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
B is the answer to your question
Answer:
salt water heterogeneous mixture , pure substance compound,element solut8on
<span><span>There is no formula. The speed of light is a fundamental constant which appears in other formulas but there’s no formula to compute the numerical value.Well, actually, that’s not quite right. The numerical value in meters per second is known exactly, because we use the speed of light to define the meter. It is: <span><span><span>c=299,792,458 m/s</span><span>c=299,792,458 m/s</span></span>
</span>. Exactly. But the thing is — this value is purely an artifact of our unit system. Other unit systems will give other values, so the number value is entirely arbitrary.</span></span>
Answer:
Equilibrium concentration of 
is 12.5 M
Explanation:
Given reaction: 
Here, ![K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5BH_%7B2%7DO%5D%7D)
where 
represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations
Here, ![[C_{2}H_{4}]=0.015M](https://tex.z-dn.net/?f=%5BC_%7B2%7DH_%7B4%7D%5D%3D0.015M)
, ![[C_{2}H_{5}OH]=1.69M](https://tex.z-dn.net/?f=%5BC_%7B2%7DH_%7B5%7DOH%5D%3D1.69M)
and 
So, ![[H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M](https://tex.z-dn.net/?f=%5BH_%7B2%7DO%5D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5Ctimes%20K_%7Bc%7D%7D%3D%5Cfrac%7B1.69%7D%7B0.015%5Ctimes%209.0%7D%3D12.5M)
Hence equilibrium concentration of is 12.5 M
