Depression of a freezing point of the solutions depends on the number of particles of the solute in the solution.
1 mol of C6H12O6 after dissolving in water still be 1 mol, because C6H12O6 does no dissociate in water.
1 mol of C2H5OH after dissolving in water still be 1 mol, because C2H5OH does no dissociate in water.
1 mol of NaCl after dissolving in water gives 2 mol of particles (ions), because NaCl is a strong electrolyte(as salt) and completely dissociates in water.
NaCl ----->Na⁺ + Cl⁻
1 mol of CH3COOH after dissolving in water gives more than 1 mol but less than 2 moles, because CH3COOH is a weak electrolyte (weak acid) and dissociates only partially.
So, most particles of the solute is going to be in the solution of NaCl,
so<span> the lowest freezing point has the aqueous solution of NaCl.</span>
Given question is incomplete. The complete question is as follows.
Balance the following equation:

Answer: The balanced chemical equation is as follows.

Explanation:
When a chemical equation contains same number of atoms on both reactant and product side then this equation is known as balanced equation.
For example, 
Number of atoms on reactant side:
H = 5
P = 1
O = 6
Ca = 1
Number of atoms on product side:
H = 6
P = 2
O = 9
Ca = 1
In order to balance this equation, we will multiply
by 2 on reactant side and we will multiply
by 2 on product side. Hence, the balanced chemical equation is as follows.

Overharvesting that would be your answer!
Explanation:
1 is a process that occurs when gases in earth's atmosphere trap the sun's heat .
2 . heat in the atmosphere to warm the planet .
I only know the first too sorry ✨
First, we need to get the value of Ka:
when Ka = Kw / Kb
we have Kb = 1.8 x 10^-5
and Kw = 3.99 x 10^-16 so, by substitution:
Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11
by using the ICE table :
NH4+ + H2O →NH3 + H+
intial 0.013 0 0
change -X +X +X
Equ (0.013-X) X X
when Ka = [NH3][H+] / [NH4+]
by substitution:
2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X
∴X = 5.35 x 10^-7
∴[H+] = X = 5.35 x 10^-7
∴PH = - ㏒[H+]
= -㏒(5.35 x 10^-7)
= 6.27