Answer:
A. The partial pressure for CH4 = 0.0925atm
B. The partial pressure for C2H6 = 0.925atm
C. The partial pressure for C3H8 = 0.346atm
D. The partial pressure for C4H10 = 0.115atm
Explanation:
Total pressure = 1.48atm
Total mole = 0.4+4+1.5+0.5=6.4
A. Mole fraction of CH4 = 0.4/6.4 = 0.0625
The partial pressure for CH4 = 0.0625 x 1.48 = 0.0925atm
B. Mole fraction of C2H6 = 4/6.4 = 0.625
The partial pressure for C2H6 = 0.625 x 1.48 = 0.925atm
C. Mole fraction of C3H8 = 1.5/6.4 = 0.234
The partial pressure for C3H8 = 0.234 x 1.48 = 0.346atm
D. Mole fraction of C4H10 = 0.5/6.4 = 0.078
The partial pressure for C4H10 = 0.078 x 1.48 = 0.115atm
Answer:
= 74.4 grams / mole. Ernest Z. The reaction will produce 15.3 g of KCl
Explanation:
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
AR 385-63/MCO 3570.1C is used in conjunction with DA PAM 385-63.
AR 385-63/MCO 3570.1C is the regulation or order which provides revised range safety policy for the Army and Marine Corps.
This order/regulation applies on the Active Army or the Army National Guard of the United states.
The answer is (3) 53.5 g/mol. The gram formula mass means that the mass of one mol compound. 1N=14, 4H=4, 1Cl=35.5. So the gram formula mass of NH4Cl=14+4+35.5=53.5 g/mol.
