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3 years ago

A solution that is able to dissolve additional solute is best described as

Chemistry

1 answer:

worty [1.4K]3 years ago

6 0

<h2>Question:</h2>

A solution that is able to dissolve additional solute is best described as _____.

<h2>Answer:</h2>

<h2>D.) Unsaturated </h2>

<u>Unsaturated</u> means the substance is poured into the solvent that can be dissolved. It is a solution (with less solute than the saturated solution) that completely dissolves, leaving with no remaining substances.

<h2>For example:</h2>

If you were to pour olive oil into a glass of water, it will dissolve, so it is <u>unsaturated</u>.

_________

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Which phase change results with increased kinetic energy?

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An increase in kinetic energy corresponds to an increase in temperature. Out of boiling, condensation, freezing, and precipitation, boiling is the only one that indicates an increase in temperature.

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Refer to your periodic table:

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Explanation:

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3 years ago

How is stoichiometry used to calculate the amount of product from amount of reactant?

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<span>Stoichiometry deals with the quantitative measurement of reactants and products in a chemical reaction. Let suppose you are given with following reaction;

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3 years ago

Read 2 more answers

Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or

natka813 [3]

Answer:

Explanation:

You need to remember that the oxidation number of H is +1, except when it is in a metal hydrites like NaH, where its oxidation number is -1. Then, the oxidation number of O is -2, but in peroxides is -1. So with these rules you just have to multiply the ox. number with the name of atoms and all the elements in the reaction must sum 0.

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3 years ago

If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a

vodomira [7]

Answer:

pH = 4.543

Explanation:

CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ <em>C</em> CH3CH2COOH = 0.048 M

⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M

mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543

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3 years ago