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2 years ago

How many molecules are there in an ideal gas with a volume of 7.27 L at STP?

Chemistry

1 answer:

Marizza181 [45]2 years ago

8 0

Answer:

n = 0.324 mol

Explanation:

PV = nRT

STP:

273.15K and 101.325kPa

n = PV/RT

n = (101.325kPa)(7.27L)/(8.314 L kPa/mol K)(273.15K)

n = 0.3243693408 mol

n = 0.324 mol

Hope that helps

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Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

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<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

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Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

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

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

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$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

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Moles of CO gas reacted = 2.25 moles

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When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

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