Answer:
See explanation below
Explanation:
In an electrochemical cell, electricity is obtained by the gradual deterioration of the anode.
Hence, surface area of the metal will affect the length of time within which the electrochemical cell works.
The greater the surface area of the metal, the longer the electrochemical cell can function and the greater the quantity of electricity produced, hence the answer above.
Answer:
0.3793 M
Explanation:
The unknown metal is zinc. So the equation of the reaction is;
Zn(s) + Cu^2+(aq) -------> Zn^2+(aq) + Cu(s)
From Nernst equation;
E = E° - 0.0592/n log Q
[Cu2+] = 0.050179 M
n = 2
[Zn^2+] = ?
E = 1.074 V
E° = 0.34 - (-0.76) = 1.1 V
Substituting values;
1.074 = 1.1 - 0.0592/2 log [Zn^2+]/0.050179
1.074 - 1.1 = - 0.0592/2 log [Zn^2+]/0.050179
-0.026 = -0.0296 log [Zn^2+]/0.050179
-0.026/-0.0296 = log [Zn^2+]/0.050179
0.8784 =log [Zn^2+]/0.050179
Antilog(0.8784) = [Zn^2+]/0.050179
7.558 = [Zn^2+]/0.050179
[Zn^2+] = 7.558 * 0.050179
[Zn^2+] = 0.3793 M
Answer:
Hb would be 78.4% saturated.
Explanation:
This problem can be solved by using simple unitary method.
At 100 mm Hg pressure of oxygen, Hb is saturated by 98%
So, at 1 mm Hg pressure of oxygen, Hb is saturated by 
%
Hence, at 80 mm Hg pressure of oxygen, Hb is saturated by 
% or 78.4%
Therefore, at 80 mm Hg pressure of oxygen in the lungs, Hb would be 78.4% saturated.
It would contain 0.105M moles of NaOH in each liter of solution
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