chemistry

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

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  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.

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  2. 28g

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  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

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  4. This type of solution gives me a lot of help I searched this question at many sites but this is the most easiest

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  5. Explained very properly!!!!Thank you

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  6. Yes it is good

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  7. Yes I understand this method

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  8. Thanks it helped me

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  9. I understand very clearly

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