Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same?

Chemistry

1 answer:

Marina CMI [18]4 years ago

8 0

Answer:

option d

Explanation:

Molecular sizes of gaseous molecules are very less. Volume occupied by the all the molecules of the gases are very less or negligible as compared to the container in which it is kept. Therefore, most of the volume occupied by gaseous molecules are negligible.

Volume occupied by the gaseous molecules are actually the volume of the container and its does not depend upon the amount, molecular mass or dipole moment of the gaseous molecules.

Therefore, the correct option is d ‘Because most of the volume occupied by the substance is empty space.’

You might be interested in

Compute the sugar content in an 8 oz sample of a soft drink. If the sugar content as per label on the product =10g per 100ml.

Nataly_w [17]

Answer:

$m_{sugar}=23.7g\ sugar$

Explanation:

Hello,

In this case, we can first compute the volume of the sample in mL from the ounces:

$8oz*\frac{29.5735mL}{1oz} =236.6mL$

Thus, with the volume of the sample, we can compute the amount of sugar given the 10 g of sugar per 100 mL of soft drink as shown below:

$m_{sugar}=236.6mL*\frac{10g\ sugar}{100mL}\\ \\m_{sugar}=23.7g\ sugar$

Best regards.

6 0

3 years ago

A car mechanic had a sealed bucket containing a substance in the gas phase. She left the bucket outside over the weekend. When s

motikmotik

A phase change from Gas to liquid means that it was cold outside and so the molecules went from being very free moving to forming bonds with one another due to the cold.

4 0

3 years ago

How many grams of barium sulfate can be produced from the reaction of 2.54 grams sodium sulfate and 2.54 g barium chloride? Na2S

Rama09 [41]

Answer: 2.796 grams

Explanation:

$Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of sodium sulphate}=\frac{2.54g}{142g/mol}=0.018moles$

$\text{Number of moles of barium chloride}=\frac{2.54g}{208g/mol}=0.012moles$

According to stoichiometry:

1 mole of $BaCl_2$ reacts with 1 mole of $Na_2SO_4$

0.012 moles of $BaCl_2$ will react with= $\frac{1}{1}\times 0.012=0.012moles$ of $Na_2SO_4$

Thus $BaCl_2$ is the limiting reagent as it limits the formation of product. $Na_2SO_4$ is the excess reagent as (0.018-0.012)=0.006 moles are left unused.

1 mole of $BaCl_2$ produces 1 mole of $BaSO_4$