Answer:
u need to put a picture so we know what your talking about
Answer:
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) = ![K_{f} *(0.859 m)](https://tex.z-dn.net/?f=K_%7Bf%7D%20%20%2A%280.859%20m%29)
=====> -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).
f(x) = 3x + 1
f(9) = 3(9) + 1
f(9) = 27 + 1
f(9) = 28
f(9) = 28. I hope this helps! I have a g00gle classroom where I help with algebra including this. If you are interested, please consider joining. I will post the class code in the comments if it would let me. Sorry it wouln't let me type g0ogle properly.