<h3>
Answer:</h3>
A. 1.4 V
<h3>
Explanation:</h3>
We are given the half reactions;
Ni²⁺(aq) + 2e → Ni(s)
Al(s) → Al³⁺(aq) + 3e
We are required to determine the cell potential of an electrochemical cell with the above half-reactions.
E°cell = E(red) - E(ox)
From the above reaction;
Ni²⁺ underwent reduction(gain of electrons) to form Ni
Al on the other hand underwent oxidation (loss of electrons) to form Al³⁺
The E.m.f of Ni/Ni²⁺ is -0.25 V and that of Al/Al³⁺ is -1.66 V
Therefore;
E°cell = -0.25 V - (-1.66 V)
= -0.250V + 1.66 V
= + 1.41 V
= + 1.4 V
Therefore, the cell potential will be +1.4 V
133.0873 g/mol
(NH4)3PO3 - molar mass
Answer:
the answer is 1 sir BC i t is BC it is BC its 2
A likely application of a radioactive isotope with a short half-life such as Technetium-99 will be as a medical tracer. It will likely be used by a doctor to check the movement of substances within a person's body.
A radioactive isotope with such a long half-life like Rubidium-87 is likely used in the determination of the age of fossils and artifacts found by archaeologists.<span />
Answer:
The answer is Ionization energy.
Explanation:
Ionization Energy. The ionization energy tends to increase as one moves from left to right across a given period or up a group in the periodic table.
