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AlexFokin [52]
3 years ago
12

When 167.2 joules of heat is added to 4.00 grams of water at 10c the resulting temperature is what

Chemistry
1 answer:
Katena32 [7]3 years ago
5 0
Q=mc \Delta T \\\\
Q=168,2J\\
m=4g\\
c=4.184\frac{J}{g^{o}C}\\\\
\Delta T=\frac{Q}{mc}=\frac{167,2J}{4g*4,184\frac{J}{g^{o}C}}=9,99^{o}C\\\\
\Delta T = T_{f}-T_{i}\\\\
9,99^{o}C=T_{f}-10^{o}C\\\\
T_{f}=9,99^{o}C+10^{o}C=19,99^{o}C\approx20^{o}C
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Answer:

n_{H_2}^{equilibrium}=0.393mol

Explanation:

Hello,

In this case, given the amounts of water and carbon dioxide we should invert the given reaction as hydrogen will be producted rather than consumed:

H_2O(g) + CO(g)\rightleftharpoons H_2(g) + CO_2(g)

Consequently, the equilibrium constant is also inverted:

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In such a way, we can now propose the law of mass action:

Kc'=\frac{[H_2][CO_2]}{[H_2O][CO]}

And we can express it in terms of the initial concentrations of the reactants and the change x due to the reaction extent:

Kc'=\frac{(x)(x)}{([H_2O]_0-x)([CO]_0-x)}=1.87

Thus, we compute the initial concentration which are same, since equal amount of moles are given:

[H_2O]_0=[CO]_0=\frac{0.680mol}{70.0L}=0.0097M

Hence, solving for x by using the quardratic equation or solver, we obtain:

x_1=0.00561M\\x_2=0.0361M

For which the correct value is 0.00561M since the other one will produce negative concentrations of water and carbon monoxide at equilibrium. Therefore, the number of moles of hydrogen at equilibrium for the same 70.0-L container turn out:

n_{H_2}^{equilibrium}=0.00561mol/L*70.0L=0.393mol

Best regards.

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