It should be 13.48%.
1130g NaCL/7250g Ice + 1130 NaCl=0.1348 x 100%= 13.48%
Answer:
(a) 22.3 torr; 5.6 torr; (b) 27.9 torr; (c) 77.7 % heptane; 23.3 % octane
(d) Heptane is more volatile than octane
Explanation:
We can use Raoult's Law to solve this problem.
It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction. In symbols,
(a) Vapour pressure of each component
Let heptane be Component 1 and octane be Component 2.
(i) Moles of each component

(ii) Total moles

(iiii) Mole fractions of each component

(iv) Partial vapour pressures of each component

(b) Total pressure

(c) Mass percent of each component in vapour

The ratio of the mole fractions is the same as the ratio of the moles.

If we have 1 mol of vapour, we have 0.799 mol of heptane and 0.201 mol of octane

(d) Enrichment of vapour
The vapour is enriched in heptane because heptane is more volatile than octane.
Answer:
the population of grasshoppers will decrease
Explanation:
a food source is becoming more and more limited thus, its predator will also start to die out due to starvation
Caffeine is more soluble in dichloromethane and the both are separated by evaporating the solvent.
Caffeine is an organic plant material which is more soluble in non-polar solvents than in polar solvents. As such, caffeine is more soluble in dichloromethane than in pure water.
In order to carry out a liquid-liquid exaction of dichloromethane from a commercial teabag, the dichloromethane is mixed with water. The caffeine is found to be more soluble in the organic dichloromethane layer than in water.
The two solvents can now be separated using a separating funnel and the solution is evaporated to obtain the caffeine.
Learn more: brainly.com/question/967776
Unit of measurement
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