Answer:
C₃H₆O₃
Explanation:
Data:
EF = CH₂O
MM = 90. g/mol
Calculations:
EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u
The molecular formula is an integral multiple of the empirical formula.
MF = (EF)ₙ
MF = (CH₂O)₃ = C₃H₆O₃
Calculate the mass of a sample of lead (cPb = 0.16 J/g℃) when it loses 200 J cooling from 75.0℃ to 42.0℃.
1 answer:
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(i) 1 mole of Ar =6.022×10^23
atoms of Ar
52 mol of Ar=52×6.022×10 ^23
atoms of Ar =3.131×10 ^25 atoms of Ar
(ii) Atomic mass of He = 4amu
Or
4amu is the mass of He atoms = 1
Therefore 52 amu is the mass of He atoms=0.25×52=13 atoms of He
(iii) Gram atomic mass of He = 4g
Or
4g of He contains =6.022×10 ^23
atoms
Therefore 52 g of He contains=
6.022×10 ^23/4×52 =7.83×10 ^24 atoms
