The current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.
<h3>What is current?</h3>
The current is given as the product of the charge with time. In the electrochemical analysis of the nickel, there will be a reduction of the nickel ion to nickel. The formation is given as:

There is the deposition of 1 mole of Ni with 2 electrons transfer. The transfer of charge for 1 mole that is 58.7 grams Nickel is:

The mass of Ni to be deposited is 1.22 grams. The charge required is given as:

The current required to transfer 4010.7 C of charge in 1800 seconds is given as:

Thus, the current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.
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0.4 g/ml.............................
Number of moles of CO2 =
Mass /Ar
= 50.2 / (12 + 32)
1.14 mols
For every 1 mol of gas, there will be
24000 cm^3 of gas
Vol. = 1.14 x 24 dm^3
= 27.36 dm^3
Answer:
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The concentration may be expressed as % m/m, this is the mass of ions in 100 mass units of solution, whose formula is:
% m/m = [mass of ions / mass of solution]*100
Then,
%m/m = [8.5*10^ -3 grams of calcium ions] / [490 grams of solution] * 100 =
% m/m = 1.74 * 10^ -5 %
Answer: 1.74 * 10 ^ -5 %