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:
Mass of heptane = 102g
Vapor pressure of heptane = 454mmHg
Molar mass of heptane = 100.21
No of mole of heptane = mass/molar mass = 102/100.21
No of mole of heptane = 1.0179
Therefore the partial pressure of heptane = no of mole heptane *Vapor pressure of heptane
Partial pressure of heptane = 1.0179*454mmHg
Partial pressure of heptane = 462.1096 = 462mmHg
the partial pressure of heptane vapor above this solution = 462mmHg
Answer:
A mole (mol) is the amount of a substance that contains 6.02 × 10 23 representative particles of that substance. The mole is the SI unit for the amount of a substance. There are, therefore, 6.02 × 10 23 water molecules in a mole of water molecules. Water (H2O) is made from 2 atoms of hydrogen and 1 atom of oxygen.
Answer:
I think the teaspoon will cool down faster.
Explanation:
The teaspoon of water will cool down faster because it is a smaller amount of water so the entire [drop] will absorb heat energy faster.
Answer:
B. Conductivity can be used to calculate the salinity of the water.
I am 100% sure this is the correct answer :)