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!
We will use the formula for freezing point depression :
but first, we need to get the molality m of the solution:
- molality m = moles of C2H5OH / mass of water Kg
when moles of C2H5OH = mass of C2H5OH/ molar mass of C2H5OH
= 11.85 g / 46 g/mol
= 0.258 moles
and when we have the mass of water Kg = 0.368 Kg
so, by substitution on the molality formula:
∴ molality m = 0.258 moles / 0.368 Kg
= 0.7 mol/Kg
and when C2H5OH is a weak acid so, there is no dissociation ∴ i = 1
and when Kf is given = 1.86 C/m
so by substitution on ΔTf formula:
when ΔTf = i Kf m
∴ ΔTf = 1 * 1.86C/m * 0.7mol/Kg
= 1.302 °C
Answer:
16.9 mol
Explanation:
8Al + 3Fe₃O₄ ⟶ 4Al₂O₃ + 9Fe
You want to convert moles of Fe₃O₄ to moles of Al.
The molar ratio is 8 mol Al:3 mol Fe₃O₄. Calculate the moles of Al
Moles of Al = 6.34 × 8/3
Moles of Al = 6.34 × 2.667
Moles of Al = 16.9 mol Al
Molten state of matter is the inner core and outer core of earth
