[CO] = 1 mol / 2L = 0.5 M
[
According to the equation:
and by using the ICE table:
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
initial 0.5 0.5 0 0
change -X -X +X +X
Equ (0.5-X) (0.5-X) X X
when Kc = X^2 * (0.5-X)^2
by substitution:
1.845 = X^2 * (0.5-X)^2 by solving for X
∴X = 0.26
∴ [CO2] = X = 0.26
Answer:
0.70 J/g.°C
Explanation:
Step 1: Given data
- Mass of graphite (m): 402 g
- Heat absorbed (Q): 1136 J
- Initial temperature: 26°C
- Specific heat of graphite (c): ?
Step 2: Calculate the specific heat of graphite
We will use the following expression.
Q = c × m × ΔT
c = Q / m × ΔT
c = 1136 J / 402 g × (30°C - 26°C)
c = 0.70 J/g.°C
Answer:
79.04 L
Explanation:
We are given;
Initial Volume; V1 = 6.24L
Initial Pressure; P1 = 760 mm Hg
Final pressure; P2 = 60.0mm Hg
To solve for final volume, we will use Boyles law;
P1•V1 = P2•V2
Let's make V2 which is the final volume the subject;
V2 = (P1•V1)/P2
V2 = (760 × 6.24)/60
V2 = 79.04 L
Answer:
1.125 moles
Explanation:
2mole of HCl produced 1mole of H2
2.25moles of HCl will produce x moles
cross multiply
2x=™1×2.25
x= 2.25÷2
x=1.125mole
