Answer:
Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.
Explanation:
The CO appears twice hence in he intermediate reaction it only forms path of the enabling reagents and it further reacts to form the final product. Accounting for the CO in the intermediate reaction that undergoes further reaction will impact on the stoichiometry of the reaction.
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
They don't share and both deal with only 1 electron.
Lithium gives away 1 electron.
F will receive that electron.
The chemical formula is LiF
Answer:
number of moles of NaCl produce = 12 mol
Explanation:
Firstly, we need to write the chemical equation of the reaction and balance it .
Na(s) + Cl2(g) → NaCl(s)
The balanced equation is as follows:
2Na(s) + Cl2(g) → 2NaCl(s)
1 mole(71 g) of chlorine produces 2 moles(117 g) of sodium chloride
6 mole of chlorine gas will produce ? mole of sodium chloride
cross multiply
number of moles of NaCl produce = 6 × 2
number of moles of NaCl produce = 12 moles
number of moles of NaCl produce = 12 mol
