Chemical Calculations and Equations

Chemical Calculations and Equations
Chapter #10 Chemical Calculations and Equations

STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process.

8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2H2O

8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2H2O 6.33 g H2

8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2H2O 6.33 g H2 mole H2 2.016 g H2

8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2H2O 6.33 g H2 Mole H2 2.016 g H2

8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2H2O 6.33 g H2 Mole H2 2 mole H2O 2.016 g H2O 2 Mole H2

8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2 H2O 6.33 g H2 Mole H2 2 mole H2O 18.02 g H2O 2.016 g H2 2 Mole H2 Mole H2O

8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2 H2O 6.33 g H2 Mole H2 2 mole H2O 18.02 g H2O = 28.3 g H2O 2.016 g H2 2 Mole H2 Mole H2O

8.5 Excess and Limiting Reactants
Reactants are substances that can be changed into something else. For example, nails and boards are reactants for carpenters, while thread and fabric are reactants for the seamstress. And for a chemist hydrogen and oxygen are reactants for making water.

8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?

8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? Yes, only one house!

8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we had enough boards?

8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards?

8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards? Yes, nails are in excess!

8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards? Yes, nails are in excess! Nine more houses if we have an adequate amount of boards.

8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?

If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.

If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O H2O 10.0 g O2

If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O H2O 10.0 g O2 mole O2 32.0 g O2

If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O H2O 10.0 g O2 mole O2 2 mole H2 32.0 g O2 mole O2

If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O H2O 10.0 g O2 mole O2 2 mole H2 2.02 g H2 32.0 g O2 mole O2 mole H2

If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O H2O 10.0 g O2 mole O2 2 mole H2 2.02 g H2 = 1.26 g H2 32.0 g O2 mole O2 mole H2

Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.

Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water. 10.0 g O2 mole O2 32.0 g O2

Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water. 10.0 g O2 mole O2 2 mole H2O 32.0 g O2 mole O2

Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water. 10.0 g O2 mole O2 2 mole H2O 18.0 g H2O 32.0 g O2 mole O2 mole H2O

Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water. 10.0 g O2 mole O2 2 mole H2O 18.0 g H2O = 11.3 g H2O 32.0 g O2 mole O2 mole H2O

8.5 Percentage Yield The percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield?

8.5 Percentage Yield The percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield? Yield (the lab amount) percent yield = X 100 Theoretical Yield (by conversions) 8.66 percent yield = X 100 = 76.6% 11.26

8.7 Thermochemistry Exothermic Reaction Endothermic Reaction Eact + ΔH

8.7 Thermochemistry When a chemical or physical change takes place energy is either lost of gained. A Thermochemical equation describes this change. Equations gaining energy are called endothermic and equations losing energy are called exothermic.

8.7 Thermochemical Equations
When a chemical or physical change takes place energy is either lost of gained. A Thermochemical equation describes this change. Equations gaining energy are called endothermic and equations losing energy are called exothermic. Examples: C3H6O (l ) O2 (g) CO2(g) H2O (g) ΔH = kj Exothermic H2O (l) H2O (g) ΔH = kj Endothermic

How many kj of heat are released when 709 g of C3H6O are burned?

How many kj of heat are released when 709 g of C3H6O are burned? C3H6O (l ) O2 (g) CO2(g) H2O (g) ΔH = kj 709 g C3H6O

How many kj of heat are released when 709 g of C3H6O are burned? C3H6O (l ) + 4O2 (g) CO2(g) H2O (g) ΔH = kj 709 g C3H6O mole C3H6O 58.1 g C3H6O

How many kj of heat are released when 709 g of C3H6O are burned? C3H6O (l ) O2 (g) CO2(g) H2O (g) ΔH = kj 709 g C3H6O mole C3H6O 58.1 g C3H6O

How many kj of heat are released when 709 g of C3H6O are burned? C3H6O (l ) O2 (g) CO2(g) H2O (g) ΔH = kj 709 g C3H6O mole C3H6O 1790 kj = kj mole C3H6O 58.1 g C3H6O

13.2 SOLUTION REVIEW Solutions are homogenous mixtures.
They consist of a larger component called the solvent and one or more smaller components called the solutes. Can be in the solid, liquid, or gaseous state.

13.2 Solution Examples Margarine Tap Water Steel 18 Carat Gold Air
Sterling Silver

13.2 Solution Examples Composition of air: Dry air contains roughly (by volume) 78% nitrogen, 21% oxygen, 0.93% argon, 0.038% carbon dioxide, and small amounts of other gases. Air also contains a variable amount of water vapor, on average around 1% What is the solvent in air ?

13.2 Solution Examples Composition of air: Dry air contains roughly (by volume) 78% nitrogen, 21% oxygen, 0.93% argon, 0.038% carbon dioxide, and small amounts of other gases. Air also contains a variable amount of water vapor, on average around 1% What is the solvent in air ? Nitrogen, N2

13.2 Solution Examples Composition of air: Dry air contains roughly (by volume) 78% nitrogen, 21% oxygen, 0.93% argon, 0.038% carbon dioxide, and small amounts of other gases. Air also contains a variable amount of water vapor, on average around 1% What is the solvent in air ? Nitrogen, N2 What is a solute in air?

13.2 Solution Examples Composition of air: Dry air contains roughly (by volume) 78% nitrogen, 21% oxygen, 0.93% argon, 0.038% carbon dioxide, and small amounts of other gases. Air also contains a variable amount of water vapor, on average around 1% What is the solvent in air ? Nitrogen, N2 What is a solute in air? Oxygen, O2

13.2 Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ?

13.2 Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? Gold

13.2 Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? Gold What are the solutes?

13.2 Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? Gold What are the solutes? Silver and Copper

13.2 Solution Examples Examples of solutions include: Salt water:
what is the solvent in salt water ?

what is the solvent in salt water ? Water, H2O

what is the solvent in salt water ? Water, H2O What is a solute in sea water?

what is the solvent in salt water ? Water, H2O What is a solute in sea water? NaCl (salt)

13.2 Solution Properties Some general properties of solutions include:
Solutions may be formed between solids, liquids or gases. They are homogenous in composition They do not settle under gravity They do not scatter light (Called the Tyndall Effect) Solute particles are too small to scatter light and therefore light will go right through a solution like is shown on the next slide.

13.3 Solution Properties Tyndall Effect
Laser light reflected by a colloid. In a solution you would not see any red light.

13.3 Solution Properties Soluble substances are those that can dissolve in a given solvent. Insoluble or immiscible substances are those that cannot dissolve in a given solvent. Which of the following are soluble in water? NaCl, sugar, cooking oil, alcohol, gasoline, motor oil

13.3 Solution Properties Soluble substances are those that can dissolve in a given solvent. Insoluble or immiscible substances are those that cannot dissolve in a given solvent. Which of the following are soluble in water? NaCl, sugar, cooking oil, alcohol, gasoline, motor oil Which of the following are immiscible in cooking oil? NaCl, sugar, alcohol, gasoline, motor oil, water

13.3 Solution Properties The maximum amount of a given solute a solvent can dissolve is called the solubility. The solubility is dependent on the temperature and pressure. Solubility is often expressed in terms of grams of solute per 100 g of solvent but may have other units. When a solvent contains the minimum amount of a solute possible the solutions is said to be unsaturated. When a solvent contains the maximum amount of a solute possible the solutions is said to be saturated. When a solvent contains more than the maximum amount of a solute possible the solutions is said to be supersaturated.

13.3 Solution Properties Solutions form when a soluble solute(s) is dissolved in a solvent. In biological systems aqueous (solutions where water is the solvent) are of particular importance. The solubility of most liquids and solids in water increases with temperature. The effect of pressure on the solubility of liquid or solid solutes in water is negligible.

13.3 Solution Properties Solubility Curves

13.3 Solution Properties By forming a solution at a high temperature then slowly cooling it we can form supersaturated solutions that contain more solute than in a saturated solution. These kinds of solutions are very unstable and tend to separate out the excess solute with the slightest disturbance.

13.3 Solution Properties The solubility of gases in water decreases with temperature. Are cold carbonated drinks bubblier than warm carbonated drinks? The solubility of many gases in water is directly proportional to the pressure being applied to the solution. i.e. double the pressure, double the solubility What happens when the cork is removed from a bottle of champagne? What is the origin of decompression sickness? Anyone heard of hyperbaric therapy?

13.3 Solution Properties When we place an ionic solid in water there will be attractive forces between the ions at the surface of the crystal and the water molecules. These attractive forces are called ion-dipole forces. Water molecules orient such that the positive end of the molecule is oriented towards the negative ions at the surface and vice versa.

13.3 Solution Properties How do solutions form?
Why do some substances leave one phase and enter the solution and others don’t? How can we use chemistry to predict solubility's? Lets first look at the formation of a solution between an ionic solute and a polar solvent such as H2O.

13.3 Solution Properties Ionic compounds are composed of oppositely charged ions arranged in a repeating 3-d arrangement. They are held together by attractive forces between oppositely charged ions.

13.3 Solution Properties Ionic compounds are composed of oppositely charged ions arranged in a repeating 3-d arrangement. They are held together by attractive forces between oppositely charged ions Why is chloride ion larger than sodium ion?

13.3 Solution Properties red is the region where electrons are found most often and blue is where electrons are rarely found

13.3 Solution Properties If the attractive force between the surface ion and the solvent is greater than the forces between the ion and the solid then the ion will enter the solution phase. K+ H2O The ion that has left the solid and becomes completed surrounded by water molecules. It has become solvated or hydrated. Strong Ion dipole forces hold the ions to water.

13.3 Solution Properties Note the different orientation of water molecules around the oppositely charged ions. Positive pole of water directed to the negative ions and the negative pole directed to the positive ions

13.3 Solution Properties In a solution of an ionic compound a solvated ion will occasionally collide with the surface of the solid. Sometimes when this happens the ion will “stick” to the surface and become part of the solid phase again. This will happen more frequently the more concentrated the solution is.

13.3 Solution Properties When the rate of ions leaving the solid equals the rate of ions going back to the solid the system is at equilibrium and the solution is saturated. When a solution is at equilibrium with its solute macroscopically there will be no change occurring. However, at the molecular level lots is happening, just in equal and opposite directions.

13.3 Solution Properties Supersaturated solutions can form because there are no sites for solute ions to collide with. When we place a “seed” crystal in a supersaturated solution this provides the needed sites and the excess solute crystallizes very quickly.

13.3 Solution Properties In the you tube video we watched you can just see the tiny seed crystals on the persons finger.

13.3 Solution Properties Polar but non-ionic solutes dissolve in water via a similar mechanism as for ionic compounds.

13.3 Solution Properties A solute will be insoluble in a solvent if:
Forces between solute particles are greater than the forces between solute particles and the solvent.

13.3 Solution Properties A solute will be insoluble in a solvent if:
Forces between the solvent particles and solute particles are stronger than forces between the solvent and the solute. e.g. The only attractive force between oil and water will is dispersion forces. These are weak compared to hydrogen bonds between water molecules.

13.3 Solution Properties In a polar solvent there will be attraction between the oppositely charged ends of the molecule. Hydrogen bonds are represented by dotted lines between the water molecules. A hydrogen bond is and intermolecular force between hydrogen of one molecule and O, N, or F of another molecule. Hydrogen must be directly attached to O, N, or F in at least one of the two hydrogen bonded molecules.

13.3 Solution Properties A good “rule of thumb” that works especially well for non-ionic compounds is: “Like dissolves like” i.e. Polar solvents dissolve polar solutes well and non-polar solvents dissolve non-polar solutes well.

13.3 Solution Properties Increasing rate
The rate of dissolution is dependent upon: The surface area of the solute. i.e. how finely divided it is. Increasing rate

13.3 Solution Properties How hot the solution is.
i.e. the kinetic energy of solute and solvent. The rate of stirring. Typically when we are preparing a solution in the lab we will both heat and stir.

13.3 Solution Properties When a solute dissolves in a solvent heat can be released or absorbed. When heat is absorbed the process is endothermic and the solution becomes cooler. This effect is used in instant cold packs for sporting injuries and first aid.

13.3 Solution Properties Endothermic Solution
Solvent temperature 22.2° Solvent temperature 11.3°

13.3 Solution Properties Exothermic Solution
More commonly dissolution is an exothermic process and heat is released when a solute is dissolved. Sometimes when we make a solution it will get so hot it boils!!

13.5 SOLUTION CONCENTRATION
The ratio of the amount of solute to amount of solution, or solvent is defined by the concentration. solute solute Concentration = = solution solvent There are various combinations of units that are used in these rations. Ratio X 102 X 103 X 106 X 109 g solute % (w/w) ppt (w/w) ppm (w/w) ppb (w/w) = g solution g solute = % (w/v) ppt (w/v) ppm (w/v) ppb (w/v) mL solution mL solute = % (v/v) ppt (v/v) ppm (v/v) ppb (v/v) mL solution

1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl

1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl 33.6g H2O g NaCl

1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl 33.6g H2O g NaCl

1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O g NaCl

1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O g NaCl 44.6 g NaCl 100 g solution

1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O g NaCl 44.6 g NaCl 333 g solution 100 g solution

1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O g NaCl 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution

1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O g NaCl 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution Mass of water?

1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O g NaCl 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution Mass of water? 333 g solution – 149 g NaCl = 184 g H2O

How many grams of NaCl are required to dissolve in 88.2 g of water to make a 29.2% (w/w) solution. A sugar solution is 35.2%(w/v) find the mass of sugar contained in a 432 mL sample of this sugar solution.

The solution concentration can also be defined using moles. The most common example is molarity (M). The molarity of a solution is defined as: “The number of moles of solute in 1 L of solution” and is given the formula: Moles solute Molarity (M) = Liters solution

A student dissolves 25.8 g of NaCl in a 250 mL volumetric flask. Calculate the molarity of this solution. (picture of volumetric flask is on the next slide) Find the mass of HCl required to form 2.00 L of a M solution of HCl. A student evaporates the water form a 333 mL sample of a M solution of NaCl. What mass of salt remains? Find the molarity of sodium ions in a solution containing 2.35g of in a one liter volumetric flask.

In the lab we would use a piece of glassware called a volumetric flask to prepare this solution.

Often we will want to make a dilute solution from a more concentrated one. To determine how to do this we use the formula : C1V1 = C2V2 Where: C1 = concentration of more concentrated solution V1 = volume required of more concentrated solution C2 = concentration of more dilute solution V2 = volume of more dilute solution We can use any units in this equation but they must be the same on both sides.

How would one prepare 50.0 mL of a 3.00 M solution of NaOH using a 7.10 M stock solution? C1V1 = C2V2 (7.10 M)V1 = (3.00 M) (50.0 mL) (7.10 M)V1 = (3.00 M) (50.0 mL) (7.10 M) (7.10 M) V1 = 21.1 mL This means that you add 21.1 mL of the concentrated stock solution to a 50.0 mL volumetric flask and add water until the bottom of the meniscus touches the line on the volumetric flask.

13.8 SOLUTION STOICHIOMETRY
Consider the following balanced equation: CaCl2 (aq) + 2 AgNO3 (aq) → 2 AgCl (s) + Ca(NO3)2 (aq) Find the mass of silver chloride formed from 33.2 mL of a M solution of silver nitrate and an excess of calcium chloride.

Consider the following balanced equation: CaCl2 (aq) + 2 AgNO3 (aq) → 2 AgCl (s) + Ca(NO3)2 (aq) Find the mass of silver chloride formed from 33.2 mL of a M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution

Consider the following balanced equation: CaCl2 (aq) + 2 AgNO3 (aq) →2 AgCl (s) + Ca(NO3)2 (aq) Find the mass of silver chloride formed from 33.2 mL of a M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution L solution 103 mL

Consider the following balanced equation: CaCl2 (aq) + 2 AgNO3 (aq) →2 AgCl (s) + Ca(NO3)2 (aq) Find the mass of silver chloride formed from 33.2 mL of a M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution 2 moles AgCl g AgCl 33.2 mL moles AgCl L solution 2 moles AgNO3 103 mL

Consider the following balanced equation: CaCl2 (aq) +2 AgNO3 (aq) → 2 AgCl (s) + Ca(NO3)2 (aq) Find the mass of silver chloride formed from 33.2 mL of a M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution moles AgCl g AgCl 33.2 mL moles AgCl L solution moles AgNO3 103 mL

Consider the following balanced equation: CaCl2 (aq) AgNO3 (aq) → 2 AgCl (s) + Ca(NO3)2 (aq) Find the mass of silver chloride formed from 33.2 mL of a M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution moles AgCl g AgCl 33.2 mL moles AgCl L solution moles AgNO3 103 mL

Consider the following balanced equation: CaCl2 (aq) AgNO3 (aq) →2 AgCl (s) + Ca(NO3)2 (aq) Find the mass of silver chloride formed from 33.2 mL of a M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution 2 moles AgCl g AgCl 33.2 mL moles AgCl L solution 2 moles AgNO3 103 mL = g AgCl

Consider the following balanced equation: CaCl2 (aq) + 2 AgNO3 (aq) →2 AgCl (s) + Ca(NO3)2 (aq) Find the mass of silver chloride formed from 33.2 mL of a M solution of silver nitrate and an excess of calcium chloride. Find the mass of silver chloride formed from 33.2 mL of a M solution of silver nitrate and mL of a M solution of calcium chloride solution. Find the volume of the excess reactant.

13.9 COLLIGATIVE PROPERTIES
Colligative properties are physical properties that depend on the number of particles, but not the nature of the particles. We will discuss three colligative properties: Boiling point elevation Freezing point depression Osmotic pressure

13.9 COLLIGATIVE PROPERTIES
The vapor pressure above a solution is lower than that above the pure solvent. This has some interesting effects: The boiling point of a solution is higher than the pure solvent (boiling point elevation). The freezing point of a solution is lower than the pure solvent (freezing point depression).

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Chemical Calculations and Equations

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