Question

calculate the weight of carbon momoxide having same number of O2 atoms as are present in 22g of carbon dioxide

Answer 1

I assume O2 is an abbreviation for oxygen and you DON'T mean O2 atoms.
22 g CO2 = 22/molar mass = 22/44 = 0.50 mole CO2. Since 1 mole CO2 contains 6.022E23 molecules of CO2, 0.50 mole will contain just half that number of CO2 molecules(3.011E23) and that number of atoms(6.022E23).
So 1 mole CO should contain the same number of oxygen atoms.
1 mole CO (28 grams) contains 6.02E23 molecules and 6.022E23 atoms oxygen.

Answer 2

28g

Answer 3

CO2
W(gram)=22g
Molecular weight= 44u
N= no. of molecules
NA= Avogadro no.

N/NA=w(gram)/molecular weight
N/6.022×10^23=22/44
N=22×6.022×10^23/44
N=3.011×10^23 molecules

No. of oxygen atoms in CO2=3.011×10^23×2
=6.022×10^23 atoms

No. of oxygen atoms in CO2=no. of CO molecules=
6.022 ×10^23 atoms

CO
W(gram)=?
Molecular weight=28u
No. of molecules=6.022×10^23
NA =Avogadro no.=6.022×10^23

W(gram)/molecular weight=no. of molecules/ NA
W(gram) /28=6.022×10^23/6.022×10^23
W(gram)=1×28
=28 grams

Answer 4

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Answer 5

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Answer 6

Yes it is good

Answer 7

Yes I understand this method

Answer 8

Thanks it helped me

Answer 9

I understand very clearly