Question

2.92 A 50.0-g silver object and a 50.0-g gold object are both added

to 75.5 mL of water contained in a graduated cylinder. What is
the new water level in the cylinder? (2.7)
ifacturing of computer chins cylinders of silicon

Answer 1

The problem involves using the density formula to determine the substance a piece of jewelry might be made of, and Archimedes' principle to find the new water level in a cylinder after submerging two objects.

Step-by-step explanation:

The question pertains to the concept of density and displacement of water when a solid object is submerged. From the information provided, we can determine the density of a piece of jewelry and infer the substance it might be made of. Additionally, we solve for the new water level in a cylinder after submerging two objects with known masses.

Firstly, to calculate the density of an object, we use the formula density (d) = mass (m) / volume (V). If the mass of the jewelry is given as 132.6 g and the water displacement from submerging it is the difference between the final and initial volumes (61.2 mL - 48.6 mL), which equals 12.6 mL, then the density of the jewelry can be calculated. For part (b), we could compare this calculated density with densities of known substances to guess the material of the jewelry.

For the student's question about the graduated cylinder, the new water level can be determined by Archimedes' principle, which states that the volume of water displaced will be equal to the volume of the objects submerged. In this case, since the densities of silver and gold are not provided, we cannot calculate their respective volumes to determine the new water level directly, but we do know that the volume of the water will increase by the sum of their volumes.

Answer 2

82.9 mL

Step-by-step explanation:

1. Volume of silver

`\begin{array}{rcl}\text{Density}&=& \frac{\text{Mass}}{\text{Volume}}\\\\\rho&=& (m)/(V)\\\\V &=& (m)/(\rho)\\\\& = & \frac{\text{50.0 g}}{\text{10.49 g$\cdot$mL}^(-1)}\\\\& = & \text{4.766 mL}\\\end{array}\\\text{The volume of the silver is $\large \boxed{\textbf{4.766 mL}}$}`

2. Volume of gold

`\begin{array}{rcl}V& = & \frac{\text{50.0 g}}{\text{19.30 g$\cdot$mL}^(-1)}\\\\& = & \text{2.591 mL}\\\end{array}\\\text{The volume of the gold is $\large \boxed{\textbf{2.591 mL}}$}`

3. Total volume of silver and gold

V = 4.766 mL + 2.591 mL = 7.36 mL

4 New reading of water level

V = 75.5 mL + 7.36 mL = 82.9 mL