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Zinaida [17]
3 years ago
9

A sample of 0.0200 mol of chlorine gas is kept at 27.0°c and 0.150 atm. what would be its pressure if the temperature was increa

sed to 227.°c and the volume kept the same
Chemistry
1 answer:
Andrews [41]3 years ago
6 0
Answer is:  0,250 atm.
T₁(chlorine) = 27°C = 300K.
T₂(chlorine) = 227°C = 500K.
T₂/T₁ = 1,66.
p1(chlorine) = 0,150 atm.
p2(chlorine) = ?
V(chlorine) = constant.
n(chlorine) = 0,0200 mol.
R - gas constant, R = 0.0821 atm·l/mol·K. 
Ideal gas law: p·V = n·R·T.
p = n·R·T÷V.
If temperature was increased 1,66 times, pressure also increase 1,66 times: 1,66·0,150 atm = 0,250 atm. 
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Consider the following reaction where K. = 154 at 298 K: 2NO(g) + Brz(9) 2NOBr(g) A reaction mixture was found to contain 2.69x1
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Explanation:

2NO(g) + Br_2(g)\rightleftharpoons 2NOBr(g)

Equilibrium constant of reaction = K=154

Concentration of NO = [NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M

Concentration of bromine gas = [Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M

Concentration of NOBr gas = [Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M

The reaction quotient is given as:

Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}

Q=328.06

Q>K

The reaction will go in backward direction in order to achieve an equilibrium state.

1. In order to reach equilibrium NOBr (g) must be produced.  False

2. In order to reach equilibrium K must decrease. False

3. In order to reach equilibrium NO must be produced. True

4. Q. is less than K . False

5. The reaction is at equilibrium. No further reaction will occur. False

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12.70 L of a gas has a pressure of 0.63 atm. What is the volume of the gas at 105kPa?
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Answer: 7.693 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

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P_2\text{ and }V_2 are final pressure and volume.

We are given:

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Putting values in above equation, we get:

0.63\times 12.70mL=1.04\times V_2\\\\V_2=7.693L

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