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3 years ago

Calculate the volume of 0.684mol of carbon dioxide at s.t.p. show working

Chemistry

1 answer:

solong [7]3 years ago

6 0

Answer: The volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = ?

n = number of moles = 0.684

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $273K$ (at STP)

$V=\frac{nRT}{P}$

$V=\frac{0.684\times 0.0821L atm/K mol\times 273K}{1atm}$

$V=15.3L$

Thus the volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L

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<h3>Dilution equation</h3>

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3 years ago

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Answer:

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