The current needed to deposit all of the nickel in 5 hours is 2.219 A
<h3>
What is Faraday's Law of electrolysis?</h3>
According to the first law of Faraday, the amount of substance liberated at an electrode is directly proportional to the quantity of charge passed through an electrolyte.
m = zq , where z = electrochemical equivalent
An electrochemical equivalent is the mass of the substance liberated at an electrode when 1 Columb of charge is passed through the electrolyte.
For the nickel,
z = 2
The molar mass of NiS0₄ = 154.75 g
154.75 g of Ni needs 2 x 96485 C
32 g of Ni needs
32 g of Ni needs 39944.79 C
We know,
q = It
Where q = charge
I = current and t = time in seconds
q = 39944.79 C and t = 5 x 60 x 60 = 18000 s
= 2.219 A
Hence, the current needed to deposit all the nickel is 2.219 A in 5 hours
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Answer:
A. The reactants are changed to form the products.
Explanation:
Chemical reactions are reactions that involves a change in the chemical composition of substances involved while a nuclear reaction is the process of fusing together or splitting the nucleus of an atom. According to this question, matter is said to undergo both types of reaction.
However, one similarity in both chemical and nuclear reactions is that substances called REACTANTS are changed to form PRODUCTS. In nuclear reaction, the atoms joined or split are the reactants while the ones formed are the products.
Answer:
It is a pure substance. Magnesium is an element.
Answer:
CH₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1 g
Mass of CO₂ = 3.14 g
Mass of H₂O = 1.29 g
Empirical formula =?
Next, we shall determine the mass of Carbon and hydrogen present in the compound. This can be obtained as follow:
For Carbon, C:
Mass of CO₂ = 3.14 g
Molar mass of CO₂ = 12 + (2×16)
= 12 + 32
= 44 g/mol
Molar mass of C = 12 g/mol
Mass of C =?
Mass of C = molar mass of C/ Molar mass of CO₂ × Mass of CO₂
Mass of C = 12/44 × 3.14
Mass of C = 0.86 g
For hydrogen, H:
Mass of C = 0.86 g
Mass of compound = 1 g
Mass of H =?
Mass of H = (Mass of compound) – (mass of C)
Mass of H = 1 – 0.86
Mass of H = 0.14 g
Finally, we shall determine the empirical formula of the cyclopropane. This can be obtained as follow:
Mass of C = 0.86 g
Mass of H = 0.14 g
Divide by their molar mass
C = 0.86 / 12 = 0.07
H = 0.14 / 1 = 0.14
Divide by the smallest
C = 0.07 / 0.07 = 1
H = 0.14 / 0.07 = 2
Thus, the empirical formula of cyclopropane is CH₂