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10 months ago

Chemistry Equilibrium

The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is:

[Assume : No cyano complex is fomed ; k_sp (AgCN) = 2.2 * 10^-16 and K_a (HCN) = 6.2 * 10^-10]

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alok kumar singh

Contributor-Level 10

$\underset{S S}{A g C N_{(S)} ? A {g^{+}}_{(a q)} + C {N^{-}}_{(a q)}}$

$C N^{-} + H^{+} ? H C N$

Before reaction S 10^-3 0

After reaction 0 10^-3 S

$∴ H C N ? H^{+} + C N^{-}$

$S^{2} = \frac{2.2 * 1 0^{- 16}}{6.2 * 1 0^{- 7}} = \frac{2.2}{6.2} * 1 0^{- 9}$

$S = \sqrt[]{\frac{2.2}{6.2} * 1 0^{- 9}} = \sqrt[]{\frac{22}{6.2} * 1 0^{- 10}}$

$= \sqrt[]{3.54 * 1 0^{- 10}} = 1.88 * 1 0^{- 5} = 1.9 * 1 0^{- 5}$

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10 months ago

Chemistry Class 11th

Five moles of an ideal gas at 293K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external pressure 4.3 MPa. The heat transferred in this process is……. kJ mol^-1. (Rounded-off to the nearest integer).

[Use R = 8.314 J mol^-1 K^-1]

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Payal Gupta

Contributor-Level 10

If $Δ E = 0$

Q = W

W = -P_ext $(Δ V) = - 4.3 n R T [\frac{1}{P_{2}} - \frac{1}{P_{1}}]$

= $- 4.3 * 5 * 8.314 * 293 [\frac{1}{1.3} - \frac{1}{2.1}]$

= 15347.70 K = 15.3 kJ

Q = 15 kJ

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10 months ago

Chemistry Class 12th

The unit cell of copper corresponds to a face centered cube of edge length 3.596 $\overset{o}{A}$ with one copper atom at each lattice point. The calculated density of copper in kg/m³ is…….

[Molar mass of Cu = 63.54g : Avogadro Number = 6.022 * 10²³]

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Payal Gupta

Contributor-Level 10

Density = $\frac{M * z}{N_{A} a^{3}}$

$d = \frac{63.5 * 4}{6.02 * 1 0^{23} * {(3.596 * 1 0^{- 10})}^{3} * 1000} = 9076 k g / m^{3}$

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10 months ago

Chemistry Chemistry Chemical Kinetics

The rate constant of a reaction increases by five times on increase in temperature from 27°C to 52°C. The value of activation energy in kJ mol^-1 is………. (Rounded-off to the nearest integer).

[R = 8.314JK^-1mol^-1]

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Payal Gupta

Contributor-Level 10

$l o g (\frac{k_{2}}{k_{1}}) = \frac{E a}{2.303 * 8.314} [\frac{1}{300} - \frac{1}{325}]$

log (5) = $\frac{E a}{2.303 * 8.314} [\frac{1}{300} - \frac{1}{325}]$

$∴ E a = \frac{0.7 * 2.303 * 8.314 * 300 * 325}{25} = 52271.7$

Ea in kJ/mole = $\frac{52271.7}{1000} = 52.2 k J / m o l$

the nearest integer is 52.

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10 months ago

Chemistry Class 12th

The major product of the following chemical reaction is:

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alok kumar singh

Contributor-Level 10

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10 months ago

Chemistry Class 12th

Which one of the following reactions will not form acetaldehyde?

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alok kumar singh

Contributor-Level 10

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10 months ago

Chemistry Chemistry Coordination Compounds

The hybridization and magnetic nature of ${[M n {(C N)}_{6}]}^{4 -} a n d {[F e {(C N)}_{6}]}^{3 -},$ respectively are:

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alok kumar singh

Contributor-Level 10

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Chemistry Class 11th

Copper reduces $N O_{3}^{-}$ into NO and NO₂ depending upon the concentration of HNO₃ in solution. (Assuming fixed $[C u^{2 +}] a n d P_{N O} = P_{N O_{2}}$ ), the HNO₃ concentration at which the thermodynamic tendency for reduction of $N O_{3}^{-} i n t o$ NO and NO₂by copper is same is 10^x M. The value of 2x is……. (rounded-off to the nearest integer).

$[G i v e n, E_{C u^{2 +} / C u}^{0} = 0.34 V, E_{N O_{3}^{-} / N O}^{0} = 0.96 V, E_{N O_{3}^{-} / N O_{2}}^{0} = 0.79 V a n d a t 298 K, \frac{R T}{F} (2.303) = 0.059]$

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Payal Gupta

Contributor-Level 10

$N O_{3}^{-} + 4 H^{+} + 3 e^{-} \to N O + 2 H_{2} O; E_{N O_{3}^{-}}^{0} = 0.96 V$

$L e t [H N O_{3}] = y M$

For same thermodynamic tendency of reduction

$E_{N O_{3}^{-} / N O} = E_{N O_{3}^{-} / N O_{2}}$

$0.96 - \frac{0.059}{3} l o g \frac{p_{N O}}{y^{5}} = 0.79 - 0.059 l o g \frac{p_{N O_{2}}}{y^{3}}$

Assume $(p_{N O} = p_{N O_{2}} = 1 b a r)$

$0.17 = - \frac{0.0591}{1} l o g \frac{p_{N O_{2}}}{y^{3}} + \frac{0.059}{3} l o g \frac{p_{N O}}{y^{5}}$

$= \frac{0.059}{3} [l o g \frac{p_{N O}}{y^{5}} - l o g \frac{{p^{3}}_{N O_{2}}}{y^{9}}] = \frac{0.059}{3} * l o g [\frac{p_{N O}}{y^{5}} * \frac{y^{9}}{{p^{3}}_{N O_{2}}}] = \frac{0.059}{3} l o g y^{4}$