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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 There are 1000 mililiters in a liter, so 1000 ml for every liter, you have 5 liters, so:
5L*1000 = 5000 mL
 
        
                    
             
        
        
        
Answer:
ΔH =  - 2020.57 kJ/mol
Explanation:
Given that :
mass of propanol = 1.685 g
the molar molar mass = 60 g/mol
Thus; the number of  moles = mass/molar mass
= 1.685 g/60 g/mol
= 0.028 g/mol
However ;
ΔH = heat capacity C × Δ T
Given that:
The temperature increases from  298.00 K to 302.16 K. 
Then ;
 Δ T = 302.16 K - 298.00 K
 Δ T = 4.16 K
heat capacity C = 13.60 kJ/K
∴
ΔH = 13.60 kJ/K × 4.16 K
ΔH =  56.576 kJ
The equation of the given reaction can be represented as :

Thus for 0.028 mol of heat liberated; ΔH =  56.576 kJ
For 1 mole of heat liberated now: 
ΔH =  56.576 kJ/0.028 mol
ΔH =  2020.57 kJ/mol
SInce , Heat is liberated, the reaction undergoes an exothermic reaction thus;
ΔH =  - 2020.57 kJ/mol
 
        
                    
             
        
        
        
A very disgusting type of lemonade 
        
             
        
        
        
Answer:
–2.23 L
Explanation:
We'll begin by calculating the final volume. This can be obtained as follow:
Initial pressure (P₁) = 1.03 atm
Initial volume (V₁) = 3.62 L
Final pressure (P₂) = 2.68 atm
Final volume (V₂) =?
P₁V₁ = P₂V₂
1.03 × 3.62 = 2.68 × V₂
3.7286 = 2.68 × V₂
Divide both side by 2.68 
V₂ = 3.7286 / 2.68
V₂ = 1.39 L
Finally, we shall determine the change in volume. This can be obtained as follow:
Initial volume (V₁) = 3.62 L
Final volume (V₂) = 1.39 L
Change in volume (ΔV) =?
ΔV = V₂ – V₁
ΔV = 1.39 – 3.62
ΔV = –2.23 L
Thus, the change in the volume of her lung is –2.23 L.
NOTE: The negative sign indicate that the volume of her lung reduced as she goes below the surface!