2. Density is a physical property of a molecule that's measured as mass per unit volume. (a) Given that density is usually expressed in the units of g/cm', calculate the density of a cube of plastic
that has a mass of 2.331 g and an edge length of 1.500 cm (volume of a cube is V = 1).
I
(b) Draw two liquids in a graduated cylinder on a separate sheet of paper. You won't submit the drawing as
part of your answer. Liquid A has a density of 1.004 g/cm and Liquid B has a density of 0.6223 g/cm”.
Assume they don't mix. Label which liquid is on top and which liquid is on bottom, and then draw the cube
where you would expect it to be if it's dropped into the graduated cylinder with these two liquids.
Explain the location for liquid A, liquid B, and the cube in the graduated cylinder.

A = 0.691g/cm³

B = liquid A at the bottom, liquid B in the middle and the cube at the top

Explanation:

Hello,

a)

Density = mass / volume

Density = ?

Mass = 2.331g

Volume = ?

Length of the side = 1.5cm

Volume of a cube = L³

Volume of the cube = 1.50³

Volume = 3.375cm³

Density = mass / volume

Density = 2.331 / 3.375

Density = 0.691g/cm³

b) since the question doesn't require submitting a diagram, I would go ahead and answer it without any but explaining the basics for you.

Densities:

Liquid A = 1.004g/cm³

Liquid B = 0.6223g/cm³

Cube = 0.691g/cm³

If the three substance don't mix and are place in a granulated cylinder,

Liquid A would be at the bottom

Liquid B would stay at the middle

The cube would be at the top

Reason is that, when two immiscble liquid are placed in a container, the liquid with the higher density will stay at the bottom while that of lower will float on top.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g