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Flauer [41]
3 years ago
10

If gas particles start colliding with the walls of their metallic container with increased force, what is their direct effect? l

ower gas pressure higher gas pressure lower volume of gas higher volume of gas
Chemistry
2 answers:
stepladder [879]3 years ago
8 0
The answer is B) Higher gas pressure.
AveGali [126]3 years ago
3 0

Answer: Higher gas pressure

Explanation:

p=\frac{Force}{Area}

Pressure\propto Force

According to force and pressure relation stated above, pressure exerted is directly proportional to the force applied.

From this we can easily predict that pressure of the gas will increase when gas particles are colliding with the walls of metallic container with increased force.


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sleet_krkn [62]

Answer:

p_{H_2O}=2.00atm

Explanation:

Hello!

In this case, according to the following chemical reaction:

2H_2+O_2\rightarrow 2H_2O

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

n_{H_2}=\frac{3.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0914molH_2 \\\\n_{O_2}=\frac{1.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0305molH_2

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

n_{H_2O}^{by\ H_2} = 0.0914molH_2*\frac{2molH_2O}{2molH_2} =0.0914molH_2O\\\\n_{H_2O}^{by\ O_2} = 0.0305molO_2*\frac{2molH_2O}{1molO_2} =0.0609molH_2O

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

p_{H_2O}=\frac{0.0609molH_2O*0.08206\frac{atm*L}{mol*K}*400K}{1.00L}\\\\ p_{H_2O}=2.00atm

Best regards!

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