Answer:
Explanation:
1. the 1/2 reaction that occurs at the cathode
3Cl2(g) +6e^- -------------> 6Cl^- (aq)
2 the 1/2 reaction that occurs at the anode
2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-
2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-
E0 = -0.59v
3Cl2(g) +6e^- -------------> 6Cl^- (aq)
E0 = 1.39v
3Cl2 (g) + 2MnO2 (s) + 8OH^(−) (aq)---------> 6Cl^(−) (aq) + 2MnO4^(−) (aq) + 4H2O (l)
E0cell = 0.80v
Answer:The formulas of ionic compounds are:
a)
b)
c)
d)
Explanation:
Formulas for the an ionic compounds is determine by:
Criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.
(a) Copper bromide :Given that it contains
ion.

(b) Manganese oxide : Given that it contains
ion.

(c)Mercury iodide :Given that it contains 

(d) Magnesium phosphate :Given that it contains 

Answer:
balanced equation mole ratio 5 2 mol NO/1 mol O2
10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2
20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO
actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2
Because the actual mole ratio of NO:O2 is larger than the balanced equation mole
ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.
Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO
0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO
Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2
Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N
Explanation:
Explanation:
They are called or known as cations