[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
its 40
I just took the test right now to get this answer
Answer:
east to west is the answer
2.168 L of air will leave the container as it warms
<h3>Further explanation</h3>
Given
V₁=2.05 L
T₁ = 5 + 273 = 278 K
T₂ = 21 + 273 = 294 K
Required
Volume of air
Solution
Charles's Law
When the gas pressure is kept constant, the gas volume is proportional to the temperature
Input the value :
V₂=(V₁.T₂)/T₁
V₂=(2.05 x 294)/278
V₂=2.168 L
Answer:
Copper ions are reduced into copper atoms.
Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎
Explanation:
During electrolysis, the positive H⁺ and Cu⁺ ions move to the negative cathode and negative OH⁻ and Cl⁻ ions move to the positive anode.
At cathode, copper ions are preferentially discharged due to the low electromotive force required to discharge them compared to the hydrogen ion. The copper ions gain the two electrons lost by the chloride ions when the are discharged. (2 Cl⁻₍aq₎ → Cl₂₍g₎ + 2e⁻)
Thus the half equation is as follows:
Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎
