Answer:
#8 : 21.5 L of CHâ‚„
#9 : 36 g of Hâ‚‚O
#9 : 5.4 x 10²² molecules of CO₂
Explanation:
#8 : Part 1.
Data given:
no. of moles of CHâ‚„ = 0.960 mol
volume of CHâ‚„ = ?
Solution
volume can be calculated by following formula
     No. of moles = Volume of gas / molar volume
Rearrange the above equation
      Volume of gas = No. of moles x molar volume . . . . . . (1)
Where
molar volume = 22.4 L/mol
Put values in equation 1
      Volume of gas = 0.960 mol x 22.4 L/mol
      Volume of gas = 21.5 L
So,
By calculation 0.960 moles have 21.5 L volume of CHâ‚„
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#9 : Part 2.
Data given:
no. of moles of Hâ‚‚O = 2.0 mol
mass of Hâ‚‚O = ?
Solution
volume can be calculated by following formula
     No. of moles = mass in grams / molar mass
Rearrange the above equation
      mass in grams = No. of moles x molar mass . . . . . . (2)
Where
molar mass of Hâ‚‚O = 2 (1) + 16
molar mass of Hâ‚‚O = 18 g/mol
Put values in equation 2
   mass in grams = 2.0 mol x 18 g/mol
    mass in grams = 36 g
So,
By calculation 2.0 moles have 36 g mass of Hâ‚‚O
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#10 : Part 3.
Data given:
volume of COâ‚‚ = 2 L
no. of molecules of COâ‚‚ = ?
Solution
First we have to find out number of moles of COâ‚‚
Following formula will be used
     No. of moles = Volume of gas / molar volume
Where
molar volume = 22.4 L/mol
Put values in above equation
      No. of moles = 2 L / 22.4 L/mol
      No. of moles = 0.0893 mol
So,
No. of moles of COâ‚‚ = 0.0893 mol
Now
we will calculate number of molecules by using following formula
     No. of moles = no. of molecules / Avogadro's number
Rearrange the above equation
      no. of molecules = No. of moles x Avogadro's number . . . . . . (3)
Where
Avogadro's number = 6.022 x 10²³
Put values in above equation 3
      no. of molecules =  0.0893 mol x 6.022 x 10²³
      no. of molecules =  5.4 x 10²²
So,
By calculation 2 L of CO₂ have 5.4 x 10²² molecules of CO₂
