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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I will use [pV/T] in the state 1 = [pV/T] in the state 2.
State 1:
p = 1.0 atm
V = 25 liter
T = 100 + 273.15 = 373.15 K
State 2:
p = 19.71 mmHg * 1.atm / 760 mmHg = 0.0259atm
V= ?
T = 25 + 273.15 = 298.15 K
Application of the formula
1.0 atm * 25 liter / 373.15 k = 0.0259 atm * V / 298.15 K =>
V = [1.0atm * 25 liter / 373.15 K]*298.15K/0.0259atm = 771 liter
Answer:
32.00 g. Hope this helps! PLEASE GIVE ME BRAINLIEST!!!!! =)
K is Potassium
Cl is Chlorine
There are 2 elements.