Question Bank Directory

Identify the wrong statement with reference to the gene 'I' that controls ABO blood groups.

For the reaction: $X_2O_4(l) \rightarrow 2XO_2(g)$ with the given values $\Delta U = 2.1 \text{ kcal}$ and $\Delta S = 20 \text{ cal K}^{-1}$ at $300 \text{ K}$, what is the value of $\Delta G$?

$2\text{Zn} + \text{O}_2 \rightarrow 2\text{ZnO}; \Delta G^\circ = -616\text{ J}$ $2\text{Zn} + \text{S}_2 \rightarrow 2\text{ZnS}; \Delta G^\circ = -293\text{ J}$ $\text{S}_2 + 2\text{O}_2 \rightarrow 2\text{SO}_2; \Delta G^\circ = -408\text{ J}$ $\Delta G^\circ$ for the following reaction is: $2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2$

Reversible expansion of an ideal gas under isothermal and adiabatic conditions are shown in the figure: $\text{AB} \rightarrow \text{Isothermal expansion}$, $\text{AC} \rightarrow \text{Adiabatic expansion}$. Which of the following options is not correct?

Match the following columns and select the correct option. Column-I: (a) Gregarious, polyphagous pest (b) Adult with radial symmetry and larva with bilateral symmetry (c) Book lungs (d) Bioluminescence. Column-II: (i) Asterias (ii) Scorpion (iii) Ctenoplana (iv) Locusta

Identify the incorrect statement.

The following two reactions are known: $\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g); \Delta H = -26.88\text{ kJ}$ $\text{FeO}(s) + \text{CO}(g) \rightarrow \text{Fe}(s) + \text{CO}_2(g); \Delta H = -16.5\text{ kJ}$ The value of $\Delta H$ for the following reaction: $\text{Fe}_2\text{O}_3(s) + \text{CO}(g) \rightarrow 2\text{FeO}(s) + \text{CO}_2(g)$ is:

Which of the following options correctly describes the free expansion of an ideal gas under adiabatic conditions?

<p class="p1" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(20, 20, 19); white-space: normal;">An em wave is propagating in a medium with</p><p class="p2" style="font-width: normal; font-size: 7px; line-height: normal; font-family: Symbol; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(0, 0, 0); white-space: normal;">$\rightarrow$</p><p class="p1" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(20, 20, 19); white-space: normal;">a velocity&nbsp;<span class="s1" style="font-width: normal; line-height: normal; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(0, 0, 0);">$\hat{v}$</span></p><p class="p3" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(0, 0, 0); white-space: normal;">$V = V_i$</p><p class="p1" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(20, 20, 19); white-space: normal;"><span class="s2" style="font-width: normal; line-height: normal; font-family: Symbol; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(0, 0, 0);">$=$</span>. The instantaneous</p><p class="p1" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(20, 20, 19); white-space: normal;">oscillating electric field of this em wave is</p><p class="p1" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(20, 20, 19); white-space: normal;">along $+y$ axis. Then the direction of oscillating</p><p class="p1" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(20, 20, 19); white-space: normal;">magnetic field of the em wave will be along</p>

If the bond energies of $\text{H}-\text{H}$, $\text{Br}-\text{Br}$, and $\text{H}-\text{Br}$ are $433$, $192$, and $364\text{ kJ mol}^{-1}$ respectively, the $\Delta H^\circ$ for the reaction $\text{H}_2(g) + \text{Br}_2(g) \rightarrow 2\text{HBr}(g)$ will be:

If the standard enthalpy of neutralization reaction of HCl and NaOH is −57.3 kJ mol⁻¹, then find out the enthalpy of neutralization of 0.25 mol of HCl by 0.25 mol of NaOH:

Hydrolysis of sucrose is given by the following reaction: $\text{Sucrose} + \text{H}_2\text{O} \rightleftharpoons \text{Glucose} + \text{Fructose}$. If the equilibrium constant ($K_c$) is $2 \times 10^{13}$ at $300 \text{ K}$, the value of $\Delta_r G^\ominus$ at the same temperature will be:

For the reaction, $X_2O_4(l) \rightarrow 2XO_2(g)$, $\Delta U = 2.1 \text{ kcal}$, $\Delta S = 20 \text{ cal K}^{-1}$ at $300 \text{ K}$. Hence, $\Delta G$ is:

<p class="p1" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(20, 20, 19); white-space: normal;">A block of mass m is placed on a smooth</p><p class="p1" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(20, 20, 19); white-space: normal;">inclined wedge ABC of inclination&nbsp;<span class="s1" style="font-width: normal; line-height: normal; font-family: Symbol; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal;"></span>&nbsp;as shown</p><p class="p1" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(20, 20, 19); white-space: normal;">in the figure. The wedge is given an</p><p class="p1" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(20, 20, 19); white-space: normal;">acceleration 'a' towards the right. The</p><p class="p1" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(20, 20, 19); white-space: normal;">relation between a and&nbsp;<span class="s1" style="font-width: normal; line-height: normal; font-family: Symbol; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal;"></span>&nbsp;for the block to</p><p class="p1" style="font-width: normal; font-size: 10px; line-height: normal; font-family: Helvetica; font-size-adjust: none; font-kerning: auto; font-variant-alternates: normal; font-variant-ligatures: normal; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-position: normal; font-feature-settings: normal; font-optical-sizing: auto; font-variation-settings: normal; color: rgb(20, 20, 19); white-space: normal;">remain stationary on the wedge is</p>

Given the following bond energies: H-H bond energy $= 431.37 \text{ kJ mol}^{-1}$ C=C bond energy $= 606.10 \text{ kJ mol}^{-1}$ C-C bond energy $= 336.49 \text{ kJ mol}^{-1}$ C-H bond energy $= 410.50 \text{ kJ mol}^{-1}$ Based on the data given above, enthalpy change for the reaction $C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)$ will be:

The correct option for free expansion of an ideal gas under adiabatic condition is:

2 mole of an ideal gas at $27^\circ\text{C}$ temp. is expanded reversibly from $2\text{ lit.}$ to $20\text{ lit.}$ Find entropy change ($R = 2\text{ cal/mol K}$):

For vaporization of water at $1$ atmospheric pressure, the values of $\Delta H$ and $\Delta S$ are $40.63\text{ kJ mol}^{-1}$ and $108.8\text{ J K}^{-1}\text{ mol}^{-1}$, respectively. The temperature when Gibbs energy change ($\Delta G$) for this transformation will be zero, is:

The entropy change involved in the conversion of 1 mole of liquid water at 373 K to vapour at the same temperature will be [$\Delta H_{vap} = 2.257\text{ kJ/g}$] [MP PET 2002]

The correct difference between first and second order reactions is that