The solubility of NaCl in water will not be affected by an increase in pressure.
We know that the density of NaCl(s) in 2.165 g/cm³ at 25 °C and we want to know how will its solubility in water be affected when the pressure is increased.
<h3>What is solubility?</h3>
Solubility is the maximum mass of a solute that can be dissolved in 100 grams of solvent at a determined temperature.
The solubility of a solid, such as NaCl, in a liquid, is mainly affected by the temperature. However, since solids are not compressible, an increase in pressure will not affect its solubility.
On the other hand, the solubility of gases in water will increase with an increase in pressure, as stated by Henry's law.
The solubility of NaCl in water will not be affected by an increase in pressure.
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When temperature increases pressure also increases.
When temperature decreases pressure decreases.
That is why you have low tire pressure when it’s cold out
Answer:
B. as a food preservative in the manufacture of detergents
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The concentration of a dextrose solution prepared by diluting 14 ml of a 1.0 M dextrose solution to 25 ml using a 25 ml volumetric flask is 0.56M.
Concentration is defined as the number of moles of a solute present in the specific volume of a solution.
According to the dilution law, the degree of ionization increases on a dilution and it is inversely proportional to the square root of concentration. The degree of dissociation of an acid is directly proportional to the square root of a volume.
M₁V₁=M₂V₂
Where, M₁=1.0M, V₁=14ml, M₂=?, V₂=25ml
Rearrange the formula for M₂
M₂=(M₁V₁/V₂)
Plug all the values in the formula
M₂=(1.0M×14 ml/25 ml)
M₂=14 M/25
M₂=0.56 M
Therefore, the concentration of a dextrose solution after the dilution is 0.56M.
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Answer:
0.42%
Explanation:
<em>∵ pH = - log[H⁺].</em>
2.72 = - log[H⁺]
∴ [H⁺] = 1.905 x 10⁻³.
<em>∵ [H⁺] = √Ka.C</em>
∴ [H⁺]² = Ka.C
∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.
<em>∵ Ka = α²C.</em>
Where, α is the degree of dissociation.
<em>∴ α = √(Ka/C) </em>= √(8.065 x 10⁻⁶/0.45) = <em>4.234 x 10⁻³.</em>
<em>∴ percentage ionization of the acid = α x 100</em> = (4.233 x 10⁻³)(100) = <em>0.4233% ≅ 0.42%.</em>
