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lapo4ka [179]
3 years ago
7

When a gas is pumped into a small rigid container the pressure inside the container increases as more particles are added. If th

e number of particles is doubled then the pressure will double. What would happen if the number of particles were decreased by half?
Chemistry
2 answers:
Sever21 [200]3 years ago
7 0

Answer:  A. Pressure would decrease by half

Explanation:

marysya [2.9K]3 years ago
3 0
Since the container of the gas is rigid, the volume of the gas will remain constant. Therefore, when the number of particles were decreased in half then the pressure will also be half of the original given they both are subjected to the same temperature.

PV = nRT

V, T and R are constants so they can be lumped together to a constant k.

P/n = k
P1/n1 = P2/n2

since n2 = n1/2
P1/n1 = P2/<span>n1/2</span>
P2 = P1/2
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Assume that all of this fossil fuel is in the form of octane (C8H18) and calculate how much CO2 in kilograms is produced by worl
AnnZ [28]

The given question is incomplete. the complete question is:

The world burns the fossil fuel equivalent of approximately 9.50\times 10^{12} kg of petroleum per year. Assume that all of this petroleum is in the form of octane. Calculate how much CO2 in kilograms is produced by world fossil fuel combustion per year.( Hint: Begin by writing a balanced equation for the combustion of octane.)

Answer: 29\times 10^{12}kg

Explanation:

Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

2_C8H_{18}+17O_2\rightarrow 16CO_2+18H_2O

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of octane}=\frac{9.50\times 10^{12}\times 10^3g}{114g/mol}=0.083\times 10^{15}moles

According to stoichiometry :

As 2 moles of octane give = 16 moles of CO_2

Thus 0.083\times 10^{15}moles of octane give =\frac{16}{2}\times 0.083\times 10^{15}=0.664\times 10^{15}moles  of CO_2

Mass of CO_2=moles\times {\text {Molar mass}}=0.664\times 10^{15}moles\times 44g/mol=29.2\times 10^{15}g=29.2\times 10^{12}kg

Thus 29\times 10^{12}kg of CO_2 is produced by world fossil fuel combustion per year.

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Explanation:

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Answer:

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Answer:

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