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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Regards!
Answer: -
Solubility of a substance depend on the balance of intermolecular forces between the solvent and solute, and the entropy change that accompanies this process.
Temperature and pressure also plays a role in solubility.
A solution having Group 1 cations like lithium, sodium, potassium etc are always soluble.
A solution having NH₄⁺ is soluble.
All salts with anion as nitrates, acetates, chlorates, and perchlorates are soluble in water.
<span> the average mass of all the different isotopes of the atom</span>
#1...Evaporation
#2....Gas
Answer:
K = 361.369
Explanation:
C2H4(g) + H2O(g) → CH3CH2OH(g)
∴ ΔG°f(298.15K) CH3CH2OH(g) = - 174.8 KJ/mol
∴ ΔG°f(298.15) C2H4(g) = 68.4 KJ/mol
∴ ΔG°f(298.15) H2O(g) = - 228.6 KJ/mol
⇒ ΔG°f(298.15) = - 174.8 - ( - 228.6 + 68.4 ) = - 14.6 KJ/mol
∴ R = 8.314 E-3 KJ/mol.K
∴ T = 298.15 K
⇒ K = e∧(-(-14.6)/((8.314 E-3)(298.15)))
⇒ K = e∧(5.889)
⇒ K = 361.369