The answers to the given questions are:
1. The mass of the gold that displaces 89.1 mL of water is 1719.63 grams.
2. The volume of water displaced by 1719.63 g of silver is 163.93 mL.
3. The percent of gold in a crown of 1719.63 g that displaces 138.4 mL of water is 34.1%.
1. We can find the mass of gold with Archimedes' principle.
[tex] m_{Au} = d_{Au}*V [/tex]
Where:
[tex]m_{Au}[/tex]: is the mass of gold =?
[tex]d_{Au}[/tex]: is the density of gold = 19.3 g/mL [/tex]
V: is the volume of water displaced = volume of the object = 89.1 mL (according to Archimedes' principle)
Hence, the mass of gold is:
[tex] m_{Au} = 19.3 g/mL*89.1 mL = 1719.63 g [/tex]
2. We can calculate now the volume of water displaced by a mass of silver as follows:
[tex] V = \frac{m_{Ag}}{d_{Ag}} [/tex]
Where [tex]d_{Ag}[/tex] is the density of silver = 10.49 g/mL
[tex] V = \frac{1719.63 g}{10.49 g/mL} = 163.93 mL [/tex]
Therefore, the volume of water displaced by 1719.63 grams of silver is 163.93 mL.
3. We know that a mass of 1719.63 g of pure gold displaces 89.1 mL and that the same mass of silver displaces 163.93 mL of water, so:
[tex] \%_{Au}V_{Au} + \%_{Ag}V_{Ag} = 100\% V_{t} [/tex]
Where:
[tex] \%_{Au}[/tex]: is the percent of gold in the crown
[tex] \%_{Ag}[/tex]: is the percent of silver in the crown
[tex] V_{Au} [/tex]: is the volume of water displaced by the mass of gold = 89.1 mL
[tex] V_{Ag} [/tex]: is the volume of water displaced by the mass of silver = 163.93 mL
[tex] V_{t}[/tex]: is the total volume of water displaced = 138.4 mL
By changing the above percent values to decimal ones, we have:
[tex] p_{Au}V_{Au} + p_{Ag}V_{Ag} = 1*V_{t} [/tex] (1)
Since the sum of the gold and silver percent must be equal to 1 (which is equal to 100%), we can express the silver percent in terms of gold percent:
[tex] p_{Au} + p_{Ag} = 1 [/tex]
[tex] p_{Ag} = 1 - p_{Au} [/tex] (2)
Now, by entering eq (2) into (1) we have:
[tex] p_{Au}V_{Au} + (1 - p_{Au})V_{Ag} = V_{t} [/tex]
After solving for [tex]p_{Au}[/tex] we have:
[tex] p_{Au} = \frac{V_{t} - V_{Ag}}{V_{Au} - V_{Ag}} [/tex]
[tex] p_{Au} = \frac{138.4 mL - 163.93 mL}{89.1 mL - 163.93 mL} = 0.341 [/tex]
The above value in percent is:
[tex] p_{Au} = 34.1 \% [/tex]
Therefore, the percent of gold in the crown is 34.1%.
You can find more about Archimedes' principle here: https://brainly.com/question/13106989?referrer=searchResults
I hope it helps you!
