Chapter 5 Answers and Explanations

Chapter 5 Answers and Explanations

Iron and carbonates are insoluble.

carbonate has a charge of -2.
To cancel out, both charges need to have a magnitude of 6.
The best representation of that is B.

When those solutions mix, the ion will be removed from the solution.

The K+ and NO - 3 remain the same as when they started.

Both NO - 3 and Na+ are spectator ion, because of the PbCl2 form.

The amount of chloride ion is twice as much as the amount of lead ion.
If equal moles of both are used, the NaCl would be the limiting reagent, and almost all of the chloride ion would be present in the solid, with very few left in the solution.
Significant Pb2+ remains in solution, as it is in excess.

The formula of Cu2CO3 is what you can determine from the information given from the rules.

In the net ionic equation, spectator ion don't show up in the final reaction.

Z must have a higher reduction potential than X because Z was able to reduce to solid Z by taking electrons from metal X.

Y must have a lower reduction potential than X because Y was unable to take electrons from metal X.

The energy must be absorbed by the reactants for the reaction to occur.

The amount of energy needed to break the bonds of the reactants is more than the amount that is released by the formation of the bonds in the products.

The oxidation number of the pure element is 0.

The oxidation number of chromium is equal to its charge of +3, so it has a charge of +2.
The total charge on the ion is -2 and the oxygen is -2.

The oxidation number is X + -2(4).

The oxidation number is +6.

The energy level of the activated complex is always higher than the energy level of the products and reactants.

The reaction will stop when either magnesium orphosphate run out.

The excess magnesium ion must have left the solution because there are no excessphosphate ion in it.

Every mole of Ca(OH)2 contains 2 moles of oxygen.

64 grams of oxygen is equal to 4 mol.

The mass of the oxygen must be calculated first.

Oxygen, carbon, and hydrogen can be converted to moles.

There are three carbon atoms and six hydrogen atoms for every oxygen atom.

You are expected to be able to do simple calculations even without a calculator.

3.36 L O2 14 is 0.1500 mol O2 x.

The amount of force with which the gas particles are hitting the container walls is called pressure.

An increase in temperature is a sign of increased velocity.
The molecule will have more energy when they hit the container walls, but they will also hit those walls more often if the molecule is moving faster.
There are two factors that contribute to the increased pressure.

The amount of bond energy present in the reactants and products is not changed by a catalyst.

The reaction can progress at a slower rate even without a catalyst.

The molecule weight of CuSO4 is 160 g/mol.

The lost mass was due to water, so 1 mole of the hydrate must have contained 90 grams of H2O.

The formula for the hydrate must be CuSO4 * 5 H2O.

One mole of CO2 is produced for every mole of CaCO3 consumed.

1.5 moles of CO2 are produced.

34 L 18 is the volume of CO2.

If the number of moles of gas changes over the course of the reaction, the pressure will not change.

The pressure will remain the same at the end of the reaction, because the number of moles of gas doesn't change in the balanced equation.

The oxidation state of NO is +2.

The nitrogen value must get more negative if it is reduced.
The oxidation state of nitrogen in N2 is 0.
The nitrogen oxidation state in the other choices is greater than +2.

The oxidation numbers of the reactants and products are listed below.

The oxidation states of oxygen and hydrogen are not changed.

Add the reduction potential for Fe2+ to the oxidation potential for Cu to get -0.7 V.

Add the reduction half-reaction to the oxidation half-reaction to get the reaction potential.

The potentials for (A)-(D) are here.

Choice is the only reaction with a positive voltage that is favored.

Find out how many electrons are provided by the current.

The Pb half-cell has a bigger reduction potential than the Cr half-cell.

To balance the electrons, the reduction half-reaction must be doubled and the oxidation half-reaction must be tripled.

The net ionic equation does not include the nitrate ion.

The Na+ from the salt bridge replaces the positive charge in the solution.

The excess Cr3+ is balanced by the Cl- anions from the salt bridge.

The reduction potential of the ion in solution must be greater than the reduction potential of the metal.

Oxidation involves the loss of electrons.

The only substance that can be reduced at the anode is the solid copper, because there is no hydrogen gas present in the solution to start.

The calculation becomes (0.00 V) + (-0.34 V) + (0.34 V) for this reaction to occur.

In the initial Cu/Zn cell, Cu2+ ion are produced and become part of the solution.

The SRP of Cu2+ is greater than the SRP of H+, so later in the reaction, Cu2+ is reduced into solid copper at the zinc cathode, which causes the copper to plate out on the zinc.
H+ has a higher reduction potential than Ni2+, which means the H+ reduction would continue at the cathode, and no nickel would be plated out of solution.

The positive ion would be attracted to the positive dipoles of the water molecule, while the negative ion would be attracted to the positive dipoles of the water molecule.

One Be2+ and two Cl- ion are produced for every mole of BeCl2 that dissociates.

The concentration of the Be2+ will be the same as the base concentration, and the concentration of the Cl- will be twice that.

The concentration of the lead nitrate solution can be used to determine how much is required.

The excess nitrate and beryllium ion would adhere to the solid, giving it a false high molar mass.

Decanting the solution into a funnel setup would allow the water to carry the spectator ion and catch all of the precipitate.

The reduction reaction must be increased by two and the oxidation reaction increased by five.

When the reactions combine, 10 of the hydrogen ion cancel out.

There will be a change in the color of the solution.

The balanced equation can be used to convert to moles of hydrogen peroxide.

The concentration of H2O2 should be converted to that.

We will use a volume of less than 50.0 mL in our calculations because we are interested in the concentration of the solution.

You can't measure out a volume in a beaker to any number of places.

The number of significant figures in your answer would be reduced if you used a beaker.

The volume of permanganate would be artificially high because the rest was air.

The permanganate's recorded volume will eventually lead to a calculated H2O2 that is also too high.

Some of the permanganate that leaves the buret might stick to the sides of the flask if you don't do this.

This would mean that the recorded volume of permanganate is too high, which would lead to an artificially high calculate concentration of hydrogen peroxide.

A triple bond is required in order to fill the octet.

It must be one of the other two reactants because there is plenty of water.

From the balanced equation, CuSO4 is consumed in a 2:1 ratio, so run out of Mg first.

The yield of H2 can be found by using the limiting reagent, Mg.

The number of moles of H2 produced will be less than the number of moles of Mg consumed.

When 0.060 moles of Mg are consumed, 0.060 moles of CuSO4 are also consumed.

All of the Mg consumed ends up in the solution.

The Cu/M cell is reduced by the higher reduction potential of copper.

The oxidation potential of metal M is different to the reduction potential.

M is the anode whenOxidation occurs.

The metal M should be flowing into the copper bar.

This will increase the cell's power.

The nickel reduction is more positive than the other one.

The top two reactions must be flipped to have the reactants on the other side.
The water oxidation has a more positive value and will occur.

Oxygen gas would form and bubble up to the surface.

The rates of chemical reactions are determined by the structure of the atom.

The initial rate of a reaction is dependent on the concentrations of its reactants.

The rate of appearance of a product or the rate of disappearance of a reactant is described.
The rate law for a reaction cannot be determined from a balanced equation and must be determined from experimental data presented on the test.

The rate of the reaction will be affected by a change in the concentration of the reactant.

When the concentration of an individual reactant is doubled, the rate can be seen.

The other reactant concentrations remain constant from experiment 3 to experiment 4.

The other reactant concentrations remain constant from experiment 1 to experiment 3.

The other reactant concentrations remain constant from experiment 1 to experiment 2.

The reaction is said to be third order because of the sum of the exponents.

The rate constant can be calculated using any of the lines of data on the table once the rate law has been determined.

It's important to carry along units throughout all rate constant calculations because the units of the rate constant are dependent on the order of the reaction.

The rate of appearance of D is equal to the rate of disappearance of A and C because the coefficients of all three are the same.

The rate at which D appears is half the rate at which B disappears.

Each rate law can be expressed as a graph that shows the rate constant, the concentration of the reactant, and the elapsed time.

The rate of a zero-order reaction does not depend on the concentration of reactants, so it will always be the same at a given temperature.

The rate of a first-order reaction depends on the concentration of a single reactant raised to the first power.

Natural logarithms are used in the rate law for a first-order reaction.

The rate law uses natural logarithms to create a linear graph comparing concentration and time.
The rate law uses natural logarithms to create a linear graph comparing concentration and time.
Intercept is given by A.

We can use slope-Intercept form to come up with an equation.

The rate of a second-order reaction depends on the concentration of a single reactant raised to the second power.

The inverses of concentrations are used in the rate law for a second-order reaction.

A linear graph comparing concentration and time is created by inverses in the rate law.

As the concentration decreases, the line moves upward.

The time it takes for half of the substance to react is called the half-life of a reactant.

The chart should start with the time at zero.

Half-life is constant for a first order reactant.

If it takes 30 seconds for 50% of a reactant to decay, then in the next 30 seconds 50% more will decay, leaving 25% of the original amount remaining.
Only 25% will remain 30 seconds later.

Half-life is not constant for zero order or second order reactants.

The half-life equation cannot be used for anything other than a first order reactant.

Zero order and second order reactants have half-lives that vary over time.
The time it takes for the amount of a zero or second order reactant to decrease from 100% to 50% will not be the same as the time it takes for the reactant to decrease from 50% to 25%.

Let's see if we can come up with an example.

The first two lines of the table will be used.

The half-life is about 15 minutes.

The half-life equation can be used to confirm this.

Half-life can be used to examine the rate of decay of a radioactive substance.

As time goes on, a radioactive substance will slowly decay into a more stable form.

Chemical reactions occur because reactants are constantly colliding with one another.

They are referred to as effective collisions because they lead to a chemical reaction.

Chemical reactions are not produced by effective collisions.
A small fraction of the reactant molecule will collide with enough energy to cause a reaction between them.

Increased concentration will increase the rate of reaction.

They are more likely to collide with more molecules moving around.
If the surface area available to react is larger, the reaction will proceed faster.
The acid can only react with the metal atoms on the surface, so dropping a large chunk of metal into acid will cause the metal to dissolution, but slowly.
The acid can react with a lot more metal atoms if you grind the metal into a powder.

Stirring is a physical factor that can affect a reaction rate.

If you were to drop sugar into water, stirring the water will make it dissolution quicker.
We need to talk about homogeneity to understand why stirring a mixture doesn't always speed up a reaction.
Some parts of the water contain more sugar than others, and the mixture would be heterogeneous.
Each portion of water will have the same number of sugar molecule in it.

Stirring does not increase the reaction rate.

Causing the solid sugar molecule to move will cause them to collide with the liquid water molecule more often, which increases the likelihood that a reaction will occur.
The number of sugar/water molecule collisions that are already happening due to the inherent motion of the molecule in their aqueous state are less than the number of sugar/water molecule collisions that are caused by stirring.

Increasing temperature means that the molecule are moving faster, which means that they have more average energy.

This can be seen on a diagram.

On the above graph, you can see that for the reaction at the higher temperature, a larger fraction of the reactant molecule have enough energy to exceed the activation energy barrier.

If the reactants collide with the correct orientation, there will be reactions.

There are many possible collision orientations in the reaction 2NO2F - 2NO2 + F2.

If the N-F bonds can break and the F-F bonds can form, the reaction will happen.

While there is no way to quantify collision orientation, it is an important part of collision theory, and you should be familiar with the basic concept underlying it.

A device called a spectrophotometer can be used to measure the concentration of a solution.

The amount of light absorbed by a solution is measured by a spectrophotometer.
As the reaction progresses, the amount of light that is absorbed will change.

You can't use Beer's Law until you are 21.

Beer's Law can be used whenever you please.
It pairs well with solutions that change color.

Beer's Law is often thought of as a correlation between absorbance and concentration of the solution, because of the constants of path length and molars.

Beer's Law can be used to determine the concentrations of reactants in solutions that are invisible to the human eye if a spectrophotometer that emits light in the ultraviolet region is used.

While studying Beer's Law, you may run into a device called a colorimeter.

The difference between a colorimeter and a spectrophotometer is that a colorimeter only emits light at specific frequencies, while a spectrophotometer can emit light anywhere.

There are many chemical reactions.

The balanced equation is the sum of a series of steps.

When adding up the various steps in a reaction, intermediates will always cancel out.

Cancel the species that appear on both sides.

The mechanism is consistent with the balanced equation if we add up all the steps.

There are either one or two reactants in the elementary steps.
In the above mechanism, step I is unimolecular and steps II and III are bimolecular.

Since the slowest step is the most important step in determining the rate of a reaction, the steps leading up to it are also used to see if the mechanism is consistent with the rate law.

The rate for an elementary step can be determined by taking the concentration of the reactants in that step and raising them to the power of any coefficient attached to that reactant.

I is an example.
There is an intermediate present in step II.

Even though intermediates can appear in rate laws, if they can be replaced with a reactant via an earlier step, we should try and do so.

As the sides are in equilibrium, we can see that [X] is equivalent to [A]2.

The rate law of elementary steps can only be determined using the coefficient method.

The coefficients from the balanced equation cannot be used to determine the overall rate law.
The only way to determine the rate law for a full reaction is by knowing which step is the fastest and applying the above method to that step.
You must use experimental data to determine the rate law if you don't know the relative speed of the elementary steps.

An energy diagram is a graphical representation that shows the energy level of the products and the reactants, as well as the required activation energy for a reaction to occur.

An energy diagram can be used to look at the energy change in each step.

The overall reaction is exothermic.

The first and second steps of the reaction are endothermic.

In the book, we mentioned that a catalyst increases the rate of a chemical reaction without being consumed in the process.

In some cases, a catalyst is necessary because the reaction would be too slow to be useful without it.

A catalyst is present before and after the reaction.

There are catalysts in the elementary steps.

An alternative reaction pathway with a lower activation energy is provided by a catalyst.

X is a catalyst and Y is a reaction intermediate.

In rate laws catalysts and intermediates can appear.

The example is the decomposition of H2O2 in the presence of the iodide ion.

I- is a catalyst and OI- is an intermediate.
As I- appears to be a reactant in the slow step, it would be part of the overall rate law.

When a catalyst is introduced to a reaction, the reaction undergoes catalysis.

There are many types of reactions.
One of the most common is surface catalysis, in which a reaction intermediate is formed.
The catalyst binding to the reactants reduces the overall activation energy of the reaction.
In biological applications, genes are very common.
In acid-base catalysis, reactants gain or lose protons in order to change the rate of reaction.
Big Idea 6 will look at bases and acids in more depth.

The following information can be used to answer questions.

C + A - D Step III.

The progress of the reaction is aided by this reaction intermediate.

The overall order of the reaction is changed by this catalyst.

The overall activation energy of the reaction is lowered.

The reactions will move faster and collide more often.

The activation energy of the reaction will be lowered.

The activation energy of a higher fraction of the molecule will be the same.

The half life will not be increased by these.

The reactants and products are in the gaseous phase.

No appears at half the rate that NOCl disappears.

The rate at which Cl2 appears is the same as the rate at which NOCl disappears.

The rate of reaction and the rate constant will not change.

The rate of reaction will change.

The rate of reaction will increase.

A red solution of Co2+ ion is seen under white light.

The light will be absorbed by the red light.

The solution matches the wavelength of red light.

A decomposition reaction took place.

The concentration of A was recorded in the chart.

A sample had to be decayed to 25 percent of the original amount.

In order to study the rate of the reaction, it was necessary to determine the order of the reaction with respect to each of the reactants.

The units should be included.

The units should be included.

No + no is a combination of N2O2 and 2NOBr.

The data was collected after the first-order reaction was allowed to proceed.

The rate constant is calculated using the values for concentration and time given in the table.

A and B are present in a closed container.

Explain why the rate and rate constant will be affected by the following changes to the reaction system.

Use your knowledge to answer the questions.

Justify your answers.

The Raschig process can be used to produce hydrzine.

Justify your answer.

The only colored species in the reaction is the crystal violet, so a spectrophotometer can be used to determine its concentration over time.

You can graph a function of [CV+] vs. time on the axes.