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gulaghasi [49]
2 years ago
13

Which of the following is the correct rank order of the diffusion rates of either Cl2 or H2S through air at a pressure of 1.0 at

m?
Chemistry
1 answer:
Radda [10]2 years ago
3 0

Answer:

Cl2 at 250 K < Cl2 at 300 K < H2S at 250 K < H2S at 300 K.

Explanation:

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If you dissolve 8.50 g of ca(no3)2 in 755 ml of distilled water what is the concentration in molarity of the calcium nitrate sol
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Molarity is defined as the number of moles of solute in 1 L of solution 
the mass of Ca(NO₃)₂ present - 8.50 g
therefore number of moles of Ca(NO₃)₂ - 8.50 g / 164 g/mol = 0.0518 mol
the volume of solution prepared is 755 mL 
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For the reaction C2H4(g) + H2O(g) --&gt; CH3CH2OH(g)
Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

Explanation :

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = -45.6 kJ = -45600 J

\Delta S^o = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)

\Delta G^o=-12666.6J=-12.7kJ

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln k

where,

\Delta G^o = standard Gibbs free energy  = -12666.6 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k

k=3.35\times 10^{2}

Therefore, the value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

3 0
3 years ago
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