Answer: the bonds in the methane and oxygen come apart, the atoms rearrange and then re-bond to form water and carbon dioxide
Explanation:^
Answer:
NaI > Na2SO4 > Co Br3
meaning that NaI has the highest freezing point, and Co Br3 has the lowest freezing point.
Explanation:
The freezing point depression is a colligative property.
That means that it depends on the number of solute particles dissolved.
The formula to calculate the freezing point depression of a solution of a non volatile solute is:
ΔTf = i * Kf * m
Where kf is a constant, m is the molality and i is the van't Hoff factor.
Molality, which is number of moles per kg of solvent, counts for the number of moles dissolved and the van't Hoff factor multipllies according for molecules that dissociate.
The higher the number of molecules that dissociate, the higher the van't Hoff, the greater the freezing point depression and the lower the freezing point.
As the question states that you assume equal concentrations (molality) and complete dissociation you just must find the number of ions generated by each solute, in this way:
NH4 I → NH4(+) + I(-) => 2 ions
Co Br3 → Co(+) + 3 Br(-) => 4 ions
Na2SO4 → 2Na(+) + SO4(2-) => 3 ions.
So, Co Br3 is the solute that generate more particles and that solution will exhibit the lowest freezing point among the options given, Na2SO4 is next and the NaI is the third. Ordering the freezing point from higher to lower the rank is NaI > Na2SO4 > CoBr3, which is the answer given.
Answer:
the change is evaporation
Explanation:
the water heats up at the surface of the water and evaporates
0.17 M is the is the molal concentration of this solution
Explanation:
Data given:
freezing point of glucose solution = -0.325 degree celsius
molal concentration of the solution =?
solution is of glucose=?
atomic mass of glucose = 180.01 grams/mole
freezing point of glucose = 146 degrees
freezing point of water = 0 degrees
Kf of glucose = 1.86 °C
ΔT = (freezing point of solvent) - (freezing point of solution)
ΔT = 0.325 degree celsius
molality =?
ΔT = Kfm
rearranging the equation:
m = 
m= 0.17 M
molal concentration of the glucose solution is 0.17 M
The concentration of AlCl3 solution if 150 ml of the solution contains 550 mg of cl- ion is 0.0344 M
calculation
concentration = moles /volume in liters
volume in liters = 150 /1000= 0.15 L
number of moles calculation
write the equation for dissociation of Al2Cl3
that is AlCl3 ⇔ Al^3+ + 3 Cl ^-
find the moles of Cl^- formed
moles =mass/molar mass
mass in grams= 550/ 1000 =0.55 grams
molar mass of Cl^- =35.5 g/mol
moles is therefore= 0.55/35.5 =0.0155 moles
by use of mole ration betweem AlCl3 to Cl^- which is 1:3 the moles of AlCl3 is =0.0155 x 1/3= 5.167 x10^-3 moles
concentration of AlCl3 is therefore= 5.167 x10^-3/ 0.15 =0.0344 M
