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bekas [8.4K]
3 years ago
10

BrainlieSTTTTTT A gas is originally stored at a pressure of 25 atm with a volume of 3 L. If the pressure were increased to 75 at

m, what would be the new volume of the gas?
a.8.333 L
b.1 L
c.75 L
d.1 L
Chemistry
1 answer:
netineya [11]3 years ago
3 0

Answer:

D. is correct.

Explanation:

0+-0&-0=kd

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Answer:

Describe what is happening within the system when it is at equilibrium in terms of concentrations, reactions that occur, and reaction rates.

Explanation:

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Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)
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<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

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\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]

For the given chemical reaction:

C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]

We are given:

\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

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Answer:

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