From 200 mg of co2 10^21 molecules are removed how many grams of co2 are left
Answers
Answered by
0
Step-by-step explanation:
C O
12. 16
weight 12+32= 44 amu
44g of co2 contains = 6 × 10^23mo.
therefore
200g contains=
=((6 × 10^23)/44)×200
= 27.2×10^23 molecules
now,
after removing 10^21 molecules
= (27.2×10^23)-10^21
= 10^21((27.2× 10^2)-1)
= 2719× 10^21
= 2.719 × 10^24
Answered by
1
Answer:
One mole of atoms =6.023×10
23
atoms = Gram atomic mass of element
44 grams of CO
2
=1moleCO
2
=6×10
23
molecules of CO
2
200mg=
44×1000
200×1
moles=
44×1000
200×10
23
×6.023 molecular of CO
2
=27.37×10
20
molecular of CO
2
=2.737×10
21
molecules of CO
2
molecules left =1.737×10
21
molecular of CO
2
moles of CO
2
left =
6.023×10
23
1.737×10
4
=0.29×10
−2
moles of CO
2
=2.9 mill moles
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