Let's see how these rules are applied. A solution of lead(II) nitrate is mixed with a solution of sodium iodide. Predict what will happen.
If both products aresoluble, the reaction would be listed as no reaction. There is a chance for a reac tion in the reaction question part of the AP exam.
The reactions involved in thebatteries are called redox reactions. The exchange of electrons from a more active species to a less active one is involved in a redox reaction. Predicting the relative activities from a table of activities or a half-reaction table is possible.
Chapter 16 goes into more detail about the chemistry.
There is a table in the back of the book.
You may want to memorize the oxidizing and reducing agents.
Suppose a piece of zinc metal is placed in a solution with a blue Cu2+ cation. The substance is copper metal. The blue color of the solution fades as the copper metal is deposited. The more active zinc metal is losing electrons to form the Zn2+ cation, while the Cu2+ ion is gaining electrons to form the less active copper metal.
The zinc metal is being degraded. The section on single displacement reactions can be found later in the chapter.
The Cu2+ ion is the oxidizing agent. The reducing agent is zinc metal. Adding oxidation and reduction together will give you the overall redox reaction.
There is a simultaneous loss and gain of electrons in these reactions. In the oxidation reaction electrons are being lost, but in the reduction half-reaction they are being gained. As reactants are being converted into products, electrons are being exchanged. This electron exchange can be direct or indirect, as when copper metal plates out on a piece of zinc.
The oxidation numbers of the reactant species can be looked at to determine what is being reduced. You may be asked to assign oxidation numbers or identify changes in oxidation numbers on the AP exam.
An increase in oxidation number is a sign ofOxidation. The Zn metal went from an oxidation state of zero to +2. A decrease in oxidation number is a reduction. The oxidation state of Cu2+ went from +2 to zero. Write the net ionic equation to figure out if a particular reaction is a redox reaction. Decrease the oxidation numbers of the elements in the reaction. A redox reaction is when one or more elements have changed oxidation number.
There are several types of reactions. In the next few pages, we will look at some of these types of reactions.
Some types of reactions are definitely redox. The reactions of metals with nonmetals to form ionic compounds and the reactions of nonmetals with other nonmetals to form covalent compounds are included.
In the first reaction, we have the combination of an active metal with an active nonmetal. Stable compound water is formed by the very active oxygen reacting with hydro Gen. Oxygen and chlorine are being reduced while the hydrogen and potassium are being oxidation.
Many decomposition reactions are redox reactions.
The chlorine is going from a less stable oxidation state to a more stable oxidation state. Oxygen is being degraded from -2 to 0.
In order to force the nonspontaneous reaction to occur, we must supply energy in the form of electricity.
Since the element becomes an ion, all of these single replacement reac tions are redox reactions.
The displaced atom is on the product side of the equation.
The free-response section of the AP Chemistry exam has the least easily oxidation Reactions.
A table of metals that have the same ease of oxidation is useful for predicting what will happen. A table of half-cell potentials is an alternative to the activity series.
The net ionic form is required on the AP exam.
There wouldn't be a reaction if a piece of copper metal was placed in a solution that was not tin.
Hydrogen will not be displaced from water by the Groups IA and IIA elements on the activity table.
As one goes from top to bottom in the periodic table, the reactivity decreases.
The sources of heat energy are very important to our society.
The highly reactive oxygen forming a stable compound is the driving force in all of these reactions. The exothermic nature of the reaction shows this.
It is important to balance the oxygen last in any of the combustion reactions.
Transition metal ion form when a salt is dissolved in water. If you want to score high reactions, you need to review the knowledge. In this case, suppose the dissolved substance is water.
If ammonia is added to the solution, the water molecule can be displaced. The ammonia ligand displaced the water ligand from the chromium complex because nitrogen is a better electron-pair donor than oxygen.
The nitrogen and oxygen in the water are donor atoms. The electrons are donated to the Lewis acid by the atoms. The coordination number is 6. Coordination numbers are usually 2, 4, or 6. Most other central ion have a coordination number of 6 and zinc, copper, nickel, and Platinum form complexes with a coordination number of 2.
The reactions between acids and bases are very common.
The hydronium ion reacts with the hydroxide ion to form water. You should review the acid-base part of the Equilibrium chapter after you finish this section.
acids may be damaging to the skin and react with bases to yield salts. Bases taste bitter, feel slippery, and react with acids to form salts.
The H+ will accept an electron pair from ammonia if the acid is NH3(g) HNH3Cl. Ammonia accepts the H+ and gives an electron pair that the H+ will bond with.
After the bond is formed, it is the same as a covalent bond formed by donation of one electron by both of the bonding atoms.
The hydrogen ion and the anions from the acid are produced by the strong acids above.
The hydroxide ion is produced by the compounds listed above, not the compound.
Unless told otherwise, assume that acids and bases not on the lists above are weak and will establish an equilibrium system when placed into water.
Acid-base properties are found in some salts. When dissolved in water, NH4 will act as a weak acid and donate a protons. We will look at the acid-base properties in more detail in the next section.
There are oxides that have acidic or basic properties. When the oxides are dissolved in water, they show their properties. Most of the time, reactions of this type are not redox reactions.
Many oxides of metals that have a +2 charge are called basic oxides because they will react with acids.
Many nonmetal oxides are called acidic oxides because they react with water to form an acidic solution. It is the reason that soft drinks have bubbles.
acids react with bases to form water and salt.
Some salts have acid-base properties. Salts containing cations of strong bases and anions of strong acids are not acidic or basic. They are neutral, reacting with no acids or bases. KNO3 is an example. The strong acid HNO3 and the strong base KOH give rise to the potassium.
Salts with anions of strong acids behave as acidic salts.
Basic salts include strong bases and anions not of strong acids. Na2CO3 would be an example.
CO2(g) + H2O(l) 2 H+(aq) + CO2 3 - H2CO3(aq) - CO2(g) + H2O(l) is the same type of reaction for acid carbonates.
The hydride of the alkali metals and of calcium, strontium, and barium have acid-base properties.
In this case, water is behaving as H+OH-.
A titration is a common laboratory application.
Suppose you wanted to know the molarity of the solution.
Adding a couple of drops of a suitable acid-base indicator to a known volume of acid in a flask would be enough. phenolphthalein is pink in a basic solution and acidic in an acidic solution, which is an indicator. You would fill the buret with a strong base solution. The base solution can be added to the acid solution in the flask with the buret. The use of a pH meter can follow the course of the titration. Since it is an acid solution, the pH of the solution will be low. The acid will be neutralized as the base is added. The base is added until it reaches the equivalence point.
A strong acid with a strong base has a rapid rise in the vicin ity of the equivalence point. The indicator turns pink when the smallest amount of base is added. The endpoint will be as close to the equivalence point as possible.
The basic solution's pH is greater than 7 after the equivalence point has been passed.
The solution for the molarity of the acid solution can be solved with reaction schihiometry.
Chapter 7 contains a discussion of solution stoichiometry.
An acid solution of known concentration can be used to measure an unknown base. The pH will be greater than 7 initially and will decrease as the titra tion proceeds. The indicator will start pink and disappear at the end.
There is a balanced chemical equation. H+ is supplied by the acid and the base accepts it. The acid can give two H+ if it is diprotic.
moles are the key to any experiment. The numbers of moles can be calculated. The mass and moles can be calculated with the formula weight. The density and volume of a substance may be combined to deter it from being mined. The final and initial readings from a buret can be used to calculate the solution's volume. The moles present can be calculated using this volume and the molarity.
The moles of a gas can be calculated using the volume, temperature, and pressure of a gas. You have to be very careful on the AP exam to distinguish between the values that you measure and those that you calculate.
The balanced chemical equation can be used to calculate the moles of any substance in a reaction. Chapter 7 contains other calculations.
The compounds ionize.
The chemical species should be shown in the ionic and net ionic equations.
Unless they are strong acids that are ionizing, don't break apart ionic and net ionic equations.
Reducing agents and oxidizers are reactants.
Carbon dioxide and water are the products of the complete combustion. If oxygen is present as well, you will also have carbon dioxide and water, as well as sulfur, if that element is present.
No carbon dioxide can be formed if a substance does not contain carbon.
No hydroxide ion, OH-, will be formed if an alcohol like methanol is dissolved in water.
It is not a strong acid.
In titration calculations, you have to consider the reaction stoichiometry.
The charges should be indicated correctly.
The coordination numbers are 2, 4, and 6.
Don't confuse measured and calculated values.
You can use the questions here to practice for the AP Chemistry exam. There are 16 multiple-choice questions that are similar to what you will see in the chemistry section of the exam. Section II of the exam has a short free-response question. You can make these review questions even more authentic by following the instructions.
You can't use a calculator. The periodic table and equation sheet can be found at the back of the book.
In the absence of air, a sample of copper metal is reacted with con Fe3(PO4)2(s) + H2O(l) centrated nitric acid.
The gas bubbles will be created by the equation for the reaction of acetic acid with addition of hydrochloric acid.
The solutions of this ion are blue.
Water will release insoluble metal.
Oxygen is the gas that is 3 + 2 H+ + 2 Cl-.
The compound will react with strong bases.
A solution is prepared.
The solution has the following ion: Co2+, cipitated with chloride ion. If Hg2+ is present.
The Dilute NH balanced net ionic equation is added.
A solution of chlorine gas bubbles through a mixture of solu 2Cl2 + 2 NH3 - Hg + HgNH2Cl + NH + tion.
The compound PbCl2 reacts with water.
The solution becomes neutral when AgNO3 is added.
The formation of a brown precipitate takes forms. What is the place after the reaction is complete?
The concentration of the 2 is very small because of the white whole number coefficients.
The formation of 2 H3PO4(aq) - Fe3(PO4)2(s) + 6 H2O (l) causes the black color.
Carbonates produce carbon dioxide gas in 3, which eliminates choices C and D. choice A is eliminated because none of the other ion arrows must be equal.
0.10 moles of chloride ion is whatqueous solutions of Cu2+ are normally. Iron ion give a variety of colors but are of chloride will react with 0.15 moles of lead in the absence of two Pb2+.
The other ion is odorless.
The limiting reagent is the HCl.
One-half the silver to half the mula should be written as PbCO3.
The doubling of the volume acid is a strong acid, so it should be written as a separate H+ and Cl- ion concentration.
H2CO3 is a carbonic acid that quickly converts to CO2 and H2O.
PbCl2 is the only possible pre charges because A cannot be correct because of the solution.
The balanced chemical equation is 3 Cu(s) + but there are two nitrate ion produced per 8 HNO3(aq) - 3 Cu(NO3)2(aq) + 2 NO(g) + solute formula. On ion, the copper is below hydrogen. H2 can't form by this equal, but some of the lead is precipitated as acid-metal reaction.
Ammonia will cause the to NO. The oxidation and reduction must be metal hydroxides.
acetic acid is a weak acid. There are nitrates from nitric acid. The solution that is least likely is a strong base, so it will separate into K+ and cause an observable change.
The K+ ion will come from any potassium compound that forms. oxidation is caused by chlorine. The sium ion is left out of the equation because it is a spectator ion.
The silver-ammonia complex will be produced if the solution reaction is strong.
The reaction of potassium to another solution. The acid-base reaction to release ammonia must be a reduction because the only species gas available is hydrogen.
27 O2 + 16 CO2(g) + 22 H2O(l) + 2 N2(g) is the balanced equation.
There are two types of free-response questions on the AP exam.
An example of a short answer question can be found here.
You have 5 minutes to answer the question. You can use the tables in the back of the book.
An iron(III) nitrate, Fe(NO3)3 solution is mixed with a K3PO4 solution.
There are four gasses present.
The starting materials are in solution. KNO3 and FePO4 might form in the reaction.
The nitrate and the K+ are not in the precipitate.
There are points that can be scored. You can get 1 point for correctly identifying the precipitate. You get 2 points for the explanation and 1 point for identifying the spectator ion.
The free-response section of the AP exam contains reaction questions. This might not be true in the multiple-choice part.
Energy can be released in a reaction or absorbed.
Chemical equations are balanced by adding coefficients in front of the chemical species until the number of each type of atom is the same on both the right and left sides of the arrow.
The balanced equation must have coefficients in the lowest whole-number ratio.
Water is the universal solvent, dissolving a wide variety of both ionic and polar substances.
Non-electrolytes do not conduct an electrical current when dissolved in water.
solvation is a process in which most ion in solution attract and bind a layer of water.
Some compounds form ionized water.
The reactants and products are shown in their un-ionized form, the ionic equation shows the strong electrolytes in the form of ion, and the net ionic equation shows only those species that are undergoing chemical transformation.
Precipitation reactions form an insoluble compound.
You should be able to apply the rules.
Reduction and oxidation take place at the same time.
Oxidation and reduction are related to the loss of electrons.
Two or more reactants combine to form one product in a combination reaction.
Decomposition reactions are when a compound breaks down into two or more simpler substances.
In a single displacement reaction, atoms of an element replace atoms of another element in a compound.
The activity series can be used to predict whether or not an element will move.
The chemical species rapidly combine with diatomic oxygen gas, emitting heat and light. Carbon dioxide and water are products of the com plete combustion of a hydrocarbon.
Substances that exhibit different colors are indicators.
Acids are electron-pair acceptors.
The bases areelectron-pair donors.
Coordinate covalent bonds are bonds in which one atom is 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609-
Strong acids and bases completely ionize/dissociate, and weak acids and bases only partially ionize/dissociate.
Acids form a salt in a neutralization reaction.
There are many hydrides that react with water.
A solution of known concentration is used in a titration to determine the concentration of an unknown solution.
The moles of H+ from the acid are equal to the moles of OH- from the base in an acid-base titration. The equivalence point is the point at which the indicator changes color.
A metal ion is bonded to two or more compounds called ligands.
In the previous chapter on chemical reactions, reactants and products were discussed. An industrial chemist is not interested in the number of molecules being produced; she or he is interested in the amount of products being formed per hour or day. The questions are of interest. A production chemist is interested in the bigger picture, not the smaller one. A chemistry student will not be weighing out individual atoms and molecules in the lab, but large numbers of them in grams. There needs to be a way to bridge the gap between the world of atoms and molecules and the world of grams and kilograms. The mole concept is one of the most important concepts in chemistry.
A mole is a number of particles, like a dozen or a pair. A mass of a chemical substance is represented by the mole. The mass is expressed in grams. The atomic mass of an element is described in Chapter 5. The mass was associated with an atom. The mass of a compound can be calculated by simply adding together the mass of the individual elements in the compound. The unit of grams is used to represent the quantity of a mole.
The way in which grams and moles are converted to particles is given by the relationship above. You can calculate the other two if you have any of the three quantities.
The coefficients of the balanced chemical equation are not only the number of individual atoms or molecules at the microscopic level, but also the number of moles at the macroscopic level.
If the formula of a compound is known, it is easy to determine the percent composition of each element in the compound. Suppose you want to calculate the percentage of hydrogen and oxygen in water.
Add the percentages together as a good check. They should be close to it.
A check is the total.
The percentage data was calculated from the chemical formula, but the empirical formula can be used if the percent compositions are known. The data could be percentage, mass, or even moles. The procedure is the same: convert each to moles, divide each by the smallest number, and use an appropriate multiplier.
The formula mass can be calculated. If the actual mass is known, dividing it by the empirical formula mass gives a rounded number that is used to calculate the subscripts in the formula.
A sample of a gas was analyzed and found to have 2.34 g of nitrogen and 5.34 g of oxygen. The mass of the gas was determined to be about 90 g/mol.
The formula can be determined by dividing the actual com pound by the empirical mass. The empirical mass is 46 g/mol.
To one significant figure, 90 g/mol is 2.
It's important to use as many digits as possible. Failure to do so may result in incorrect ratios and formulas.
The balanced chemical equation shows the relative ratio of reactants and products as well as the species of reactants.
The number of grams or molecule can be calculated if the number of moles is known. The relationship between the reactants and products and the conversion from moles of one chemical species in the reaction to another is defined by the coefficients in a balanced chemical equation.
Consider the process above.
The balanced chemical equation shows the ratio of 2 mol NH3 to 1 mol N2
How many moles of hydrogen would it take to react with 20.0 mol of nitrogen?
The balanced chemical equation is where the new ratio came from.
If you wanted an answer in grams, you would have to use moles.
You will need a balanced chemical equation.
We will convert from grams of hydrogen to moles of hydrogen to moles of ammonia using the correct ratio.
40.0 g of H2 and Cl2 are combined. It will be produced.
One reactant was present in excess. Some of the other reactant was left over. The amount of product being formed is determined by this reactant. You can't assume it's the reactant in the smallest amount.
Each reactant is assumed to be the limiting reactant and the amount of formation is calculated. The advantage of this method is that you can practice your calculation skills; the disadvantage is that you have to do more calculations.
The moles of the reactants in the balanced chemical equation are calculated. The limit reactant has the smallest mole-to-coefficient ratio. Many use this method.
Let's consider the reaction again. If 50.0 g of nitrogen and 40.0 g of hydrogen were allowed to react, what would happen?
It is most likely an L.R. when the quantities of more than one reactant are given.
Let's look at another case. 125 g of P2O5 and 50.0 g of H2O were supplied.
The balanced chemical equation gives the 1mol and 3mol. This is the L.R., because the 0.880 is smaller.
Finish by using the number of moles.
The maximum amount of product that could be formed from the given amount of reactants is calculated in the preceding problems. There are many reasons for this, but the main reason is that most reactions don't go to completion; they establish an equilibrium system. Not as much as expected is not formed. The percent yield can be used to judge the efficiency of the reaction.
Consider the problem in which 60.8 g NH3 could be formed. 52.3 g NH3 was formed, if that reaction was carried out.
A 25.0-g sample of calcium oxide is heated with excess hydrogen chloride to produce water and calcium chloride.
The yield is calculated using the theoretical yield.
This includes canceling g CaCl2 with g.
We discuss solutions further in Chapter 13, but solu tion is so common on the AP exam that we will discuss it here briefly also.
The NaCl is the solute and the water the solvent if it is dissolved in water.
The only concentration unit we will use in Chapter 13 is molarity.
An example of calculating molarity can be found in the mol solute/L solution. A solution of NaCl has 39.12 g of this compound.
The moles or grams of solute present can be calculated by knowing the volume of the solution and the molarity.
We can use molarity to calculate moles.
The volume and molarity of the solution must first be converted to moles before the solution can be worked in the same fashion as before.
moles must be involved in tichiometry experiments. Initial and final volumes are measured.
The difference between the initial and final values may be included in a calculation. moles can be calculated using the mass in grams and the mass in mula. The volume of a solution and its molarity can be used to calculate moles.
The experiment will be based on further calculations using the moles that have been calculated.
The number is 6.022 x 1023, not 10-23.
It is important to know the difference between moles and molecules.
It is important to get the lowest ratio of whole numbers in empirical formula problems.
The mol/coefficient ratio is used to determine the limiting reactant in limiting-reactant problems.
The limit reactant is a chemical species to the left of the arrow.
Percent yield is the actual yield of a substance divided by the theoretical yield of the same substance.
When you need or have the number of atoms, be careful with Avogadro's number.
To practice for the AP Chemistry exam, use these questions to review the content of this chapter. There are 18 multiple-choice questions that are similar to what you will see in the chemistry section of the exam. Section II of the exam has a long free-response question. You can make these questions even more authentic by following the instructions.
You can't use a calculator. The periodic table and equation sheet can be found at the back of the book.
The acid needs to be completely neutralized by NaOH and the acid needs to be 30.0 mL.
This reaction is used in the titration of iron.
There is a 1+ charge for silver. All silver compounds are insoluble in this reac.
According to the above equation, copper metal reacts with nitric acid.
Au2O3 can be used to make gold metal, Au, and oxygen gas. The gold(III) oxide mass is 442 g/mole.
A solution containing 0.20 mole of KBr and How many moles of MnSO4 are produced when 0.20 mole of MgBr2 in 2.0 L of water is pro 1.0 mole of KMnO4 and vided.
Calcium reacts with water according to the mula mass, which is the same for both the empirical and above reaction.
The equivalent expression is 1,000 mL.
The reaction is H2SO4(aq) + 2 KOH(aq) - K2SO4(aq) + 2 H2O(l).
The reaction is H2C2O4 and 2 NaOH.
To get an estimate, round the values and pick the closest answer.
There is enough base to react with only one hydrogen ion from the acid. H 2AsO4 is left. Answer D is not correct because it is a cation.
Simplistic is important in this example.
Pick the closest answer that is slightly lower since the 45.20 was rounded up.
Ag+ is not a possible product because all silver compounds are insoluble, according to the question. The formula for silver carbonate should be Ag2CO3. Hydrochloric acid is a strong acid, so it should be writ ten. H2CO3 is a carbonic acid that quickly converts to CO2 and H2O.
It's important to have a whole-number ratio. The smallest value leads to a whole number and the value 1.33 is too far from a whole number to round. 3 V and 4 O is given by multiplying them by 3.
halving the concentrations will be achieved by doubling the volume. After the reaction, some of the magnesium is left, the rest is precipitated, and there are two nitrates.
The numerator gives the mass of the water and the denominator is the mass of the formula. The percentages for the other compounds are A, B, and D. rounding will simplify the work.
The limit of the reactant is 0.80 mole of acid.
There is a 2:1 relationship between the mass of reactant and the formula mass.
The equation is HNO2 + NaOH - NaNO2 + H2O.
It should remain as HNO2 because it is a weak acid. It will separate into Na+ and OH- ion if it is a strong base. Any compound that forms will yield Na+ ion. The net ionic equation does not include the spectator ion.
The limiting reagent is KMnO4. The mole of KMnO4 will produce a mole of MnSO4.
The balanced equation is 2 C8H18(g) + 25 O2(g) - 16 CO2(g) + 18 H2O(g).
An equal number of moles of SnO2 and CrO2 are present.
The larger coefficient in the balanced reaction is the limiting reactant.
The volume of water is irrelevant. 0.10 mole of Pb(NO3)2 will be required for 0.20 mole of KBr. To get the final answer, you need to total the two yields.
For the empirical formula to be the same as the molec ular formula, the two formulas must be the same. An empirical formula that is different from the molecular formula can be simplified.
You have 15 minutes to answer the question. You can use the tables in the back of the book.
A sample of a monoprotic acid contained 40.0% C and 6.71% H.
The sample was placed in a hot water bath. 1.18 g of vapor was found to be 300.0 mL at 100degC and 1.00 atm. The weight of the acid can be determined.
The grams of each element are equivalent to the percent if there are 100 g of sample. The moles of each element are given by dividing the grams by the molar mass.
The empirical formula is CH2O.
If you show your work, you get 1 point for correctly determining any of the elements and 1 point for the complete empirical formula correct.
If you show your work, you will get 1 point for the correct number of moles and 1 point for the correct answer.
The ideal gas equa tion is used here. Do not forget, you have to change the temperature.
If you show your work, you will get 1 point for getting any part of the calculation correct and 1 point for getting the correct final answer.
You get 1 point for each correct formula.
The one formula is double the other formula if you review the knowledge you need to score high. The larger molecule can be produced by the smaller molecule.
You can point to any relationship between the two formulas. If you note the second formula is two lighter molecules added together, you get 1 point.
There is a chance for 10 points.
The mole is the amount of substance that contains the same number of particles.
Avogadro's number is the number of particles per mole.
A mole is a mass expressed in grams.
You can calculate the others if you have any of the three.
The lowest whole-number ratio is indicated by the empirical formula.
The formula tells which elements are present and which ones are not.
The empirical formula can be calculated from percent composition data or chemical analysis.
The amount of one substance in a chemical equation can be calculated by using another substance.
It will be necessary to convert to moles sooner or later in working problems that involve a quantity other than moles.
The limit reactant is used first.
The limiting reactant can be calculated using the mol/coefficient ratio.
Percent yield is how much was formed in the reaction, divided by the theoretical yield to get the maximum amount of product formed.
A solution is a mixture of a solute and a solvent.
The number of moles of solute per liter is called molarity.
You have to be able to work reaction stoichiometry problems using molarity.
gases are the best understood and have the best model. The four main physical properties of gases are volume, pressure, temperature and amount. You will spend a lot of time working on problems in this chapter because gas laws show up on the AP exam a lot. Before we look at the gas laws, let's look at the model that scientists use to represent the state of gas.
Gases are composed of particles.
We assume that the volume of the gas particles themselves is insignificant because they are tiny compared to the distances between them.
The gas particles are moving in straight lines and colliding with each other inside the container. The inside container walls contain the pressure of the gas.
The gas particles are not thought to repel each other. They are assumed to be elastic if they collide. The energy from one gas molecule to another is retained.
There are no ideal gases and only gases that approach ideal behavior. We know that there are interactions between real gas particles and that they occupy a finite volume. The factors cause gases to deviate from the ideal behavior. A non-polar gas at a low pressure and high temperature would be close to ideal behavior. Modification of equations to account for non-ideal behavior will be shown later in this chapter.
Let's quantify a couple of the postulates of the KMT before we start looking at the gas law relationships. The motion of the gas particles is described by postulate 3.
The average speed of a gas particle is what it is called.
The root mean square speeds are very high. The value of hydrogen gas is 2,000 m/s.
The average energy of the gas particles is calculated by Postulate 5.
The relationship between the average energy of the gas particles and the temperature is shown.
The physical properties of volume, pressure, temperature, and moles are related to each other by the gas laws. We will look at the individual gas relationships law first. The use of the individual equation is not required in order to know these relationships for the AP exam. We will combine the relationships into a single equation that you will need to apply. We need to describe a few things about pressure.
The two types of pressure are measured in different ways.
The open end of the evacuated hollow tube is immersed in a pool of mercury. The weight of the atmospheric gases on the surface of the mercury pool will force the liquid up into the tube. The column of mercury inside the tube will be stable as the opposing forces bal ance each other. The atmospheric pressure is the height of the mercury column above the mercury pool. The column is at sea level. The pressure is called 1 atmosphere. The unit torr is used to calculate atmospheric pressure at sea level. The pascal is the SI unit of pressure and it is 101,325 Pa. 1 atm is 14.69 pounds per square inch in the United States.
The pressure of the gas is balanced against a column of mercury.
If you know any of the three quantities, you can calculate the fourth.
If a sample of gas is heated, the volume must increase to keep the pressure constant.
The temperature must be expressed in kelvin in any gas law calculation.
As the temperature increases, the volume also increases.
As the gas is heated, the particles move with greater force, hitting the inside walls of the container more often. The pressure of the gas increases.
You should be able to solve for the sixth unknown in this equation if you give any five.
A bottle of gas with a pressure of 2.50 atm is heated to 80 degrees. The combined gas law can be used to calculate the new pressure. Let's do some reasoning before we start working. The number of moles of gas inside the bottle hasn't changed.
If the amount and volume are constant, the pressure should increase if the temperature increases. If the new pressure is less than 2.50 atm, you've made an error. K and degC are the same as 20 and 80degC.
The answer is reasonable because the new pressure is greater than the original pressure. All the units were canceled except atm, which you wanted.
There are two conditions in this situation. A balloon has a volume at sea level of 10.0 L at 760.0 torr and 20degC. The pressure of the balloon is 450.0 torr and the temperature is -20degC.
The new volume of the balloon is calculated. You have to tell the temperature in K in the calculations. It's fine to leave the pressures in torr. It doesn't matter what pressure or volume unit you use, as long as they are consistent in the problem. The volume should increase because the pressure is decreasing. The volume should decrease because the temperature is decreasing. It is difficult to predict the end result because you have two competing factors. You have to do the calculations.
Substitute the quantities into the equation.
The units canceled and left the desired volume unit of liters. In this case, the pressure decrease had a greater effect than the tem perature decrease, because the volume did increase. Looking at the numbers, this seems reasonable. There is a relatively small change in the temperature compared to a much larger change in the pressure.
The amount of gas has not changed. Avogadro's law comes into play there.
If a container is kept at constant pressure and temperature, and you increase the number of gas particles in that container, the volume will have to increase in order to keep the pres sure constant.
The amount of gas is related to the other physical properties through another relationship that Avogadro developed, so we could work this into the combined gas law.
The simplest way to calculate an ideal gas constant is to use the 0.0821 and convert the volume to liters and the pressure to atm.
Let's see if we can use the ideal gas equation. If you wanted to know what volume 20.0 g of hydrogen gas would be, you could ask. The ideal gas equation requires the conversion of grams of hydrogen gas to moles of hydrogen gas before you can use it. H2 is a diatomic gas.
You have a lot of gas. It would occupy about 10mol x 22.4 L/mol by Avogadro's relationship. The pressure is slightly less than standard pressure of 1 atm, which would tend to increase the volume, and the temperature is greater than standard temperature of 0degC, which would also increase the volume. You can expect a volume greater than 224 L, that's what you found.
The collection of a gas over water and the displacement of water by oxygen gas are both examples of calculations using the Dalton's law. There is a mix of gasses in this situation. The total pressure in this case is usually atmospheric pressure, and the partial pressure of the water is determined by looking up the vapor pressure of water at the temperature of the water in a reference book. The partial pressure of the oxygen is generated by simple subtraction.
The mole fraction of gas A is the same as the moles of gas A.
The rate of effusion is determined by the lighter the gas is.
The gases 2 and 1 have the same mass.
Suppose you wanted to calculate the ratio of effusion rates for hydrogen and nitrogen gases. The diatomic mass of H2 is 2.016 g/mol and the diatomic mass of N2 is 28.02 g/mol.
As fast as nitrogen gas, hydrogen gas would effuse through a pinhole 3.728 times.
The answer is reasonable since the lower the mass, the faster the gas is moving.
Graham's law can be used to calculate the mass of an unknown gas.
The gas law relationships can be used. The knowledge you need to score high grams of KClO3 is present. You heat the mixture.
displacement of water collects the oxygen gas that is formed. It has a volume of over 500 mL. The pressure is 755.0 torr. The water has a vapor pressure of 26.7 torr.
You need to know the pressure of the oxygen gas.
You have 542 mL of oxygen gas at 728.3 torr and 300.K.
The ideal gases are gases that obey the postulates of the theory. Remember, a couple of those postulates were shaky.
There were no attractive forces between the gas particles. The ideal gas equation works well when approximations are fine. It would be great to have a more accurate model for doing precise work or when a gas has a large attractive force. The volume of the gas is less than the ideal gas because of the finite volume of gas molecule. The more moles of gas there are, the bigger the real volume is.
The attraction of the gas particles for each other reduces the force of gas particles colliding with the container walls. Concentration of gas particles and mag nitude of the particles' intermolecular force determine the amount of attraction. The higher the attraction is, the less real pressure there is.
The larger, more concentrated, and stronger the intermolecular forces of the gas, the more deviation from the ideal gas equation one can expect and the more useful the van der Waals equation becomes.
Gas law experiments usually involve pressure, volume, and temperature.
Gas law type information and calculations for non-gaseous materials are common errors.
Water Vapor, H2O(g), is a common consideration. Vapor pressure varies with the temperature. In these cases, the pressure of a gas sample is adjusted for the presence of water vapor. The total pressure is the pressure of the gas or gases being collected. Vapor pressure of water is subtracted from the total pressure when a gas is needed. To find the vapor pressure of water, you have to measure the temperature and use a table to show it.
Time is measured in experiments on Graham's law. The amount of time required is the measurement. The rate of effusion is the amount of material effusing divided by the time.
The ideal gas equation is used in most gas law experiments.
Many of the experiments involve gas. The combined gas law can be used to generate moles of a gas by adjusting the volume and Avogadro's relationship.
It is possible to determine the temperature using a thermometer. Normally, the temperature measurement is in the degree C range. The temperature must be converted to aKelvin temperature.
A barometer is used to measure pressure. The pressure needs to be adjusted to compensate for the presence of water vapor. The knowledge you need to score high is reviewed in step 4. Subtract the value from the measured pressure. Values from tables are not used for an experiment.
A measurement of the volume of a container can be made.
Knowing the number of moles of gas is needed to calculate the volume.
The number of moles can be determined using stoichiometry. A quantity of something is converted to something else. There are a number of ways in which this conversion can be accomplished. volatilization or reaction is the most common method of stoichiometric methods. The simplest method is volatilization. A weighed quantity is used to convert a substance to moles. The mole-to-mole ratio is used to convert the quantity of one of the substances to moles.
The gas law must be used to adjust the volume to the law. The molar volume of a gas is determined by combining these values with the STP conditions.
Values found in tables and conversions from one unit to another are not experimental measurements.
If you are using any of the gas laws, make sure you are dealing with gases.
We don't know how many times people have applied gas laws in situations where no gases were involved.
It's a common mistake to fail to do so.
You need to be sure that you are calculating the volume, pressure, etc.
Don't just write a number down from your calculator, analyze the problem. You should check your number of significant figures.
The combined gas equation is used if you have a gas at a certain set of conditions. The ideal gas equation is useful if moles of gas are involved.
Group all initial-condition quanti ties on one side of the equals sign and all final-condition quantities on the other side of the equation.
Be sure to use the correct mass for the gases that are diatomic.
To practice for the AP Chemistry exam, use these questions to review the content of this chapter. There are 20 multiple-choice questions that are similar to what you will see in the chemistry section of the exam. Section II of the exam has a long free-response question. You can make these questions even more authentic by following the instructions.
You can't use a calculator. The periodic table and equation sheet can be found at the back of the book.
A sample of gas is in a container.
The container's volume is doubled. What happens to the gas if it begins by heating a solid to produce a pressure?
The temperature is halved.
The ideal barometric pressure in the room gases is A, B, and C. The volume of a real gas is larger than is known. The additional informa was calculated from the ideal gas equation.
There are two balloons in a flask. Ammonia will be under a lot of pressure. One contains 14 g of nitrogen, and the ideal under which of the following sets contains 20.0 g of argon.
An additional 0.13 atm of argon gas is pumped into a steel tank at constant temperature.
A sample of methane gas was collected. There is no change in the number of gas in the sample. The partial pressure atoms can be determined.
There is an increase in the volume of gas.
A student has three identical 2.0 L flasks.
flasks B and C contain hydrogen, H and He. The three flasks should be ranked in order of decreasing pressure.
If it has a density of 5.47 g/L, you should use the formula for a gaseous Silane.
An ideal gas sample weighing 1.28 g at 127degC (B) Si2H6 and 1.00 atm has a volume of 0.250 L.
Increasing the temperature of an ideal gas from (B) The moving outside air pushes the balloon 50degC to 75degC at constant volume will cause higher.
Each container has a temperature and gas samples in it. Each container has 16 g of methane, CH4; another container has a small amount of gas.
The containers are filled with the same average speed.
The pressure on the gas may be removed from the temperature.
The identity of the gas is irrelevant.
To volume, temperature, and pressure, it is necessary to convert moles to mass and eliminate gas during the experiment.
The ideal gas equation can be used to calculate the total pressure.
The number of moles is determined by the gas.
If you want to score high, you need to review the knowledge you have.
The gases effuse faster. It is possible to delay looking up among the lighter choices because of the methane vapor pressure of water.
Real gases are different from ideal gases because of two factors: the rate difference and the molar mass of methane.
A steel tank has a constant volume. The problem states that the temperature is subtracted from the observed volume constant. Adding gas to the tank will increase in volume for a real gas, and the number of moles of the gas and the pressure pressure has a term added to compensate for. Correcting for a stant temperature means there will be a constant smaller pressure. There are only two average speeds.
The answer is C.
A large polar molecule has ideal behavior at deviation.
The only polar molecule listed is the partial pressure of any gas.
The mass of gas must be determined.
0.10 1.00 L mol partial pressure of CO is 0.24 atm.
The vapor pres 1 L is called B. You could find moles from the ideal gas if you didn't recognize the conditions as sure of water.
The gases have a 2.0 g/mole molar mass.
The hot air balloon has a CH4. The surrounding air has a lower density than an 8.0-g sample.
The "lighter" objects will float on volume and the "heavier" objects will float on temperature.
The molar mass can be obtained by dividing it by the number of moles, which is the ideal gas equation.
The temperature needs to be converted tokelvin.
The number of moles, the temperature, and the not allowed by the problem are required for choice B. Choice C requires the same volume. The average change in the composition of the gas will be the same as with choice A. The heavier molecule doesn't need a decrease in volume.
The gas is moving slower for this reason. The speeds are not the same.
The deviation from ideal gas behavior is caused by all three variables changing. The number of moles decreases when the escaping gas, SO2, is the only polar gas.
The pressure remaining species should decrease if fewer moles are present. The gas mol is moving slowly.
The behavior is due to the speed of the molecule.
This would result in a lower average speed. The species are in a container. There will be a lower temperature if the atomic mass is heavier than average.
You have 20 minutes to answer the question. You can use the tables in the back of the book.
A sample containing 1/3 mol of KClO3 is heated until it starts decomposing. The oxygen is collected in a bottle. This information can be used to answer the following questions.
The sample volume is slightly larger.
calculate the number of moles of gas formed with a balanced chemical equation.
If the equation is balanced correctly, you will get 1 more point.
You get 1 point for the correct answer and 1 point for the work.
The volume of 1 mol of O2 should be 22.4 L because oxygen was not the only gas in the sample. Water was present. There is more gas that leads to a larger volume.
You get 1 point for discussing the pres ence of water vapor.
You get a point for the equation and a point for the math. If you use an incorrect number of moles of O2 from a balanced equation, you can get a total point.
This will get you a point.
Total your points with SO3(g) + H2O(l) There is a chance for 10 points.
Gases are small particles of negligible volume moving in a random straight-line motion, colliding with the container walls and each other. No energy is lost, but energy can be transferred from one particle to another. There is no attraction between the particles.
Know the different units used in atmospheric pressure and how a barometer works.
The volume and pressure of a gas are related to the amount and tempera ture.
If the amount and pressure are constant, the volume and temperature of a gas are proportional.
If the amount and volume are constant, the pressure and temperature of a gas are proportional.
If the pressure and temperature are constant, the number of moles and volume of a gas are proportional. If the gas is at STP, you should not use the 22.4 L.
The lower the mass of a gas, the faster it will effuse.
Know how to apply the gas laws to reactions.
Gas laws are useful for gases, but not for liquids. If you are dealing with a gas, be sure to apply a gas law.
Energy changes will be described in this chapter. There are two types of energy. Unless a nuclear reaction occurs, energy cannot be created or destroyed. Energy exchanges between a system and surroundings will be discussed. It could be a beaker or it could be Earth.
The joule and the calories are the most common units of energy used in the study of thermodynamics.
Cal is a kilocalorie, which is 1,000 cal.
Units of J/K are the most common units of heat capacity.
The amount of energy absorbed can be calculated if the specific heat capacity, mass, and temperature are known.
A calorimeter is a laboratory instrument. The change in temperature is measured when a process is carried out with known amounts of substances.
These could be reactions that happen in open beakers. This type of calorimeter is used to measure the specific heat of a substance. A mass of solid is heated to a certain temperature and then added to a calorimeter with a mass of water.
The heat capacity of the solid can be solved with this equation.
The calorimeter is used to measure the energy changes that occur during a fire. A weighed sample of the substance being investigated is placed in the calorimeter. The sample is set afire by a hot wire and a calorimeter is used to measure a mass of water. Sometimes the heat capacity of the calorimeter/water system is known.
A bomb calorim eter was used to ignite a 1.5886 g sample of glucose. The temperature went up by 3 degrees. The heat capacity of the calorimeter was 3.562kJ/degC, and the calorimeter contained 1.000 kg of water.