Assuming all gas is removed from the tank when filling balloons, how many 0.75 L balloons can be filled from a tank that contains 375 g CO2 at 1.0 atm of pressure at a temperature of 298 K

A tank that contains 375 g of CO₂ at 1.0 atm and 298 K can be used to fill 277 0.75 L-balloons.

First, we will convert 375 g of CO₂ to moles using its molar mass (44.01 g/mol).

1 mole of CO₂ at 1.0 atm and 298 K occupies 24.4 L. The volume occupied by 8.52 moles of CO₂ is:

8.52 moles of CO₂ at 1.0 atm and 298 K occupies 208 L. The number of 0.75 L-balloons that can be filled is:

A tank that contains 375 g of CO₂ at 1.0 atm and 298 K can be used to fill 277 0.75 L-balloons.

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278 balloons can be aired.

Explanation:

We apply the Ideal Gases Law to determine the total available volume for each baloon of 0.75L. So we need to divide the volume we obtain by the volume of each baloon.

We determine the moles of CO₂

375 g. 1 mol / 44g = 8.52 moles

V = (n . R . T) / P → (8.52 mol . 0.082 . 298K) / 1 atm = 208.3 L

That's the total volume from the tank, so we can inflate (208.3 / 0.75) = at least 278 balloons

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.