b. canola oil
c. peanut oil
d. sunflower oil
Answer:
Canola Oil
Explanation:
The data given is as;
Oil Melting Point / Freezing Point
Canola Oil 10 °C
Corn Oil -11 °C
Sunflower Oil -17 °C
Peanut Oil -2 °C
When the mixture of these oils is cooled to 5 °C only the Canola oil will solidify because it has a melting point of 10 °C and exists in solid state below 10 °C (Freezing Point). Therefore, it can easily be separated from remaining mixture which is present in liquid state.
(3) a hard sphere with positive particles uniformly embedded
(4) a hard sphere with negative particles uniformly embedded
Answer:
a
Explanation:
A Brønsted-Lowry base is a species that accepts a proton (hydrogen ion) from another species. NH3, OH-, and even water itself are examples of Brønsted-Lowry bases, denoting they accept protons.
A Brønsted-Lowry base is a species that can accept a proton (a hydrogen ion) from another species. For instance, in a reaction between water and ammonia, NH3 is the Brønsted-Lowry base because it accepts a proton from water. This means that any species capable of accepting a proton, such as hydroxide ion (OH-), ammonia (NH3), or water itself can be considered a Brønsted-Lowry base.
For example, think about the dissociation of water:
H2O + H2O ⇌ H3O+ + OH-
In this reaction, water is acting as both a Brønsted-Lowry acid and a Brønsted-Lowry base. One water molecule donates a proton and becomes a hydroxide ion (the conjugate base), while the other accepts a proton to become hydronium (the conjugate acid).
Another example would be the ionization of ammonia in water:
NH3 + H2O ⇌ NH4+ + OH-
Here, ammonia (NH3) is the Brønsted-Lowry base as it accepts a proton from water to become ammonium (NH4+).
#SPJ12
Intermolecular forces.
The boiling point of standard water is 100 degree Celsius, with the addition of solute the boiling point is elevated. The freezing point of the solution will be -18.04 degree Celsius.
The boiling point is the temperature at which the liquid is converted to vapor. The change in boiling point of the aqueous solution gives the molality of the solution as:
The depression in freezing point from molality is given as;
The freezing point of aqueous water is zero degree Celsius. The freezing point of the solution will be:
The freezing point of the solution is -18.04 degree Celsius.
Learn more about boiling point, here:
Answer:
T°fussion of solution is -18°C
Explanation:
We have to involve two colligative properties to solve this. Let's imagine that the solute is non electrolytic, so i = 1
First of all, we apply boiling point elevation
ΔT = Kb . m . i
ΔT = Boiling T° of solution - Boiling T° of pure solvent
Kb = ebuliloscopic constant
105°C - 100° = 0.512 °C kg/mol . m . 1
5°C / 0.512 °C mol/kg = m
9.7 mol/kg = m
Now that we have the molality we can apply, the Freezing point depression.
ΔT = Kf . m . i
Kf = cryoscopic constant
0° - (T°fussion of solution) = 1.86 °C/m . 9.76 m . 1
- (1.86°C /m . 9.7 m) = T°fussion of solution
- 18°C = T°fussion of solution