You have a partially filled party balloon with 2.00 g of helium gas. you then add 2.74 g of hydrogen gas to the balloon. assuming constant temperature and pressure, how many times bigger is the party balloon - comparing before and after the hydrogen gas has been added?

First, we need to get moles of He:

moles of He = mass/molar mass of He

when the molar mass of He = 4 g/mol
                     
and when the mass = 2 g 
 
by substitution:

  moles of He = 2 g / 4 g/mol 

                        = 0.5 moles

and when V = nRT/P and n is the number of moles

so, V1 = 0.5RT/P

then, we need moles of H2 = mass / molar mass 

                                               = 2.74g / 2g/mol

                                               = 1.37 moles

∴ moles of He + moles of H2 = 0.5 moles + 1.37 moles 

                                                 = 1.87 moles

so, V2= 1.87RT/P

from the V1 and V2 formula:

∴V2/V1 = 1.87 / 0.5

              = 3.74 

∴ the party balloon is 3.74 times bigger

The total number of electrons is 14

A.(nh4)(2co3)is the correct answer