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olchik [2.2K]
3 years ago
12

Determain the number of moles in 2.24l of ch4 at stp

Chemistry
1 answer:
valkas [14]3 years ago
6 0

Answer:

0.1 mole of CH₄

Explanation:

From the question given above, the following data were obtained:

Volume of CH₄ = 2.24 L

Number of mole of CH₄ =?

The number of mole of CH₄ can be obtained as follow:

Recall:

1 mole of a gas occupy 22.4 L at stp. This implies that 1 mole of CH₄ occupies 22.4 L at stp.

22.4 L = 1 mole of CH₄

Therefore,

2.24 L = 2.24 × 1 mole of CH₄ / 22.4

2.24 L = 0.1 mole of CH₄.

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What is the volume of 0.80 grams of o2 gas at stp? (5 points) group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l
Vladimir [108]

Answer:

0.56L

Explanation:

This question requires the Ideal Gas Law:  PV=nRT where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.

Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:R=0.0821\frac{L\cdot atm}{mol\cdot K}

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means T=273.15K and P=1atm

Lastly, we must calculate the number of moles of O_2(g) there are.  Given 0.80g of O_2(g), we will need to convert with the molar mass of O_2(g).  Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2:  32g\text{ }O_2=1mol\text{ }O_2

Thus, \frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n

Isolating V in the Ideal Gas Law:

PV=nRT

V=\frac{nRT}{P}

...substituting the known values, and simplifying...

V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}

V=0.56L \text{ } O_2

So, 0.80g of O_2(g) would occupy 0.56L at STP.

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Answer:

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Explanation:

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1 mol ------ 58, 5 g

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Explanation:

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