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lyudmila [28]
4 years ago
5

A fixed amount of gas occupies a volume of 7.25 l at a pressure of 4.52 atm. what will be the volume occupied if the pressure is

decreased to 1.21 atm at constant temperature?
Chemistry
1 answer:
Eddi Din [679]4 years ago
6 0
P1V1=P2V2
SO
7.25X4.52=1.21XV2
V2=27.1 l
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\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]

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\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol

Putting values in above equation, we get:

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Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

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