Answer:
0.2M NaOh
Explanation:
there are 0.2 mol of NaOH in 8.0 g. (8.0/40) =0.2. Molarity = mol/L = 0.2M.
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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Answer:
Everything around you can be broken down into smaller particles called atoms. The particles of one substance are all the same and different substances are made up of different particles.
Explanation:
Answer:
3
Explanation:
If oxygen reacts with iron, then both must be reactants and rust the product of that reaction
Answer:
2 mol of SO3 produces 1 mol O2
3 mol SO3 produces 3/2 mol of O2
so O2 produced = 1.5(32) =48 gm
Explanation:
