Answer:
The freezing point will be -2.046°C.
Explanation:
The freezing point depression equation is 
Where;
= The temperature depression of the freezing point
= The constant of freezing point depression which is solvent dependent = 1.86°C/m
i = The number of particles the substance decomposes into in solution = 1 for sugar (a covalent compound)
m = The molality of the solution = 1.1
Therefore, we have;
Therefore the freezing point will be 0 - 2.046°C = -2.046°C.
6CO2 + 7H2O = C6H12O6 + 6O2 + H2O
Explanation:
CO2 + H2O = C6H12O6 + O2 + H2O
Starting carbon atoms,
1C on reactants and 6C on products,
therefore you should start by putting a 6 on CO2
6CO2 + H2O = C6H12O6 + O2 + H2O
Proceed with Hydrogen,
Reactants → 2H
Products → 14H
You should multiply H2O with 7 .
6CO2 + 7H2O = C6H12O6 + O2 + H2O
the we need to proceed for Oxygen,
Reactants → 12 + 7= 19O
Products → 6 O in C6H12O6, 2O in O2 and 1O in H2O
we need to balance the equation without disrupting the balance of other atoms, so we need to Multiply O2 with 6
The required balanced equation is
6CO2 + 7H2O = C6H12O6 + 6O2 + H2O
The separation of a mixture by passing it in solution through a medium in which the components move at different rates
Answer:
Freezing T° of solution = - 7.35 °C
Explanation:
This is about the freezing point depression, a colligative property which depends on solute.
The formula is: Freezing T° pure solvent - Freezing T° solution = m . Kf . i
Freezing T° of pure solvent is -3.1°C
At this case, i = 1. As an organic compound the urea does not ionize.
We determine the molality (mol/kg of solvent)
We convert mass to moles:
12.3 g . 1mol / 60.06 g = 0.205 moles
0.205 mol / 0.3 kg = 0.682 mol/kg
We replace data in the formula:
-3.1°C - Freezing T° of solution = 0.682 mol/kg . 6.23 kg°C/mol . 1
Freezing T° of solution = 0.682 mol/kg . 6.23 kg°C/mol . 1 + 3.1°C
Freezing T° of solution = - 7.35 °C
