Answer:
This is an oxidation-reduction (redox) reaction:
2 Ni0 - 4 e- → 2 NiII
(oxidation)
2 O0 + 4 e- → 2 O-II
(reduction)
Ni is a reducing agent, O2 is an oxidizing agent.
The balanced neutralization reaction here is:
Ca(OH)2 + 2HBr --> 2H2O + CaBr2
Notice that two moles of Her are required to neutralize every one mole of Ca(OH)2. This means that for however many moles of Her reacted, HALF as many moles of Ca(OH)2 reacted as well.
Moles of HBr reacted = 0.75 M x 0.345 L = 0.259 mol
Moles of Ca(OH)2 reacted = 0.259 mol / 2 = 0.130 mol
Concentration of Ca(OH)2 = 0.130 mol / 0.250 L = 0.52 M
A variable is not consistent or having a fixed pattern; liable to change.
