Answer:
3.47 ×10^-10
Explanation:
The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)
A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.
E°cell = E°cathode - E°anode
E°cathode = -0.41 V
E°anode = -0.13 V
E°cell = -0.41 -(-0.13) = -0.28 V
From
E°cell = 0.0592/n log K
n= 2, K= the unknown
-0.28 = 0.0592/2 log K
log K = -0.28/0.0296
log K = -9.4595
K = Antilog ( -9.4595)
K= 3.47 ×10^-10
Answer:
2Cl——>Cl2+2e-
Explanation:
It shows an electron loss or gain
Answer:
pH = 7.8
Explanation:
The Henderson-Hasselbalch equation may be used to solve the problem:
pH = pKa + log([A⁻] / [HA])
The solution of concentration 0.001 M is a formal concentration, which means that it is the sum of the concentrations of the different forms of the acid. In order to find the concentration of the deprotonated form, the following equation is used:
[HA] + [A⁻] = 0.001 M
[A⁻] = 0.001 M - 0.0002 M = 0.0008 M
The values can then be substituted into the Henderson-Hasselbalch equation:
pH = 7.2 + log(0.0008M/0.0002M) = 7.8
Answer:
B
Explanation:
Noble gases are in group 18 (neon, argon, etc)
