Haryana State Board Class 11 Chemistry Solutions For Chapter 6 Thermodynamics

Question 1. Discuss whether the difference between the heat of the reaction at constant pressure and constant volume will depend on the temperature of the following reaction.
Answer: 

2CO(g) + O2(g)→+2CO2(g)

We know, AH = AU+AnRT

For the given reaction, An = 2-(2 + 1) = -1

ΔH-ΔU = ΔnRT = -RT

So, the value of (ΔH-ΔU) depends on the temperature.

Question 2. Calculate the change in internal energy of the gas when it expands from 2L to 8L at a constant pressure of 2 atm absorbing 400 J of heat in the process. (1 L-atm = 101.35 J)
Answer:

We know, w = -Pex (V2-V1)

∴ ω=_2(8- 2)=-12 L.atm=-12 x 101.3 J =-1215.6 J.

According to the 1st law of thermodynamics, A U=q + ω

Given that q = 400 J

∴ ΔU= (400- 1215.6)J = -815.61

∴ Change in internal energy of the gas =-815.61.

thermodynamics class11 chemistry

Question 3. At 25 °C the standard heat of formation of liquid H20 is -286.0 kJ mol-1. Calculate the change in standard internal energy for this formation reaction.
Answer:

Formation reaction of \(\mathrm{H}_2 \mathrm{O}(l): \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

For this rections \(\Delta n=0-\left(1+\frac{1}{2}\right)=-\frac{3}{2} \text {. }\)

We know, ΔH° = ΔU° + ΔnRT

∴ \(-286.0 \mathrm{~kJ}=\Delta U^0+\left(-\frac{3}{2}\right) \times 8.314 \times 10^{-3} \times(273+25) \mathrm{kJ}\)

or, ΔU° = -282.28 kj

Therefore, the change in internal energy for the formation reaction of H20(/) is -282.28 kJ.

Question 4. The temperature of 4 mol of a gas decreases from 40°C to -60 °C on adiabatic reversible expansion. The molar-specific heat of the gas at a constant volume being 12 J.K-1 mol-1, determines the change in internal energy and work done in this process
Answer:

For an adiabatic expansion

\(w=\Delta U=n C_{V, m}\left(T_2-T_1\right)\left[T_2<T_1\right]\)

∴ ΔU = 4 X 12 X (213- 313) J = -4.8 kj and w = Δt = -4.8 kj

∴ In the process, the change in internal energy = -4.8 kJ, and the amount of work done = 4.8 kJ.

Question 5. Write the thermodynamic relation generally used to predict whether a reaction is spontaneous or not For exothermic and endothermic reactions with their change in entropies being positive and independent of temperature variations, comment on the spontaneity of the reactions in both cases for temperature variations. Compute ΔH° at 298 K for: OH(g)→+H(g) + O(g)
Answer:

The thermodynamic relation that is generally used to predict the spontaneity of a reaction is AG = AH- TAS, where AG, AH, and AS are the changes in free energy, enthalpy, and entropy in the reaction at a given pressure and temperature of T K. At a given pressure and temperature, for a spontaneous reaction AG < 0. In exothermic reaction, AH < 0.

In such a reaction, if AS> 0, then AG = AH-TAS = -ve-T(+ve). As T is always positive, AG will always be negative at any value of T. Hence, an exothermic reaction will always be spontaneous at any temperature if AS > 0.

In an endothermic reaction, AH > 0. In such a reaction, if AS >0, then AG = AH-TAS =+ve-T(+ve) So, AG will be -ve only when |TASl > |AH|. It happens at high temperatures. So, an endothermic reaction with AS > 0 will be spontaneous at high temperatures.

chapter 6 chemistry class 11

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{OH}(\mathrm{g}) ; \Delta H^0=10.06 \mathrm{kcal} \quad \cdots[1]\)

⇒ \(\begin{aligned}
& \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) ; \Delta H^0=104.18 \mathrm{kcal} \\
& \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{O}(\mathrm{g}) ; \Delta H^0=118.32 \mathrm{kcal}
\end{aligned}\)

By dividing each of the equations [2] and [3] by 2, and then adding them together, we have

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \text {; }\)

ΔHº= (52.09 + 59.161) kcal = 111.25 kcal

Subtracting equation [1] from equation [4], we have

OH(g)→H(g) + O(g) ; ΔHº = (111.25 -10.06) kcal = 101.19 kcal

So, ΔHº for the given reaction is 101.19 kcal.

Question 6. For which of the following reactions, AS > 0 —

  1. \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)
  2. \(\mathrm{HCl}(g)+\mathrm{NH}_3(g) \rightleftharpoons \mathrm{NH}_4 \mathrm{Cl}(s)\)
  3. \(\mathrm{NH}_4 \mathrm{NO}_3(s) \rightleftharpoons \mathrm{N}_2 \mathrm{O}(g)+2 \mathrm{H}_2 \mathrm{O}(g)\)
  4. \(\mathrm{MgO}(s)+\mathrm{H}_2(g) \rightleftharpoons \mathrm{Mg}(s)+\mathrm{H}_2 \mathrm{O}(g)\)

Answer: 3. \(\mathrm{NH}_4 \mathrm{NO}_3(s) \rightleftharpoons \mathrm{N}_2 \mathrm{O}(g)+2 \mathrm{H}_2 \mathrm{O}(g)\)

⇒ \(\mathrm{NH}_4 \mathrm{NO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{N}_2 \mathrm{O}(\mathrm{g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

Question 7. The heat of combustion of benzene is x J-mol-1. The heat of the formation of carbon dioxide and water are y J.mol-1 and k J .mol-1 respectively. Calculate the heat of the formation of benzene.
Answer: \(\mathrm{C}_6 \mathrm{H}_6(l)+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l) ;\)

With increasing the number of species in the system, the entropy system increases for which AS > 0.

ΔH = -x J-mol-1

C(s) +O2(g) CO2(g) ; AH = -y J. mol-1

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) ; \Delta H=-z \mathrm{~J} \cdot \mathrm{mol}^{-1}\)

6 x eqn.(2) + 3 x eqn.(3)- eqn.(1) we get,

6C(s) + 3H2(g)→CgH6(l) ;

= (x- 6y- 3z) J .mol-1

ch 6 chemistry class 11

Question 8…… The combustion reaction for methane is—

⇒ \(\mathrm{CH}_4(g)+2 \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)\)

For this reaction, An = 1- (1 + 2) = -2

we know ΔH0 = ΔU° + ΔnRT

For the above combustion reaction, ΔH° = – X- 2RT

As T is positive, ΔH0 < ΔU0

Question 9. The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter, and A U was found to be -742.7 kj.mol-1 at 298 K. Calculate the enthalpy change for the reaction at 298 K:

⇒ \(\mathrm{NH}_2 \mathrm{CN}(s)+\frac{3}{2} \mathrm{O}_2(g) \rightarrow \mathrm{N}_2(g)+\mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l)\)

thermodynamics exercises

Answer: For the given reaction. Δn \(=(1+1)-\frac{3}{2}=+\frac{1}{2}\)

We know, ΔH = ΔU+ΔnRT

∴ \(\Delta H=\left(-742.7+\frac{1}{2} \times 8.314 \times 10^{-3} \times 298\right)\)

= -741.64kJ . mol-1

∴ The change in enthalpy for the reaction is -741.46kJ.mol-1

Question 10. Calculate the number of kj necessary to raise the temperature of 60g of aluminum from 35°C to 55°C. The molar heat capacity of A1 is 24J.mol-1. K-1.
Answer: Heat (q) required to raise temp, of m g of a substance (specific heat capacity=c) from T1 to T2 is: q=mc(T2-T1)

⇒ \(c=\frac{\text { Molar heat capacity }}{\text { Molar mass }}=\frac{24 \mathrm{~J} \cdot \mathrm{K}^{-1}}{27 \mathrm{~g}}=\frac{24}{27}\left(\mathrm{~J} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\right)\)

⇒ \(\text { So, } q=60 \times \frac{24}{27}[(273+55)-(273+35)] \mathrm{J}=1.066 \mathrm{~kJ}\)

Hence, 1.066kJ heat is required.

Question 11. Calculate enthalpy change on freezing 1 mol water at 10°C to ice at -10°C. A = 6.03kJ. mol1 at 0°C. Cp[H20(J)]=75.3J.mol-1.K-1 , Cp[H20(s)]=36.8J. mol-1.K-1
Answer: The given process can be considered as the sum of the following two processes—

Water(10°C) -4 water(0°C) 4 ice(0°C) 4 ice (-10°C)

To calculate enthalpy changes in steps (1) and (3), we use the relation, qp=AH=Cp(T2-T1) [At constant P, q = AH].

Step 1: AHj =Cp[H2O(l)](T2-T1)

= 75.31. mol-1.K-1 [273- (273 + 10)]K

=-0.753 kl-mol-1

Step 2: Freezing of water takes place. In this step,

AH2 = -AfusH = -6.03kj.mol-1

Step 3: AH3 =Cp[H20(s)](T2- TJ)

= 36.8[(273—10)—(273+0)] =-0.368 kj-mol-1

Step 2: Freezing of water takes place. In this step

ΔH2 = -ΔfusH = -6.03kj. mol-1

Step 3: AH3 =Cp[H20(s)](T2- TJ)

= 36.8[(273—10)—(273+0)] =-0.368 kl-mol-1

So, the total change in enthalpy in the entire process,
AH =AHJ+AH2 + AH3

=[- 0.753- 6.03- 0.368]kJ. mol-1

=-7.151kj.mol-1

Note: The answer to this problem is given as -5.65kJ.mol-1. The process may be: water at -10°C -> ice at -10°C,

solutions thermodynamics

Question 12. Enthalpy of combustion of C to CO2: -393.5 kj.mol-1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.
Answer:

Given:C(s) + O2(g)→CO2(g); AH0 = -393.5 kj.mol-1

According to this equation, 44 g of CO2 = 12 g of C

∴ \(35.2 \mathrm{~g} \text { of } \mathrm{CO}_2 \equiv \frac{12}{44} \times 35.2 \equiv 9.6 \mathrm{~g} \text { of } \mathrm{C} \equiv 0.8 \mathrm{~mol} \text { of } \mathrm{C}\)

From above equation, 1 mol C = 393.5 kJ of heat released

∴ 0.8 mol of C = 393.5 x 0.8kJ = 314.8kJ of heat released Therefore, the heat released on the formation of 35.2 g of CO2 from C and O2 is 314.8 kj

Question 13. Enthalpies of formation of CO(g), CO2(g) , N2O(g) and N2O4(g) are -110, -393, 81 and 9.7kj-mol-1 respectively. Find the value of ArH for the reaction: N2O4(g) + 3CO(g)→N2O(g) + 3CO2(g)
Answer:

⇒ \(\begin{aligned}
& \Delta_r H=\Sigma \Delta_f H \text { (Products) }-\Sigma \Delta_f H \text { (Reactants) } \\
&=3 \Delta_f H\left[\mathrm{CO}_2(g)\right]+\Delta_f H\left[\mathrm{~N}_2 \mathrm{O}(g)\right] \\
&-\Delta_f H\left[\mathrm{~N}_2 \mathrm{O}_4(g)\right]+3 \Delta_f H[\mathrm{CO}(g)] \\
&= {[3(-393)+1(81)-1(9.7)-3(-110)]=-777.77 \mathrm{~kJ} }
\end{aligned}\)

Question 14. Calculate the standard enthalpy of the formation of CH3OH(l) from the following data:

⇒ \(\begin{aligned}
\mathrm{CH}_3 \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \\
\Delta_r \mathrm{H}^0=-726 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{C}(s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta_c H^0=-393 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\
& \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta_f H^0=-286 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)
Answer: To get the formation reaction for CH3OH(l), we combine three thermochemical equations provided in the given way:

\(\begin{array}{r}
\left.\mathrm{C}(s)+\mathrm{O}_2(\mathrm{~g})+2\left[\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\right]-\mathrm{CH}_3 \mathrm{OH}(l)-\frac{3}{2} \mathrm{O}_2(\mathrm{~g})\right\rceil \\
\mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)-\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{H}_2 \mathrm{O}(l) ;
\end{array}\)

AH0 =[- 393 + 2(—286)- (-726)]kJ.mol-1 or, C(s) + 2H2(g) + O2(g)→CH3OH(Z); AH0 = -239 kj-mol-1 Eq. [1] represents formation reaction of CH3OH(Z). So, standard enthalpy offormation is -239 kl.mol-1.

Question 15. Calculate the enthalpy change for the process CCl4(g)→C(g) + 4Cl(g) & calculate the bond enthalpy of C—Cl in CCl4(g). Given: AvapH°(CCl4) = 30.5 kj-mol1, Af H°(CCl4) = -135.5 kl.mol-1, AaH°(C)=715.0 kj. mol-1, AaH°(Cl2) =242 kj-mol-1 where AaH° is enthalpy of atomisation. m For the reaction 2Cl(g)→Cl2(g) , what are the signs of AH and AS?
Answer:

Based on the given information, we can write the following thermochemical equations-

⇒ \(\begin{aligned}
& \mathrm{CCl}_4(l) \rightarrow \mathrm{CCl}_4(g) ; \Delta H^0=30.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\
& \mathrm{C}(s)+2 \mathrm{Cl}_2(g) \rightarrow \mathrm{CCl}_4(l) ; \Delta H^0=-135.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

C(s)-C(g) ; ΔH0 = 715.0 kl. mol-1

Cl2(g)→2Cl(g) ; ΔH0 = 242 kl .mol-1

By, eq. [3]- eq. [2]- eq. [1] + 2 x eq. [4] we have,

⇒ \(\begin{gathered}
\mathrm{C}(s)-\mathrm{C}(s)-2 \mathrm{Cl}_2(g)-\mathrm{CCl}_4(l)+2 \mathrm{Cl}_2(g) \\
\mathrm{C}(g)-\mathrm{CCl}_4(l)-\mathrm{CCl}_4(g)+4 \mathrm{Cl}(g)
\end{gathered}\)

ΔH° = [715.0- (-135.5)- 30.5 + 2 x 242] kj

Or, CCl4(g)C(g) + 4Cl(g); ΔH° = 1304 k

In the CC14 molecule, there are four C—Cl bonds. To break all these bonds, 1304 kj of energy is required. Hence, C—Cl bond enthalpy in CCl4(g) = 1/4x 1304 = 326 kj.mol-1.

Haryana Board Class 8 Maths Solutions

  • Chapter 1 Rational Numbers
  • Chapter 2 Linear Equation in One Variable
  • Chapter 3 Understanding Quadrilaterals
  • Chapter 4 Practical Geometry
  • Chapter 5 Data Handling
  • Chapter 6 Square and Square Roots
  • Chapter 7 Cube and Cube Roots
  • Chapter 8 Comparing Quantities
  • Chapter 9 Algebraic Expressions and Identities
  • Chapter 10 Visualising Solid Shapes
  • Chapter 11 Mensuration
  • Chapter 12 Exponents and Powers
  • Chapter 13 Direct and Indirect Proportions
  • Chapter 14 Factorisation
  • Chapter 15 Introduction to Graphs
  • Chapter 16 Playing with Numbers

Class 10 Heredity and Evolution Questions and Answers

Heredity Questions and Answers

Question 1.

  1. What is genetics?
  2. Give the common name of the plant on which Mendel performed his experiments.
  3. What for did Mendel use the term factors and what are these factors called now?
  4. What are genes? Where are the genes located?

Heredity and Evolution, Class 10 question answer

Answer:

  1. Genetics: It is the study of genes, genetic variations and laws governing the passage of heredity from generation to generation.
  2. Garden Pea, Edible Pea.
  3. Factors: They are particulate entities which function as units of inheritance that are transferred from generation to generation without being deleted or lost. The factors are now known as
  4. Genes: Genes are units of genetic material that have particular functions in inheritance. They are linear segments of DNA/chromosomes.

Question 2. How do Mendel’s experiments show that the

  1. Traits may be dominant or recessive.
  2. Traits are inherited independently.

Answer:

Traits are Dominant or Recessive: This can be discovered by performing a monohybrid cross. A between pure round-seeded and pure wrinkled-seeded plants yields only round-seeded plants in the F1 generation. Self-pollination of F1 plants yields both round-seeded and wrinkled-seeded plants in the ratio of 3:1. TIIIH to poHHiblo only If (lie allele lor wrinkled seed to present in F1 generation but is unable to express its effect, Therefore, the trail of wrinkled seed is recessive while the trait of round seed which is expressed in a generation is dominant.

class 10 heredity questions

Independent Inheritance of Traits: This can be found by performing a dihybrid cross. A cross between pure round and yellow seeded (RRYY) Pea plant with pure wrinkled and green seeded (rryy) plant yields rounded and yellow seeded F1 generation. Self-polIination of F1 plants gives rise to four types of plants—round yellow (9/16), round green (3/16), wrinkled yellow (3/16) and wrinkled green (1/16). Here round yellow and wrinkled green are parental types. Round green and wrinkled yellow are recombinations which can develop only if the factors for these traits are inherited independently of one another.

Question 3.

  1. How many pairs of chromosomes are present in human beings? Out of these how many arc sex chromosomes 7 How many types of sex chromosomes are found in human beings?
  2. “The sex of a newborn child is a matter of chance and none of the parents may be considered responsible for it”. Draw a flowchart showing the determination of the sex of the newborn to justify the statement.

Answer:

Human beings have 23 pairs of chromosomes. One pair of chromosomes are sex chromosomes or allosomes. In females both the sex chromosomes are similar. They are called XX chromosomes. In males, the two sex chromosomes are different (heteromorphic). One is a normal X chromosome (as found in females). The second is a shorter Y chromosome.

Question 4.

  1. What are dominant and recessive traits?
  2. “Is it possible that a trait is inherited but may not be expressed in the next generation ?’’ Give a suitable example to justify this statement.

Answer:

Dominant Trait: It is that trait which expresses its effect in both homozygous and heterozygous states, Example TT, Tt.

Recessive Trait: It is that trait which is unable to express its effect in the heterozygous state, i.e., in the presence of its alternate trait. The trait expresses its effect in only the homozygous state, Example tt.

Question 5. If we cross purebred tail (dominant) Pea plants with purebred dwarf (recessive Pea plants) we get Pea plants of F1 generation. If we now self-cross the Pea plants of the F1 generation, then we obtain Pen plants of the F2 generation

  1. What do the plants of the F1 generation look like?
  2. What is the ratio of tall plants to dwarf plants in F2 generation?
  3. State the types of plants not found in the F1 generation but appeared in the F2 generation, mentioning the reason for the same.
  4. What are homologous structures? Give an example. Is it necessary that homologous structures always have common ancestors?

Answer:

  1. F1 Generation. All tall p
  2. F2 Gcncmtlon: 3 tall; I dwarf.
  3. Dwarf plant: The trait of dwarfness is recessive and cannot express its effect in the presence of an alternate dominant trait.
  4. Homologous Structures, See question 5 above. Homologous structures always have common ancestors,

Question 6.

  1. What Is the law of dominance? Explain with an example.
  2. What is genetics?
  3. What are genes? Where are the genes located?

Answer:

Law of Dominance. Out of the two contrasting factors or alleles of a character if present together as in a hybrid, only one factor called dominant expresses its effect. The other factor called recessive docs not express its effect.

Example Cross a pure tall Rea plant with a pure dwarf Pea plant. All the individuals of the F1 generation are tall though they carry the factor of dwarfness as well.

Heredity Genes Located

Question 7. 

  1. What does the F1 progeny of tall plants with round seeds and short plants with wrinkled seeds look like?
  2. What are recessive traits?
  3. Mention the type of new combination of plants obtained in F2 progeny along with their ratio, if F1 progeny was allowed to self-pollinate.
  4. If 1600 plants were obtained in F2 progeny, write the number of plants having traits
    1. Tall with round seeds
    2. Short with wrinkled seeds. Write the conclusion of the above experiment.

Answer:

  1. Round and tall (both the traits are dominant over wrinkled seed and short height).
  2. Recessive traits are those traits which are unable to express their effect in the presence of their alternate traits, as in the F1 generation.
  3. New Combinations
    1. Tall and wrinkled seeded = 3/16
    2. Short and round seeded = 3/16
  4. Tall with round seeds—1600 x 9/16 = 900
    1. Short with wrinkled seeds—1600 x 1/16 = 100.

Conclusion: The two factors of a character separate independently of the two factors of another character, coming together randomly at the time of fertilization. It is called the law of independent assortment.

Heredity Of Heredity

Class 10 Biology Respiration Question and Answers

Respiratory System Questions And Answers

Question 1. What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration?
Answer:

Respiratory System Questions

The advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration

Aquatic organisms obtain oxygen for their respiration from water. Water has about 1% oxygen in a dissolved state. Terrestrial organisms obtain oxygen directly from the air. The oxygen content of air is 21%. So terrestrial organisms are in a favourable position to obtain the required oxygen without too much work.

On the other hand, aquatic animals which extract oxygen from water have to perform more vigorous and rapid breathing to obtain the required oxygen. A fish will breathe 100 times/ min as compared to 14-20/min in humans.

Question 2. What are the different ways in which glucose is oxidized to provide energy in various organisms?
Answer:

The different ways in which glucose is oxidized to provide energy in various organisms

There are three common methods by which glucose is catabolised to release energy-aerobic respiration, lactic acid anaerobic respiration and alcoholic anaerobic respiration.

Respiratory Questions

Aerobic Respiration. Glucose is completely oxidised with oxygen as a terminal oxidant. It yields carbon dioxide and energy. The energy yield is 686 kcal/2870 kJ per mole of glucose. About 38 ATP molecules are formed for storing energy. The process occurs partly in cytoplasm and partly in mitochondria.

Respiration Aerobic Respiration In Various Organisms

Lactic Acid Anaerobic Respiration. Glucose is incompletely catabolised to form two molecules of lactic acid without involving the use of oxygen. It occurs in many bacteria and skeletal muscles during vigorous exercise. Only about 50 kcal/210 kJ of energy is released from which 2 ATP molecules are formed. The process occurs exclusively in the cytoplasm.

Respiration Lactic Acid Anaerobic Respiration

Ethyl Alcohol Anaerobic Respiration. A molecule of glucose is catabolised to form two molecules of ethyl alcohol. Two molecules of carbon dioxide are evolved. The energy yield is about 50 kcal/210 kJ, equivalent to 2ATP molecules. Oxygen is not required. Respiration occurs inside the cytoplasm. Mitochondria are not required.

Respiration Ethyl Alcohol Anaerobic Respiration

Question 3. How are oxygen and carbon dioxide transported in human beings?
Answer:

Transport of Oxygen. It occurs through blood. 97% of oxygen combines with haemoglobin to form oxyhaemoglobin (Hb4O8). The remaining 3% dissolves in blood plasma. In the tissues, oxyhaemoglobin breaks down to release oxygen. The deoxygenated blood travels to the lungs for reoxygenation.

Transport of Carbon Dioxide. Carbon dioxide produced in cellular respiration passes into the blood. 70% of carbon dioxide forms sodium bicarbonate inside the plasma. 23% of carbon dioxide combines with haemoglobin to produce carbaminohaemoglobin. The remaining 7% gets dissolved in blood plasma. Blood-carrying carbon dioxide reaches alveoli where carbon dioxide diffuses into alveolar air.

Question 4. How are the lungs designed in human beings to maximize the area for the exchange of gases?
Answer:

Lungs designed in human beings to maximize the area for the exchange of gases as follows

Human Respiratory System, Class 10 pdf

After entering a lung, each bronchus divides and redivides repeatedly to form segmental bronchi and then each segmental bronchus produces different ranks of bronchioles. The terminal bronchioles form respiratory bronchioles.

The latter form alveolar sacs that are studded with bunches of alveoli. It is estimated that a human lung has about 300 million alveoli. The total alveolar surface of the two lungs is about 80 m2 or roughly 40 times the surface area of the human body. Such a large surface area is most suitable for rapid gaseous exchange between the blood and the alveolar air.

Human Health And Disease Class 12 Multiple Choice Questions

Question 1. Rabies is also called:

  1. Infantile paralysis
  2. Lock Jaw
  3. Hydrophobia
  4. Amoebiasis.

Answer: 3. Hydrophobia

Question 2. Rubeola virus causes:

  1. Smallpox
  2. Measles
  3. Chickenpox
  4. Influenza.

Answer: 2. Measles

Question 3. Black Death or Great Mortality is the name given to:

  1. Plague
  2. Malaria
  3. Cholera
  4. Cancer.

Answer: 1. Plague

human health and disease pyq 

Question 4. A person with blood group ‘O’ can receive blood transfusion from persons with blood groups:

  1. O and A B
  2. O, A and B
  3. O only
  4. AB only.

Answer: 3. O only

Question 5. After which animal is the ‘Rh’ factor named?

  1. Dragon fly
  2. Monkey
  3. Man
  4. Rat.

Answer: 2. Monkey

Human Health And Disease Class 12 Multiple Choice Questions

Question 6. The universal recipient in blood transfusion belongs to the group:

  1. A
  2. O
  3. AB
  4. B.

Answer: 3. AB

Question 7. Rh factor is connected with:

  1. Carbohydrate metabolism in the body
  2. Initiation of protein synthesis
  3. Blood clotting
  4. Blood of man.

Answer: 4. Blood of man

Question 8. Haemophilia is a disease caused by deficiency of:

  1. Water in plasma
  2. RBCs
  3. WBCs
  4. Thromboplastin.

Answer: 4. Thromboplastin.

Question 9. Rh negative individuals are:

  1. Homozygous dominant (RR)
  2. Heterozygous (Rr)
  3. Homozygous recessive (rr)
  4. Possess only one gene (R or r).

Answer: 3. Homozygous recessive (rr)

Question 10. Which of the following is a genetic trait?

  1. Albinism
  2. Leucoderma
  3. Tuberculosis
  4. Diphtheria.

Answer: 1. Albinism

Question 11. The Rh factor was discovered by:

  1. William Harvey
  2. Landsteiner and Weiner
  3. James Watson
  4. Robert Hooke.

Answer: 2. Landsteiner and Weiner

Question 12. Universal donor in blood transfusion belongs to the blood group:

  1. A
  2. O
  3. AB
  4. B.

Answer: 2. O

Question 13. A woman with blood group O marries a man with blood group A. The child conceived to such a woman will:

  1. Survive with normal health
  2. Survive with slight anaemic conditions
  3. Will not survive long
  4. Will die soon after birth.

Answer: 1. Survive with normal health

Question 14. Which of the following is biologically incompatible marriage because of the danger of erythroblastosis foetal?

  1. Rh man and Rh woman
  2. Rh+ man and Rh+ woman
  3. Both Rh positive
  4. Both Rh negative.

Answer: 2. Rh+ man and Rh+ woman

Question 15. Down’s Syndrome has karyotype make-up as:

  1. 44 + 3x
  2. 44 + xxy
  3. 45 + xx in female
  4. 44 + xy or 44+ xx in female.

Answer: 3. 45 + xx in female

Question 16. If a man is Rh+ and the lady is Rh then:

  1. The first child will die
  2. The first child will survive
  3. No child will be bom
  4. None of these.

Answer: 2. The first child will survive

Question 17. If a human mother has blood group ‘O’, the foetus in the womb would die, the blood group of the foetus is

  1. A
  2. B
  3. AB
  4. No effect in any case.

Answer: 4. No effect in any case.

Question 18. To a person of blood group AB blood groups of A and B are given separately, what will happen:

  1. Clumping occurs with A
  2. Clumping occurs with B
  3. Clumping occurs with both
  4. Clumping does not occur.

Answer: 4. Clumping does not occur.

Question 19. Which of the following diseases is genetically similar to haemophilia?

  1. Albinism
  2. Colour blindness
  3. Phenylketonuria
  4. None of the above.

Answer: 2. Colour blindness

Question 20. The disease Erythroblastosis foetal in human embryos is caused by to:

  1. Disadjustment of blood groups
  2. Disadjustment of Rh factor
  3. Both of these
  4. None of these.

Answer: 2. Disadjustment of Rh factor

Question 21. ABO blood groups were discovered by:

  1. Landsteiner
  2. Darwin
  3. Harvey
  4. Weiner.

Answer: 1. Landsteiner

Question 22. The second Rh+ child conceived to an Rh the woman will:

  1. Die immediately
  2. Surely survive
  3. Survive with serious malformations
  4. Usually die at an early stage.

Answer: 1. Die immediately

Question 23. Persons with blood group ‘O’ lack:

  1. Antigens
  2. Antibody ‘a’
  3. Antibody ‘b’
  4. Haemophilia.

Answer: 1. Antigens

Question 24. Foetal sex can be determined by examining cells from amniotic fluid looking for

  1. Barr bodies
  2. Chiasmata
  3. Sex chromosome
  4. Kinetochores.

Answer: 1. Barr bodies

Question 25. The causes of diseases like peptic ulcers, hypertension, coronary heart diseases, and liver cirrhosis are:

  1. Certain viruses and bacteria
  2. Excess of fats in food
  3. Excessive use of alcoholic drinks
  4. Not yet identified.

Answer: 3. Excessive use of alcoholic drinks

Question 26. Trisomic condition of Down’s syndrome arises due to:

  1. Triploidy
  2. Non-disjunction
  3. Translocation
  4. Dicentric bridge formation.

Answer: 2. Non-disjunction

Question 27. Condition of sex chromosomes in male children with Down’s syndrome will be:

  1. XY
  2. XXY
  3. XX
  4. XO.

Answer: 1. XY

Question 28. Phenylketonuria (PKU) is:

  1. A nutritional disorder due to protein deficiency
  2. An inborn error of metabolism due to a recessive autosomal gene mutation
  3. A sex-linked disorder which is more common in males than in females
  4. None of the above.

Answer: 2. An inborn error of metabolism due to a recessive autosomal gene mutation

Question 29. Chromosomes of Klinefelter’s syndrome usually have:

  1. One X
  2. XXY
  3. 2X only
  4. No Y.

Answer: 2. XXY

Question 30. XO chromosome abnormality is:

  1. Turner’s syndrome
  2. Klinefelter’s syndrome
  3. Criminal syndrome
  4. Down’s syndrome.

Answer: 1. Turner’s syndrome

Question 31. Tay Sach’s disorder is:

  1. A chromosomal disorder in young children
  2. The result of severe protein deficiency in diet damages the muscle development in growing children
  3. An inherited disorder in which the spinal cord and brain are severely damaged within a few months of birth leading to paralysis and mental retardation, and finally the infant’s death in 3 or 4 years
  4. A genetic abnormality which appears at the age of 25-30 years and damages the bones.

Answer: 3. An inherited disorder in which the spinal cord and brain are severely damaged within a few months of birth leading to paralysis and mental retardation, and finally the infant’s death in 3 or 4 years

Question 32. Active immunity is due to:

  1. Memory cells
  2. Killer T-cells
  3. Helper T-cells
  4. Suppressor T-cells.

Answer: 1. Memory cells

Question 33. Passive immunity is provided through:

  1. Exogenous supply of antigens
  2. Exogenous supply of antibodies
  3. Endogenous supply of antigens
  4. Endogenous supply of antibodies.

Answer: 2. Exogenous supply of antibodies

Question 34. B-cells produce antibodies in response to instruction received from:

  1. Killer T-cells
  2. Suppressor T-cells
  3. B-lymphocytes
  4. All of the above.

Answer: 4. All of the above.

Question 35. Opsonisaiion is carried out by:

  1. IgM
  2. IgG
  3. IgA
  4. IgD and IgE.

Answer: 2. IgG

Question 36. Lysis of foreign cells is mediated through:

  1. IgM and IgG
  2. IgG and IgA
  3. IgA and IgD
  4. IgD and IgE.

Answer: 1. IgM and IgG

Question 37. T-cells have a life of:

  1. 4-5 days
  2. 4-5 weeks
  3. 4-5 months
  4. 4-5 years.

Answer: 4. 4-5 years.

Question 38. Antigenic determinant sites bind to which portions of an antibody molecule?

  1. Light chains
  2. Heavy chains
  3. Intermediate chains
  4. Both (1) and (2).

Answer: 4. Both (1) and (2).

Question 39. The cells that actually release the antibodies are:

  1. Helper T-cells
  2. Cytotoxic T-cells
  3. Mast cells
  4. Memory cells.

Answer: 3. Mast cells

Question 40. Types of T-cells are

  1. Killer cells and helper cells
  2. Killer cells and suppressor cells
  3. Killer, helper and suppressor cells
  4. Killer, helper and depressor cells.

Answer: 3. Killer, helper and suppressor cells

Question 41. Interferon is a protein that:

  1. Inactivates a virus
  2. Protects unattacked cells from virus
  3. Prevents viruses from taking over the cellular machinery
  4. All of the above.

Answer: 2. Protects unattacked cells from virus

Question 42. The most modern technique of vaccine preparation is:

  1. Extraction of antigens from pathogens
  2. Multiplication of antigens through DNA technique
  3. Weakening of pathogens through chemical treatment
  4. Attenuation of the pathogen through heat.

Answer: 2. Multiplication of antigens through DNA technique

Question 43. AIDS testing on normal individuals is done by:

  1. Separation of virus
  2. Reduction in immunity of the individual
  3. Identification of antibodies
  4. Identification of antigen-toxin.

Answer: 3. Identification of antibodies

Question 44. When people are born without B-cells or T-cells, they suffer from:

  1. Autoimmune diseases
  2. Immunodeficiency diseases
  3. Both of these
  4. None of these.

Answer: 2. Immunodeficiency diseases

Question 45. Which of the following is incorrect:

  1. Skin is a tough outer layer, so it prevents entry of bacteria and viruses
  2. Skin secretes oil and sweat by the glands which makes the surface basic and thus kills the microbes
  3. Skin secretes lysozyme along with sweat which prevents infection
  4. Some bacteria are present on the skin which release acids and other metabolic wastes that inhibit the growth of pathogens.

Answer: 2. Skin secretes oil and sweat by the glands which makes the surface basic and thus kills the microbes

Question 46. Which one of the following diseases is due to an allergic reaction?

  1. Enteric fever
  2. Hay fever
  3. Skin cancer
  4. Goitre.

Answer: 2. Hay fever

Class 11 Chemistry Hydrogen Question and Answers

Hydrogen Question and Answers

Question 1. Explain why concentrated HCl is not used in the laboratory preparation of H2 gas.
Answer:

Concentrated HCl is not used for the laboratory preparation of dihydrogen because HCl, being highly volatile, gets mixed with dihydrogen.

Question 2. Write down the name and formula of a compound that on electrolysis produces dihydrogen at the anode.
Answer:

Sodium hydride(NaH).

thermodynamics class11 chemistry

Question 3. What is syngas?
Answer:

syngas

All mixtures of CO and H2 irrespective of their composition are called synthesis gas or syngas.

Question 4. Which isotope of hydrogen is used as a tracer in organic reactions?
Answer:

Deuterium D is usually used as a tracer in determining the mechanism of organic reactions.

Question 5. Explain why dihydrogen is not suitable for balloons.
Answer:

As dihydrogen, the lightest substance known, is a highly inflammable gas, it is not suitable for balloons.

Question 6. Which bond between two atoms has the highest bond dissociation enthalpy?
Answer:

Bond dissociation enthalpy of the H —H bond is highest.

Question 7. Explain why H2 is more reactive than D2
Answer:

This is because the H —H bond dissociation enthalpy is less than the D —D bond dissociation enthalpy.

Question 8. What change is expected to take place when vegetable oils are hydrogenated?
Answer:

Carbon-carbon double bonds are converted to carbon-carbon single bonds.

Question 9. Which isotope of hydrogen is used in nuclear reactors?
Answer:

Deuterium(H or D).

Question 10. Why are ionic hydrides used as solid fuels?
Answer:

When heated, ionic hydrides decompose to evolve dihydrogen gas which ignites readily.

chapter 6 chemistry class 11

Question 11. The densities of ionic hydrides are greater than that of the metal from which they are formed—why?
Answer:

The densities of ionic hydrides are greater than that of the metal from which they are formed

This is because hydride ions (H) occupy the holes in the lattice of the metal without distorting the metal lattice.

Question 12. Give examples of two interstitial hydrides.
Answer:

CuH and FeH.

Question 13. Which gaseous compound on treatment with dihydrogen produces methanol?
Answer:

Carbon monoxide(CO).

Question 14. Give the chemical reaction that occurs when hydrogen is used as a rocket fuel.
Answer:

\(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l)+286 \mathrm{~kJ}\)

hydrogen solutions

Question 15. A sample of water containing KCl does not behave as hard water, but a sample of water containing CaCI2 or MgCl2 behaves as hard water—why?
Answer:

A sample of water containing KCl does not behave as hard water, but a sample of water containing CaCI2 or MgCl2 behaves as hard water

The potassium salt of soap is soluble in water and forms a lather while calcium or magnesium salt of soap is insoluble in water and does not form a lather.

Question 16. What is EDTA, a compound used to determine the hardness of water?
Answer:

EDTA

EDTA is the disodium salt of ethylenediamine tetraacetic acid \(\left[\mathrm{NaO}_2 \mathrm{C}(\mathrm{COOH}) \mathrm{NCH}_2 \mathrm{CH}_2 \mathrm{~N}(\mathrm{COOH}) \mathrm{COONa}\right]\)

 

ch 6 chemistry class 11

Question 17. Can distilled water be called deionized water?
Answer:

Distilled water can be called deionized water because it does not contain any cations and anions.

Question 18. What is the difference between the water softened by the permit process and the water softened by the organic ion exchangers?
Answer:

The difference between the water softened by the permit process and the water softened by the organic ion exchangers

Although the water softened by the permit process contains no cation, it contains various anions (for example Cl-, SO|- etc.). However, the water softened by the organic ion exchangers contains no cations and anions.

chapter 9 Chemistry, class  11 

Question 19. What will be the hardness of a sample of water, 106 g of which contains| mol A12(SO4)3?
Answer:

50 ppm.

Question 20. What is Calgon?
Answer:

Calgon

Sodium hexametaphosphate, Na2[Na4(PO3)6]

Question 21. Give the chemical formula of permtit.
Answer:

Na2Al2Si2O8 xH8O.

Question 22. What is the main source of heavy water?
Answer:

The main source of heavy water

Ordinary water is the main source of heavy water.

Question 23. Can sea animals survive in distilled water?
Answer:

Sea animals cannot survive in distilled water because distilled water contains no salt or dissolved oxygen.

Question 24. Although D20 resembles H20 chemically, it is a toxic substance—why?
Answer:

D2O is toxic because D+ ions react at a much slower rate than H+ ions in enzyme-catalyzed reactions.

Question 25. Which compound is used to color hair golden?
Answer:

Dilute solution of H2O2.

Question 26. What is the trade name of hydrogen peroxide used as an antiseptic?
Answer:

Perhydrol.

Question 27. What is the strength in the normality of an ‘11.2 volume’ H2O2 solution?
Answer:

2(N) H2O2 solution.

Question 28. Name a compound that suppresses the decomposition of H2O2.
Answer:

Acetanilide (PhNHCOCH3).

Question 29. H2O2 molecule has an open-book-like structure. What is the angle between the two pages of the book in the gas phase?
Answer:

111.5°.

Question 30. Name an organic compound without peroxo bond which is used to manufacture H2O2.
Answer:

2-ethylanthraquinol.

Question 31. Why do most of the reactions of H2 occur at much higher temperatures?
Answer:

Due to the much higher bond dissociation enthalpy of the H —H bond, dihydrogen is quite stable and relatively inert at temperature. It dissociates into atoms at about 5000K. For this reason, most of the reactions of dihydrogen occur at much higher temperatures.

ch 6 chemistry class 11

Question 32. What characteristics do you expect from electron-deficient hydrides with respect to their structure and chemical reactivity?
Answer:

Electron-deficient hydrides have less number of electrons in the valence shell of the central atom and so, their mononuclear units do not satisfy the usual Lewis octet rule. Due to a deficiency of electrons, these hydrides act as Lewis acids and form complex entities with Lewis bases such as NH3, H- ion, etc.

⇒ \(\mathrm{B}_2 \mathrm{H}_6+2 \ddot{\mathrm{N}}_3 \rightarrow\left[\mathrm{BH}_2\left(\mathrm{NH}_3\right)_2\right]^{+}\left[\mathrm{BH}_4\right]^{-}\)

⇒ \(\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{LiH} \rightarrow 2 \mathrm{Li}^{+}\left[\mathrm{BH}_4\right]^{-} \text {(Lithium borohydride) }\)

Question 33. Explain whyit is harmful to bathe in heavy water and use it for drinking purposes.
Answer:

Being a very hygroscopic substance, heavy water (D2O) absorbs water from the body and thereby damages body cells. Also, it retards some cellular processes such as mitosis, cell division, and various enzyme-catalyzed reactions. For these reasons, it is harmful to bathe in heavy water and use it for drinking purposes.

Question 34. Explain why the thermal stability of H2O2 is very low.
Answer:

The bond dissociation enthalpy of the O —O bond presenting H2O2 molecule is very low (35kcal . mol-1) i.e., the bond is very weak. For this reason, the thermal stability of H2O2 is extremely low.

Question 35. How the presence of H- ions be confirmed in ionic hydrides?
Answer:

In the molten state, ionic hydrides conduct electricity with the liberation of dihydrogen at the anode. This confirms the presence of hydride (H ) ions in them.

Question 36. How do you separate 2 allotropic forms of hydrogen?
Answer:

Ordinary hydrogen is a mixture of 75% of ortho and 25% of para-isomer at room temperature. On passing through activated charcoal kept at 20K, the para-isomer is adsorbed leaving behind the ortho-isomer. From the charcoal surface, para-hydrogen can be released by reducing pressure.

thermodynamics exercises

Question 37. Mention the difference in chemical characteristics of the two hydrides obtained when hydrogen combines with two elements having atomic numbers 17 and 20.
Answer:

The highly electronegative Cl having atomic number 17 combines with hydrogen to form the covalent hydride H —Cl. On the other hand, the highly electropositive Ca atom having atomic number 20 combines with hydrogen to form the ionic hydride CaH2.

Question 38. Two samples of hard water contain the same cations, Ca2+ & Mg2+. One is marked as temporary and the other as permanent. In which respect do they differ?
Answer:

The two samples of water differ with respect to the anions of calcium and magnesium salts present. the sample watermarked as temporary hard water contains bicarbonates of Ca2+ and Mg2+ ions while the sample watermarked as permanent hard water contains chlorides and sulfates of Ca2+ and 2+ ions.

Question 39. Tube-well water, if left for some time, assumes brownish turbidity—explain.
Answer:

Tube-well water, if left for some time, assumes brownish turbidity

Tube-well water sometimes contains soluble ferrous bicarbonate [Fe(HCO3)2]. This compound, on aerial oxidation, is converted into brown ferric hydroxide, Fe(OH)3, which remains in water as colloidal suspension, and because of this, water assumes a brownish turbidity.

⇒ \(4 \mathrm{Fe}\left(\mathrm{HCO}_3\right)_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \rightarrow \underset{\text { (brown) }}{4 \mathrm{Fe}(\mathrm{OH})_3}+8 \mathrm{CO}_2\)

Question 40. Write the reactions for:

  1. Preparation of H2O2 from two sodium salts and
  2. Preparation of D2O2 from potassium persulphate.

Answer:

Hydrogen Sodium dihydrogen

⇒ \(\begin{aligned} & \mathrm{K}_2 \mathrm{~S}_2 \mathrm{O}_8+2 \mathrm{D}_2 \mathrm{O} \rightarrow \mathrm{D}_2 \mathrm{O}_2+2 \mathrm{KDSO}_4 \\ & \text { Potassium } \\ & \text { persulphate } \end{aligned}\)

Question 41. Between deionised water and distilled water which one is more pure and why?
Answer:

Both deionized and distilled water are free from ions. Yet distilled water is superior to deionized water in terms of purity. This is because of the fact that deionized water contains a small amount of dissolved silica and CO2 along with some germs and organic impurities. However, distilled water prepared in a glass apparatus does not contain any impurities other than trace amounts of silica and CO2.

Question 42. Why is Na2O2 used for purifying air in submarines and in crowded places?
Answer:

Na2O2 reacts with CO2 of air to evolve O2 (2Na2O2 + 2CO2 —> 2Na2CO3 + O2t). For this reason, Na2O2 is used for the purification of air in submarines and in crowded places.

solutions thermodynamics

43. Comment on the reactions of dihydrogen with

  1. Chlorine
  2. Sodium And
  3. Copper
  4. Oxide.

Answer:

  1. Dihydrogen reduces chlorine (Cl) to chloride ion (Cl-) and itself gets oxidized to form H+ ions. These two ions (H+ and Cl-) share an electron pair between themselves to form a covalent molecule of hydrogen chloride (HCl)
    ⇒ \(\mathrm{H}_2(g)+\mathrm{Cl}_2(g)\rightarrow 2 \mathrm{HCl}(g)\)
  2. Sodium reduces dihydrogen to hydride ion (H~) and itself gets oxidized to sodium ion (Na+). In this reaction, an electron gets completely transferred from Na thereby forming ionic sodium hydride (NaH).
    \(2\mathrm{Na}(s)+\mathrm{H}_2(s)\stackrel{\Delta}{\longrightarrow} 2\mathrm{Na}^{+} \mathrm{H}^{-}(s)\)
  3. Dihydrogen reduces copper(II) oxide to metallic copper while it gets oxidized to form a covalent molecule of H2O.
    ⇒ \(\stackrel{+2}{\mathrm{CuO}}\stackrel{-2}{\mathrm{O}}(s)+\stackrel{0}{\mathrm{H}_2}(s)\stackrel{\Delta}{\longrightarrow}\stackrel{0}{\mathrm{Cu}}(s)+\stackrel{+1}{\mathrm{H}_2}\stackrel{-2}{\mathrm{O}}(l)\)

Question 45. An ionic alkali metal hydride has a covalent character to some extent and it does not react with oxygen and chlorine. This hydride is used in the synthesis of another hydride. Write the formula of the hydride and what happens when it reacts with AI2CI6.
Answer:

Since the alkali metal hydride possesses sufficient covalent character, the hydride is of the smallest alkali metal, Li, i.e. the hydride is LiH. As LiH is quite stable, it does not react with oxygen and chlorine. Lithium hydride reacts with AI2CI6 to form lithium aluminum hydride (LiAlH4) which is extensively used as a reagent in the synthesis of different organic compounds.

⇒ \(8\mathrm{LiH}+\mathrm{Al}_2\mathrm{Cl}_6\longrightarrow2\mathrm{LiAlH}_4+6\mathrm{LiCl}\)

Question 46. Sodium reacts with dlliydrogcn to form a crystalline ionic solid. It is non-volatile and a non-conductor of electricity. It also reacts vigorously with water to liberate II2 gas. Write the formula of the Ionic solid and give a reaction between the lids solid & water. What happens when the ionic solid in its molten stater Is electrolysed?
Answer:

Sodium reacts with dihydrogen to form sodium hydride which is a crystalline ionic solid.

⇒ \(2\mathrm{Na}+\mathrm{H}_2\stackrel{\Delta}{\longrightarrow}2\mathrm{Na}^{+}\mathrm{H}^{-}\)

Sodium hydride reacts with water as follows:

⇒  \(2 \mathrm{NaH}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+2 \mathrm{H}_2\)

Sodium hydride, in its solid state, does not undergo electrolysis. However, in its molten state, undergoes electrolysis. However, in its molten state, NaH undergoes electrolysis to liberate dihydrogen (H2) at the anode and metallic sodium at the cathode.

⇒ \(\mathrm{Na}^{+} \mathrm{H}^{-}(\text {molten }) \stackrel{\text { Electrolysis }}{\longrightarrow} \underset{\text { Cathode }}{2 \mathrm{Na}(\text { molten })}+\underset{\text { anode }}{\mathrm{H}_2(g)}\)

Question 47. Why is seawater not used in boilers?
Answer:

  1. Sea water is hard water. It contains Mg(HCO3)2 and Ca(HCO3)2, which on boiling, form a hard heat-insulating thick layer or scale of MgCO3 and CaCO3 on the inner surface of the boiler. Consequently, much heat is required to raise the temperature of the boiler, and thus, fuel economy is adversely affected.
  2. Again at much higher temperatures, the boiler scales and the metal of the boiler expand unequally. Due to such uneven expansion, cracks are formed on the scales. When hot water comes in contact with the hot metal surface of the boiler through these cracks, it is suddenly converted into steam. Due to the high pressure, thus developed, the boiler may burst leading to accidents. Again, MgCl2 and MgSO4 present in seawater undergo hydrolysis to form HCl and H2SO4 which degrade the metallic component of the boiler. So, seawater is not used in boilers.

Question 48. The values of melting point, enthalpy of vaporization, and viscosity of H2O and D2O are given below:

  1. Melting point(K)     373.0             374.4
  2. Enthalpy of
    vaporisation
    (kJ • mol-1, 373K)     40.66         41.61
  3. Viscosity(centipoise) 0.8903      1.107

From the given data, determine which liquid has a greater magnitude of intermolecular forces of attraction.

Answer:

The magnitude of intermolecular forces of attraction depends on the magnitudes of melting point, enthalpy of vaporization, and viscosity of liquid. As, these parameters have higher values in the case of D2O, the magnitude of intermolecular forces of attraction is greater for D2O than for H2O.

Question 49. How will you prepare heavy water from ordinary water? Explain its principle.
Answer:

Heavy water (D2O) is prepared by prolonged electrolysis of ordinary water. As water is not a good conductor of electricity, an alkaline solution of water [~0.5(N) NaOH] is used for electrolysis. The bond dissociation energy of the O —H bond is less than that of the O —D bond. So, electrolysis of H2O occurs at a faster rate and more easily than D2O. Consequently, the amount of D2O in ordinary water increases. Pure D2O is obtained when the amount of residual liquid decreases.

Question 50. Can phosphorus form PHg with its outer electronic configuration of 3s23p3?
Answer:

Phosphorus cannot form PH5 although it shows +3 and +5 oxidation states. Dihydrogen acts as a weak oxidizing agent due to the high bond dissociation enthalpy of H—Hbond (435.88 kJ-mo-1) and slightly negative electron-gain enthalpy (-73 kJ mol-1). So, dihydrogen can oxidize phosphorus to a +3 oxidation state but not to its highest oxidation state of +5. Therefore, phosphorous can form only PH3 and not PH5.

Question 51. How will you prepare dinitrogen from HNO3?
Answer:

Magnesium and manganese react with a very dilute solution of HN03(2%) to form dihydrogen.

⇒ \(\mathrm{Mg}+2\mathrm{HNO}_3\rightarrow\mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \uparrow\)

⇒ \(\mathrm{Mn}+2\mathrm{HNO}_3\rightarrow\mathrm{Mn}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \uparrow\)

Question 52. Do you think the hydrides of N, O, and F will have lower boiling points than the hydrides of their corresponding group members? State reasons.
Answer:

The hydrides of the elements N, O, and F are NH3, H2O, and HF respectively. These hydrides are expected to have lower boiling points than that of their corresponding group members (PH3, H2S & HCl) when their masses are considered.
However, because of the high electronegativity of N, O, and F, their hydrides undergo extensive hydrogen bonding (intermolecular). As a result, boiling points of NH3, H2O, and HF are much higher than the hydrides of their corresponding group members, i.e., PH3, H2S, and HCl respectively.

Question 53. KF reacts with HF to form the compound, KF-2HF. Discuss the probable structure of the compound.
Answer:

The h-bond in the HF molecule is very strong. When KF gets added to HF, one F ion forms an H — bond with two HF molecules to form the H2F3 ion [F—-H —F—H —F —].

Question 54. Calculate the amount of energy liberated due to combustion of 4g dihydrogen.
Answer:

Amount of energy liberated due to combustion of1 mol, i.e., 2g dihydrogen = 242 kJ • mol-1

⇒ \(\left[\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(g) ; \Delta \mathrm{H}=-242 \mathrm{~kJ}\cdot\mathrm{mol}^{-1}\right]\)

Amount of energy liberated due to 4g dihydrogen = 242×2 = 484 kJ-mol-1

Question 55. Under what conditions, water reacts with calcium cyanamide and what are the products formed due to this reaction?
Answer:

Superheated steam reacts with calcium cyanamide under high pressure to form ammonia gas and calcium carbonate as a result of hydrolysis.
\(\text { CaNCN }(\text{Calciumcyanamide })+3\mathrm{H}_2\mathrm{O}\rightarrow\mathrm{CaCO}_3+2\mathrm{NH}_3\)

Question 56. Why H2O2 is a better oxidant than water?
Answer:

H2O2 is a better oxidant than water because H2O2 being unstable readily dissociates to form stable water molecules along with the evolution of O2 gas. \(2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2+\text { Heat }\)

Solved Questions

Question 1. What is heavy water? Why is it so-called?
Answer:

Heavy water

Deuterium oxide is commonly known as heavy water because its density is higher than that of ordinary water, i.e., it is heavier than ordinary water.

Question 2. A water sample contains 1 millimole of Mg2+ ion per liter. Calculate the hardness of the water sample in ppm units.
Answer:

1 millimol of Mg2+=1 millimole of MgCl2 = 0.095 g of MgCL2.

1 1. or 1000 g or 103 g of water contains 0.1 g of MgCl2.

106g of water contains 0.095 x 103 = 95g of MgCl2

The degree of hardness of water is 95 ppm.

Question 3. BaO2 is a peroxide but MnO2 is not a peroxide explain.
Answer:

There is peroxide linkage in the Ba02 molecule ( —O —O—). But there is no such bonding present in MnO, molecule (O —Mn=0). Thus Mn02 is not a peroxide.

 

Class 11 Biology Human Growth Development Multiple Choice Questions

Biology Human Growth Development Multiple Choice Questions

Question 1. Which breaks the dormancy of potato tuber?

  1. Gibbercllin
  2. IAA
  3. ABA
  4. Zcalin.

Answer: 1. Gibberellin

mcq of biology

Question 2. The hormone responsible for senescence is:

  1. ABA
  2. Auxin
  3. GA
  4. Cytokinin.

Answer: 1. ABA

Question 3. Which of the following prevents the fall of fruits?

  1. GA3
  2. NAA
  3. Ethylene
  4. Zcatin.

Answer: 2. NAA

Question 4. The plants respond to photoperiods due to (the presence of:

  1. Stomata
  2. Phytochromes
  3. Phytohormones
  4. Enzymes.

Answer: 2. Phytochromes

multiple choice biology questions

Question 5. Which of the following types of phytohormones resemble nucleic acids in some structural aspects?

  1. Cytokinins
  2. Auxins
  3. Gibberellins
  4. Abscisic acid.

Answer: 2. Auxins

Question 6. Ripening of fruits can be accelerated by :

  1. Heating
  2. Ethylene
  3. Nitrogen
  4. IAA.

Answer: 2. Ethylene

Class 11 Biology Human Growth Development Multiple Choice Questions

 

biology multiple choice questions

Question 7. Coconut milk factor is:

  1. abscisic acid
  2. cytokinins
  3. an auxin
  4. a gibberellin.

Answer: 2. cytokinins

Question 8. In arithmetic growth, following mitotic cell division, only one daughter cell continues to divide while others differentiate by plotting the length of the organ against time, a linear curve is obtained. It is expressed as :

  1. Lt = rt – LO
  2. Lt = tR + LO
  3. LO = Lt + rt
  4. Lt = LO + rt.

Answer: 4. Lt = LO + rt.

Question 9. The stress-related hormone abscisic acid also :

  1. aids in the development of fruit
  2. helps in the closure of stomata
  3. stimulates cell division
  4. accelerates fruit ripening.

Answer: 2. helps in the closure of stomata

Question 10. Apical dominance is due to a high concentration of:

  1. auxin in terminal bud
  2. gibberellin in the apical bud
  3. auxin in lateral bud
  4. auxin and gibberellin in the lateral bud.

Answer: 1. auxin in terminal bud

Question 11. Regulation of flowering by length of day and night is :

  1. phototropism
  2. photoperiodism
  3. nyctinasty
  4. none of the above.

Answer: 2. photoperiodism

Question 12. Pruning of plants promotes branching due to sensitization of axillary buds by :

  1. Ethylene
  2. Cytokinin
  3. Gibberellin
  4. IAA.

Answer: 4. IAA.

Question 13. One set of plants was grown at 12 hours day and 12 hours night period cycles and it flowered while in other set night phase was interrupted by a flash of light and did not produce flowers. Under which one of the following categories will you place this plant?

  1. Long day plant
  2. Day-neutral plant
  3. Short day plant
  4.  Darkness neutral plant.

Answer: 3. Short day plant

Question 14. Cell elongation in internodal regions of green plants takes place due to :

  1. Cylokinm
  2. Gibberellin
  3. Indolcacctic acid
  4. Ethylene,

Answer: 2. Gibberellin

Question 15. Senescence is inhibited by :

  1. Ethylene
  2. Gibberellic acid
  3. Abscisic acid
  4. Cytokinin.

Answer: 4. Cytokinin

Question 16. To lest any chemical substance by using the living system is called

  1. Grafting
  2. Cloning
  3. Biopsy
  4. Bioassay.

Answer: 4. Bioassay.

Question 17. Which hormone is the maximum in coconut milk?

  1. Gibberellin
  2. Ethylene
  3. Cytokinin
  4. Auxin.

Answer: 3. Cytokinin

Question 18. Gibberellins can promote seed germination because they influence

  1. the rate of cell division
  2. production by hydrolyzing enzyme
  3. synthesis. abscisic acid
  4. absorption of water through the hard seed coat.

Answer: 2. production of hydrolyzing enzymes

Question 19. Avena curvature lest is a bioassay for examining the activity of:

  1. auxins
  2. gibberellins
  3. cytokinins
  4. ethylene,

Answer: 1. auxins

Question 20. An enzyme that can stimulate the germination of barley seeds is:

  1. Inverlasc
  2. Protease
  3. a-amylase
  4. Lipase.

Answer: 3. a-amylase

Question 21. In which of the following fruits is the edible part of the aril?

  1. Orange
  2. Litchi
  3. Pomegranate
  4. Custard apple.

Answer: 2. Litchi

Question 22. Treatment of seed at low temperature under moist conditions to break its dormancy is called :

  1. Chelation
  2. Stratification
  3. Scarification
  4. Vernalization.

Answer: 4. Vernalization

Question 23. Parthenocarpic tomato fruits can be produced by :

  1. raising the plants from vernalized seeds
  2. treating the plants with phenylmercuric acetate
  3. removing androecium of flowers before pollen grains are released
  4. treating the plants with low concentrations of gib-acrylic acid and auxins.

Answer: 4. Treat the plants with low concentrations of gib- bicyclic acid and auxins.

Question 24. How does pruning help in making the hedge dense?

  1. The apical shoot grows faster after pruning
  2. It releases wound hormones
  3. It induces the differentiation of new shoots from the rootstock
  4. It frees axillary buds from apical dominance

Answer: 4. It frees axillary buds from apical dominance

Question 25. Which one of the following pairs, is not correctly matched?

  1. Gibberellic Acid-Leaf fall
  2. Cytokinin-Cell division
  3. IAA-Ccll wall elongation
  4. Abscisic Acid-Stomatal closure.

Answer: 1. Gibberellic Acid-Leaf fall

Question 26. The wavelength of light absorbed by the Pr form of phytochrome is:

  1. 680 nm
  2. 720 nm
  3. 620 nm
  4. 640 nm

Answer: 1. 680 nm

Question 27. “Foolish Seedling” disease of rice led to the discovery of:

  1. ABA
  2. 2, 4 – D
  3. IAA
  4. GA.

Answer: 4. GA.

Question 28. One hormone helps in the ripening of fruits while the other stimulates the closure of stomata. These are respectively :

  1. abscisic acid and auxin
  2. ethylene and abscisic acid
  3. abscisic acid and ethylene
  4. ethylene and gibberellic acid
  5. gibberellic acid and abscisic acid.

Answer: 2. ethylene and abscisic acid

Question 29. Pick out the correct statements :

A. Cytokinins especially help in delaying senescence

B. Auxins are involved in regulating apical dominance

C. Ethylene is especially useful in enhancing seed germination

D. Gibberellins are responsible for the immature falling of leaves.

  1. (A) and (C) only
  2. (A) and (D) only
  3. (B) and (C) only
  4. (A) and (B) only
  5. (B) and (D) Only.

Answer: 4. (B) and (D) Only.

Question  30. Match List I and List II and select the correct option.

Growth And Development Match The Correct Answer

  1. (A)-3 (B)-1,  (C)-5,  (D)-4,  (E)-2
  2. (A)-4 (B)-5,  (C)-1  (D)-3,  (E)-2
  3. (A)-2. (B)- 1, (C)-5,  (D)-3.  (E)-4
  4. (A)-3, (B)- l  (C)-5.  (D)-2,  (E)-4
  5. (A)-4. (B)- 1,  (C)-5  (D)- 3,  (E)-2.

Answer: 1. A)-3 (B)-1,  (C)-5,  (D)-4,  (E)-2

Question 31. Senescence is an active developmental cellular process in the growth and functioning of a flowering plant, is indicated in:

  1. annual plants
  2. floral parts
  3. vessels and tracheid differentiation
  4. leaf abscission.

Answer: 2. floral parts

Question 32. The importance of day length in the flowering of plants was first shown in

  1. cotton
  2. Petunia
  3. Lemna
  4. tobacco.

Answer: 4. tobacco.

Question 33. One hormone is used to speed up the malting process in barley, another is used to promote flowering in pineapple, and the third helps in the delay of leaf senescence. These are respectively.

  1. auxin, gibberellin, and cytokinin
  2. gibberellin, cytokinin, and auxin.
  3. gibberellin, auxin, and cytokinin
  4. cytokinin, auxin, and gibberellin
  5.  auxin, cytokinin, and gibberellin.

Answer: 3. gibberellin, auxin, and cytokinin

Human Health And Disease MCQ With Answers For Class 12 Biology

Question 1. Growth of the population of an area depends upon:

  1. Natality rate
  2. Immigration
  3. Environmental resistance
  4. All of these.

Answer: 4. All of these.

Question 2. In India, there is a decline in the female population as compared to males. This is because of:

  1. Female infanticide
  2. Fewer females reach reproductive age
  3. The number of female children born is less
  4. All the above.

Answer: 1. Female infanticide

Question 3. Woman plays a greater role in procreation than man (women are biologically superior to men) because they:

  1. Nurses the foetus in her womb
  2. Feeds the infant on her milk
  3. Cleans infant’s excreta
  4. Does all the above.

Answer: 4. Does all the above?

Question 4. Which of the following causes the storage of fat in the liver?

  1. Amphetamines
  2. Barbiturates
  3. All synthetic drugs
  4. Alcohol.

Answer: 4. Alcohol.

Question 5. Depression in the activities of central nervous system may be caused by:

  1. Narcotics
  2. Stimulants
  3. Both narcotics as well as stimulants
  4. Neither narcotics nor stimulants.

Answer: 1. Narcotics

Question 6. Which of the following is not a derivative of opium?

  1. Morphine
  2. Heroin
  3. Cocaine
  4. Codein.

Answer: 3. Cocaine

Human Health And Disease MCQ With Answers For Class 12 Biology

Question 7. Which of the following is not obtained from Cannabis indica?

  1. Marijuana
  2. Bhang
  3. Charas
  4. Ganja.

Answer: 1. Marijuana

Question 8. The drug extracted from Argot fungus is:

  1. Pethidine
  2. Methadone
  3. L.S.D.
  4. Opium.

Answer: 3. L.S.D.

Question 9. Which of the following drugs is responsible for the production of an abnormal foetus?

  1. L.S.D.
  2. Cocaine
  3. Heroin
  4. Alcohol.

Answer: 1. L.S.D.

Question 10. Match the words in Column A with those in Column B

Drugs And Drug Dependence Mental Health Growth Of Human Populations Match The Column

  1. (1-E), (2-B), (3-A) and (4-C)
  2. (1-D), (2-A), (3-C), and (4-E)
  3. (1-D), (2-A), (3-E) and (4-B)
  4. (1-D), (2-E), (3-A) and (4-B).

Answer: 2. (1-D), (2-A), (3-C), and (4-E)

Question 11. Consumption of alcohol causes:

  1. Dilation of blood vessels
  2. Constriction of blood vessels
  3. Both of the above
  4. None of the above.

Answer: 1. Dilation of blood vessels

Question 12. LSD is obtained from:

  1. Papaver somniferum
  2. Cannabis sativa
  3. Claviceps
  4. Solanum nigrum.

Answer: 3. Claviceps

Question 13. Morphine is obtained from:

  1. Papaver somniferum
  2. Cannabis sativa
  3. Claviceps
  4. None of the above.

Answer: 1. Claviceps

Question 14. Barbiturates and valium area

  1. Stimulants
  2. Sedatives
  3. Hallucinogens
  4. Carcinogens.

Answer: 2. Sedatives

Question 15. Which one of the following is the most harmful derivative of opium?

  1. Codein
  2. Pethidine
  3. Methadone
  4. Heroin.

Answer: 4. Heroin.

Question 16. Tobacco smoke contains

  1. CO2, hydrocarbon and tar
  2. CO2 and tar
  3. CO2 polycyclic aromatic hydrocarbon and tar.
  4. CO2, CO and tar.

Answer: 3. CO2, polycyclic aromatic hydrocarbon and tar.

Question 17. Neuritis is an inflammation of:

  1. Only axon
  2. Entire neuron
  3. Only cyton
  4. Only dendrites.

Answer: 1. Only axon

Question 18. Mental health is influenced by:

  1. Biological factors
  2. Social factors
  3. Both (1) and (2)
  4. Economic factors.

Answer: 3. Both (1) and (2)

Question 19. Barbiturates are extensively used as:

  1. Antiseptics
  2. Stimulants
  3. Disinfectants
  4. Sedatives.

Answer: 4. Sedatives.

Question 20. The most important organ damaged by alcohol is:

  1. Heart
  2. Brain
  3. Lung
  4. Liver.

Answer: 4. Liver.

Question 21. A severe type of mental illness in which patients lose touch with reality is known as:

  1. Psychosis
  2. Neurosis
  3. Epilepsy
  4. All of the above.

Answer: 1. Psychosis

Question 22. Bhang, charas, marijuana and hashish are obtained from:

  1. Nicotiana plant
  2. Papaver somniferum
  3. Cannabis sativa
  4. None of the above.

Answer: 3. Cannabis sativa

Question 23. The continuous and excessive intake of alcohol causes:

  1. Fatty liver syndrome
  2. Cirrhosis
  3. Hypertension
  4. All of the above.

Answer: 4. All of the above.

Question 24. Which of the following diseases are included under the global immunisation programme launched by WHO?

  1. Diphtheria, whooping cough, tetanus, polio, tuberculosis and measles
  2. Diphtheria, pertussis, tetanus, measles, polio and mumps
  3. Pertussis, tuberculosis, tetanus, mumps, malaria and measles
  4. Diphtheria, tetanus, tuberculosis, smallpox and measles.

Answer: 1. Diphtheria, whooping cough, tetanus, polio, tuberculosis and measles

Question 25. Which of the following compounds can alter a person’s thoughts, feelings and perceptions?

  1. LSD
  2. Mescalin and psilocybin
  3. ganja
  4. All of the above

Answer: 4. All of the above

Question 26. Seizure is a condition of:

  1. Mood fluctuation
  2. Epileptic attack
  3. Trembling caused by mental sickness
  4. Tremor of limbs in Parkinson’s disease.

Answer: 2. Epileptic attack

Question 27. Paralysis agitans was discovered by:

  1. Parkinson
  2. Huntington
  3. Waller
  4. Einthoven.

Answer: 1. Parkinson

Question 28. ECTis:

  1. Electro-cardio-traction
  2. Electro convulsive therapy
  3. Emergency cardiac treatment
  4. Effects of chewing tobacco.

Answer: 2. Electroconvulsive therapy

Question 29. Mental diseases can be treated by:

  1. Shock treatment
  2. Psychotherapy
  3. Social therapy
  4. All of the above.

Answer: 4. All of the above.

Question 30. The universal immunisation programme was launched in:

  1. 1947
  2. 1990
  3. 1985
  4. 1975.

Answer: 3. 1985

Question 31. Global Immunisation Programme was started in:

  1. May, 1974
  2. May, 1984
  3. August, 1985
  4. May, 1963.

Answer: 1. May, 1974

Question 32. Project Pulse Polio is concerned with

  1. Eliminating polio by 2000
  2. Widespread attack of polio in certain pockets
  3. Segregation of polio-prone areas
  4. Repeated vaccination for polio.

Answer: 1. Eliminating polio by 2000

Question 33. Any drug which is taken intravenously is called:

  1. 1-drug
  2. 2-drug
  3. 3-drug
  4. 4-drug.

Answer: 4. 4-drug.

Question 34. Select the vitamin deficiency reported in an alcoholic:

  1. A
  2. D
  3. E
  4. K.

Answer: 1. A

Question 35. Ciliotoxin present in tobacco smoke is:

  1. Acrolein
  2. Formaldehyde and acetaldehyde
  3. Nitrogen oxides
  4. All the above.

Answer: 4. All the above.

Question 36. The most serious form of mental illness is:

  1. Neurosis
  2. Psychosis
  3. Epilepsy
  4. Cancer.

Answer: 2. Psychosis

Question 37. Nicotine derived from tobacco plants is:

  1. Steroid
  2. Glycoproteins
  3. Lipoproteins
  4. Alkaloid.

Answer: 4. Alkaloid.

Question 38. Any chemical which causes loss of sensation is:

  1. Sedative
  2. Analgesic
  3. Anaesthetic
  4. Stimulant.

Answer: 3. Anaesthetic

Question 39. Tobacco smoke causes inflammation of lung alveoli and causes:

  1. Bronchitis
  2. Lung cancer
  3. Pulmonary TB
  4. Emphysema.

Answer: 4. Emphysema

Question 40. In an alcoholic, urine is generally:

  1. Hypo-osmotic
  2. Hyper-osmotic
  3. Iso-osmotic
  4. None of these.

Answer: 2. Hyper-osmotic

Question 41. Which of the following chemicals is derived from urea malonic acid and used in inducing sleep?

  1. Diazepam
  2. Benzodiazepine
  3. Barbiturate
  4. Amphetamine.

Answer: 3. Barbiturate

Question 42. Select the correct statement from the ones given below:

  1. Morphine is often given to persons who have undergone surgery as a painkiller
  2. Cocaine is given to patients after surgery as it stimulates recovery)
  3. Chewing tobacco lowers blood pressure and heart rate
  4. Barbiturates when given to criminals make them tell the truth.

Answer: 1. Morphine is often given to persons who have undergone surgery as a painkiller

Class 11 Chemistry Classification of Elements and Periodicity in Properties Question and Answers

Question 1. What will be the name (IUPAC) and symbol if the element with atomic number 119 is discovered? Write its electronic configuration. Also, write the formulas of the stable chloride and oxide of this element.
Answer:

IUPAC name: Ununennium, Symbol: Uue

  • Atomic number ofthe element =119 = 87 + 32
  • It is known that the element with atomic number 87 is francium (Fr). Fr belongs to group-1 in the 7th period of the periodic table.
  • So, the element with atomic number 119 will take its position in group-1 and 8th period just below francium(Fr).
  • The electronic configuration of this element will be [UuojBs1, (where Uuo = Ununoctium, Z = 118). It will be an alkali metal with valency=1
  • If the symbol ofthe element is ‘M’ then the formulas of its stable chloride and oxide will be MCI and M20 respectively.

periodic table question answer 

Question 2. Formulas of oxide and chloride of an element A are A205 and AClg respectively. Which group of Mendeleev’s periodic table will the element belong to? State whether it is a metal or a non-metal.
Answer:

Hiking oxygen as standard, the valency ofthe element A is 5, and taking chlorine as standard, the valency of A is 3. Since oxygen-based valency (8-chlorine-based valency) is the same as that ofthe group number of the element, the element will be of group Mendeleev’s periodic table. Thus, itis a non-metal.

Question 3. A and B are two elements with atomic numbers 9 and 17 respectively. Explain why the element A is a more powerful oxidizing agent than the element B.
Answer:

Given that A and B are two elements with atomic numbers 9 and 17 respectively.

Electronic configuration of gA: ls22s22p5. Electronic configuration of B: ls²2s²2p63s23p5. Both A and B can accept electrons to form the anions of A¯ and B¯ having gas electronic configuration. However, the anions A¯ is more stable than B¯ because of the smaller atomic size of A. So A is a stronger oxidizing agent than B.

Question 4. Elements A, B, and C have atomic numbers (Z- 2), Z, and (Z +1) respectively. Of these, B is an inert gas. Which one of these has the highest electronegativity? Which one of these has the highest ionization potential? What is the formula of the compound formed by the combination of A and C? What is the nature of the bond in this compound?
Answer:

Since element B (atomic no =Z) is an inert gas, the element ‘A’ with atomic number (Z- 2) is included in group VIA. On the other hand, the element C, having an atomic number (Z + 1) must belong to groups (alkali metal). Hence, the electronegativity of the element A is maximum.

The element B, being an inert gas, has the highest value of ionization potential.

The valency of the element A, belonging to group (8- 6) = 2, and that ofthe element C, being an element of group IA, is 1. Therefore, the formula of the compound formed by A and C will be C2A.

A being a strongly electronegative element and C being a strongly electropositive element complete their octet through gain and loss of electrons respectively. So, the nature of the bond formed between C and A in C2A isionic or electrovalent bond.

Question 5. The atomic radius of 10Ne is more than that of gF —why?
Answer:

The atomic radius of 10Ne is more than that of gF

Fluorine forms diatomic molecules, thus the atomic radius of fluorine is a measure of half of the internuclear distance in its molecule (i.e., half of covalent bond length of F2 molecule) but neon being an inert gas, its atoms are incapable of forming covalent bonds by mutual combination amongst themselves.

The only force that comes into play between the atoms is the weak van der Waals force. So a measure of the atomic radius of Ne is equal to its van der Waals radius but the van der Waals radius is always greater than the covalent radius.

Thus, the atomic radius of neon is larger than that of fluorine. Furthermore, due to an increase in the number of electrons in the outermost 2p -orbital of Ne, there occurs an increase in electron-electron repulsion. So 2p-orbital of Ne suffers expansion leading to its increased atomic radius.

Question 6. The first electron affinity ofoxygen is negative but the second electron affinity is positive—explain.
Answer:

The first electron affinity ofoxygen is negative but the second electron affinity is positive

When an electron is added to the valence shell of an isolated gaseous O-atom in its ground state to form a negative ion, energy is released.

Because a neutral oxygen atom tends to complete its octet with electrons. So, the electron affinity of oxygen is an exothermic process and its value is negative.

When an extra electron is added to the O¯ ion, that second electron experiences a force of repulsion exerted by the negative charge ofthe anion. So, first, this process requires a supply of energy from an external source. This accounts for the endothermic nature of second electron affinity and has a positive value.

Question 7. The electron affinity of sodium is negative but magnesium has a positive value—why?
Answer:

The electron affinity of sodium is negative but magnesium has a positive value

Electronic configuration of 11Na: ls22s22p63s1 Electronic configuration of 12Mg: ls22s22p63s2. The addition of one electron to the 3s -orbital of Na leads to a comparatively stable electronic configuration with a fulfilled orbital.

So, this process of the addition of electrons to Na is an exothermic, process. So, the electron affinity of Na is negative. On the other hand, Mg has fulfilled 3s orbital and has a stable electronic configuration. So the addition of an electron to the 3p -orbital destabilizes the electronic configuration of Mg.

Additional energy is required for addition of electrons to the outermost shell of Mg i.e., this process is endothermic and thus the value of electron affinity of Mg is positive.

Question 8. If the electron affinity of chlorine is 350 kJ.moI-1, then what is the amount of energy liberated to convert 1.775 g of chlorine (existing at atomic state) to chloride ions completely (in a gaseous state)
Answer:

Atomic mass of chlorine = 35.5 g .mol 1

The energy liberated in the conversion of 35.5 g of Cl- to Clion =350 kJ

∴ The energy liberated in the conversion of 1.775 g of Cl to Cl-

\(\text { ion }=\frac{350}{35.5} \times 1.775=17.5 \mathrm{~kJ}\)

Question 9. The second ionization enthalpy of Mg is sufficiently high and the second electron-gain enthalpy of O has a positive value. How do you explain the existence of Mg2+02- rather than Mg+O-?
Answer:

The lattice energy of an ionic crystal depends on the force of attraction between the cations and anions \(\left(F \propto \frac{q_1 q_2}{r^2}\right)\)

So, the magnitude of lattice energy increases as the charges on the cation and anion increase. Consequently, the lattice energy of Mg²+O² is very much greater than that of Mg+O¯.

The lattice energy of Mg²+O²¯ is so high that it exceeds the unfavorable effects of the second ionization enthalpy of Mg and the second electron-gain enthalpy of O.

So, Mg²+O²¯ is a stable ionic compound and its formation is favored over Mg+O¯.

class 11 chemistry periodic classification solutions

Question 10. The atomic numbers of some elements are given below. Classify them into three groups so that the two elements in each group exhibit identical chemical behavior: 9, 12, 16, 34, 53, 56.
Answer:

Elements with atomic numbers 9 and 53 belong to the -block and they have similar outer electronic configurations (ns2np5). So they will exhibit similar chemical properties. Their group number = (10 + 2 + 5) = 17. Elements with atomic numbers 12 and 56 belong to s -block and theyhave similar outer electronic configurations (ns2).

class 11 chemistry chapter 3 exercise

So they will exhibit similar chemical properties. Their group number = 2. Elements with atomic numbers 16 and 34 belong to the -block and they have similar outer electronic configurations (ns2npA).

So, they will exhibit similar chemical properties. Their group number = (10 + 2 + 4) = 16. So, based on the similarity in chemical properties the given elements are divided into three groups :

Important Questions for Class 11 Physics For Work and Energy

Question 1. An electron and a proton are detected In a cosmic ray experiment. Both of them have a kinetic 100 keV. Which one is faster?
Answer:

Given

An electron and a proton are detected In a cosmic ray experiment. Both of them have a kinetic 100 keV.

Kinetic energy, K = \(\frac{1}{2} m v^2 \text {, i.e., } v=\sqrt{\frac{2 K}{m}}\)

For the same kinetic energy, \(v \propto \frac{1}{\sqrt{m}}.\)

The mass of an electron is less than that of a proton.

So, the electron has a higher velocity, i.e., the electron is faster.

Question 2. Is air resistance a conservative force?
Answer:

Air resistance is not a conservative force

To displace any object against air resistance, work has to be done. This work can never be restored, because to return the object to its initial state, further work has to be done to overcome the air resistance again. Hence air resistance is not a conservative force.

Question 3. A box Is lifted vertically by 6 m in 3 s.

  1. If the box is lifted to the same height in a zig-zag way, or
  2. If the box is lifted in 5 s instead of in 3s, what would be the change in the work done?

Answer:

  1. Gravitational force is a conservative force. Hence, work done to lift a body to the same height will be the same, whatever may the nature of the path be.
  2. Work done is independent of time. Hence, the same work is done in lifting the box in 5 s instead of in 3 s.

Question 4. A hydrogen gas-filled balloon of mass m can rise to a maximum height of h above the earth’s surface. On the earth’s surface, the potential energy of the balloon = kinetic energy = 0; at the height h, the potential energy and kinetic energy of the balloon are mgh and 0 respectively. Is the law of conservation of energy violated here?
Answer:

Given

A hydrogen gas-filled balloon of mass m can rise to a maximum height of h above the earth’s surface. On the earth’s surface, the potential energy of the balloon = kinetic energy = 0; at the height h, the potential energy and kinetic energy of the balloon are mgh and 0 respectively.

The potential energy of an object of mass m raised the inclination. to a height h, is mgh. Here, mg = weight of the object = net downward force. But for a balloon of mass m, the net downward force is not mg.

When it is stationary at a height of h, the upward buoyancy balances the weight; the net downward force is zero. So the potential energy of the balloon at that height = 0, not mgh. This means that the total mechanical energy is zero, on the ground as well as at the height h. Thus, the law of conservation of energy is not violated.

Question 5. What is the percentage increase in the momentum of a body when its kinetic energy increases by 69%?
Answer:

Kinetic energy, K = \(\frac{p^2}{2m}\), where p = momentum.

∴ \(\frac{K_1}{K_2}=\frac{p_1^2}{p_2^2} \quad \text { or, } \frac{p_2}{p_1}=\sqrt{\frac{K_2}{K_1}}=\sqrt{\frac{100+69}{100}}=\frac{13}{10}=\frac{130}{100}\)

∴ Increase in momentum = 30%.

Question 6. A particle of mass m, moving with a constant acceleration, acquires a velocity v0 in time t0. Initially, the particle was at rest. Find the average power and the instantaneous power of the applied force.
Answer:

Given

A particle of mass m, moving with a constant acceleration, acquires a velocity v0 in time t0. Initially, the particle was at rest.

Work done on the particle in time t0 (W) = kinetic energy gained in \(t_0 s=\frac{1}{2} m v_0^2\)

Hence, average power, \(\frac{\frac{1}{2} m v_0^2}{t_0}=\frac{m v_0^2}{2 t_0}\)

If the velocity after a time t is v, then instantaneous power at a time t,

P = \(\frac{d W}{d t}=\frac{d}{d t}\left(\frac{1}{2} m v^2\right)=m v \frac{d v}{d t}=\frac{m v^2}{t}\)

(as for a constant acceleration \(\frac{d v}{d t}\) = \(\frac{v}{t}\))

Question 7. Is it possible to increase the kinetic energy of a body without applying an external force?
Answer:

The potential energy stored in a body can be transformed into its kinetic energy. Hence, it is possible to increase the kinetic energy of a body without applying an external force.

Question 8. The height of the vertex of an inclined plane is h. From the vertex, a body is released along the inclined plane. Explain whether the kinetic energy of the body at the base of the inclined plane depends on the inclination.
Answer:

Given

The height of the vertex of an inclined plane is h. From the vertex, a body is released along the inclined plane.

Let m = mass of the body.

Hence, the kinetic energy of the body at the base of the inclined plane = its potential energy at the vertex of the inclined plane = mgh.

Hence, the kinetic energy at the base depends on the height of the vertex h of the inclined plane, but not on the inclination.

Question 9. A body tied to a thread is made to revolve in a horizontal plane with a definite velocity and the thread does not snap. But when the body Is made to revolve in a vertical plane, the thread snaps. How is It possible?
Answer:

Given

A body tied to a thread is made to revolve in a horizontal plane with a definite velocity and the thread does not snap. But when the body Is made to revolve in a vertical plane, the thread snaps

Let the length of the thread be l, the mass of the body m, the velocity of revolution v, and tension in the thread T when the body is rotated in the horizontal plane.

Work And Energy A Body Tied To A Thread Is Made To Revolve In A Horizontal Plane With A Define Velocity

Required centripetal force for revolution, \(\frac{m v^2}{l \sin \theta}=T \sin \theta\)…(1)

and \(m g=T \cos \theta\)….(2)

From equations (1) and (2) we get, \(\sin ^2 \theta+\cos ^2 \theta=\frac{m v^2}{l T}+\left(\frac{m g}{T}\right)^2\)

or, \(T^2=\frac{m v^2}{l} T+m^2 g^2\)

or, \(T^2-\frac{m v^2}{l} T-m^2 g^2=0\)

∴ T = \(\frac{\frac{m v^2}{l} \pm \sqrt{\left(\frac{m v^2}{l}\right)^2+4 m^2 g^2}}{2}\)

∴ T>0 and \(\frac{m v^2}{l}<\sqrt{\left(\frac{m v^2}{l}\right)^2+4 m^2 g^2}\),

T = \(\frac{1}{2}\left[\frac{m v^2}{l}+\sqrt{\left(\frac{m v^2}{l}\right)^2+4 m^2 g^2}\right]\)

or, \(T=\frac{1}{2}\left[\frac{m v^2}{l}+\sqrt{\left(\frac{m v^2}{l}+2 m g\right)^2-2 \cdot \frac{m v^2}{l} \cdot 2 m g}\right]\)…(3)

Now, \(\frac{m v^2}{l}+2 m g>\sqrt{\left(\frac{m v^2}{l}+2 m g\right)^2-2 \cdot \frac{m v^2}{l} \cdot 2 m g}\)

∴ From equation (3), it can be inferred that

T< \(\frac{1}{2}\left[\frac{m v^2}{l}+\left(\frac{m v^2}{l}+2 m g\right)\right]\)

or, \(T<\left(\frac{m v^2}{l}+m g\right)\)….(4)

Again, when the body is revolved in the vertical plane, tension in the thread at the lowest point of the circular path,

T’ = \(\frac{m v^2}{l}+m g\)…(5)

From equations (4) and (5) we get, T < T’

So, the tension in the thread at the lowest point in the second case is greater than that in the first case, and hence, in the first case though the thread does not snap, in the second case the thread may snap.

Question 10. Two bodies of different masses have the same momentum. Compare their kinetic energies.
Answer:

Given

Two bodies of different masses have the same momentum.

Let K1, P1, and K2, p2 be the kinetic energies and momenta of the two bodies of masses m1 and m2, respectively.

Now, \(K=\frac{1}{2} m v^2=\frac{m^2 v^2}{2 m}=\frac{p^2}{2 m}\)

∴ \(K_1=\frac{p_1^2}{2 m_1}, K_2=\frac{p_2^2}{2 m_2}\)

∴ \(\frac{K_1}{K_2}=\frac{m_2 p_1^2}{m_1 p_2^2}=\frac{m_2}{m_1} \quad \text { as } p_1=p_2\)

Hence, the body with the lighter mass has greater kinetic energy, i.e., the kinetic energy is inversely proportional to the mass, if the momentum remains constant.

Question 11. An unbalanced system of forces can produce acceleration as well as deformation in a body, but ‘a balanced system of forces produces deformation only.
Answer:

Given

An unbalanced system of forces can produce acceleration as well as deformation in a body, but ‘a balanced system of forces produces deformation only.

In case of an unbalanced system of forces acting on a body, a resultant force acts at the center of mass. The body accelerates, as per the equation, F = ma. But for a balanced system of forces, the resultant force is zero and hence the acceleration of the centre of mass is zero.

However, in both situations, different parts of the body may have relative velocities with respect to one another. Hence, deformation may take place.

For example, if a rubber rope is pulled on both ends with equal forces, the rope extends in length, i.e., it is deformed, but its center of mass remains steady at one place. When the forces applied at the ends are unequal, the rope not only extends but also moves in the direction of the greater force.

Question 12. What are the characteristics of an elastic collision between two bodies?
Answer:

The characteristics of an elastic collision between two bodies

In an elastic collision,

  1. Total kinetic energy is conserved and
  2. Total momentum is conserved.

Question 13. Two protons are drawn near each other. Will the potential energy of the system increase or decrease? What will happen if a proton and an electron are drawn near each other?
Answer:

Given

Two protons are drawn near each other.

Since the protons are positively charged particles, they will repel each other. To bring them nearer, work has to be done against the force of repulsion and this will be stored up in the system as potential energy. So the potential energy of the system will increase. The potential energy decreases when a proton and an electron are brought nearer.

Question 14. What is the condition for the collision of two bodies to be one-dimensional?
Answer:

The speeds of the centers of mass of the two particles before the collision must be along the same straight line.

Question 15. A shot is fired from a gun. What will be the changes in momentum and kinetic energy?
Answer:

Given

A shot is fired from a gun.

The momentum of the system will be conserved. However the kinetic energy will increase because the chemical energy of the explosives used in firing will be converted into kinetic energy.

Question 16. A molecule in a gas container hits a horizontal wall with a speed of 200 m · s-1 at an angle of 30° with the normal and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
Answer:

Given

A molecule in a gas container hits a horizontal wall with a speed of 200 m · s-1 at an angle of 30° with the normal and rebounds with the same speed.

As there is no external force, linear momentum is conserved. The collision is elastic because the magnitude of the molecular velocity remains the same.