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Ksenya-84 [330]
2 years ago
10

Is milk a solution, suspension, compound , colloid ?

Chemistry
1 answer:
kakasveta [241]2 years ago
3 0

Answer:

Colloids (heterogeneous)

The difference between a colloid and a suspension is that the particles will not settle to the bottom over a period of time, they will stay suspended or float. An example of a colloid is milk. Milk is a mixture of liquid butterfat globules dispersed and suspended in water.

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6 0
3 years ago
Which of the following is not a type of fault?
bekas [8.4K]

normal is the answer


8 0
3 years ago
HELP I DONT KNOW HOW TO DO THIS !!
BlackZzzverrR [31]

Answer:

you divide 71.5 by 2

Explanation:

it's may look intimidating but the question is actually really simple.

5 0
2 years ago
An aqueous potassium carbonate solution is made by dissolving 5.51 moles of K 2 CO 3 in sufficient water so that the final volum
ad-work [718]

Answer:

1.67mol/L

Explanation:

Data obtained from the question include:

Mole of solute (K2CO3) = 5.51 moles

Volume of solution = 3.30 L

Molarity =?

Molarity is simply the mole of solute per unit litre of the solution. It can be expressed mathematically as:

Molarity = mole of solute /Volume of solution

Molarity = 5.51 mol/3.30 L

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Therefore, the molarity of K2CO3 is 1.67mol/L

3 0
3 years ago
Read 2 more answers
1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction
nikklg [1K]

Answer:

H_{comb}=-4406kJ/mol

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol

Best regards.

4 0
3 years ago
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