# 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

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

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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