This force is opposing in nature .it resists object velocity .
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:
The answer might be 4, been a while since I've done third grade math though.
To find out how many grams are in 4.65 moles of Al(NO₂)₃
Find out what the molar mass of Al(NO₂)₃ is
Al = 26.98 g/mol Al
N = 14 g/mol N
O = 16 g/mol O
Next, you have to look at the subscripts and figure out which they belong to, in this case:
Al = 26.98 g/mol Al
N₃ = 42 g/mol N₃
O₆ = 96 g/mol O₆
Finally, add the numbers together, so:
26.98 g/mol Al + 42 g/mol N₃ + 96 g/mol O₆ =
164.98 g/mol Al(NO₂)₃
Now, you have 4.65 mol Al(NO₂)₃ so
164.98 g/mol Al(NO₂)₃ × 4.65 mol Al(NO₂)₃ =
767.157 grams of Al(NO₂)₃
mm = 1500 x 10
m