The volume of water in a transfer pipette is 15.23 ml. A 6.7 mL volume of water was then transferred out. Considering
significant figures, what is the new volume?

The new volume of water in the pipette is 8.53 mL

Explanation:

The given information are;

The initial volume of water in the transfer pipette = 15.23 mL

The amount of water transferred from the pipette = 6.7 mL

The new volume of water in the transfer pipette is given as follows;

New volume of water in the pipette =  Initial volume of water in the pipette - Volume of water transferred out

Therefore, we have;

New volume of water in the pipette = 15.23 mL - 6.7 mL = 8.53 mL

The new volume of water in the pipette = 8.53 mL

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%