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3 years ago

Calculate the molar solubility of silver chloride in 0.15 M sodium chloride.

Chemistry

2 answers:

vova2212 [387]3 years ago

7 0

Answer:

S AgCl = 1.066 E-9 M

Explanation:

NaCl(s) → Na+(aq) + Cl-(aq)

0.15 M 0.15 M 0.15 M

AgCl(s) → Ag+(aq) + Cl-(aq)

S S S + 0.15

∴ Ksp AgCl = 1.6 E-10 = [Ag+]*[Cl-]

⇒ Ksp = 1.6 E-10 = ( S )*( S + 0.15 )

⇒ S² + 0.15S - 1.6 E-10 = 0

⇒ S = 1.066 E-9 M

Alex777 [14]3 years ago

6 0

Answer:

S = 1.1 × 10⁻⁹ M

Explanation:

NaCl is a strong electrolyte that dissociates according to the following expression.

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

Given the concentration of NaCl is 0.15 M, the concentration of Cl⁻ will be 0.15 M.

We can find the molar solubility (S) of AgCl using an ICE chart.

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

I 0 0.15

C +S +S

E S 0.15+S

The solubility product (Ksp) is:

Ksp = 1.6 × 10⁻¹⁰ = [Ag⁺].[Cl⁻] = S (0.15 + S)

If we solve the quadratic equation, the positive result is S = 1.1 × 10⁻⁹ M

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3 years ago

The half-life of tritium (H-3) is 12.3 years. If 48.0mg of tritium is released from a nuclear power plant during the course of a

Rudiy27

Answer:

The amount left after 49.2 years is 3mg.

Explanation:

Given data:

Half life of tritium = 12.3 years

Total mass pf tritium = 48.0 mg

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Solution:

First of all we will calculate the number of half lives.

Number of half lives = T elapsed/ half life

Number of half lives = 49.2 years /12.3 years

Number of half lives = 4

Now we will calculate the amount left after 49.2 years.

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6 0

3 years ago

What compound is formed when methyloxirane (1,2-epoxypropane) is reacted with ethylmagnesium bromide followed by treatment with

Pavlova-9 [17]

Answer:

Pentan-2-ol

Explanation:

On this reaction, we have a Grignard reagent (ethylmagnesium bromide), therefore we will have the production of a carbanion (step 1). Then this carbanion can attack the least substituted carbon in the epoxide in this case carbon 1 (step 2). In this step, the epoxide is open and a negative charge is generated in the oxygen. The next step, is the treatment with aqueous acid, when we add acid the hydronium ion ( $H^+$ ) would be produced, so in the reaction mechanism, we can put the hydronium ion. This ion would be attacked by the negative charge produced in the second step to produce the final molecule: "Pentan-2-ol".