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Q1M.1

Calculate the oxidation number for nitrogen in the following substances:

A. NH3

B. N2

C. NO2

D. NO3−

S1M.1

This problem is asking us to figure out the oxidation number of nitrogen and we can do this by usually calculating the oxidation number of the other element in the compound most of the time.

A. It is given that any element in the first element is going to have an oxidation charge of +1. Since Hydrogen falls into the first group it causes the charge to be +1. On top of that if we look down at the subgroup of Hydrogen we can see there are three Hydrogen, so we multiply +1 * 3 to get an overall charge of +3 for hydrogen. Now to calculate the oxidation number we can automatically assume that the charge is neutral unless given. Thus to make it a neutral charge nitrogen has to be -3 for oxidation number.

B. This problem only consists of the sole element Nitrogen and no other element. It is a given rule that if there is just a self-element without any given charge we can assume that Nitrogen is going to equal 0.

C. It is a given that Oxygen has a charge of -2 unless in peroxide. If we look down at the subgroup of Oxygen we can see there are two hydrogens, so we multiply -2 * 2 to get an overall charge of -4 for Oxygen. Now to calculate the oxidation number we can automatically assume that the charge is neutral unless given. Thus to make it a neutral charge nitrogen has to be +4 for oxidation number.

D. It is a given that Oxygen has a charge of -2 unless in peroxide. If we look down at the subgroup of Oxygen we can see there are two hydrogens, so we multiply -2 * 3 to get an overall charge of -6 for Oxygen. Now to calculate the oxidation number we can automatically see that the overall compound charge is -1. Thus to make the compound of charge -1, nitrogen has to be +5 for oxidation number.

Q1M.2

In the oxidation-reduction reaction:

Br₂(l)+2I⁻(aq)→2Br⁻(aq)+I₂(s)

A. which substance(s) is being reduced?

B. which element(s) is increasing in oxidation number?

C. which element(s) is gaining electrons?

D. which substance(s) is the oxidizing agent?

E. which substance(s) is the reducing agent?

S1M.2

A. A reduced substance is a substance that is gaining an electron. If we look at the element Br as a reactant it has a neutral charge but as a product, it has an electron charge of -2. Since the substance is gaining an electron the substance that is being reduced is Br.

B. If we look at the element charges we can see that Br gained an oxidation number from 0 to -2 and I went from -2 to 0. In this case, the increasing oxidation number for oxidation numbers would be I.

C. Br is the element that is gaining electrons because it charge went from 0 to -2, this negative charge is only associated when gaining an electron.

D. I is the oxidizing agent because it is losing its electron, and is giving electrons to Br for oxidation to occur.

E. Br₂is the reducing agent because it is gaining the electrons, and is taking electrons from I for reduction to occur.

Q1M.3

Classify each of the following substances as an oxidizing agent, reducing agent or both. List the oxidizing agents in order of decreasing strength; list the reducing agents in order of decreasing strength:

Ni(s), H⁺(aq), Au(s), Cl₂(g), Sn²⁺(aq), Mg(s), Fe²⁺(aq)

S1M.3

Some ways to help in classifying an oxidizing agent or reducing agent is by looking at the standard reduction potentials. These data allow us to compare the oxidative and reductive strengths of a variety of substances. The half-reaction for the standard hydrogen electrode (SHE) lies more than halfway down the list. All reactants that lie below the SHE in the table are stronger oxidants than H⁺, and all those that lie above the SHE are weaker. Using that knowledge the best reducing agent would be from highest strength to lowest Mg(s), Fe²⁺(aq), Ni(s), Sn²⁺(aq), H⁺(aq). Using that knowledge the best oxidizing agent would be from highest strength to lowest Cl₂(g), Au(s), H⁺(aq).

Q3M.1

An electric current is passed through an aqueous solution of lithium bromide.

A. What is produced at the cathode?

B. What is produced at the anode?

S3M.1

In an electrolytic cell, the positive terminal of the battery is connected to the anode and the negative terminal of the battery is connected to the cathode. Here LiBr or lithium bromide acts as the electrolyte. Lithium is positively charged ( +1 ) and Bromine is negatively charged ( -1 ), and so after passing current, they split into their respective ions and are attracted towards oppositely charged electrodes. Lithium will be attracted towards the negatively charged electrode, which is Cathode, and Bromide ions will be attracted towards the positively charged electrode, which is Anode.

A. Lithium ions (Which in this case is negative) are supplied with electrons to replace the ones they lost when they became ions at the cathode. Lithium atoms are more active than hydrogen so they react with water replacing hydrogen in water molecules. The displaced hydrogen atoms join in pairs to make H₂, molecular hydrogen. The water molecules that have lost H become OH- ions. In the end what is produced is H₂and OH^- .

B. At the anode, Bromide ions are stripped of their extra electrons and oxidized to bromine atoms which join together to make Br₂ .

Q3M.2

An electric current is passed through an aqueous solution of lead(II) chloride.

A. Write the half-reaction that takes place at the cathode.

B. Write the half-reaction that takes place at the anode.

S3M.2

A. At the cathode Lead is going through reduction and is gaining two electrons from Chlorine and can be seen as: Pb²⁺+ 2e^-→ Pb

B. At the anode Chlorine is going through oxidation and is losing two electrons and can be seen as: 2Cl^-→ Cl₂+ 2e^-

Q3M.3

“White gold” it is plated with rhodium (Rh) in order to make it look more like silver. How many coulombs of electricity must be pumped through an rhodium(III) solution in order to plate 1 gram of solid rhodium?

S3M.3

This problem works a little bit like a dimensional analysis problem using this chemical half reaction: Rh³⁺+ 3e^- → Rh.

We do most of the work by taking the 1 gram amount of solid rhodium and converting it into the correct moles and multiplying it to the Farhad's principle which is a given of 96485 Coulomb.

1 g Rh * (1 mol Rh/103 g Rh) * (3 mol e-/1 mol Rh) * (96485 C/1 mol e-) = 2810 coulomb

Q8B.29

What is the approximate half-life for each of the following three reactions?

A1→B1		A2→B2		A3→B3
Time, s	[A][A], M	Time, s	[A][A],M	Time, s	[A][A], M
0	1.00	0	1.00	0	1.00
25	0.80	25	0.77	25	0.82
50	0.64	50	0.52	50	0.70
75	0.52	75	0.27	75	0.60
100	0.40	100	0.00	100	0.51
150	0.24			150	0.42
200	0.15			200	0.35
250	0.07			250	0.29

S8B.29

To calculate the approximate half-life for this problem, it would be best to start off by calculating the initial rate constant and then plugging that into a half-life calculation for first order. To calculate initial rate we can do that by dividing change in concentration over change in time. It would be correct to go up to the point closest to the half of the decay reaction far all three reactions so the next steps is calculating the intial rate:

A1→B1: -(.52M - 1M)/(75s - 0s) = .0064 M/s

A2→B2: -(.52M - 1M)/(50s - 0s) = .0096 M/s

A3→B3: -(.51M - 1M)/(100s - 0s) = .0049 M/s

With these constant rates we can place them into our first order half reaction that goes with ln(2)/k to give us the final half life for each of these equations:

A1→B1: .693/.0064 = 77 seconds

A2→B2: .693/.0096 = 52 seconds

A3→B3: .693/.0049 = 102 seconds

Q8B.33

The reaction A+B→C+D is in second order in A and first order in B. The value of k is 0.0205 M^-2 min^-1. What is the rate of this reaction when [A] is 0.005M and [B] is 3.02M.

S8B.33

When writing a rate law, orders are taken into serious consideration. Orders are always put to the power of the concentration so for a second order it would be to the power of two and for the first order it is only to the power of one, giving us the result of this: k[A]²[B].

Plug in the concentrations and k to find result: .0205 * [.005]²* [3.02] = .000001548

Q8B.35

The reaction: H₂O₂(aq)→H₂O(l)+12O₂(g). Yields the following data when decomposed at 600 K was obtained.

Time(s)	[H2O2][H2O2] (M)
0	2.00
100	1.80
200	1.62
300	1.48
400	1.36
500	1.26

What are the average rate of reaction over the first 500 seconds and reaction order?

S8B.35

To calculate the average rate of a reaction it is the change of concentration divided over the change of time. Since we are trying to calculate the average of 500 seconds we would compare the initial concentration and time to 500 seconds.

it can be seen as: -((1.26M-2M)/(500s- 0s)) = .00148

Q9.5

Balance each of the following nuclear equations and indicate the type of nuclear reaction (a-emission, b-emission, fission, fusion, or “other”).

A. Pu⁹⁴₂₃₉+N⁰₁→Sn⁵⁰₁₃₀+?+3N⁰₁

B. ?+Li³₆→2H²₄

C. Po⁸⁴₂₁₀→He²₄+?

D. U⁹²₂₃₅+₁⁰N→?³⁰₇₂+?+4N₁⁰

E. I⁵³₁₂₅→I⁵³₁₂₅+?

F. U⁹²₂₃₈→?+Th^?₂₃₄

G. U⁹²₂₃₅+₁⁰N→Br^?₈₆+?^?₁₄₇+?N₁⁰

H.Th⁹⁰₂₃₄→?+?⁹¹₂₃₄

S9.5

There are a few forms of radioactive emissions that happen to occur and they all have different characteristics so when balancing the nuclear equations it makes it easier to see which ones they are.

A. Pu⁹⁴₂₃₉+N⁰₁→Sn⁵⁰₁₃₀+Ru⁴⁴₁₀₉+3N⁰₁To calculate the answer we added up the sum of atomic number and mass on both sides and subtracted it to calculate a atomic number of 44 and atomic mass of 109. We then found from the periodic table that the element is Ru.

We can tell from the reaction equation that this is a fission reaction which involves an Nitrogen and eventually the product of nitrogen also are forming which easily give off the fact that it is a fission reaction.

B. ¹₂H+Li³₆→2H²₄To calculate this answer we added up the sum of atomic number and mass on both sides and subtracted it to calculate the atomic numbers and atomic mass. Furthermore, we can go onto show that this problem is a fusion reaction.

C. Po⁸⁴₂₁₀→He²₄+Pb⁸²₂₀₆ To calculate the answer we added up the sum of atomic number and mass on both sides and subtracted it to calculate a atomic number of 82 and atomic mass of 206. We then found from the periodic table that the element is Pb.

We can tell from the reaction equation that this is an a-emission equation since it involves a He emission which gives the indication.

D. U⁹²₂₃₅+₁⁰N→Zn³⁰₇₂+Sm⁶²₁₆₀+4N₁⁰

To calculate the answer we added up the sum of atomic number and mass on both sides and subtracted it to calculate an atomic number of 62 and an atomic mass of 160. We then found the periodic table that the element is Sm and Zn.

We can tell from this equation that this is a Fission reaction because one of the reactants is N.

E. I⁵³₁₂₅→I⁵³₁₂₅+ ve This n goes onto just gain an antineutrino which is consisting of a massless and chargeless particle. This reaction under those circumstance leads to an "other" equation.

F. U⁹²₂₃₈→H²₄+Th⁹⁰₂₃₄ To calculate the answer we added up the sum of atomic number and mass on both sides and subtracted it to calculate an atomic number of 2 and atomic mass of 4. This reaction is an a-emission reaction as it goes on to indicate the alpha particle on the particle.

G. U⁹²₂₃₅+₁⁰N→Br³⁵₈₆+La⁵⁷₁₄₇+3N₁⁰To calculate this answer we added up the sum of atomic number and mass on both sides and subtracted it to calculate the atomic numbers and atomic mass. This reaction is a fission reaction as it undertakes n and also has a release of 3 N particles.

H. Th⁹⁰₂₃₄→e-¹₀+Pa⁹¹₂₃₄To calculate this answer we added up the sum of atomic number and mass on both sides and subtracted it to calculate the atomic numbers and atomic mass. This reaction is considered an beta decay reaction because it goes onto a form Pa which is a incremented equation.

Q9.6

The isotope Cs⁵⁵₁₃₇ undergoes beta emission with a half-life of 30 years.

A. Write a balanced nuclear equation for this reaction.

B. What fraction of Cs-137 remains in a sample of the isotope after 60 years?

C. What mass of Cs will be left in a 24.0 g sample of Cs⁵⁵₁₃₇ after 90 years?

D. What fraction of Cs-137 has decayed after 120 years?

S9.6

A. Cs⁵⁵₁₃₇→ B + Xe⁵⁴₁₃₇+ ve. To calculate this answer we added up the sum of atomic number and mass on both sides and subtracted it to calculate the atomic numbers and atomic mass, beta particles tend to have a -1 charge.

B. We know that it takes 30 years for decay to occur so we divide 60 by 30 like this which equals n: 60/30 = 2. We then plug this equation into the fraction left seen as (1/2)ⁿ. We have to plug this into the equation and calculate: (1/2)²= .25 left.

C. We know that it takes 30 years for decay to occur so we divide 90 by 30 like this which equals n: 90/30 = 3. We then plug this equation into the fraction left seen as (1/2)ⁿ. We have to plug this into the equation and calculate: (1/2)³= .125 left. We then take this value of what is left and plug it into a total mass which consist of 24g like this: 24 * .125 = 3 grams.

D. We know that it takes 30 years for decay to occur so we divide 90 by 30 like this which equals n: 120/30 = 4. We then plug this equation into the fraction left seen as (1/2)ⁿ. We have to plug this into the equation and calculate: (1/2)⁴= .0625 left after that we place it over a left equation of 15/16 it goes onto calculate a fraction decay.

Q9.7

What is the half-life of an isotope that is 75% decayed after 16 days?

S9.7

Since 75% is decayed only 25% is left and we plug that into a fraction left equation. 25 = (1/2)ⁿ; we then go on to do basic algebra to indicate that the goes onto calculate n to equal to 2. Then we have to calculate using time past divided by n and can be seen as:

16/2 = 8 days.

Q10.13

Balance each of the following nuclear equations and indicate the type of nuclear reaction (a-emission, b-emission, fission, fusion, or “other”).

A. Pu⁹⁴₂₃₉+N⁰₁→Sn⁵⁰₁₃₀+?+3N⁰₁

B. ?+Li³₆→2H²₄

C. Po⁸⁴₂₁₀→He²₄+?

D. U⁹²₂₃₅+₁⁰N→?³⁰₇₂+?+4N₁⁰

E. I⁵³₁₂₅→I⁵³₁₂₅+?

F. U⁹²₂₃₈→?+Th^?₂₃₄

G. U⁹²₂₃₅+₁⁰N→Br^?₈₆+?^?₁₄₇+?N₁⁰

H.Th⁹⁰₂₃₄→?+?⁹¹₂₃₄

S10.13

There are a few forms of radioactive emissions that happen to occur and they all have different characteristics so when balancing the nuclear equations it makes it easier to see which ones they are.

We can tell from the reaction equation that this is an a-emission equation since it involves a He emission which gives the indication.

D. U⁹²₂₃₅+₁⁰N→Zn³⁰₇₂+Sm⁶²₁₆₀+4N₁⁰

We can tell from this equation that this is a Fission reaction because one of the reactants is N.

Q10.14

The isotope Cs⁵⁵₁₃₇ undergoes beta emission with a half-life of 30 years.

A. Write a balanced nuclear equation for this reaction.

B. What fraction of Cs-137 remains in a sample of the isotope after 60 years?

C. What mass of Cs will be left in a 24.0 g sample of Cs⁵⁵₁₃₇ after 90 years?

D. What fraction of Cs-137 has decayed after 120 years?

S10.14

Q10.15

What is the half-life of an isotope that is 75% decayed after 16 days?

S10.15

16/2 = 8 days.

Q10.16

Explain what makes an isotope radioactive. Why do radioactive isotopes undergo radioactive decay? How does the energy released by nuclear reactions compare to that released by ordinary chemical reactions? Why?

S10.16

Isotopes are radioactive due to an unstable nucleus. Nuclei can be unstable because they are too large or if their neutron-to-proton ratio is “incorrect”. Radioactive decay enables a nucleus to become more stable. Alpha-decay makes a nucleus smaller and beta-decay reduces the neutron-to-proton ratio. The energy released in nuclear reactions is several orders of magnitude greater than the energy released in ordinary chemical reactions. This enormous amount of energy reflects the loss in mass that accompanies nuclear reactions.

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