Hey there!
It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.
In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:
Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:
Thus, the density of the calcite sample will be:
This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL
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Answer:
Do to half of the mnairals this can not be made into a lab there is an error
Explanation:
Mass was conserved, because it was losing all shape, weight, and matter. It really has no mass.
Answer:
Final temperature is 302 K
Explanation:
You can now initial volume with ideal gas law, thus:
V = 
Where:
n are moles: 2 moles
R is gas constant: 0,082 
T is temperature: 300 K
P is pressure: 1 atm
V is volume, with these values: <em>49,2 L</em>
The work in the expansion of the gas, W, is: 1216 J - 34166 J = <em>-32950 J</em>
This work is:
W = P (Vf- Vi)
Where P is constant pressure, 1 atm
And Vf and Vi are final and initial volume in the expansion
-32950 J = -1 atm (Vf-49,2L) × 
Solving: <em>Vf = 49,52 L</em>
Thus, final temperature could be obtained from ideal gas law, again:
T = 
Where:
n are moles: 2 moles
R is gas constant: 0,082
P is pressure: 1 atm
V is volume: 49,52 L
T is final temperature: <em>302 K</em>
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I hope it helps!
