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Zanzabum
3 years ago
10

What are the two components of a solution? Write two properties of a solution

Chemistry
1 answer:
BARSIC [14]3 years ago
3 0
The two components of a solution are solvent and solute.

A solution is a homogenous mixture, stable, and the particles are very small.


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Can sedimentation and decantation be used for all types of mixtures? Explain
Bumek [7]

Explanation:

1. Sedimentation and decantation cannot be used for all types of mixtures.

Decantation is a separation technique in which is used to separate immiscible liquids or mixtures containing liquid and solids within them.

In decantation, gravity is used to bring the denser materials to settle at the bottom.

For homogenous mixtures, it is not possible to use decantation. A solution of sugar and water will not decant.

2. Yes, mass of an object reduces the settling time of such object in a mixture.

The higher the mass, the faster the rate of settling. Also, as we know, mass is directly proportional to density. A body with a high density will settle faster in solution.

4 0
3 years ago
(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct
Zarrin [17]

Explanation:

The given data is as follows.

   \Lambda^{o}_{m}(NaCl) = 1.264 \times 10^{-2}

   \Lambda^{o}_{m}(H-O=C-ONO) = 1.046 \times 10^{-2}

   \Lambda^{o}_{m}(HCl) = 4.261 \times 10^{-2}

Conductivity of monobasic acid is 5.07 \times 10^{-2} S m^{-1}

     Concentration = 0.01 mol/dm^{3}

Therefore, molar conductivity (\Lambda_{m}) of monobasic acid is calculated as follows.

                 \Lambda_{m} = \frac{conductivity}{concentration}

                                  = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}

                                 = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}

                                 = 5.07 \times 10^{-3} S m^{2} mol^{-1}

Also, \Lambda^{o}_{m} = \Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}

                            = 4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}

                            = 4.043 \times 10^{-2} S m^{2} mol^{-1}

Relation between degree of dissociation and molar conductivity is as follows.

               \alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}

                             = \frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}

                             = 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

Putting the values into the above formula we get the following.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

                        = \frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}

                        = 0.017973 \times 10^{-2}

                       = 1.7973 \times 10^{-4}

Hence, the acid dissociation constant is 1.7973 \times 10^{-4}.

Also, relation between pK_{a} and K_{a} is as follows.

                 pK_{a} = -log K_{a}

                              = -log (1.7973 \times 10^{-4})

                              = 3.7454

Therefore, value of pK_{a} is 3.7454.

                             

3 0
2 years ago
Are HF CH4 and CO2 acids or bases>
Vlad [161]
Huh I do not understand this
4 0
3 years ago
Functions of the following ingredients in the production of dairy products.ii. Enzyme in cheese​
Serjik [45]

Answer:

sugar in melk

Explanation:

cow eat sugar then make sugar :)

5 0
3 years ago
Which one of the following is an alkane?<br><br> A. C2H4<br> B. C5H10<br> C. C3H8<br> D. C4H6
kondaur [170]
Hello

Answer :

C. C3H8
3 0
3 years ago
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