Answer:
D. 108 grams of KNO3(Potassium nitrate)
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basic or acidic conditions and the reactants must be heated.
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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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<span>When two metals touch in the mouth, a small shock is created. this is known as a </span>galvanic action
3.37 x 10¹⁰ molecules
Explanation:
Given parameters:
Volume of water = 1pL = 1 x 10⁻¹²L
Density of water = 1.00g/mL = 1000g/L
Unknown:
Number of water molecules = ?
Solution:
To solve this problem, we first find the mass of the water molecule in the inkjet.
Mass of water = density of water x volume of water
Then, the number of molecules can be determined using the expression below:
number of moles = 
Number of molecules = number of moles x 6.02 x 10²³
Solving:
Mass of water = 1 x 10⁻¹² x 1000 = 1 x 10⁻⁹g
Number of moles:
Molar mass of H₂O = 2 + 16 = 18g/mol
Number of moles = 
= 5.6 x 10⁻¹⁴moles
Number of molecules = 5.6 x 10⁻¹⁴ x 6.02 x 10²³ = 33.7 x 10⁹
= 3.37 x 10¹⁰ molecules
Learn more:
Number of molecules brainly.com/question/4597791
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