Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2
Answer:
5 moles of Fe
Explanation:
The equation of the reaction is;
2 Al(s) + Fe 2O 3(s) --> 2Fe (s) + Al 2O 3 (s)
Now;
1 mole of Fe2O3 require 2 moles of Al
3 moles of Fe2O3 requires 3 × 2/1 = 6 moles of Al
Hence Al is the limiting reactant.
If 2 moles of Al yields 2 moles of Fe
5 moles of Al yields 5 × 2/2 = 5 moles of Fe
Answer:
1552.83J Released
Explanation:
1. mass/m=225
Initial temp:86C, final:32.5C
Changed Temp: 32.5-86= -53.5C
s=0.129 J/gC
Formula: q= m times s times changed Temp.
q=(225)(0.129)(-53.5)
q= -1552.83 J
q=1552.83 J Released
9 is B
6 is A
i’m not sure about the rest
