Calculate the freezing point (kf = 1.86 °c/m) of a 2.1m aqueous solution of kcl.

7.8 °c

3.9 °c

-7.8 °c

-3.9 °c

The freezing point of the solution is - 3.90 °C.

Explanation:

We can solve this problem using the relation:

ΔTf = (Kf)(m),

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = - 1.86 °C/m,

m is the molality of the solution (m = 2.1 m).

∴ ΔTf = (Kf)(m) = (-1.86 °C/m)(2.1 m) = - 3.90 °C.

∵ The freezing point if water is 0.0 °C and it is depressed by - 3.90 °C.

∴ The freezing point of the solution is - 3.90 °C.

The answer is b i think

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer