Answer:
<h3>3.Ba=2</h3><h3>4.Ni=2...PO4=3</h3><h3>5.Al=3...CrO4=2</h3><h3>6.NH4=1...SO4=2</h3>
Answer:
3.1 * 10^-14
Explanation:
Note that E°cell = 0.0592/n log K
We can obtain E°cell from the standard reduction potentials of cadmium and hydrogen
Anode reaction
H2(g) ----> 2H+ + 2e
Cathode reaction
Cd^2+(aq) + 2e -----> Cd(s)
E°cell = E°cathode - E°anode
E°cathode = –0.40 V
E°anode = 0 V
E°cell = –0.40 V - 0 V
E°cell = –0.40 V
E°cell = 0.0592/n log K
Where n=2 electrons transferred
–0.40 = 0.0592/2 log K
–0.40 = 0.0296 log K
log K = –0.40/0.0296
log K = -13.5135
K = Antilog ( -13.5135)
K = 3.1 * 10^-14
The best formula that represents compounds only is <span>CO2, H2O, NH3.</span>
