Consider the following data for air trapped in a flask: pressure = 0.988 atm room temperature = 23.5°c volume of the flask = 1.042 l for this calculation, assume air is 78.5% nitrogen and 21.5% oxygen (by number of moles). r = 0.0821 l•atm•mol-1•k-1 n = 14.01 g/mol o = 16.00 g/mol what is the mass of air in the flask

total mass will be =  = 1.207g

Explanation:

First what is given  

Pressure P= 0.988 atm                  Room TemperatureT = 23.5°C= 296.5 K

Volume V= 1.042 L

Nitrogen in air is 80 % (moles number) = 0.8                

Ideal gas constant R = 0.0821 L atmmol-1K-1

Given mass of N = 14.01 g/mol

Given mass of oxygen O = 16.00 g/mol

Total number of moles = ?

So first we have to find the total number of moles by using formula  

Total number of moles  

n = PV/ RT  

adding the values  

moles n= 0.988atm x 1.042L / (0.0821L-atm/mole-K x 296.6K)  

  = 1.0294 / 24.35

= 0.042 moles (total number of moles)

So by using Nitrogen percentage  

Moles of nitrogen = total moles x 80/100

                            = 0.042moles x 0.8  

So moles of O2= Total moles – moles of N2  

                       =   0.042moles - 0.034moles

    Moles of O2 = 0.008moles

Now for finding the mass of the N2 and oxygen  

Mass of Nitrogen N2 = no of moles x molar mass

                           = 0.034 x 28

                           = 0.952 g

Mass of oxygen O2 = no of moles x molar mass

                           = 0.008 x 32

                           =  0.256 g

total mass will be = Mass of Nitrogen N2 + Mass of oxygen O2  

                             =0.952 g  + 0.256 g

                               =1.207g

- "the acute form of diverticulitis can manifest itself with one or more severe attacks of infection and inflammation.

the inflammation of diverticulitis can eventually result in a bowel obstruction, which may cause constipation, thin stools, diarrhea, abdominal swelling or bloating, and abdominal pain."

d

explanation:

rutherford's experiment showed that the atom does not contain a uniform distribution of charge.