[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
KOH is a compound containing two ions, K+ and OH-.
<span>The polyatomic ion present is OH- which is called hydroxide. </span>
<span>The compound is named potassium hydroxide.</span>
Answer:
Q = 4.056 J
Explanation:
∴ m = 406.0 mg = 0.406 g
∴ <em>C </em>= 1.85 J/g.K
∴ T1 = 33.5°C ≅ 306.5 K
∴ T2 = 38.9°C = 311.9 K
⇒ ΔT = 311.9 - 306.5 = 5.4 K
⇒ Q = (0.406 g)(1.85 J/gK)(5.4 K)
⇒ Q = 4.056 J
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:
Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:
Best regards.
H is 4*10^6 M , OH is 2.5*10^-9 M
