Answer:
Explanation:
Your goal here is to figure out the mass of iron and the mass of sulfur present in exactly
100 g
of this compound, which is what you need in order to know the compound's percent composition.
The problem provides you with the mass of iron and with the mass of sulfur present in
27.9 g
of this compound, so you must use this information to scale up your sample from
27.9 g
to
100 g
.
To do that, set up the known composition of the compound as a conversion factor.
To find the number of grams of iron present in
100 g
of compound, set up the conversion factor like this
17.6 g iron
27.9 g compound
You will end up with
100
g compound
⋅
17.6 giron
27.9
g compound
=
63.1 g iron
To find the mass of sulfur present in
100 g
of compound, you can either use a conversion factor set up in a similar manner
100
g compound
⋅
10.3 g sulfur
27.9
g compound
=
36.9 g sulfur
or use the fact that the compound contains only iron and sulfur to say that the mass of sulfur will be equal to
100 g compound
−
63.1 g iron = 36.9 g sulfur
You now know the mass of iron and the mass of sulfur present for every
100 g
of compound, so you can say that the compound has the following percent composition
63.1% iron
36.9% sulfur
Here
%
means per
100 g
of compound.
Answer:
D.
Explanation:
I'm not 100 percent but I went to three different website articles that said so
Hope this helps have a great day!
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Answer is: mass oxygen is 129,28 grams.
Chemical reaction: 2C₁₈H₃₈ + 55O₂ → 36CO₂ + 38H₂O.
m(C₁₈H₃₈) = 37,5 g.
n(C₁₈H₃₈) = m(C₁₈H₃₈) ÷ M(C₁₈H₃₈).
n(C₁₈H₃₈) = 37,5 g ÷ 254 g/mol.
n(C₁₈H₃₈) = 0,147 mol.
From chemical reaction: n(C₁₈H₃₈) : n(O₂) = 2 : 55.
n(O₂) = 4,04 mol.
m(O₂) = 4,04 mol · 32 g/mol.
m(O₂) = 129,28 g.
