You get the north hemisphere and the south hemisphere.
Cl2 + 2Cu --> 2CuCl Oxidation reaction is Cu --> Cu+ + 1e Reduction reaction is Cl + 1e --> Cl- Redox reaction is Cu + Cl --> Cu+ + Cl-
82,653,950,016
0.5 cu decimetres or 500 cu cm or half a million cu mm
In this case, zinc will undergo oxidation and copper ions will experience reduction. The reduction half-reaction is Cu^2+ (aq) + 2e^- → Cu (s), and the oxidation half-reaction is Zn (s) → Zn^2+ (aq) + 2e^-. Overall, the reaction is Zn (s) + Cu^2+ (aq) → Zn^2+ (aq) + Cu (s).
1 mg
Zn(s) --> Zn2+(aq) + 2e : Oxidation Cu+(aq) + 1e --> Cu(s) : Reduction
Copper-64 (Cu-64) has a half-life of approximately 12.7 hours. After one half-life (12.7 hours), half of the original sample would remain. Therefore, from a 2 mg sample, after 12 hours, approximately 1 mg of Cu-64 would remain, as it has not yet fully completed one full half-life.
54 feet 4 inches
The half equation for the reduction of copper oxide by carbon is: CuO + C -> Cu + CO
(all figures approximate)Mass of earth = 5.97 * 1024 kilogramsRadius of earth = 6.371 * 106 metresVolume of earth = 4/3 * pi * radius3 = 1.0832 * 1021 cu metresSo:(average) density = mass / volume = 5,510 kgs / cu metre
1 cubic yard = 27 cubic feet1 cubic foot = 1,728 cubic inches1 gallon = 231 cubic inches(0.5 cubic yard) x (27 cu-ft/cu-yd) x (1,728 cu-in/cu-ft) / (231 cu-in/gal) = 100.987 gallons (rounded)
Cu(s) + 2AgNO3(aq) ---> Cu(NO3)2(aq) + 2Ag(s) So you need half as many moles of Cu. Thus 5.8/2 = 2.9 moles of Cu are needed.