Answer:
0.92 kg
Explanation:
The volume occupied by the air is:

The moles of air are:

The heat required to heat the air by 10.0 °C (or 10.0 K) is:

Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:

Answer: 241.6 grams of CO2
Explanation: you take 84.3 grams C5H12 and divide it by 72.15 grams of C5H12(which is the molar mass) you take that answer and calculate the mols of CO2 by multiplying the 1.168 you got before and multiply it by 5. You take the answer you get from that and multiply it by the molar mass of CO2 and get the theoretical yield and then you just plug it in. 94= (x/257.02)x100 and solve to find x which is the actual yield.
Answer:
The square root of the molar mass of B ÷ the square root of the molar mass of A
Explanation:
Graham’s Law applies to the effusion of gases:
The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).
If you have two gases A and B, the ratio of their rates of effusion is

Answer:
Option A. It has stayed the same.
Explanation:
To answer the question given above, we assumed:
Initial volume (V₁) = V
Initial temperature (T₁) = T
Initial pressure (P₁) = P
From the question given above, the following data were:
Final volume (V₂) = 2V
Final temperature (T₂) = 2T
Final pressure (P₂) =?
The final pressure of the gas can be obtained as follow:
P₁V₁/T₁ = P₂V₂/T₂
PV/T = P₂ × 2V / 2T
Cross multiply
P₂ × 2V × T = PV × 2T
Divide both side by 2V × T
P₂ = PV × 2T / 2V × T
P₂ = P
Thus, the final pressure is the same as the initial pressure.
Option A gives the correct answer to the question.
Answer:
diluted or stronger
Explanation:this is caused by the addition of oxygen and it is normally overbearing of ends up changing the whole molecule