For each of the -3.3 trials, and in the same sequence in which you entered the freezing points, enter your calculated values for the freezing point depression constant of water. Be sure to use the average value for the freezing point of pure water and the value for the molality of the solution that you entered earlier. Report your values to 3 significant figures.

The morality of the solution is calculated as 0.859 m. We are required to determine the freezing point depression constant of pure water. The freezing point depression of the solution is given as

* (morality of solution)

and are the freezing points of the pure solvent (water, 0°C) and = freezing point depression constant of water. Therefore,

*(0.859 m)

=====> (0°C) – (3.00°C) =

=====> -3.00°C =

Ignore the negative sign (since is positive) and get

= (3.00°C) / (0.859 m) = 3.492°C/m

The freezing point depression constant of the solvent is 3.5°C kg/mole  

3.5 K.kg/mole (temperature differences are the same in Celsius and Kelvin scales).

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse