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CHEMISTRY 20 Unit 2 Gases and solutions C = n or C = m V MV P 1 V 1 = P 2V 2 T1 T2 1

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CHEMISTRY 20

Unit 2

Gases and solutions

C = n or C = m V MV

P 1 V 1 = P2V2

T1 T2

PV = m R T M

Prepared by Daniel VeraartCIS Abu Dhabi

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Chemistry 20Lab #10

Date: ___________________ Name: ____________________________

Balloon in a bottle

You cannot see most gases, but you can learn about some of the properties of gases by creating situations that you van visualize. In this lab, you will deduce some properties of gases by blowing up a balloon inside a bottle.

Materials: 1 L clean plastic soft drink Bunsen burner and an ice pick Round balloon

Procedure:

1. Insert the balloon in the bottle as shown in the picture.

2. Predict how much you will be able to inflate the balloon inside the bottle, relative to the size of the bottle. Record your prediction in the table of observations.

3. Inflate the balloon inside the bottle as much as you can. How large did get? Record your observations. Allow the balloon to deflate.

4. Using a hot ice pick, puncture a hole in the middle of the bottom of the plastic bottle. Inflate the balloon again. Record your observations.

Prediction Observation #1 Observation #2

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Analysis:

1. a) Was your prediction in step 2 verified when you carried out step 3? ___________

b) What happened to the air already inside the bottle when you blew up the balloon inside the bottle

__________________________________________________________________

__________________________________________________________________

__________________________________________________________________

__________________________________________________________________

2. a) Was there a difference in the volume to which you were able to inflate the balloon before and after _____________you punctured a hole in the bottle?

b) If there was a difference, explain why?

__________________________________________________________________

__________________________________________________________________

__________________________________________________________________

__________________________________________________________________

3. Describe one property of gases that was demonstrated in this activity?

__________________________________________________________________

__________________________________________________________________

__________________________________________________________________

__________________________________________________________________

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Chemistry 20Day 11

Date: ___________________ Name: ____________________________

Properties of gases

What are gases? Gases are defined as fluids that have no shape or volume

of their own, but take on the shape and volume of their container in which they are confined.

Macroscopic Properties of gases

Gases have the following macroscopic properties (properties that can be observed):

1. Gases are compressible

2. Gases expand as the temperature is increased (if pressure remains constant)

3. Gases have very low resistance to flow (air flows very easily)

4. Gases have low densities. (Density of H2O(g) = 0.001 g/mL)

5. Gases mix evenly and completely when put in the same container.

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Gases and Pressure

Atmospheric pressure

Most of the Earth's atmosphere lies within 11 km of its surface.

Pressure is defined as force per unit area., thus atmospheric pressure as the force that a 1 m2 column of air exerts on Earth's surface.

Because for many years, mercury barometers were used to measure atmospheric pressure, the common units of pressure became the mmHg.

At sea level, the atmospheric pressure 760 mmHg = 1 atmosphere = 101.325 kPa (N/m2)

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Boyle’s LawRelationship between Pressure and volume of gases

Irish scientist Robert Boyle (1627-1691) studied the relationship between pressure and volume of gases at constant temperature.

By making careful measurement between pressure and volume of a trapped gas, he was able to describe what happened when the pressure exerted on a gas was increased.

The relationship that we saw in this demonstration is known as the Boyle's Law. As you observed, when the pressure on the gas was increased, the volume of the gas decreased.

Mathematically, this relationship is called an inverse proportion (Volume is inversely proportional to pressure)

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Boyle’s law For a constant mass of gas at constant temperature, the volume varies inversely with the pressure.

=> If the pressure decreases, the volume increases=> If the pressure increases, the volume decreases

Therefore,P1 . V1 = P2 . V2

Example 1

A weather balloon with a volume of 2000 L at a pressure of 96.3 kPa rises to a height of 1000 m, where the atmospheric pressure is measured to be 60.8 kPa. Assuming there is no change in temperature, what is the final volume of the weather balloon?

Example 2

A 2.5 L container is filled with helium gas at a pressure of 3.5 atm. If the temperature remains constant, and the volume increases to 9.0 L, what is the final pressure on the gas?

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Chemistry 20Homework #11

Name: ____________________________ Date: __________________________

Exercises

1. A sample of gas in a flexible container has a volume of 6.9 L after its pressure has been increased from 1.0 atm to 3.5 atm. What was the initial volume of the gas?

Answer: ________________

2. A flexible container holding 3.50 L of hydrogen gas at standard atmospheric pressure has to be compressed into a volume of 2.75 L. If there is no change in temperature, what pressure is required?

Answer: ________________

3. A sample of neon gas at room temperature is collected in a 2.5 L balloon at standard atmospheric pressure. The balloon is then submerged into a tub of water, also at room temperature, so that the external pressure is increased to 112.5 kPa. What will be the final volume?

Answer: ________________

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Chemistry 20Day 12

Date: ___________________ Name: ____________________________

Gases and temperature

Charles’ Law

More than a century after Boyle had determined the relationship between the pressure and volume of a gas, French physicist Jacques Charles determined the relationship between the temperature and volume of a gas.

Charles became interested in the effect of temperature on gas volume after observing the hot air balloons that had become popular as flying machines.

In addition, Charles observed another very consistent result in their experiments; when they extrappolated their linear plots down to zero volume, all the lines converged at one value of temperature, -273.15C.

It was Scottish physicist Lord Kelvin, who in 1848, interpreted te significance of this temperature of -273.15C.

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He suggested that this was the theoretically lowest possible temperatue or absolute zero.

He then established a new temperature scale (The Kelvin scale) based on absolute zero starting point on the scale.

The Kelvin scale

The size of a unit of temperature on the Kelvin scale is the same as the size of a degree on the Celsius scale.

Only the starting points for the temperature scales are different.

The units on the Kelvin scale is Kelvin (K)

Celsius to Kelvin --> Add 273.15

Kelvin to Celsius --> Subtract 273.15

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Charles’ law For a constant mass of gas at constant pressure, the volume varies directly with its temperature (in Kelvin).

If the temperature increases, the volume increases (and vice-versa)

Therefore, V1 = V2 T1 T2

Note Temperature has to be in Kelvin.

Example:

1. A birthday balloon is filled to a volume of 1.50 L of helium gas in a 21.00C room. The balloon is then taken outside on a warm sunny day. The volume expands to 1.53 L. Assuming the pressure remains constant, what is the Celsius temperature outdoors?

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Chemistry 20Homework #12

Name: ____________________________ Date: __________________________

Exercises

1. Convert the following Celsius temperatures to the Kelvin scale;

a) 27.3°C : ____________ b) 37.8°C: ___________

c) 122.4°C: ___________ d) -25°C: ____________

2. Convert the following Kelvin temperatures to the Celsius scale;

a) 373.2°K : ____________ b) 275°K : ___________

c) 173°K : _____________ d) 23.5°K: ____________

3. A 75 mL balloon immersed in liquid nitrogen at -196°C is lifted out and left in a room at 22.3°C. What is the final volume of the balloon?

Answer: ________________

4. A sealed syringe contains 37.0 mL of trapped air. The temperature of the air in the syringe is the same as room temprature (25°C). The Sun shines on the syringe causing the temperature of the air inside to increase. If the volume increase to 38.6 mL, what is the new temperature of the air in the syringe?

Answer: ________________12

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5. The volume of a 1.5 L balloon at room temprature increases by 25 percent of its volume when is is placed in a hot-water bath. How does the temperature of the water bath compare with room temperature?

Answer: ________________

6. A 50g sample of dry air in a large advertising balloon occupies a voume of 27L at 20°C. What volume will the balloon occupy when the temperature is increased to 35°C at constant pressure?

Answer: ________________

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Chemistry 20Day 13

Date: ___________________ Name: ____________________________

Combined Gas Law

When Charle’s law and Boyle’s law are combined, the resulting combined gas law states the relationships among the volume, temperature, and pressure of any fixed quantity of gas.

Combined Gas Law

P1V1 = P2V2 T1 T2

Note : The temperature has to be in Kelvin.

Finding Volume using the Combined Gas Law

A small balloon contains 275 mL of helium gas at a temperature of 25.0C and a pressure of 350.0 kPa. What volume in mL would this gas occupy at 10.0C and 85.0 kPa?

Finding Temperature using the Combined Gas Law

A volume of 25.0 mL of gas is produced in a lab experiment at a temperature of -15.0C and a pressure of 700 mmHg. Predict the Celsius temperature of the gas when its volume is reduced to 20 mL and the pressure is increased to 820 mmHg.

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Avogadro's Law

Amadeo Avogadro (1776-1856) came up with this Law that relates the number of moles of gas with the volume of gas. His principle stated that equal volumes of gases at the same temperature and pressure contain the same number of molecules regardless of their chemical nature and physical properties!

Avogadro's Law

Equal volumes of all ideal gases at the same temperature and pressure contain the same number of molecules.

n1 = n2 V1 V2

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Chemistry 20Homework #13

Name: ____________________________ Date: __________________________

Exercises

1. A sample of methane gas at 105.0 kPa pressure has a volume of 25.0 L. The pressure is decreased to 95.0 kPa. What is the new volume?

Answer: ______________ L

2. A sample of neon gas has a volume of 25.0 L at 25.0°C. What is the volume at 45.0°C?

Answer: _____________ L

3. A sample of argon gas has a volume of 38.0 L at 12.0°C. What is the volume at 112.0°C?

Answer: _____________ L

4. A gas sample has a volume of 235 mL at a pressure of 115 kPa. What is the volume at 85.5 kPa ?

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Answer: ______________ L

5. At 17.0°C a sample of hydrogen gas has a volume of 45.0 L . To what temperature must this gas be heated to change the volume to 65.0 L?

Answer: _____________ °C

6. A 250 mL sample of fluorine gas at 110.0 kPa and 25.0°C is changed to 95.0 kPa and 45.0°C. What is the new volume?

Answer: _____________ mL

7. A storage tank is designed to hold a fixed volume of butane gas at 150 kPa and 35.0°C. to prevent dangerous pressure buildup, the tank has a relief valve that opens at 250 kPa. At what temperature (in °C) does the valve open?

Answer: _______________ °C

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Chemistry 20Day 14

Date: ___________________ Name: ____________________________

Molar volume of Gases

According to Avogadro, equal volumes of any gas at the same temperature and pressure contains an equal number of partcles.Therefore for all gases at each specific pressure and temperature, there must be a certains volume that contains exactly one mole of particles. This is referred as the molar volume of a gas.

Molar volume of a gas The molar volume of a gas is the volume occupied by one mole at standard ambient temperature and pressure (SATP)

This molar volume for all gases is 24.8 L at SATP.SATP = Standard ambient temperature is 25C (298.15 K) and standard pressure is 100

kPa. ***Not to be confused with STP which is 0C and pressure of 101.325 kPa!

Knowing the molar volume of gases allows scientists to work with easily measured volumes of gases when specific masses of gases are needed.

Measuring the volume of a gas is much more convenient than measuring its mass. Molar volume can be used as a conversion factor to convert amounts in moles (or grams) to volume, and vice-versa.

Examples

1. What amount in moles of oxygen is available in a volume of 5.60 L at SATP?

2. What mass of carbon dioxide is present in a 600 mL container at SATP?

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The ideal Gas Law

The gas laws discussed so far are exact only for an ideal gas. An ideal gas is a hypothetical gas that obeys all the gas laws perfectly under all conditions; that is, it does not condense into a liquid when cooled. Theoretically, an ideal gas is composed of particles of zero size that have no attraction to each other. Real gases do condense and do have particles that attract each other, and therefore real gases do not follow the gas laws exactly.

For the purpose of our course, all gases are dealt with as if they were ideal. A single, ideal-gas equation describes the interrelationship of pressure, temperature, volume, and amount of matter — the four variables that define a gaseous system.• According to Boyle's law, the volume of a gas is inversely proportional to the pressure.• According to Charles' law, the volume of a gas is directly proportional to the Kelvin temperature.• According to Avogadro, the volume of a gas is directly proportional to the amount of matter.

Combining these three statements produces the following relationship:

V = (a constant) x 1 x T x n or PV = nRT PThis last equation is known as the ideal gas law; the constant R is known as the universal gas constant. The value for the universal gas constant can be obtained by substituting SATP conditions for one mole of an ideal gas into the ideal gas law and solving for R.

R = P. V = 100 kPa x 24.79 L = 8.314 L.kPa/mol.K n . T 1.00 mol x 298.15 K

The ideal Gas Law

P . V = n . R . T

P = Pressure in kPa V = Volume in Ln = Number of moles in mol T = Temperature in K

R = universal gas constant = 8.314 kPa.L/mol.K

Finding Volumes using the Ideal Gas Law

Find the volume of 100.0 g of oxygen at SATP.

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Finding Temperature using the Ideal Gas Law

Find the Celsius temperature when 2.50 moles of a gas occupies a volume of 56.5 L under a pressure of 121.59 kPa.

Finding Molar mass using the Ideal Gas Law

Find the molar mass of a gas if a 1.58 g sample occupies a volume of 500.0 mL at SATP.

Finding Mass of gas using the Ideal Gas Law

What mass of neon gas should be introduced into a 880 mL tube to produce a pressure of 90.0 kPa at 30.0°C?

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Chemistry 20Homework #14

Name: ____________________________ Date: __________________________

1. What mass of sulfur dioxide gas is contained in 50.0 mL of the gas at SATP?

Answer: ______________ g

2. What is the volume of 66.4 g of carbon dioxide at 32.0°C and 115 kPa?

Answer: ______________ L

3. What is the volume of 44.0 g of butane gas ( C4H10(g) ) at 45.0°C and 95.0 kPa?

Answer: ______________ L

4. What is the mass of neon inside a 160.0 L sign if the temperature is -10.0°C and the pressure is 300.0 kPa ?

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Answer: ______________ g‘

5. 0.575 g of an unknown gas has a volume of 350 mL at 105 kPa and 24.0°C. What is its molar mass?

Answer: _____________ g/mol

9. A student is trying to identify a pure gas sample. She decides to determine the molar mass of the gas, and obtains the following evidence.

Mass of container = 7.02 g Mass of container + gas = 9.31 gVolume of container = 1.25L Temperature of gas = 23.4 °CPressure of gas = 102.2 kPa

Find the molar mass of the gas.

Answer: _____________ g/mol

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Chemistry 20Day 15

Date: ___________________ Name: ____________________________

Stoichiometry problems with gases

Many chemical reactions have gases as reactants and/or products. When a gas is produced (or needed) in a chemical reaction, we can calculate the volume of the gas by using the ideal gas law in the stoichiometry calculations.

Stoichiometric calculations with gases

1. Write a balanced chemical equation for the reaction and determine the two substances involved in the reaction, the given and the unknown.

2. Find the number of moles of the given substance using the following formulas:

n = m or n = P.V (If a gas is given) M R.T 3. Find the molar ratio between the given and unknown substances using the chemical

equation. This will give you the number of moles of the unknown substance.

4. Answer the question by using the number of moles found in step #3 and one of the following formulas:

m = n.M or V = n.R.T P

Examples:

1. A teacher burns 5.00 g of propane and collected the carbon dioxide produced by the reaction. Calculate the volume of carbon dioxide produced at SATP.

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2. Hydrogen sulfide reacts with sulfur dioxide to give H2O and S,

2H2S(g) + SO2(g) → 2H2O(l) + 3S(s)

If 6.0 L of H2S gas at 100.4 kPa produced 3.2 g of sulfur, calculate the temperature in C.

3. How many liters of NH3 will be produced if 4.3 grams of N2(g) are consumed at SATP according to the following reaction:

N2(g) + 3H2(g) → 2NH3(g)

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Do Homework #15

Chemistry 20Homework #15

Name: ____________________________ Date: __________________________

1 A student put a piece of tin (in excess) in 10.0 g of potassium hydroxide in solution. The products of this reaction are hydrogen gas and K2SnO2(s). At SATP, what volume of gas was collected?

Answer: _____________ L

2. A piece of sodium (in excess) reacts with 5.00 L of Chlorine gas at 50.0°C and 95 kPa. Calculate the mass of sodium chloride produced.

Answer: _____________ g

3. What volume of oxygen at SATP is needed to burn 25.0 L of methane?

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Answer: _____________ L

4. A student drops a 1.50 g piece of magnesium in excess hydrochloric acid. What volume of gas would be produced if the temperature is 20.0 °C and the pressure is 100 kPa.?

Answer: _____________ L

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Lab #13The Volume of a gas

Problem: Find the volume of gas produced when a piece of magnesium is added to hydrochloric acid.

Design: A known mass of magnesium ribbon reacts with excess hydrochloric acid in a closed system.

Material: - 100 mL cylinder - Stopper - Stopper - Thermometer-Beaker - Magnesium ** - HCl (2 mol/L) ** - Copper wire

Questions

1. Make a table of observations.

2. Calculate the volume of gas using stoichiometry.

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Chemistry 20Day 17

Date: ____________________ Name: _________________________

Types of solutionsA solution is a homogeneous (uniform) mixture of two or more substances.

Examples of solutions: Vinegar, Sea water, Air (Oxygen in nitrogen)

The substance that is present in larger quantity is usually referred as the solvent.The substance present in smaller quantity, which is dissolved in the solvent, is usually referred to as the solute.

Table representing the types of solutions

Original State of solute State of solvent Examples

Gas Gas

Gas Liquid

Gas Solid

Liquid Liquid

Liquid Solid

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Solid Liquid

Solid Solid

Dissolving and Forces of Attraction

When a solid dissolves in water, it may appear as if a chemical reaction occurs, since the solute seems to disappear.

When a solute dissolve, the bonds holding the molecules together (intermolecular bonds) are broken and the molecules or ions of solute become attracted to the water.

The solute particles disperse throughout the solution, and your unaided eye cannot distinguish solute from solvent.

Processes of Dissolving

Step 1: The intermolecular forces between the particles in the solute break = Endothermic process (needs energy)

Step 2: The intermolecular forces between the water molecules break = Endothermic process (needs energy)

Step 3: Formation of new chemical bonds = Exothermic process (releases energy)

Dissolving and Forces of Attraction

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Net energy of dissolving = Energy Step 1 + Energy Step 2 - Energy step 3

If net energy is positive, the overall energy change is endothermic

If net energy is negative, the overall energy change is exothermic

Exothermic change: A change in which energy is released during the change.

Example: Dissolving Sodium hydroxide in water

Endothermic change: A change in which energy is absorbed during the change.

Example: Reaction in cold packs

Lab #5Energy Changes in dissolving ionic compounds

Problem: To determine if the dissolving of different ionic compound is endothermic or exothermic.

Material:Beakers Scale ThermometerCalcium chloride Ammonium nitrate Lithium chlorideAmmonium chloride Potassium chloride Sodium hydroxide

Procedure:

1. Mesure 50 mL of distilled water into your beaker.

2. Mesure a moderate amount of solute.

3. Add the solute to your solvent.

4. Using a stir stick, ensure that all the solute is dissolved.

5. Mesure the temperature of your solution.

6. Dispose of your solution in the disposal bin,

7. Rinse your beaker with water.

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8. Repeat step 1-7 with your next solute.

Analysis:

1. Complete the table of observations.

SoluteTemperature

change of solvent + solute

Exothermic? Endothermic?

Calcium chloride

Ammonium nitrate

Lithium chloride

Ammonium chloride

Potassium chloride

Sodium hydroxide

2. Order your solute in order of increasing heat released.

_____________ _____________ _____________ _____________ _____________ _____________

3. Name the variables of the experiment (Manipulated, Responding and Controlled)

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Chemistry 20Day 17

Date: ____________________ Name: _________________________

Solutions of Electrolytes and Non-Electrolytes

What is an electrolyte?

Electrolytes : An electrolytes is a compound that dissolves in water to produce ions in solution, this solution will conduct an electric current.Examples of electrolytes are soluble ionic compounds and strong acids.

Example: NaCl(s) Na+(aq) + Cl-(aq) HCl(g) H+

(aq) + Cl-(aq)

Note: The conductivity of a solution is tested with a conductivity apparatus.

Acidic solution vary in their electrical conductivity:- Acids that are extremely good conductors are called strong acids.- Acids that are poor conductors are called weak acids.

There are only 6 strong acids: List of the strong acid in your data booklet (P. 8)

Non-electrolytes: Non-electrolytes contain solutes which form neutral molecules in solution. No current flows and the solution are nonconductors.

Molecular compounds are all non-electrolytes.

Example: C12H22O11(S) C12H22O11(aq)

When non-electrolytes dissolve in water, the particles of the solute are separated from each other, but their structure remains unchanged. When electrolytes dissolve in water, the particles undergo a structural change.

Examples:

1. Are the following compound electrolytes or non-electrolytes. Justify your choice.

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a) CuSO4(s) : _________________________ b) HBr(g) :_______________________________

c) C3H8(g) : ___________________________ d) Zn(s) : ________________________________

e) CaCO3(s) : _________________________ f) CH3OH(l) : _____________________________

g) Na2SO4(s) : ________________________ h) CH3COOH(g) : __________________________

i) MgCl2(s) : ________________________ j) CuI(s) : __________________________

Dissociation equations for electrolytes

Compounds that are electrolytes (soluble ionic compounds and strong acids) dissociate into ions when dissolved in water.We can represent the change of an electrolyte into ions by an equation known as dissociation equation.The dissociation equation shows the change from the pure compound to the aqueous ions.

Examples:

1. i) Classify the following substances as ionic soluble – ionic non-soluble – molecular soluble – molecular non-soluble - strong acids – weak acids.ii) Classify as electrolyte or not.iii) Write the dissociation equations for the electrolytes.

a) CaCl2(s) : ________________________________________________________________

b) Na3PO4(s) : ______________________________________________________________

c) Ca(NO3)2(s): ______________________________________________________________

d) PbI2(s): ___________________________________________________________________

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e) H2SO4(g) : ________________________________________________________________

Entities in waterWhen we add a substance to water, certain entities (compound, ions,…), will be present in that mixture. The entities present depend on the solubility of the substance and also if the substance is an electrolyte or not.

Examples: When calcium carbonate is added to water, the calcium carbonate does not dissolve and the major entities present are water and calcium carbonate.

When calcium chloride is added to water, the calcium chloride will dissolve and dissociate into ions (because it’s an electrolyte). The major entities present are water calcium ion and chloride ions.

To understand the properties of aqueous solutions and the reactions that take place in solutions, it is necessary to know the major entities present when any substance is in water.

Major entities present in waterType of substance Solubility in

waterTypical Pure substance

Major entities present if placed in water

Ionic compounds soluble

Ionic compounds non soluble

Molecular compounds

Strong acids

Weak acids

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Elements

\

ExamplesEach of the following substances is placed in water. Classify the substances and for each mixture, list the formulas of the major entities present in water environment.

Substance Classification Entities present in water

Potassium chloride

Hydrobromic acid

C2H5OH

Copper (II) nitrate

Calcium carbonate

Phosphoric acid

NH4OH

Ammonium phosphate

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Silver

C8H18

Do Homework #17

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Chemistry 20Homework #17

Name: ____________________________ Date: __________________________

Exerci ses

1. Are the following compound electrolytes or non-electrolytes. Justify your choice.

Substance Electrolytes or not JustificationCu(NO3)2

H3PO4(g)

CH4(g)

HNO3(g)

H2CO3(g)

N2(g)

H2O2(l)

HClO4(g)

2. i) Classify the following substances as ionic soluble – ionic non-soluble – molecular soluble – molecular non-soluble - strong acids – weak acids.ii) Classify as electrolyte or not. iii) Write the dissociation equations for the electrolytes.

a) Lead (IV) sulfate: i) _________________________ ii) __________________________

iii) _______________________________________

b) Nitrogen gas: i) _________________________ ii) ______________________

iii) _______________________________________

c) Sulfuric acid: i) _________________________ ii) ______________________

iii) _______________________________________

d) Zinc chloride: i) _________________________ ii) __________________________

iii) _______________________________________ e) Sodium sulfate: i) _________________________ ii) __________________________

iii) _______________________________________

h) Butanol (C4H9OH(l)) : i) _________________________ ii) ______________________

iii) _______________________________________

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3. Each of the following substances is placed in water. Classify the substances and for each mixture, list the formulas of the major entities present in water environment.

Substance Classification Entities present in water

Zinc

Acetic acid

Copper (II) sulfate

Calcium phosphate

Wax (C25H52)

Aluminum sulfate

Methanol (CH3OH)

Sulfuric acid

Benzoic acid

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Chemistry 20Day 18

Date: ____________________ Name: _________________________

SolubilityImportant definitions

Solubility: The maximum quantity of solute that dissolves in a given quantity of solvent, at a given temperature.

Example: the solubility of sodium chloride, NaCl(s), in water at 20°C is 36 g/100 mL.

Saturated: A solution with solute that dissolves until it is unable to dissolve anymore, leaving the undissolved substances at the bottom.

Example: 100 mL of saturated NaCl(aq) would contain 36 g of dissolved NaCl at 20°C. If more sodium chloride is added to the solution, it will not dissolve.

Unsaturated: A solution (with less solute than the saturated solution) that completely dissolves, leaving no remaining substances.

Example: a solution containing 20 g of dissolved NaCl(s) in 100 mL of water at 20°C is unsaturated. This solution has the potential to dissolve an additional 16 g of NaCl(s) before it becomes saturated.

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Representing Equilibrium for a Saturated SolutionAs you learned earlier, a dissociation equation can be written for the dissolving of an ionic compound, such as Copper (II) sulfate.

CuSO4(s) → Cu2+(aq) + SO4

2-(aq)

Crystallization is the reverse of dissolving and can also be written as an equation:

Cu2+(aq) + SO4

2-(aq) → CuSO4(s)

An equilibrium reaction can be represented by combining both the forward and the reverse processes into one equation with a double-headed arrow, as shown below:

CuSO4(s) Cu2+(aq) + SO4

2-(aq)

You will learn more about equilibrium in Chemistry 30.

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Concentration of solutions

In chemistry, it is important to know the concentration of a solute in a solvent.

Concentration: The concentration of a solution is defined as the amount of solute (in moles or grams) per volume of solution (usually in Liter)

The concentration of solutions can be referred to as concentrated or dilute.

A concentrated solution contains a large amount of solute. A dilute solution contains a small amount of solute.

Expression concentration

There are three main ways of expressing Concentration:

1. Concentration as Percent by Mass

2. Concentration as Parts per m illion

3. Concentration as Molar Concentration

Concentration as Percent by Mass

The concentration of a solute can be described as a mass of solute dissolved in a mass of solution.

This is usually expressed as a percentage

The percent by mass gives the mass of a solute divided by the mass of solution.

Percent by mass = Mass of solute (g) X 100% Mass of solution (g)

Sample problem

Calcium chloride, CaCl2(s) , can be used to melt ice on roads in the winter. To determine how much calcium chloride had been used on a road, a student took a sample of slush to analyze. The sample had a mass of 23.47 g. When the solvent was evaporated, te residue had a mass of 4.58 g. What was the percent by mass of calcium chloride in the slush?

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Concentration as Parts per Million or Parts per Billion

Very small concentration of a substance are often expressed as Parts per Million (ppm) or Parts per Billion (ppb).

For example, air that contains 30 ppm carbon monoxide contains 30 g of CO(g) per 1 000 000 g of air.

1 ppm = 1 g /1 000 000 g or 1 ppm = 1 g / 1000 kg or 1 ppm = 1 mg / kg

Concentration in ppm = Mass of solute (g) X 106 ppm Mass of solution (g)

Concentration in ppb = Mass of solute (g) X 109 ppb Mass of solution (g)

Sample problem

Hard water contains calcium carbonate in very small concentration. Hard water in Alberta contain up to 300 ppm of calcium carbonate. How much calcium carbonate would be present in 20 L (20 000g) of hard water?

Do Homework #18

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Chemistry 20Homework #18

Name: ____________________________ Date: __________________________

Exerci ses

1. The solubility of sodium nitrate is 87.6 g/100 mL in water at 20°C. Using the terms saturated and unsaturated, how would you describe the following systems at 20°C?

a) 20.0 g of sodium nitrate in 100 mL of water : ______________

b) 100 g of sodium nitrate in 100 mL of water : _______________

2. Write the equations that represent saturated solutions of the following compounds:

a) LiCl(s) : ___________________________________________________________

b) NH4F(s) : ___________________________________________________________

c) Na2S(s) : ___________________________________________________________

3. Calculate the percent by mass for each of the following solutes in solution;

a) An aqueous solution with a mass of 82.0 g contains 17.0 g of sulfuric acid, H2SO4(aq)

b) a benzene solution with a mass of 85.4 g contains 12.9 g of carbon tetrachloride, CCl4(l)

4. Symptoms of mercury poisoning become apparent after a person has accumulated 20 mg of mercury in his of her body;a) Express this concentration as parts per million for a 60 kg person

b) Express this amount as a percent by mass for the same person

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Chemistry 20Day 19

Date: ____________________ Name: _________________________

Concentration as Molar Concentration

The most commonly used unit of concentration in chemistry is the Molar Concentration.

Important formula

Chemists like to use the molar concentration of a substance as a measure of concentration since represents the amount of moles of substance per liter of water. (mol/L)

Formula to calculate the molar concentration:

C = n V

We can combine this formula with the formula n=m/M to form a combination that gives:

C = m or m = C.V.M M.V

C = molar concentration m = mass of substance dissolvedM = Molar mass of substance dissolved V = Volume of water

Examples:

1. 0.900 mol of NaCl(s) is dissolved in 500 mL of water, what is the molar concentration of salt in this solution?

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2. If 5.00 g of KOH(s) is dissolved in 1.50 L of water, what would be the concentration of the solution?

3. A student dissolves 15.0 g of sodium hydrogen carbonate in 250 mL of water. Determine the molar concentration of this solution.

4. A sample of water contains 0.240 mol/L of dissolved iron. What mass of iron is present in 250 mL of water?

5. What mass of sodium carbonate is needed to make 0.400 L of a 0.0500 mol/L solution?

6. A 0.245 mol/L Ba(OH)2(aq) solution contains 26.5 g of solute . What is the volume of this solution?

Do Homework #19

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Chemistry 20Homework #19

Name: ____________________________ Date: __________________________

Exercises

1. We dissolve 156 g of pure ammonia in water to make 2.00 L of solution. What is the molar concentration?

Answer: _________________

2. Calculate the number of moles of solute needed to make 100 mL of 0.660 mol/L NaHCO3

Answer: _________________

3. Calculate the mass of glucose needed to make 250 mL of 0.550 mol/L solution .

Answer: _________________

4. What volume of a 0.800 mol/L NaOCl(aq) could be made if we dissolve 119.2 g of solute?

Answer: _________________

5. Calculate the molar concentration of the following solution : 12.7 g of Ca(OCl)2 in 500 L of water.

Answer: _________________

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6. A 0.700 mol/L Na3PO4 solution contains 126 g of solute . What is the volume of this solution?

Answer: _________________

7. Calculate the mass of Na2S2O3 needed to prepare 10.0 L of a 0.00500 mol/L solution .

Answer: _________________

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Chemistry 20Day 20

Date: ____________________ Name: _________________________

Preparing and diluting Solutions

Dilution formula

The concentration of a solution may be changed in several ways, for various reasons. You can increase the concentration by simply adding more solute. You can also decrease the concentration of a solution by diluting it with water.

Formula for dilution Dilution is the process of decreasing the concentration of a solution by adding more solvent.The number of moles of solute remains the same in both solutions

Then, ni = nf ( i = initial f = final)

And we know that n = C . V , so our formula becomes :

Ci . Vi = Cf . Vf

Notes: Final volume is usually greater than initial volume.Initial concentration is usually greater than final concentration.

Examples:

1. Concentrated hydrochloric acid has a concentration of 12.4 mol/L. What volume of concentrated acid should we used to make 2.00 L of 0.250 mol/L HCl(aq) ?

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2. Concentrated acetic acid has a concentration of 17.4 mol/L. If you add 200 mL of this acid to make 4.00 L of solution, what would be the concentration of the diluted solution?

3. A student prepares a solution by dissolving 5.00 g of KOH in 250 mL of water. After the solution is prepared, his good friend messed up his solution by adding 50 mL of water. What is the new concentration of the solution (after the water was added).

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Lab #6Dilution of a solution

Problem : Make a 0.135 mol/L of a colored ionic compound and dilute the solution to a concentration of 0.0564 mol/L and again to a concentration of 0.02 97 mol/L

Procedure:

1. Make 100 mL of a 0.135 mol/L of a colored solution. Pour the solution in a beaker.2. Calculate the volume of the concentrated solution you will need. Transfer this volume in your flask.3. Add water to 100 mL.4. Repeat step 1 to 3 for the second dilution.

Questions :

1. Calculate the mass of the solute needed for the concentrated solution.

2. Calculate the volume needed for the first dilution.

3. Calculate the volume needed for the second dilution.

4. If you evaporate all the water from the third solution, what would be the mass of the solid left.

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Chemistry 20Homework #20

Name: ____________________________ Date: __________________________

Exercises

1. What is the concentration of the solution if 25.0 mL of 2.00 mol/L NaCl(aq) is diluted to a volume of 125 mL?

2. What volume of 12.1 mol/L HCl(aq) is needed to make 300.0 mL of 1.25 mol/L solution?

3. Determine the concentration of the solution when 35.0 mL of 6.00 mol/L KOH(aq) is diluted to 125 mL.

4. Calculate the volume of the original solution needed to make the new solution when 10.0 mol/L silver nitrate solution is used to make 865 mL of 3.45 mol/L solution.

5. Diluted methanol is used as winter windshield washer. A student wants to prepare 8.00 L of a 1.00 mol/L methanol solution. What volume of concentrated methanol (24,7 mol/L) should he be using?

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6. What volume of concentrated ammonia (14.8 mol/L) should a student use to prepare 5.00 L of 0.700 mol/L ammonia?

Chemistry 20Day 21

Date: ____________________ Name: _________________________

Stoichiometry with Solutions

Many chemical reactions have solutions as reactants and/or products. When a solution is produced (or needed) in a chemical reaction, we can calculate the concentration (or volume) of the solution using stoichiometry calculations.

Stoichiometric calculations with solutions

1. Write a balanced chemical equation for the reaction and determine the two substances involved in the reaction, the given and the unknown.

2. Find the number of moles of the given substance using the following formulas:

n = m or n = C x V (If a solution is given) M 3. Find the molar ratio between the given and unknown substances using the chemical

equation. This will give you the number of moles of the unknown substance.

Number of moles = Unknown Given

4. Answer the question by using the number of moles found in step #3 and one of the following formulas:

m = n.M or C = n/V or V = n/C

Example:

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1. A student wants to reacts 10.0 g of magnesium with 500 mL of hydrochloric acid. What would be the required concentration of the hydrochloric acid to react the entire piece of magnesium?

2. What is the minimum volume of 0.250 mol/L MgCl2(aq) needed to precipitate all the silver ions in 60 mL of 0.30 mol/L AgNO3(aq)?

3. What mass of zinc will react with 25 mL of 4.0 mol/L nitric acid to produce H2 gas? What quantity of hydrogen gas will be produced?

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Chemistry 20Homework #21

Name: ____________________________ Date: __________________________

Exercises

1. How much 0.80 M HCl would be needed to dissolve a CaCO3 pearl which weighs 4.0 grams?CaCO3(s) + 2 HCl(s) → CaCl2(aq) + H2O(l) + CO2(g)

2. Throwing some scrap iron in a gold nitrate solution causes the gold metal to precipitate and iron (III) nitrate to form. How much 0.50 M gold nitrate solution would react with 224 grams of iron metal?

3. Sea water is about 0.50 M NaCl. To produce Cl2 gas, a company evaporates sea water, melts the NaCl, and runs electricity through it.

2 NaCl(aq) → 2 Na(s) + Cl2(g)

How many liters of sea water are needed to fill a tank car with 1,120,000 liters of chlorine gas at SATP?

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4. If 36.0 mL of H3PO4 reacts exactly with 80.0 mL of 0.500 M NaOH, what is the concentration of the phosphoric acid?

Lab #7Determination of salinity by chemical titration

Name: ___________________ Date: __________________

In this lab we will use a silver nitrate solution to precipitate out all of the chloride in a known volume of artificial seawater. Although natural seawater contains a much wider array of ions than a mixture of NaCl and fresh water, the analytical technique is the same. When we add a solution containing silver nitrate (that is, Ag+ ions and NO3- ions) to a solution containing sodium chloride (that is, Na+ ions and Cl- ions), the chloride ions in the solution will react with the silver ions to produce a solid precipitate, silver chloride (and leaving the sodium and nitrate ions in solution):

Hypothetically we could then dry and weigh the precipitate AgCl, and from the weight determine the amount of chloride in the sample. Rather than drying and weighing the silver chloride, we can use the same reaction to measure the salinity by knowing exactly how much AgNO3 was used to precipitate all of the chloride.

We can do this by adding another chemical reagent, potassium chromate (K2CrO4), which acts as an indicator to tell us exactly when all of the chloride in the seawater has reacted to form AgCl. The potassium chromate indicator remains colorless as long as there is Cl- present, but the instant the last of the Cl- is bound up as silver chloride, the solution turns orange.

Question

What is the concentration of NaCl in salt water?

Materials • Salt• Scale• Stir stick• Small funnel• 250 mL Erlenmeyer beaker• 150 mL beaker• 40 mL disposal beaker• Graduated cylinder• Deionized water• Burette• Single use syringe• AgNO3 solution (11.6 g of AgNO3 dissolved in 350 mL of deionized water)• K2CrO4 solution

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Procedure

1. Measure a certain amount of salt (NaCl). (Between 1 and 2 grams)2. Mix the salt in 100 mL of water. Use a stir stick to ensure the salt is completely dissolved.3. Using a graduated cylinder, pour 50 mL of your salt water sample into a 250 mL

Erlenmeyer beaker.4. Add ~5 mL of the K2CrO4 solution to the same beaker. (Figure below).5. Fill the 50 ml burette with the AgNO3 solution, pouring it carefully down a funnel into the

top of the burette. This can be done once for this entire set of three titrations. Ensure that the solution fills all the way to the tip of the burette. Record the starting point.

6. Add the AgN03 solution slowly from the burette and continuously stir it with the help of your partner circulating the beaker solution.

7. Once you start to notice the solution turning orange. Record the new burette reading (the end point).

Observations

Trial 1 Trial 2 Trial 3

Starting point

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Trial 1 Trial 2 Trial 3

Ending point

Total volume added

Analysis

1. Write the balanced chemical equation for the reaction of sodium chloride (NaCl) with silver nitrate (AgNO3). The products are silver chloride (AgCl) and sodium nitrate (NaNO3). Calculate the experimental value for the concentration of sodium chloride.

2. Calculate the theoretical value for the concentration of your salt solution.

3. Calculate your percentage of error using the experimental and theoretical values of aqueous sodium chloride.

4. Explain why the color change from yellow to orange of the saline solution indicates the exhaustion of all the chloride ions.

5. The Dead Sea in Jordan is visited by thousands of people every year due to its incredible ability to make any person buoyant in its waters. Using the concentration of solutions, describe this phenomenon.

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