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3 years ago

At STP, what volume of container would be needed to hold 3.2 moles of gas?

Chemistry

1 answer:

Rzqust [24]3 years ago

3 0

78.4 L volume of container is required to hold 3.2 moles of gas.

Explanation:

STP is defined as the standard temperature and pressure of a gas in room temperature conditions. At STP, one mole of the gas which has Avogadro's number of molecules in it will occupy a volume of 22.4 L.
So, one mole of a substance or gas will occupy a volume of 22.4 L then the volume of the container needed for 3.2 moles of gas is calculated by multiplying 22.4 L, standard volume with the moles of the gas 3.2 moles.
Hence, the answer would be 78.4 L.

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A chemist must prepare of 800.0 ml potassium hydroxide solution with a pH of 13.00 at 25Â°.

ArbitrLikvidat [17]

Answer:

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

Explanation:

The pH of the solution = 13.00

pH + pOH = 14

pOH = 14 - pH = 14 - 13.00 = 1.00

$pOH=-\log[OH^-]$

$1.00=-\log[OH^-]$

$[OH^-]=10^{-1.00} M=0.100 M$

$KOH(aq)\rightarrow K^+(aq)+OH^-(aq)$

$[KOH]=[OH^-]=[K^+]=0.100 M$

Molariy of the KOH = 0.100 M

Volume of the KOH solution = 800 mL= 0.800 L

1 mL = 0.001 L

Moles of KOH = n

$Molarity=\frac{Moles}{Volume(L)}$

$0.100 M=\frac{n}{0.800 L}$

n = 0.0800 mol

Mass of 0.0800 moles of KOH :

0.0800 mol × 56 g/mol = 4.48 g

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

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3 years ago

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