Which substance is the least likely to dissolve in water?

Chemistry

2 answers:

Neko [114]4 years ago

5 0

Answer:

Vegetable oil

Explanation:

because it is nonpolar object

katrin2010 [14]4 years ago

3 0

The answer would be D

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What do genotype will appear in boxes 2 and 3?

VMariaS [17]

Tt is the genotype that will appear in boxes two and three.

If you look at the column and row that intersect to form boxes two and three, you will see that they are T and t. That is the best way I can describe it, sorry if it’s confusing.

3 0

3 years ago

Read 2 more answers

7. Diethyl ether burns in air according to the following equation. C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l) If 7.15 L of CO2 is

iren [92.7K]

From the stoichiometry of the combustion reaction, we can see that 7.4 L of oxygen is consumed.

<h3>What is combustion?</h3>

Combustion is a reaction in which a substance is burnt in oxygen. The equation of the reaction is; C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l)

We can obtain the number of moles of CO2 from;

PV = nRT

n = 1.02 atm * 7.15 L/0.082 atm LK-1mol-1 * (125 + 273) K

n = 7.29 /32.6

n = 0.22 moles

If 6 moles of oxygen produces 4 moles of CO2

x moles of oxygen produces 0.22 moles of CO2

x = 0.33 moles

1 mole of oxygen occupies 22.4 L

0.33 moles of oxygen occupies 0.33 moles * 22.4 L/ 1 mole

= 7.4 L of oxygen

Learn more about stoichiometry: brainly.com/question/13110055

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6 0

3 years ago

When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr

konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.