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bezimeni [28]
4 years ago
12

Write net ionic equations for the reactions that take place when aqueous solutions of the following substances are mixed: a. sod

ium cyanide and nitric acid, b. ammonium chloride and sodium hydroxide, c. sodium cyanide and ammonium bromide, d. potassium hydrogen sulfate and lithium acetate, e. sodium hypochlorite and ammonia.
Chemistry
1 answer:
strojnjashka [21]4 years ago
8 0

Answer:

A. H₃O⁺ + CN⁻ → H₂O + HCN

B. OH⁻ + NH₄⁺ → NH₃ + H₂O

C. CN⁻ + NH₄⁺ → HCN + NH₃

D. HSO₄⁻ + CH₃COO⁻ → SO₄²⁻ + CH₃COOH

E. ClO⁻ + NH₃ → NO REACTION

Explanation:

Part A.

Sodium cyanide consist of Na⁺ and CN⁻ ions. CN⁻ is a conjugate base and accept H⁺ ion from nitric acid to form HCN (hydrogen cyanide).

Part B.

Ammonium chloride consist of NH₄⁺ and Cl⁻ ions. Sodium hydroxide is a strong base dissociate completely in solution convert into ammonium hydroxide and sodium chloride.

Part C.

Sodium cyanide consist of counter ion Na⁺ . CN⁻ combined with ammonium bromide to form ammonium cyanide and sodium bromide.

Part D.

Potassium hydrohen sulphate consist of potassium counter and hydrogen sulphate ion. Hydrogen sulphate combined with acetate ion to form acetic acid and sulphate ion.

Part E.

There have no reaction because hypochlorite and ammonia both are bases.

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The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

<u>Answer:</u> The concentration of copper fluoride in the solution is 4.90\times 10^{-3}mol/L

<u>Explanation:</u>

To calculate the molarity of solute, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

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Putting values in above equation, we get:

\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L

Hence, the concentration of copper fluoride in the solution is 4.90\times 10^{-3}mol/L

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