How many moles are in 68.5 liters of oxygen gas at STP?

2 answers:

7 0

We can solve this problem when we use the conditions of a gas at standard temperature and pressure. It has been established that at STP where the temperature is 0 degrees Celsius and the pressure is 101.325 kPa, the volume of 1 mole of gas is 22.4 L. We will use this data for the calculations.

68.5 L ( 1 mol O2 / 22.4 L O2 ) = 3.06 mol O2

harina [27]3 years ago

5 0

Explanation:

It is known that at STP, there are 22.4 L present in one mole of a substance.

Therefore, in 68.5 liters there will be 1 mol divided by 22.4 L times 68.5 L.

Mathematically, $68.5 L \times \frac{1 mol O_{2}}{22.4 L}$

= 3.05 mol

Hence, we can conclude that there are 3.05 moles present in 68.5 liters of oxygen gas at STP.

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Carbonic acid is an important agent involved in rock and mineral dissolution. How is carbonic acid formed?

Vitek1552 [10]

Answer:

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Explanation:

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6 0

3 years ago

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Aleks04 [339]

Answer: The molar mass of the gas is 9.878 g/mol.

Explanation:

According to Graham's law, the rate of diffusion is inversely proportional to square root of molar mass of gas.

$Rate = \frac{1}{\sqrt{M}}$

where,

M = molar mass of gas

As given gas diffuses 1/7 times faster than hydrogen gas. So, its molar mass is calculated as follows.

$\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\\$

where,

$M_{1}$ = molar mass of hydrogen gas

$M_{2}$ = molar mass of another given gas

$R_{1}$ = rate of diffusion of hydrogen

$R_{2}$ = rate of diffusion of another given gas = $\frac{1}{7}R_{1}$

Substitute the values into above formula as follows.

$\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\\\frac{R_{1}}{\frac{1}{7}R_{1}} = \sqrt{\frac{M_{2}}{2}}\\7 \times 1.414 = M_{2}\\M_{2} = 9.878 g/mol$