Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.

Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.

We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:

$81.6gFe_2(SO_4)_3(\frac{1mol}{399.88g})$

= $0.204molFe_2(SO_4)_3$

$18.4gH_2O(\frac{1mol}{18.02g})$

= $1.02molH_2O$

Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.

$Fe_2(SO_4)_3=\frac{0.204}{0.204}$ = 1

$H_2O=\frac{1.02}{0.204}$ = 5

There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is $Fe_2(SO_4)_3.5H_2O$ and the name of the hydrate is Iron(III)sulfate pentahydrate.

3 0

3 years ago

A wood pole is 12 metres long and is cut into 5 equal sections. If each cut wastes 10 cm of pole, what length would each section

Sophie [7]

Each section would be 230 centimeters or 2.30 meters long.

<h3>Sectional division</h3>

The original wood pole is 12 meters long.

12 meters = 12 x 100 = 1200 centimeters

It is to be cut into 5 equal sections: 1200/5 = 240 centimeters

But each cut wastes 10 cm of the pole.

240 - 10 = 230 centimeters.

Thus, each section would be 230 centimeters or 2.30 meters long.

More on sectional divisions can be found here: brainly.com/question/15381501

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6 0

2 years ago