DO NOT CLICK THE LINK THE OTHER PERSON COMMENTED PLS
Answer:
b. unsaturated
.
Explanation:
Hello there!
In this case, according to the given information, it turns out necessary for us to bear to mind the definition of each type of solution:
- Supersaturated solution: comprises a large amount of solute at a temperature at which it will be able to crystalize upon standing.
- Unsaturated solution: is a solution in which a solvent is able to dissolve any more solute at a given temperature.
- Saturated solution can be defined as a solution in which a solvent is not capable of dissolving any more solute at a given temperature.
In such a way, since 20 grams of the solute are less than the solubility, we infer this is b. unsaturated, as 33.3 grams of solute can be further added to the 100 grams of water.
Regards!
Colligative
properties calculations are used for this type of problem. Calculations are as
follows:<span>
</span>
<span>ΔT(freezing point)
= (Kf)m
ΔT(freezing point)
= 1.86 °C kg / mol (0.705)
ΔT(freezing point) = 1.3113 °C
</span>
<span>
</span>
<span>Hope this answers the question. Have a nice day.</span>
Answer:
0.681 atm
Explanation:
To solve this problem, we make use of the General gas equation.
Given:
P1 = 785 torr
V1 = 2L
T1 = 37= 37 + 273.15 = 310.15K
P2 = ?
V2 = 3.24L
T2 = 58 = 58+273.15 = 331.15K
P1V1/T1 = P2V2/T2
Now, making P2 the subject of the formula,
P2 = P1V1T2/T1V2
P2 = [785 * 2 * 331.15]/[310.15 * 3.24]
P2 = 515.715 Torr
We convert this to atm: 1 torr = 0.00132 atm
515.715 Torr = 515.715 * 0.00132 = 0.681 atm
