Answer:
The freezing point will be 
Explanation:
The depression in freezing point is a colligative property.
It is related to molality as:

Where
Kf= 
the molality is calculated as:




Depression in freezing point = 
The new freezing point = 
Hey there!:
Molar mass Ca(NO2)2 = 132.089 g/mol
Mass of solute = 120 g
Number of moles:
n = mass of solute / molar mass
n = 120 / 132.089
n = 0.0009084 moles of Ca(NO2)2
Volume in liters of solution :
240 mL / 1000 => 0.24 L
Therefore:
Molarity = number of moles / volume of solution
Molarity = 0.0009084 / 0.24
Molarity = 0.003785 M
Hope that helps!
The temperature that would the volume of a gas be 0.550l if it had a volume of 0.432 L at -20.0 c is calculated using the Charles law formula
that is v1/T1=V2/T2
V1=0.550 l
t1=?
T2= -20 c +273 = 253 K
v2= 0.432 l
by making T1 the subject of the formula T1= V1T2/V2
T1= (0.55lL x253)/ 0.432 l = 322.11 K or 322.11-273 = 49.11 C
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
Fluorine 20 (F - Atomic number 9 and atomic mass 20). Firstly we need to know what is beta decay. Beta decay occurs when one neutron changes into a proton and an electron therefore the atomic mass will remain the same as even though we loose a neutron it is replaced by a proton, the atomic number is always raised by 1 when one beta decay occurs. The produced electron is shot out of the nucleus at an incredible speed. This speedy electron we call a beta particle.
Ok now the reaction.
20 20 0
F -> Ne + e
9 10 -1
Remember the atomic number determines the nature of the element ( i.e what elemnt it is).
Hope this helps :).