How many fluorine atoms are in 750.0 mL of a 0.500M HF solution?

Explanation:

Step 1: Given data

Step 2: Convert "V" to liters

We will use the conversion factor 1 L = 1000 mL.

750.0 mL × 1 L/1000 mL = 0.7500 L

Step 3: Calculate the moles of HF

We will use the following expression.

n = M × V

n = 0.500 mol/L × 0.7500 L = 0.375 mol

Step 4: Calculate the atoms of F in 0.375 moles of HF

We will use the following relationships:

b i think it’s a water wheel turning in a stream

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%