Question
An irregularly shaped piece of gold was lowered into a graduated cylinder holding a volume of water equal to 17 ml. The height of the water rose to 20 ml. If the mass of the gold was 27 g, what was its density?
Responses
6 g/mL
6 g/mL
9 g/mL
9 g/mL
10 g/mL
10 g/mL
7 g/mL
Responses
6 g/mL
6 g/mL
9 g/mL
9 g/mL
10 g/mL
10 g/mL
7 g/mL
Answers
GPT-4o mini
To find the density of the gold, you can use the formula for density:
\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\]
From the problem:
- The mass of the gold is 27 g.
- The volume of the gold can be determined from the change in water level in the graduated cylinder. The water level increased from 17 mL to 20 mL, so the volume of the gold is:
\[
\text{Volume of gold} = \text{Final Volume} - \text{Initial Volume} = 20 \, \text{mL} - 17 \, \text{mL} = 3 \, \text{mL}
\]
Now plug the values into the density formula:
\[
\text{Density} = \frac{27 \, \text{g}}{3 \, \text{mL}} = 9 \, \text{g/mL}
\]
Thus, the density of the gold is **9 g/mL**.
\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\]
From the problem:
- The mass of the gold is 27 g.
- The volume of the gold can be determined from the change in water level in the graduated cylinder. The water level increased from 17 mL to 20 mL, so the volume of the gold is:
\[
\text{Volume of gold} = \text{Final Volume} - \text{Initial Volume} = 20 \, \text{mL} - 17 \, \text{mL} = 3 \, \text{mL}
\]
Now plug the values into the density formula:
\[
\text{Density} = \frac{27 \, \text{g}}{3 \, \text{mL}} = 9 \, \text{g/mL}
\]
Thus, the density of the gold is **9 g/mL**.