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3 years ago

What is the molarity of a KCl solution made by diluting 75.0 mL of a 0.200 M solution to a final volume of 100.0 mL?

Chemistry

1 answer:

Ipatiy [6.2K]3 years ago

4 0

Answer:

0.150 M

Explanation:

The molarity of the solution can be calculated using the formula;

C1V1 = C2V2

Where; C1 = initial concentration of solution

C2 = final concentration of solution

V1 = volume of initial solution

V2 = volume of final solution

According to this question, C1 = 0.200M, C2 = ?, V1 = 75mL, V2 = 100.0mL

Hence,

C1V1 = C2V2

0.200 × 75 = C2 × 100

15 = 100C2

C2 = 15/100

C2 = 0.150M

Therefore, the molarity of the final KCl solution is 0.150M

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Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs

saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of $Ag^+$ when CuBr just begins to precipitate is, $1.34\times 10^{-6}M$

Percent of $Ag^+$ remains is, 0.0018 %

Explanation :

$K_{sp}$ for CuBr is $4.2\times 10^{-8}$

$K_{sp}$ for AgBr is $7.7\times 10^{-13}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

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The solubility equilibrium reaction will be:

$AgBr\rightleftharpoons Ag^++Br^-$

The expression for solubility constant for this reaction will be,

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Number of neutrons=41, and it is metalloid

<h3>Further explanation</h3>

Group 14⇒valence electron = 4(ns²np²)

Period 4⇒n=4

So electron configuration of the element :

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p² = 32 electron=atomic number

The element with atomic number 32, which is in period 4 and group 14 is Ge-Germanium

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