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2 years ago

Which one of the following compounds will be least soluble in water?

2 answers:

8 0

Water is a polar solvent. Since, 'Like dissolves like' a polar compound will be most soluble in water whereas a non-polar compound will be the least soluble.

The increasing order of polarity for the given organic compounds is:-

CH3OCH3 < CH3NH2 < CH3CHO < CH3CH2OH

Ether is the least polar and hence least soluble in water.

Juliette [100K]2 years ago

7 0

Answer: The correct answer is (A).

Explanation :

$CH_3OCH_3$ that is dimethyl ether will be least soluble in water because extent of hydrogen bonding interaction with water molecules are weakest in comparison to other listed options.

Where as :

In $CH_3CH_2OH$ that is ethanol,in which extent of inter-molecular hydrogen bonding with water molecules will be highest due to the presence of polar (-OH) group.Where as ( $-NH_2$ ) group is less polar than that of the (-OH) group.
in $CH_3NH_2$ that is methyl amine in which extent of hydrogen bonding will higher than than that of the $CH_3CHO$ that is ethanal. This is due to presence of two hydrogen atoms in primary amine which are available for inter-molecular hydrogen bonding with water molecules.

So, the order of solubility from least soluble to highly soluble compound in water will be:

$CH_3OCH_3$

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3 years ago

Which element does not have the same number of electrons in its outermost shell as the other elements in its group?(1 point)

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2 years ago

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Answer:

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Explanation:

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1 year ago

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2 years ago

Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co

amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as $p_{d}$ and the proportion of air that is by-passed as $p_{bp}$ , being $p_{d}+p_{bp}=1$ .

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

$0.004 = 0.016*p_{bp}+ 0.002*p_{d}$

Replacing the first equation in the second one we have

$0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143$

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

$100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW$

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

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2 years ago