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jeyben [28]
3 years ago
7

A cylinder contains 3.1 L of oxygen at 300 K and 2.7 atm. The gas is heated, causing a piston in the cylinder to move outward. T

he heating causes the temperature to rise to 610 K and the volume of the cylinder to increase to 9.4 L.
How many moles of gas are in the cylinder?
Express your answer using two significant figures.
Chemistry
1 answer:
zubka84 [21]3 years ago
6 0

Answer: The moles of gas present in the cylinder is 0.34 moles.

Explanation:

Given: P_{1} = 2.7 atm,   V_{1} = 3.1 L,     T_{1} = 300 K

P_{2} = ?,      V_{2} = 9.4 L,       T_{2} = 610 K

Formula used to calculate the final temperature is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{2.7 atm \times 3.1 L}{300 K} = \frac{P_{2} \times 9.4 L}{610 K}\\P_{2} = \frac{5105.7}{2820} atm\\= 1.81 atm

Now, moles present upon heating the cylinder are as follows.

P_{2}V_{2} = n_{2}RT_{2}\\1.81 atm \times 9.4 L = n_{2} \times 0.0821 L atm/mol K \times 610 K\\n_{2} = \frac{17.014}{50.081} mol\\= 0.34 mol

Thus, we can conclude that moles of gas present in the cylinder is 0.34 moles.

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Answer:

The correct statement is:

E - The entropy of the products is greater than the entropy of the reactants.

Explanation:

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150ml

Explanation:

For this question,

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So we have this equation

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M2 = 2m

V1= ??

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When we input these into equation 1, we have:

2m x v1 = 6m x 50ml

V1 = 6m x 50ml/2

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Thank you

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The _______ geometry of a water molecule is tetrahedral even though the molecular geometry is bent
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What is the mass of oxygen in 3.34 g of potassium permanganate?
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Answer:

1.35 g

Explanation:

Data Given:

mass of Potassium Permagnate (KMnO₄) = 3.34 g

Mass of Oxygen: ?

Solution:

First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)

So,

Molar Mass of KMnO₄ = 39 + 55 + 4(16)

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Since the percentage of compound is 100

So,

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for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be

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