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solong [7]
4 years ago
14

95.8 L of fluorine gas is being held at a temperature of 24.5oC. If the temperature were raised to 46.9 oC, what would the new v

olume be?
Chemistry
1 answer:
Xelga [282]4 years ago
6 0

Answer: 103 litres

Explanation:

Given that,

Original volume of fluorine gas (V1) = 95.8L

Original temperature of fluorine gas (T1) = 24.5°C

[Convert 24.5°C to Kelvin by adding 273

24.5°C + 273 = 297.5K]

New volume of gas (V2) = ?

New temperature of gas (T2) = 46.9°C

[Convert 46.9°C to Kelvin by adding 273

46.9°C + 273 = 319.9K]

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

95.8L/297.5K = V2/319.9K

To get the value of V2, cross multiply

95.8L x 319.9K = 297.5K x V2

30646.42L•K = 297.5K•V2

Divide both sides by 297.5K

30646.42L•K/297.5K = 297.5K•V2/297.5K

103L = V2

Thus, the new volume of fluorine gas will be 103 litres

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Answer: The enthalpy of formation of SO_3 is  -396 kJ/mol

Explanation:

Calculating the enthalpy of formation of SO_3

The chemical equation for the combustion of propane follows:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

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

\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]

We are given:

\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-297kJ/mol\\\Delta H^o_{rxn}=-198kJ

Putting values in above equation, we get:

-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol

The enthalpy of formation of SO_3 is -396 kJ/mol

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