23 grams of Acetaminophen is equal to how many moles?

In part A, when you added water to the sodium acetate in the flask, some of the chemical dissolved into the water. Would you cal

viktelen [127]

Answer: The solution is a SATURATED solution.

Explanation:

Although most substances are soluble in water, some are more soluble than others,that is , their solubilities differ. SOLUBILITY is a means of comparing the extent to which different solutes can dissolve in a particular solvent at a definite temperature.

From the question above, when water was added to the sodium acetate in the flask, SOME of the chemical dissolved into the water, meaning that some remained undissolved. This is because a given volume of water can only dissolve a certain amount of chemical in it at room temperature. If more chemical is added to such a solution, the chemical will remain undissolved. Such a chemical solution is said to be a SATURATED SOLUTION.

A saturated solution of a solute at a particular temperature is on which contains as much solute as it can dissolve at that temperature in the presence of undissolved solute particles.

Unsaturated solution is a type of solution that dissolves all its solutes with no presence of undissolved solute.

Supersaturated solution is one which contains more of the solute than it can normally hold at that temperature. It is an unstable solution which crystallizes out when disturbed.

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3 years ago

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A 10.0 g sample of an unknown liquid is vaporized at 120.0°C and 5.0 atm. The volume of the vapour is found to be 568.0 mL. The

Minchanka [31]

Answer:

molecular formula of liquid = C₈H₁₈

Explanation:

First we determine the empirical formula of the liquid:

Number of moles of each element present in the liquid = % mass / molar mass

For Carbon, (molar mass = 12.01 g/mol) : 84.2/12.01 =7.011 moles

For Hydrogen (molar mass = 1.01 g/mol) : 15.8/1.01 = 15.643

Simplest mole ratio of the elements, C : H is given by:

C = 7.011/7.011 = 1.0

H = 15.643/7.011 = 2.23

Multiplying through with 5, C:H = 5:11

Therefore, empirical formula is C₅H₁₁

The molecular mass of the liquid is next determined:

Using PV = nRT to find the number of moles of the liquid present

P = 5.0 atm; V = 568.0 mL = 0.568 L; R = 0.082 L*atmmol⁻¹ K⁻¹; T = 273 + 120 = 393 K

n = PV/RT = (5*0.568)/0.082*393

n = 0.088 moles

Molar mass of liquid = mass/no of moles = 10.0 g/ 0.088 moles = 113.63 gmol⁻¹

Molecular formula = n(empirical formula)

Molar mass of empirical formula, C₅H₁₁ = 71 gmol⁻¹

n = molecular mass/empirical mass = 113.63/71 = 1.6

Therefore, molecular formula = 1.6*(C₅H₁₁) = C₈H₁₈

6 0

3 years ago

8. Fill in the blanks underneath the wave spectrum to indicate the relative positions of each type of

anzhelika [568]

Ultraviolet as ultraviolet light waves are around 420 nm like stated in the photo

7 0

3 years ago

A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo

Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>

Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

= (2.684 - 2.156) g

= 0.528 g

Find the number of moles of the elements in the compound

The number of moles is given by

$n = \frac{m}{M}$

where m = given mass and

M = Molar mass

Number of moles of Zinc = $\frac{2.156}{65}$ = 0.033 moles

Number of moles of Oxygen = $\frac{0.528}{16}$ = 0.033 moles

Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

Learn more about the empirical formula:

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1 year ago

Сгорело 6 г угля С. Вычислите массу образовавшегося оксида углерода (IV)<br>просто скажите ответ

skad [1K]

English<span>Сгорело 6 г угля С. Вычислите массу образовавшегося оксида углерода (IV)
просто скажите ответ</span>

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3 years ago