The amount of current required to produce 75. 8 g of iron metal from a solution of aqueous iron (iii)chloride in 6. 75 hours is 168.4A.
The amount of Current required to deposit a metal can be find out by using The Law of Equivalence. It states that the number of gram equivalents of each reactant and product is equal in a given reaction.
It can be found using the formula,
m = Z I t
where, m = mass of metal deposited = 75.8g
Z = Equivalent mass / 96500 = 18.6 / 96500 = 0.0001
I is the current passed
t is the time taken = 75hour = 75 × 60 = 4500s
On subsituting in above formula,
75.8 = E I t / F
⇒ 75.8 = 0.0001 × I × 4500
⇒ I = 168.4 Ampere (A)
Hence, amount of current required to deposit a metal is 168.4A.
Learn more about Law of Equivalence here, brainly.com/question/13104984
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There are 3.98 × 10^23 atoms of oxygen in the sample.
Given that;
1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen
x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen
x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms
x = 0.0368 moles
The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18
Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.
Learn more: brainly.com/question/9743981
Molar mass
C₂H₄O₂ = 60.0 g/mol
n = mass / molar mass
3.00 = mass / 60.0
m = 3.00 * 60.0
m = 180 g of <span>C₂H₄O₂
hope this helps!</span>
Astatine. Because it has the smaller shell of electrons. I believe
