Volume of 12.4M HCl would you need to make 500mL of a 3.5 M HCl

141 mL

Explanation:

We can use the simple equation of:

1st Concentration x 1st Volume = 2nd Concentration x 2nd Volume

Input the values in:

12.4M x (1st Volume) = (3.5M)x(500mL)

1st Volume = 141.129

The correct significant figures will make it 141. 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%

your answer is c) plasma

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