[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:
oh I'm here thanks for your points
Answer:
a
Explanation:
im thinking because the water is a room temperature there shouldnt be anm immence amount og heat energy for it to have a good amount of energy tho i could be wrong because its not moving it could have no energy.
Answer:
Ksp = 2.74 x 10⁻⁵
Explanation:
The solubility equilibrium for Ca(OH)₂ is the following:
Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)
I 0 0
C + s + 2s
E s 2s
According to the ICE table, the expression for the solubility product constant (Kps) is:
Ksp = [Ca²⁺] x ([OH⁻])² = s x (2s)² = 4s³
Then, we calculate Ksp from the solubility value (s):
s = 0.019 M
⇒ Ksp = 4s³ = 4 x (0.019)³ = 2.74 x 10⁻⁵
Convection is when cool air sinks and warm air rises. i am not 100% sure though
