When molecules of the reactant hit each other, only a certain percentage of the collisions cause a chemical change.If the molecules are moving too slowly, they will bounce off each other. There will be no reaction.If the molecules are moving rapidly but do not collide with the correct orientation, they will again bounce off each other. There will be no reaction.
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I hope this helps
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Zane
<span>Ag+(aq) + e– → Ag(s) Eo = +0.80 V
Pb2+(aq) + 2e– → Pb(s) Eo = –0.13 V
=>
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<span>2Ag+(aq) + 2e– → 2Ag(s) Eo = +0.80 V
Pb(s) → Pb2+(aq) + 2e– - Eo = – (-0.13 V) = +0.13 V
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2Ag(+) (aq) + Pb(s) -------> 2Ag(s) + Pb(2+)
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<span>E cell = E0 cathode – Eo anode = 0.80V - (- 0.13V) = 0.80V + 0.13V = 0.93V.
Answer: + 0.93V
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PH of acidic buffer = pKa + log [CH₃COONa - HCl] / [CH₃COOH + HCl]
pKa of CH₃COOH = 4.74
Concentration of acetic acid in buffer = 2.0 M
Concentration of sodium acetate = 1.0 M
Concentration of HCl must add = x
pH = 4.74 + log (1-x) / (2+x) = 4.11
x = concentration of HCl must be added = 0.43 M
number of moles of HCl = M * V = 0.43 * 1 = 0.43 mol
mass of HCl must be added = 0.43 * 36.5 = 15.7 g
