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masya89 [10]
3 years ago
11

Rank the following substances in order from most soluble in water to least soluble in water: ethane, C2H6; 2-butanol, C4H9OH; co

pper(II) sulfate, CuSO4; and propane, C3H8.
Chemistry
1 answer:
myrzilka [38]3 years ago
6 0

Answer: CuSO4>2-butanol> C4H9OH>C2H6>C3H8

Explanation:

Ionic substances are highly soluble in water. CuSO4 is an ionic salt composed of Cu2+ and SO42- ions. These ions are solvated in water hence the salt easily dissolves in the surrounding water molecules. this accounts for the fact that it is the most soluble of the five substances in water. 2-butanol contains the -OH group which interacts with water by hydrogen bonding. This accounts for the miscibility of alkanols with water in all proportions. however, this solubility in water decreases as the alky backbone increases. hence C4H9OH is less soluble in water that 2-butanol. Alkanes are usually not quite soluble in water but again solubility decreases with increase in the alky backbone hence propane is less soluble in water than ethane.

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Answer : The mole fraction of NaCl in a mixture is, 0.360

Explanation : Given,

Moles of NaCl = 7.21 mole

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Moles of LiCl = 3.42 mole

Now we have to calculate the mole fraction of NaCl.

\text{Mole fraction of }NaCl=\frac{\text{Moles of }NaCl}{\text{Moles of }NaCl+\text{Moles of }KCl+\text{Moles of }LiCl}

Now put all the given values in this formula, we get:

\text{Mole fraction of }NaCl=\frac{7.21}{7.21+9.37+3.42}=0.360

Therefore, the mole fraction of NaCl in a mixture is, 0.360

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43. A stock glucose standard has a concentration of 1,000 mg/dL. A 1/5 dilution of this standard is made. What would be the fina
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The final concentration of the diluted standard is 0.2 mg/dL.

<h3 /><h3>What is concentration of glucose standard after 1/5 solution?</h3>

Using the dilution formula:

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Learn more about dilution at: brainly.com/question/24881505

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