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blondinia [14]
3 years ago
6

What is the difference between a suspension, a colloid, a solution, and an emulsion?

Chemistry
2 answers:
Len [333]3 years ago
5 0

Suspension is defined as the heterogeneous mixture in which solute particles suspended throughout the bulk of the particles. The particle size is more than 100 nm. In suspension, particles don't pass through filter paper. Sand in water is an example of suspension.

Colloid is defined as a mixture (heterogeneous and homogeneous) in which one substance of dispersed insoluble particles get suspended throughout other substance. The particle size is 1 to 100 nm. In colloid, particles are small, thus pass through filter paper. The particles of air which is dispersed in solid stone is an example colloid.

Emulsion is a mixture of two or more substance which are immiscible in nature. It is a part of colloid. Milk is an example of emulsion.

Solution is a homogeneous mixture with clear or transparent appearance. The particle size in solution is 10^{-7}-10^{-8}cm i.e. molecule in size. There is no effect of light occurs in the solution and solution can't filtered but can separated by the physical technique i.e. distillation.

Lostsunrise [7]3 years ago
4 0

Elumsion is a colloidal suspension of one liquid in another.


Hope this helps


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1-How many moles of NazCOs are in 10.0 ml of a 2.0 M solution?
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A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles
alexgriva [62]

Answer : The pH of the solution is, 3.41

Explanation :

First we have to calculate the moles of HF.

\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}

\text{Moles of HF}=0.250M\times 1.50L=0.375mol

Now we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (6.8\times 10^{-4})

pK_a=4-\log (6.8)

pK_a=3.17

The reaction will be:

                             HF+OH^-\rightleftharpoons F^-+H_2O

Initial moles     0.375     0.100   0.375

At eqm.   (0.375-0.100)      0     (0.375+0.100)

                     = 0.275                    = 0.475

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[F^-]}{[HF]}

Now put all the given values in this expression, we get:

pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]

pH=3.41

Thus, the pH of the solution is, 3.41

8 0
3 years ago
The volume of an object is 50cm3 and its mass is 150g. What is its density?
labwork [276]

Answer:

3g/cm3

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wlad13 [49]

Calorie (cal) or small calorie, is the amount of energy needed to heat one gram of water by one degree Celsius.

One small calorie is approximately 4.2 joules.

A calorie is a unit of energy.

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Q = 150000 J; released energy.

Convert Jules in calories:

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Q = 35833.73 calorie.

The large calorie or nutritional calorie (symbols: Cal, kcal) is the heat energy needed to raise the temperature of one kilogramof water by one degree Celsius.  

A nutritional calorie, or kilocalorie, is equal to 1000 calories.

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For example:

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E = 10460 kJ.

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