<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
In terms of workplace we need our safety measures to have a safe and to ensure the awareness of health and safety of every personnel, waste disposal and the security. Speaking of security every laboratories have their own security, well organized and the most important to store information safely and the other data. The information can be stored in LIMS (laboratory information management system) by registration all the samples. You can any kind of device to upload the information.
C. 14
you would subtract the mass number from the protons or the atomic number because in order to find the mass number you would add the protons and neutrons :)
It’s oxidation number is +6
The balanced equation that illustrates the reaction is:
2C4H6 + 11O2 ......> 8CO2 + 6H2O
number of moles = mass / molar mass
number of moles of oxygen = 2.1 / 32 = 0.065625 moles
Now, from the balanced equation, we can note that:
11 moles of oxygen are required to produce 6 moles of water.
Therefore:
0.065625 moles of oxygen will produce:
(0.065625*6) / 11 = 0.03579 moles of water
number of moles = mass / molar mass
mass = number of moles * molar mass
mass of water = 0.03579 * 18 = 0.644 grams
