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gavmur [86]
3 years ago
7

A sample of argon has a volume of 6.0 cm 3 and the pressure is 0.87 atm. If the final temperature is 35 degrees celsius, the fin

al volume is 10 cm 3 and the final pressure is 0.52 atm, what was the initial temperature of the argon
Chemistry
1 answer:
____ [38]3 years ago
3 0

Answer:

= 36.185 °C

Explanation:

Using the combined gas law;

P1V1/T1 = P2V2/T2

In this case;

P1 = 0.87 atm

V1 = 6.0 cm³

T1 = ?

P2 = 0.52 atm

V2 = 10 cm³

T2 = 35 °C + 273 = 308 K

Therefore;

T1 = P1V1T2/P2V2

   = ( 0.87 × 6.0 × 308)/( 0.52 ×10)

   = 309.185 K

Therefore; Initial temperature = 309.185 K - 273 = 36.185 °C

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Answer:

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Explanation:

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

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

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

\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]

We are given:

\Delta H_f_{(N_2O)}=81.6 kJ/mol\\\Delta H_f_{(H_2O)}=-241.8 kJ/mol\\\Delta H_f_{(N_2H_4)}= 50.6 kJ/mol\\\Delta H_f_{(N_2O_4)}=9.16 kJ/mo

Putting values in above equation, we get:

\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ

Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.

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Answer : The heat your body transfer must be, 25.1 kJ

Explanation :

Formula used :

Q=m\times c\times \Delta T

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Therefore, the heat your body transfer must be, 25.1 kJ

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