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Serjik [45]
3 years ago
7

Describe one way to prove that a mixture of sugar and water is a solution and that a mixture of sand and water is not a solution

.
Chemistry
1 answer:
Llana [10]3 years ago
7 0

Answer:

The solution is always homogeneous mixture and transparent through which the light can travel. The mixture of water and sugar is a solution because sugar is soluble in water and form homogeneous mixture while the sand can not dissolve in water and sand particles scatter the light.

Explanation:

Solution:

"The solution is always homogeneous mixture and transparent through which the light can travel"

The mixture of water and sugar is a solution because sugar is soluble in water and form homogeneous mixture. The solubility of sugar is high as compared to the sand in water because the negative and positive ends of sucrose easily dissolve into the polar solvent i.e, water

Suspension:

"Suspension is the heterogeneous mixture, in which the solute particles settle down but does not dissolve"

The mixture of water and sand is suspension. The sand can not dissolve in water because it is mostly consist of quartz. The nonpolar covalent bonds of  sand are too strong and cannot be break by water molecules.

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Answer:

Explanation:

Complete the table below and determine the solubility product constant.

[Zn(NO₃)₂] 0, M Initial 0.226, 0.101 0.0452, 0.0118

[KIO₃] 0, M Titrant 0.200, 0.200, 0.200, 0.200

V₀, mL of Zn(NO₃)₂100.0, 100.0, 100.0, 100.0

V, mL of KIO₃ titrant 12.9, 12.4, 13.0, 18.3

What needs to be determined are the following for each column/sample

V0 + V, mL

[Zn²⁺]  + [IO³⁻]    ⇄  [Zn²⁺][IO³⁻]2

log [Zn²⁺][IO³⁻]2

\frac{\sqrt{Zn^{2+}}}{1 + \sqrt{Zn^{2+}}}

Then there's the determination of Ksp itself

log Ksp =  Ksp of Zn(IO3)2 =

Considering this, which of the ion products is closest to Ksp, and why?

Finally, the titration volumes for the first three samples don't vary greatly. However the last sample is considerably larger. Why is this to be expected?

The attached figures shed more light on the solution to this problem

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What is the ph of a solution of 0.50 m acetic acid?
frosja888 [35]
You need to use the Ka for the acetic acid and the equilibrium equation.

Ka = 1.85 * 10^ -5

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Molar concentrations at equilibrium

CH3COOH         CH3COO-     H+

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Ka = x*x / (0.50 - x) = x^2 / (0.50 - x)

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Answer:

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