Answer:
The number of moles of O atom in 
mol of 
= 1.6
Explanation:
1 molecule of contains 2 atoms of O
So, 
molecules of contains 
atoms of O.
We know that 1 mol of an atom/molecule/ion represents 
numbers of atoms/molecules/ions respectively.
So, molecules of is equal to 1 mol of .
atoms of O is equal to 2 moles of O atom.
Hence, 1 mol of contains 2 moles of O atom.
Therefore, mol of contains 
moles of O atom or 1.6 moles of O atom.
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: 24 gram is a final Answer.
Explanation:1 mol of carbon weight is 12 gram then
12.044×1023 contain 2 mole carbon so
The smallest atomio radius is helium
The excess reactant is Aluminum.
<u>Explanation:</u>
We have to write the balanced equation as,
4 Al+ 3 O₂ → 2 Al₂O₃
According to the molar ratio 4: 3, from the given balanced equation, we can say that 4 atoms of Al reacted with 3 molecules of oxygen.
Given that 10 atoms of aluminum reacts with 6 molecules of oxygen, as per the ratio only 8 atoms of Aluminum is required to react with 6 molecules of oxygen, so excess reactant is Aluminum.
