Hello i need help with this question

0.14 M

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V₁) = 35 mL

Molarity of stock solution (M₁) = 3 M

Volume of diluted solution (V₂) = 750 mL

Molarity of diluted solution (M₂) =?

The molarity (i.e concentration) of the diluted solution can be obtained as follow:

M₁V₁ = M₂V₂

3 × 35 = M₂ × 750

105 = M₂ × 750

Divide both side by 750

M₂ = 105 / 750

M₂ = 0.14 M

Therefore, the concentration of the diluted solution is 0.14 M

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%

6.4 cm (max) - 25 cm (min)