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viva [34]
3 years ago
11

For each of the salts on the left, match the salts on the right that can be compared directly, using Ksp values, to estimate sol

ubilities. (If more than one salt on the right can be directly compared, include all the relevant salts by writing your answer as a string of characters without punctuation, e.g, ABC.) 1. aluminum phosphate A. Mn(OH)2 2. magnesium hydroxide B. AgBr C. CuS D. MgCO3 Write the expression for K in terms of the solubility, s, for each salt, when dissolved in water. aluminum phosphate magnesium hydroxide Ksp
Chemistry
1 answer:
tigry1 [53]3 years ago
7 0

Answer:

1.AD

2.BC

Explanation:

For aluminum phosphate AlPO4 (s) <------> Al^3+(aq) + PO4^3-(aq)

Ksp= [Al^3+] [PO4^3-]

For magnesium hydroxide

Mg(OH)2(s) <--------> Mg^2+(aq) +. 2OH^-(aq)

Ksp= [Mg2+] [2OH^-]^2

For

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You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.2750 M NaOH sol
oksano4ka [1.4K]

Answer:

a. 0.393M CH₃COOH.

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35.75mL = 0.03575L × (0.2750mol / L) = <em>9.831x10⁻³ moles of NaOH</em>

As 1 mole of acid reacts per mole of NaOH, moles of CH₃COOH in the acid solution are 9.831x10⁻³ moles.

a. As the volume of the acetic acid solution is 25.00mL = 0.02500L, the molarity of the solution is:

9.831x10⁻³ moles / 0.02500L =

<h3>0.393M CH₃COOH</h3>

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9.831x10⁻³ moles ₓ (60g / mol) = <em>0.590g of CH₃COOH</em>.

As volume of the solution is 25.00mL, the percentage of acetic acid is:

(0.590g CH₃COOH / 25.00mL) ₓ 100 =

<h3>2.360% of acetic acid in the solution</h3>

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This is an example of a double displacement type of reaction where the ions of the reactants displaces each other forming two new substances. Hope this answers the question.
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3 years ago
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