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notsponge [240]
3 years ago
10

The density of liquid oxygen at its boiling point is 1.14 kg/L , and its heat of vaporization is 213 kJ/kg . How much energy in

joules would be absorbed by 2.0 L of liquid oxygen as it vaporized?
Chemistry
1 answer:
-Dominant- [34]3 years ago
5 0
<span>2.28 kg x 213 kJ/kg = 486 kJ = 4.86E+05 J</span>
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Consider the reaction CH4(g) 2O2(g)CO2(g) 2H2O(g) Using standard thermodynamic data at 298K, calculate the entropy change for th
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the entropy change for the surroundings when 1.62 moles of CH4(g) react at standard conditions is −8.343 J/K

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The balanced chemical equation of the reaction in the question given is:

CH_{4(g)}  + 2O_{2(g)} \to CO_{2(g)} + 2 H_2O _{(g)}

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The entropy of each compound above are listed as follows in a respective order.

Entropy of (CH4(g)) = 186.264 J/mol.K

Entropy of (O2(g)) = 205.138 J/mol.K

Entropy of (CO2(g)) = 213.74 J/mol.K

Entropy of (H2O(g)) = 188.825 J/mol.K

The change in Entropy (S) of the reaction is therefore calculated as follows:

=1*S(CO2(g)) + 2*S(H2O(g)) - 1*S( CH4(g)) - 2*S(O2(g))

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=  -5.15  J/mol.K

Given that :

the number of moles = 1.62 of CH4(g) react at standard conditions.

Then;

The change in entropy of the rxn = 1.62 \ mol * -5.15 \  J/mol.K

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The particular reaction that represents the oxidation of Mg metal would be the following:

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3 years ago
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