Answer:
1.71x10²⁷
Explanation:
If we sum 1/2 of (3) + 1/2 of (1):
1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷ = 4.58x10²³
1/2 (1) 1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8
C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>
K' = 4.58x10²³ * 11.8 = 5.42x10²⁴
+1/2 (2):
<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2
C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)
K'' = 5.42x10²⁴* 316.2 =
<h3>1.71x10²⁷</h3>
Answer:
a) cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)
b) 0.068 V.
Explanation:
A) Cu2+ + 2e- euilibrium cu (s)
Hg2Cl2 + 2e- equilibrium 2Hg (l) + 1cl-
Cell Reaction: cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)
B) To calculate the cell voltage
E = E_o Cu2+/Cu - (0.05916 V / 2) log 1/Cu2+
putting values we get
= 0.339V + (90.05916V/2)log(0.100) = 0.309V
E_cell = E Cu2+/Cu - E SCE = 0.309 V - 0.241 V = 0.068V.
The motivation to abstain from adding water to concentrated acids is that, with a few acids, amid weakening, a considerable measure of warmth is discharged, by adding the corrosive to the water, the generally extensive measure of water will retain the warmth. On the off chance that you added water to concentrated corrosive when you initially beginning pouring the water, it could get sufficiently hot for the little measure of water that was filled all of a sudden bubble and splatter corrosive on you. Concentrated sulfuric corrosive is most famous for doing this, not all acids get that hot on weakening, but rather in the event that you make a propensity for continually adding the corrosive to water for every one of them, you can't turn out badly.
The reduction reaction is the gain of electrons while oxidation reaction is the loss of electrons. For potassium ion(K+), the reaction should be K+ + e- ==> K. So the answer is (1).
