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3 years ago

If a gas at a pressure of 600 mmHg and a volume of 200 mL is changed to 780 mmHg, then what will be the new volume?

Chemistry

1 answer:

svlad2 [7]3 years ago

8 0

P1V1=P2V2
(600/760)(.200)=(780/760)x
x=.1538L or 153.8mL

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How many joules would be absorbed for the water in Problem 1

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3 years ago

or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und

beks73 [17]

Answer:

The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

$C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)$

$\Delta H^{o}_{rxn}= -5104.1kJ/mol$

The expression for the entropy change for the reaction is as follows.

$\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]$

$\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol$

$\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol$

$\Delta H^{o}_{f}(O_{2})= 0kJ/mol$

Substitute the all values in the entropy change expression.

$-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]$

$-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})$

$\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol$

$=-220.1kJ/mol$

Therefore, The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

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3 years ago