SOMEONE ANSWER THIS FAST ITS DUE TODAY A miner finds a small minerallsragment with a volume of 11.5 cm3 and a mass of 57.4 g.
What is the density of the mineral fragment?
0.20 g/cm3
5.00 g/cm
5.74 g/cm
164 g/cm3

Density = mass/ volume

Explanation:

57.4g / 11.5cm3 = 4.991304(g/cm)

round to sig figs = 5.00(g/cm)

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%