1.
n = m/2.016 g/mol = V/22.4 l
V = 22.4 l
Hence,
m = 2.016 g ≈ 2.02 g
B.) 2.02 g
2.
Balancing the equation, we get,
Mg + 2HCl → MgCl2 + H2
PV = nRT
V = nRT/P
V = 1.5 mol × 8.314 atm/(mol.K) × 298 K/(0.96 atm)
V = 3871.206
V ≈ 38.2 l
I need the following option
Answer: [N2]₀ = 10M and [H2]₀ = 11M
Explanation: To calculate the initial concentration, you would have to set up an ICE table, which is an organized way of tracking known quantities or the ones you want to find. ICE stands for:
I is initial amount;
C is change in concentration;
E is for equilibrium concentration;
For the mixture,
N2 3H2 2NH3
I [N2]₀ [H2]₀ 0
C - x -3x +2x
E [N2]₀ - x =8 [H2]₀ - 3x =5 2x =4
With the product, we can find "x":
2x=4
x=2M
With x=2, find the concentrations:
[N2]₀ - x = 8
[N2]₀ = 10M
[H2]₀ - 3x = 5
[H2]₀ = 11M
The initial concentrations of nitrogen gas [N2] is 10.0 M and of hydrogen gas [H2] is 11.0 M.
Answer:
C. It decreases by a factor of 4
Explanation:
F1 = kq1*q2/r²
F2 = kq1*q2/(2r)² = kq1*q2/(4r²) = kq1*q2/(r²*4) = F1/4
