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Rudik [331]
4 years ago
7

For which salt, k2so4, li2so4, kclo4, or nh4cl, will increasing the temperature of the water have the greatest change in solubil

ity per 100 g solvent
Chemistry
1 answer:
olasank [31]4 years ago
8 0

The correct answer is NH₄Cl.  

With the increase in temperature, endothermic reactions are favored. The dissolution of NH₄Cl in water is an example of an endothermic reaction. When the temperature is increased more dissolution of the salt takes place, thus, the solubility of the salt increases.  

Hence, it can be said that NH₄Cl will exhibit the greatest change in the solubility per 100 g solvent with an elevation in temperature.  

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A solution of 2-propanol and 1-octanol behaves ideally. Calculate the chemical potential of 2-propanol in solution relative to t
andrew-mc [135]

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The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

Explanation:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:

\mu (l) = \mu ^{\circ} (l) + R*T*ln(x)

<u>Where:</u>

<em>μ (l): is the chemical potential of 2-propanol in solution    </em>

<em>μ° (l): is the chemical potential of pure 2-propanol   </em>

<em>R: is the gas constant = 8.314 J K⁻¹ mol⁻¹ </em>

<em>T: is the temperature = 82.3 °C = 355.3 K </em>

<em>x: is the mole fraction of 2-propanol = 0.41 </em>

\mu (l) = \mu ^{\circ} (l) + 8.314 \frac{J}{K*mol}*355.3 K*ln(0.41)

\mu (l) = \mu ^{\circ} (l) - 2.63 \cdot 10^{3} J*mol^{-1}

\mu (l) - \mu ^{\circ} (l) = - 2.63 \cdot 10^{3} J*mol^{-1}  

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

I hope it helps you!    

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4 years ago
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