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Hello I need help with number 3.
1. An unknown mass of gold displaces 89.1 mL of water in a container. What is the mass of the gold.
-My answer is 1719.63g
2. Using identical conditions, what volume of water will the identical mass of silver displace?
-My answer is 164.2mL
3. Using an identical system, a supposedly gold crown with the same mass is measured and found to displace 138.4 mL. What percent gold is it really? (This is an algebra problem that requires two equations with two unknowns and the answers of questions 1 and 2.)

Sagot :

whitneytr12

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!

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