What type of solvent would be required to dissolve nonpolar compounds?

2 answers:

4 0

For example, ionic compounds, which are very polar, are often soluble in the polar solvent water. Nonpolar substances are likely to dissolve in nonpolar solvents. For example, nonpolar molecular substances are likely to dissolve in hexane, a common nonpolar solvent.

AnnZ [28]2 years ago

3 0

Answer:

Solvents composed of polar molecules, such as water, dissolve other polar molecules, such as table salt, while nonpolar solvents, such as gasoline, dissolve nonpolar substances such as wax.

Explanation:

In chemistry, a common rule for determining if a solvent will dissolve a given solute is "like dissolves like." Solvents composed of polar molecules, such as water, dissolve other polar molecules, such as table salt, while nonpolar solvents, such as gasoline, dissolve nonpolar substances such as wax.

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What would be the major product obtained from hydroboration–oxidation of the following alkenes?

zmey [24]

Answer:

a. 3-methylbutan-2-ol

b. 2-methylcyclohexan-1-ol

Explanation:

For this reaction, we must remember that the hydroboration is an "anti-Markovnikov" reaction. This means that the "OH" will be added at the least substituted carbon of the double bond.

In the case of 2-methyl-2-butene, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore the "OH" will be added to carbon three producing 3-methylbutan-2-ol.

For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore the "OH" will be added to carbon 2 producing 2-methylcyclohexan-1-ol.

See figure 1

I hope it helps!

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3 years ago

How does a field experiment differ from a natural experiment?

Marina CMI [18]

Answer : behavior in a field experiment is more likely to reflect real life because of its natural setting, i.e. higher ecological validity than a lab experiment

Explanation:

8 0

2 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$