Answer:
C.0.28 V
Explanation:
Using the standard cell potential we can find the standard cell potential for a voltaic cell as follows:
The most positive potential is the potential that will be more easily reduced. The other reaction will be the oxidized one. That means for the reactions:
Cu²⁺ + 2e⁻ → Cu E° = 0.52V
Ag⁺ + 1e⁻ → Ag E° = 0.80V
As the Cu will be oxidized:
Cu → Cu²⁺ + 2e⁻
The cell potential is:
E°Cell = E°cathode(reduced) - E°cathode(oxidized)
E°cell = 0.80V - (0.52V)
E°cell = 1.32V
Right answer is:
<h3>C.0.28 V
</h3>
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Pka = - log ka
10.36 = - log ka
Ka = .......
The answer is accuracy.
That is when a test instrument is calibrated, its accuracy is improved. That is the result comes more close to what it is.
The other factor will not be improved that is its reliability and precision remains the same.
So the answer is accuracy is improved when a test instrument is calibrated.
Answer:
0.017 mol
Explanation:
CH3COOH - acetic acid - we can write as C2O2H4
From formula C2O2H4 we can see that ratio C: O = 2:2 = 1:1, so number of moles C and O are the same.
So if we have 0.017 mol C, we should have 0.017 mol O.
It's called "Ionic compound"
