All categories

3 years ago

It is logical to conclude that water cannot dissolve vegetable oil because the oil is

Chemistry

2 answers:

marin [14]3 years ago

6 0

Answer:

The oil is insoluble with water.

Explanation:

The water and oil do not mix, they are illustrated as immiscible. The molecules of water are polar, that is, they exhibit a small positive charge at one terminal and a small negative charge at the other terminal, and they attach with each other. The molecules of oil are non-polar, and they possess no charge. Due to this, the molecules of oil are more fascinated with each other than to the molecules of water, and the molecules of water are more fascinated towards each other than to the molecules of oil.

Gekata [30.6K]3 years ago

3 0

In chemistry, there is a common note that says, "Like dissolves like".

This pertains to the concept that polar substances can dissolve only other polar substances. Also, nonpolar substances are also only able to dissolve nonpolar substances.

Polarity of the substance depends primarily on the type of bond and the difference in electronegativity.

Water is a polar substance while vegetable oil is not. From the concept presented above, it may be concluded that water will not be able to dissolve the vegetable oil and the assumption is logical.

You might be interested in

Strong acids are assumed 100% dissociated in water. True As a solution becomes more basic, the pOH of the solution increases. Fa

Kruka [31]

Answer:

Strong acids are assumed 100% dissociated in water- True

As a solution becomes more basic, the pOH of the solution increases- false

The conjugate base of a weak acid is a strong base- true

The Ka equilibrium constant always refers to the reaction of an acid with water to produce the conjugate base of the acid and the hydronium ion- True

As the Kb value for a base increases, base strength increases- true

The weaker the acid, the stronger the conjugate base- true

Explanation:

An acid is regarded as a strong acid if it attains 100% or complete dissociation in water.

The pOH decreases as a solution becomes more basic (as OH^- concentration increases).

Ka refers to the dissociation of an acid HA into H3O^+ and A^-.

The greater the base dissociation constant, the greater the base strength.

The weaker an acid is, the stronger , its conjugate base will be.

7 0

3 years ago

How many moles of H2SO4 are needed to prepare 200 mL of a 1.0 M solution?

jok3333 [9.3K]

The answer is 0.2moles

4 0

3 years ago

In a balanced equation, the same number of each kind of atom is shown on each side of the equation. calculate the number of iron

Juliette [100K]

There is a missing portion of this question which shows the reaction that needs balancing:

"In a balanced equation, the same number of each kind of atom is shown on each side of the equation. Calculate the number of iron (Fe), oxygen (O), and carbon atoms (C).

Fe2O3+ 3CO --> 2Fe + 3CO2

Based on these values, is the equation balanced?"

To check if this equation is balanced we simply compare the number of each element on each side of the equation.

On the reactant side of the equation we have:

2 Fe atoms
6 O atoms
3 C atoms

On the product side of the equation we have:

2 Fe atoms
6 O atoms
3 C atoms

Therefore, both side of the reaction have the correct and equal number of each atom, so the equation is balanced.

7 0

3 years ago

Read 2 more answers

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

2 years ago

Explain why beryllium is produced when potassium is heated with beryllium

Vladimir [108]

Explanation:

The more reactive element replaces less reactive element during chemical reaction.

Since, potassium is more reactive than beryllium. When potassium reacts with beryllium choride, it replaces beryllium and forms potassium chloride and produces beryllium.

3 0

2 years ago