Answer:
yes you can by using filtration and evaporation
Sr = 1 mol
H = C = 2 mol each
O = 6 mol
Answer:
Take one example: Zinc is extracted from copper sulphate if a little bit of zinc is dissolved in copper sulphate solution. Copper sulphate fades in blue and the solution of zinc sulphate is coloured. Zinc is therefore more reactive than copp
Explanation:
<span>Ag+(aq) + e– → Ag(s) Eo = +0.80 V
Pb2+(aq) + 2e– → Pb(s) Eo = –0.13 V
=>
</span>
<span>2Ag+(aq) + 2e– → 2Ag(s) Eo = +0.80 V
Pb(s) → Pb2+(aq) + 2e– - Eo = – (-0.13 V) = +0.13 V
----------------------------------
2Ag(+) (aq) + Pb(s) -------> 2Ag(s) + Pb(2+)
</span>
<span>E cell = E0 cathode – Eo anode = 0.80V - (- 0.13V) = 0.80V + 0.13V = 0.93V.
Answer: + 0.93V
</span>
Answer:
At the nucleus point of a substance, the particles have enough kinetic energy to break free from their fixed positions.
Answer : nucleus
