3 ..............................
<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
Temperature. Water is an example. When water is at room temp. its liquid. When water is at boiling temp. It is a gas. And when water is at freezing temp. Its a solid.
I will use this conversion:
1m = 1000 mm => 1 = 1m / 1000mm
1 m = 100 cm=> 1 = 1m / 100cm
1m = 10 dm=> 1 = 1m / 10 dm
So,
B) 3.8 * 10^7 cm^3 = 3.8 * 10^7 cm^3 * [1m / 100cm]^3 = 38 m^3
C) 3.8 * 10^5 dm^3 * [1m / 10 dm]^3 = 380 m^3
D) 3.8 * 10^8 mm^3 * [1m / 1000 mm]^3 = 0.38 m^3
Now you can compare the four volumes and conclude which is the largest.
Answer: option C) 3.8 * 10^5 dm^3