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Ilya [14]
4 years ago
6

How does decreasing the temperature affect the volume of a gas?

Chemistry
1 answer:
Art [367]4 years ago
6 0
1) If the volume of a container is increased, the temperature increases. 2) If the volume of a container is decreased, the temperature decreases. This means that the volume of a gas is directly proportional to its temperature. ... If the amount of gas in a container is increased, the volume increases.
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Please help!! The answer that shows work and does not send link will get brainiest!
Elden [556K]

Answer:

5 mol H₂. this is the answer bro all the best for your test/exam

7 0
3 years ago
NEED HELP ASAP; PLEASE SHOW YOUR WORK.
Sindrei [870]

Answer:

n = 0.3 mol

Explanation:

Given data:

Volume of gas = 8.0 L

Temperature of gas = 45 °C (45+273 = 318 K)

Pressure of gas = 0.966 atm

Moles of gas present = ?

Ideal gas constant = R = 0.021 atm.L/mol.K

Solution:

Formula:

PV = nRT

P = Pressure

V = volume

n = number of moles

R = ideal gas constant

T = temperature

Now we will put the values:

0.966 atm × 8 L = n × 0.0821 atm.L/mol.K × 318 K

7.728 atm.L = n × 26.12  atm.L/mol

7.728 atm.L / 26.12  atm.L/mol = n

n = 0.3 mol

5 0
4 years ago
Show ur working:
AleksAgata [21]

Answer:

120 g SO3

Explanation:

128 g SO2  -> 160 g SO3

 96 g SO2  -> x

x= (96 g SO2 * 160 g SO3)/128 g SO2

x= 120 g SO3

8 0
4 years ago
A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr
s344n2d4d5 [400]

Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

P=p_{N_2}+p_{O_2}+p_{Ar}

p_{N_2}=P\times \chi_{N_2}

p_{O_2}=P\times \chi_{O_2}

p_{Ar}=P\times \chi_{Ar}

where,

P = total pressure = 3.9 bar

p_{N_2} = partial pressure of nitrogen gas  

p_{O_2} = partial pressure of oxygen gas  

p_{Ar} = partial pressure of argon gases  

\chi_{N_2} = Mole fraction of nitrogen gas  = 0.25

\chi_{O_2} = Mole fraction of oxygen gas  = 0.65

\chi_{Ar} = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar

Partial pressure of oxygen gas :

p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar

Partial pressure of argon gas :

p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar

7 0
4 years ago
The reaction 2a + b --> c + d has an activation energy of 80.0 kj/mol. at 320°c, the rate constant k = 1.80 x 10-2 l mol-1 s-
Naddik [55]
To determine the k for the second condition, we use the Arrhenius equation which relates the rates of reaction at different temperatures. We do as follows:

ln k1/k2 = E / R (1/T2 - 1/T1) where E is the activation energy and R universal gas constant.

ln 1.80x10^-2 / k2 = 80000 / 8.314 ( 1/723.15 - 1/593.15)

k2 = 0.3325 L / mol-s
5 0
4 years ago
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