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3 years ago

Which solution is a homogeneous mixture? sand and water salt and water salad dressing soil

Chemistry

2 answers:

Over [174]3 years ago

8 0

Answer: Option (b) is the correct answer.

Explanation:

A solution whose constituent particles are uniformly distributed in the solution is known as a homogeneous solution.

In a homogeneous solution, particles of one solute completely dissolves in the solvent. This solution is also known a true solution.

When sand and water are mixed together then after a certain interval of time sand particles will settle at the bottom. Therefore, they do not dissolve in water. So, it is not a homogeneous solution.

When salt and water are mixed together then the salt particles will dissolve in water. Therefore, it is a homogeneous solution.

Salad dressing is not a homogeneous solution as it contains different solutions like oil, lemon juice etc which does not dissolve when mixed together.

On the other hand, soil is also not a homogeneous solution because we can visualize the distinct layers present in a soil. These layers when mixed in a solution does not dissolve so they did not form a homogeneous solution.

Thus, we can conclude that out of the given options, salt and water is a homogeneous mixture.

Ket [755]3 years ago

7 0

Sand and Water Salt is deh answer my good sur

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2 years ago

For full credit, show the details of each calculation below. Use more paper if needed.

vredina [299]

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Explanation:

1) a, b) A solution is a homogeneous mixture of two or more substances. The solute is the substance present in a smaller amount, and the solvent is the substance present in a larger amount.

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2 years ago

In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti

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Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

$\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }$ ⇄ $\mathtt{3SO_{2(l)}}$

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

$K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}$

However, since we are dealing with liquids solutions;

$K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}$ since the activity of $a_{so_3}$ is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

$K = \dfrac{1}{Pso_2Po_2^{1/2}}$

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

$\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\ \simeq -68 \ kJ/mol$

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

$\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}$

$\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K } \\ \\ K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\ K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }$