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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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A 6.0 M HCI solution is diluted to make 500 mL of a 2.5 M HCI solution. What volume of the 6.0 M HCI solution is required to mak

rodikova [14]

Answer:

V₁ = 208.3 mL

Explanation:

Given data:

Initial molarity of HCl = 6.0 M

Final volume = 500 mL

Final molarity = 2.5 M

Volume of initial solution required = ?

Solution:

Formula:

M₁V₁ = M₂V₂

Now we will put the values in formula.

6.0 M × V₁ = 2.5 M ×500 mL

6.0 M × V₁ = 1250 M.mL

V₁ = 1250 M.mL / 6.0 M

V₁ = 208.3 mL

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3 years ago

A certain chemical reaction releases of heat energy per mole of reactant consumed. Suppose some moles of the reactant are put in

m_a_m_a [10]

To relate the quantity of a substance to the amount of heat transferred and heat capacity, temperature, number of moles is given by

q=ncpΔT

where

n is number of moles

cp is molar heat capacity

ΔT=Tfinal−Tinitial is the temperature difference

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3 years ago

A chemistry student is given 600. mL of a clear aqueous solution at 27.° C. He is told an unknown amount of a certain compound X

DiKsa [7]

Answer: No

Explanation:

Firstly, the molar mass of the dissolved solid is not given. This is necessary to calculate the number of moles present in solution. Secondly, solubility always has to do with temperature and the specified temperature is 27°c and not 21°c. This makes it impossible to calculate the solubility at 21°c. Further information must supplied before the solubility at 21°c can be accurately calculated.

8 0

3 years ago

what is the boiling point of the solution resulted from the dissolving of 32.5g of NaCl in 250.0g of water?

suter [353]

Answer:

The boiling point of this solution is 102.28 °C

Explanation:

<u>Step 1:</u> Data given

Mass of Nacl = 32.5 grams

Molar mass of NaCl = 58.45 g/mol

Mass of Water = 250 grams

Boiling point of water = 100°C

<u>Step 2: </u>Calculate number of moles

Number of moles = mass of NaCl / Molar mass of NaCl

Number of moles = 32.5 grams /58.45 g/mol = 0.556 moles

<u>Step 3:</u> Calculate molality

Molality = Number of moles / mass of water

Molality = 0.556 moles / 0.250 kg of water

Molality = 2.224 molal

NaCl releases twice as many moles of ions, the total molality of this solution is also twice: 4.448 molal

Step 4: Calculate boiling point

dT =( 0.512 C / molal)*4.448 molal)

dT = 2.28

The boiling point of this solution is 100 °C + 2.28 °C = 102.28 °C

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3 years ago

Number of moles in 145.7 g of squaric acid

Nesterboy [21]

To find moles : moles= Mass (C₄H₂O₄) / RFM (C₄H₂O₄)

so moles = 147.7 / 114 = <span>1.2956mol

hope that helps </span>

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3 years ago