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Elis [28]
2 years ago
5

The density of NaCl( s) is 2.165 g cm 3 at 25 C. How will the solubility of NaCl in water be affected by an increase in pressure

Chemistry
1 answer:
sammy [17]2 years ago
7 0

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.

Learn more about solubility here: brainly.com/question/11963573

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If 27.3% of a sample of silver-112 decays in 1.52 hours, what is the half-life (in hours to 3 decimal places)?
ICE Princess25 [194]

<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.

<u>Explanation:</u>

All radioactive decay processes undergoes first order reaction.

To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}

where,

k = rate constant = ?

t = time taken = 1.52 hrs

[A_o] = Initial concentration of reactant = 100 g

[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g

Putting values in above equation, we get:

k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}

To calculate the half life period of first order reaction, we use the equation:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life period of first order reaction = ?

k = rate constant = 0.2098hr^{-1}

Putting values in above equation, we get:

t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs

Hence, the half life of the sample of silver-112 is 3.303 hours.

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