Answer:- Formula of the hydrate is 
and it's name is Iron(III)sulfate pentahydrate.
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:
= 
= 
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.
= 1
= 5
There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is and the name of the hydrate is Iron(III)sulfate pentahydrate.
There are 237. 5 g of Sulfur,S in 475 g of SO2?
<h3 />
<h3>Calculation of grams of Sulfur</h3>
From the question, we can say that
- The molar mass of sulfur = 32 g/mol
- The molar mass of oxygen = 16 g/mol
Therefore,
The molar mass for SO2 = 32 + (16 × 2) g/mol = 64 g/mol
Now,
If 1 mole of SO2 contains 1 mole of S
Then 64 g of SO2, will contain 32g of S;
Such that
475 g of SO2 will give { 
} = 237. 5 g of Sulfur.
Learn more about molar mass here :brainly.com/question/18291695
<span>A. Mechanic agitations</span>
Answer:
<h2>The answer is 9 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 216 g
volume = 24 cm³
The density is
We have the final answer as
<h3>9 g/cm³</h3>
Hope this helps you
<span>Grams of solute per 100 grams of water</span>
