Assume that an object has been placed in the pan of a triple beam balance. The scales of the balance are shown below. What is the mass of the object?

190.4 g

Explanation:

To obtain the mass of the object, we must be able to obtain the reading of the triple beam balance.

Thus, the reading of the balance can be obtained by simply adding the various values as indicated by the pointer in each scale.

This can be obtained as follow:

Scale 1 = 90 g

Scale 2 = 100 g

Scale 3 = 0.4 g

Reading of the balance = Scale 1 + Scale 2 + Scale 3

Reading of the balance = 90 + 100 + 0.4

Reading of the balance = 190.4 g

Thus, the mass of the object as indicated by the triple beam balance is 190.4 g.

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%