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Rina8888 [55]
3 years ago
13

One of these two substances is a liquid at room temperature and the other one is a gas. Which one is the liquid and why?

Chemistry
1 answer:
Fofino [41]3 years ago
8 0

Answer:

B. CH3CH2OH

Explanation:

Ethanol has a chemical formula of CH3CH2OH, it is the second member of the series in the alkanol family. Ethanol is a colourless, volatile liquid with a characteristic smell and taste. It is readily soluble in water in all proportions. It has a boiling point of 78° C. The physical properties such as the solubility of alkanols are affected by the presence of hydrogen bonding. The hydroxyl group is capable of bonding to other alkanol molecules. The boiling points rise with increasing molecular mass.

    Hydrogen bonding helps the molecules to stick together. For example comparing the boiling point of pentane ( 36° C) with that of butan-1-ol (118° C) , the boiling point of alkanol is much higher even though the two compound are of similar relative molecular mass. This is due to the presence of hydrogen bonds in butanol.

    Hydrocarbons are not soluble in water but alkanols are soluble in water because of the hydroxyl groups in the molecules can form hydrogen bond with water. Solubility of alkanol in water decreases as the number of carbon atom increases. Primary alcohol with more than five carbon atoms are insoluble in water.

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a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill
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Answer:

1223.38 mmHg

Explanation:

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

Also,  

Moles = mass (m) / Molar mass (M)

Density (d)  = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

PM=dRT

Given that:-

d = 1.80 g/L

Temperature = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (32 + 273.15) K = 305.15 K

Molar mass of nitrogen gas = 28 g/mol

Applying the equation as:

P × 28 g/mol  = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

⇒P = 1223.38 mmHg

<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>

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