How would you prepare 3.5 L of a 0.9M solution of KCl?

1 answer:

4 0

$V=3,5L\\
Cm=0,9M\\
M_{KCl}=74\frac{g}{mol}\\\\
C_{m}=\frac{n}{V}\\\\
n=\frac{m}{M}\\\\
C_{m}=\frac{m}{MV} \ \ \ \Rightarrow \ \ \ m=C_{m}MV\\\\
m=0,9\grac{mol}{L}*74\frac{g}{mol}*3,5L=233,1g$

B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.

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Is the osmotic pressure of a 0.10 M solution of NaCl greater than, less than, or equal to that of a 0.10 M solution of KBr?a. eq

jonny [76]

Answer:

a. equal to

Explanation:

The osmotic pressure is calculated by the formula:

π = i * M * R * T

Where π is the osmotic pressure, M is the concentration, R is a constant, T is temperature and i is the van't Hoff's factor (the number of ions a compound forms when dissolved in water, for both NaCl and KBr is 2).

Because R is always the same, and Temperature and Concentration are equal between the two solutions, the osmotic pressure of both solutions are also equal.

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A weather system moving through the American Midwest produced rain with an average pH of 5.02. By the time the system reached Ne

bulgar [2K]

Answer:

The rain falling in New England is 2.29 times more acidic than the one in the American Midwest.

Explanation:

The acidity of a solution depends on the concentration of H⁺ ions ([H⁺]). We can calculate this concentration from the pH using the following expression.

pH = -log ([H⁺])

American Midwest

pH = -log ([H⁺])

5.02 = -log ([H⁺])

[H⁺] = antilog (-5.02) = 9.55 × 10⁻⁶ M

New England

pH = -log ([H⁺])

4.66 = -log ([H⁺])

[H⁺] = antilog (-4.66) = 2.19 × 10⁻⁵ M

The ratio of concentrations is:

$\frac{2.19 \times 10^{-5} M }{9.55 \times 10^{-6} M} =2.29$