I think its B because if u increase the mass itll have more force which will increase the momentum
Gay-Lussac's law gives the relationship between pressure and temperature of gas. For a fixed amount of gas, pressure is directly proportional to temperature at constant volume.
P/T = k
where P - pressure , T - temperature and k - constant

parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting the values in the equation

T = 4342 K
initial temperature was 4342 K
16 g. The mass of 0.60 mol Al is 16 g.
Molar mass of Al = 26.98 g/mol
Mass of Al = 0.60 mol Al x (26.98 g Al/1 mol Al) = 16 g Al
Answer:
The volume of the stock solution needed is 1L
Explanation:
Step 1:
Data obtained from the question. This include the following:
Concentration of stock solution (C1) = 6M
Volume of stock solution needed (V1) =?
Concentration of diluted solution (C2) = 1M
Volume of diluted solution (V2) = 6L
Step 2:
Determination of the volume of the stock solution needed.
With the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:
C1V1 = C2V2
6 x V1 = 1 x 6
Divide both side by 6
V1 = 6/6
V1 = 1L
Therefore, the volume of the stock solution needed is 1L
Answer: E=∆H*n= -40.6kj
Explanation:
V(CO) =15L=0.015M³
P=11200Pa
T=85C=358.15K
PV=nRT
n=(112000×0.015)/(8.314×358.15)
n(Co)= 0.564mol
V(Co)= 18.5L = 0.0185m³
P=744torr=98191.84Pa
T= 75C = 388.15k
PV=nRT
n= (99191.84×0.0185)/(8.314×348.15)
n(H2) = 0.634mol
n(CH30H) =1/2n(H2)=1/2×0.634mol
=0.317mol
∆H =∆Hf{CH3OH}-∆Hf(Co)
∆H=-238.6-(-110.5)
∆H = 128.1kj
E=∆H×n=-40.6kj.