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

Answer:

82.9 mL  

Explanation:

1. Volume of silver

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

2. Volume of gold

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

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