A chemist has a sample of hydrated Li2SiF6 and it weighs 0.4813 grams. He heats it strongly to drive off the water of hydration, and after heating, he found that the anhydrous compound has a mass of 0.391 grams. What is the percent of water in the hydrate?

percentage of water = 18.76%

Explanation:

A chemist has a sample of hydrated Li2SiF6  and it weighs 0.4813 grams.

The overall weight of the compound is 0.4813 grams.

weight of hydrated sample = 0.4813 grams.

weight of anhydrous compound = 0.391 grams

percentage weight of water = mass of water/mass of the hydrated compound × 100

mass of water = mass of hydrated compound - mass of anhydrous compound

mass of water = 0.4813 - 0.391  = 0.0903 grams.

percentage of water = 0.0903/0.4813 × 100

percentage of water = 9.03/0.4813

percentage of water = 18.761687097

percentage of water = 18.76%

18.8%

Explanation:

Step 1:

The following data were obtained from the question.

Mass of hydrated Li2SiF6 = 0.4813 g

Mass of anhydrous Li2SiF6 = 0.391 g

Step 2:

Determination of the mass of the water in the hydrate.

Mass of water = (Mass of hydrated Li2SiF6) - (Mass of anhydrous Li2SiF6)

Mass of water = 0.4813 - 0.391

Mass of water = 0.0903 g

Step 3:

Determination of the percentage of water in the hydrate.

The percentage of water = Mass of water/ mass of hydrate x 100

The percentage of water = 0.0903/0.4813 x 100

The percentage of water = 18.8%

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !