<u>Answer:</u> The mass of iron in the ore is 10.9 g
<u>Explanation:</u>
We are given:
Mass of iron (III) oxide = 15.6 g
We know that:
Molar mass of Iron (III) oxide = 159.69 g/mol
Molar mass of iron atom = 55.85 g/mol
As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.
To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:
In 159.69 g of iron (III) oxide, mass of iron present is 
So, in 15.6 g of iron (III) oxide, mass of iron present will be = 
Hence, the mass of iron in the ore is 10.9 g
I believe the answer is D
What we are give: Concentration of base (CB) = 3.4 ×
Then convert all volume in ml to L.
Volume of base (VB) 25.0ml = 0.025L
Volume of acid (VA) 16.6ml = 0.0166L
Now that we have everything we use the formula CAVA=CBVB.
Make 'CA' the subject then solve.
CA=
We can solve the equation and show the solution below:
Oxygen atomic number is 16.
Phosphorus atomic number is 32.
We have the molecular weight:
Molecular weight = (31*4) + (16*10)
Molecular weight = 284 grams/mol
Solving for the grams:
0.4 mole (for P4) * (1 mol P4O10/1 mol P4) * (284 grams P4O10/1 mole P4O10)
Total grams = 113.6
The answer is 113.6 grams.
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