Answer:
7.16 × 10⁻³ M
Explanation:
Let's consider the reduction reaction of copper during the electroplating.
Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)
We can calculate the moles of Cu²⁺ present in the solution using the following relations.
- 1 mole of electrons has a charge of 96486 C (Faraday's constant).
- 1 mole of Cu²⁺ is reduced when 2 moles of electrons are gained.
The moles of Cu²⁺ reduced are:
of Cu²⁺ are in 0.300 L of solution.
[Cu²⁺] = 2.15 × 10⁻³ mol/0.300 L = 7.16 × 10⁻³ M
no of moles in 4 grams of H2=4/2=2
2H2+O2............2H2O
the reaction shows that 2 moles of H2 react with 1 mole of O2 to produced 2 moles H2O
now no of moles in 36grams of H2O=36/18=2
therefore he needed 1 mole of O2 to produce 36 grams of H2O and 1 mole of oxygen is equal to 32 grams of O2
In the given reaction potassium metal is placed in water and when a reactive metal is placed in water, it reacts with water to form metal hydroxide and hydrogen gas, as a result, potassium hydroxide and hydrogen gas are produced in the reaction. A redox reaction is a reaction in which a substance is oxidized during the reaction whereas some other substance is reduced during the reaction, simultaneously.
The given chemical reaction is:
2K(s) + 2H20(l) arrow 2KOH(aq) + H2(g)
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Answer:
The percent yield of the reaction is 82.5%
Explanation:
Let's work with moles to get the percent yield.
Mass / Molar mass =
652.5 g / 158.03 g/m = 4.13 moles
If the theoretical yield of the reaction is 5 moles but we only made 4.13 moles, the percent yield will be:
(Produced yield / Theoretical yield) . 100 =
(4.13 / 5) . 100 = 82.5 %