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Citrus2011 [14]
3 years ago
15

Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation

Choose... color Choose... freezing point depression Choose... vapor pressure lowering Choose... density Choose...
Chemistry
1 answer:
sleet_krkn [62]3 years ago
7 0

Answer:

boiling point elevation - colligative property

color - non-colligative property

freezing point depression - colligative property

vapor pressure lowering - colligative property

density - non-colligative property

Explanation:

A colligative property is a property that depends on the number of particles present in the system.

Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.

Colour and density do not depend on the number of particles present hence they are not colligative properties.

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Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03%
nlexa [21]

<u>Answer:</u> The empirical and molecular formula of chrysotile is Mg_3Si_2H_3O_4 and Mg_6Si_4H_6O_{16}

<u>Explanation:</u>

We are given:

Percentage of Mg = 28.03 %

Percentage of Si = 21.60 %

Percentage of H = 1.16 %

Percentage of O = 49.21 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of Mg = 28.03 g

Mass of Si = 21.60 g

Mass of H = 1.16 g

Mass of O = 49.21 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Magnesium = \frac{\text{Given mass of Magnesium}}{\text{Molar mass of Magnesium}}=\frac{28.03g}{24g/mole}=1.17moles

Moles of Silicon = \frac{\text{Given mass of Silicon}}{\text{Molar mass of Silicon}}=\frac{21.06g}{28g/mole}=0.752moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.16g}{1g/mole}=1.16moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{49.21g}{16g/mole}=3.07moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.752 moles.

For Magnesium = \frac{1.17}{0.752}=1.5

For Silicon = \frac{0.752}{0.752}=1

For Hydrogen = \frac{1.16}{0.752}=1.5

For Oxygen = \frac{3.07}{0.485}=4.08\approx 4

To convert the mole ratios into whole numbers, we multiply individual mole ratios by 2

Mole ratio of Magnesium = (2 × 1.5) = 3

Mole ratio of Silicon = (2 × 1) = 2

Mole ratio of Hydrogen = (2 × 1.5) = 3

Mole ratio of Oxygen = (2 × 4) = 8

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of Mg : Si : H : O = 3 : 2 : 3 : 8

The empirical formula for the given compound is Mg_3Si_2H_3O_8

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 520.8 g/mol

Mass of empirical formula = [(24 × 3) + (28 × 2) + (1 × 3) + (16 × 8)] = 259 g/mol

Putting values in above equation, we get:

n=\frac{520.8g/mol}{259g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

Mg_{(3\times 2)}Si_{(2\times 2)}H_{(3\times 2)}O_{(8\times 2)}=Mg_6Si_4H_6O_{16}

Hence, the empirical and molecular formula of chrysotile is Mg_3Si_2H_3O_4 and Mg_6Si_4H_6O_{16}

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A 7.06% aqueous solution of sodium bicarbonate has a density of 1.19g/mL at 25°C what is the molarity and molality of the soluti
lilavasa [31]

<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg

<em>Molar concentration </em>

Assume you have 1 L solution.

Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)

= 1190 g solution

Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)

= 84.01 g NaHCO3

Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)

= 1.14 mol NaHCO3

<em>c</em> = 1.14 mol/1 L = 1.14 mol/L

<em>Molal concentration</em>

Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg

<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg

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Answer:

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Explanation:

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Based on the solubility chart, which of the listed salts is the most soluble at 25 degrees Celsius?
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I think it's Potassium iodide because it has the highest solubility at 25°C.
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