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3 years ago

A balloon has an internal pressure of 1.95 atm and a volume of 5.0 L. If the temperature where the balloon is

Chemistry

1 answer:

patriot [66]3 years ago

5 0

Answer:

13.1 L

Explanation:

Using the combined gas law:

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

Based on the information provided in this question;

P1 = 1.95 atm

P2 = 0.65 atm

V1 = 5.0 L

V2 = ?

T1 = 20 °C = 20 + 273 = 293K

T2 = -15 °C = -15 + 273 = 258K

Using P1V1/T1 = P2V2/T2

1.95 × 5/293 = 0.65 × V2/258

9.75/293 = 0.65V2/258

0.0333 = 0.00252V2

V2 = 0.0333 ÷ 0.00252

V2 = 13.1 L

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nirvana33 [79]

Answer:

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Explanation:

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7 0

3 years ago

A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp

vichka [17]

Answer:

The balanced chemical equation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat capacity of calorimeter = C = 11.66 kJ/°C

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Final temperature of the calorimeter = $T_2= 26.37^oC$

Heat absorbed by calorimeter = Q

$Q=C\times \Delta T$

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

$\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g$

Molar mass of phenol = 94 g/mol

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$\frac{Q'}{\frac{1.800 g}{94 g/mol}}=\frac{-58.4166 kJ\times 94 g/mol}{1.800 g}=3,050 kJ/mol$

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