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Diano4ka-milaya [45]
2 years ago
14

At what temperature would 0.500 moles of gas particles stored in a 100.0 mL container reach a pressure of 15.0 atm?

Chemistry
2 answers:
pychu [463]2 years ago
8 0
We use the ideal gas equation in order to calculate the temperature of the system. It is expressed as follows:

PV=nRT

where P is pressure, V is volume, n is the number of moles, R is the universal gas constant and T is the temperature.

15 atm (.1 L) = 0.5 mol (0.08206 L-atm / mol-K) T
T = 36.56 K

Hope this answers the question. Have a nice day.
sergeinik [125]2 years ago
4 0

Ideal gas law: p·V = n·R·T.

p = 15.0 atm; pressure of gas.

V = 100 mL ÷ 1000 mL/L.

V = 0.100 L; volume of gas.

n = 0.500 mol; amount of substance.

R = 0,08206 L·atm/mol·K; universal gas constant.

T = ?; temperature of gas.

T(gas) = p · V ÷ (R · n).

T(gas) = 15.0 atm · 0.1 L ÷ (0,08206 L·atm/mol·K · 0.5 mol).

T(gas) =  36,500 K; calculation is correct.


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How many moles of NaCl are produced if 239.7 grams of Na2S reacts with plenty of AlCl3?
lana66690 [7]

Answer:

6.142 moles of NaCl

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2AlCl3 + 3Na2S —> Al2S3 + 6NaCl

Next, we determine the number of mole in 239.7 g of Na2S. This is illustrated below:

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Mass of Na2S = 239.7g

Number of mole Na2S =..?

Mole = Mass /Molar Mass

Number of mole Na2S = 239.7/78.048 = 3.071 moles

Finally, we can obtain the number of mole of NaCl produced from the reaction as follow:

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3 0
3 years ago
Chamber 1 and Chamber 2 have equal volumes of 1.0L and are assumed to be rigid containers. The chambers are connected by a valve
vitfil [10]
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas  than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)

=> p1 = 2 p2

Which is easy to demonstrate using ideal gas equation:

p1 = nRT/V = 2.0 mol * RT / 1 liter

p2 = nRT/V = 1.0 mol * RT / 1 liter

=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2

2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.

So, the pressure in both chambers (which form one same vessel) is:

p = nRT/V = 3.0 mol * RT / 2liter

which compared to the initial pressure in chamber 1, p1, is:

p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1

So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.

You can also see how the pressure in chamber 2 changes:

p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
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2 years ago
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Answer:

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