Extra Credit 19
- Page ID
- 82987
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Question 17.2.8
For each of the following reactions, list the substance reduced, the substance oxidized, the reducing agent, and the oxidizing agent.
- 6H+ + MnO4- + 5SO32- → 5SO42- + 2Mn2+ + 3H2O
- 8H+ + Cr2O72- + 6HI → 2Cr3+ + 3I2 + 7H2O
- 3Cl2 + 6OH- → ClO3- + 5Cl- + 3H2O
Answer 17.2.8
When a substance is reduced, it's oxidation number is lowered as the substance is gaining electrons. The reduced substance gains electrons from the oxidized substance which loses its electrons. When a substance is oxidized, it's oxidation number increases as the substance is losing electrons.The substance being oxidized is the reducing agent as it reduces another species in the equation. The substance being reduced is the oxidizing agent because its accepting electrons allowing the other substance to become oxidized.
1. Mn is the substance reduced. MnO4- is the oxidizing agent. S is the substance oxidized. SO32- is the reducing agent.
2. Cr is the substance reduced. Cr2O72- is the oxidizing agent. I is being oxidized. HI is the reducing agent.
3. Cl is the substance reduced. Cl2 (gas) is the oxidizing agent. Cl is also the substance oxidized (Cl2-->ClO3-). Cl2 is the reducing agent.
Question 19.1.17
Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)
- Fe(s)+H2SO4(aq)⟶
- FeCl3(aq)+NaOH(aq)⟶
- Mn(OH)2(s)+HBr(aq)⟶
- Cr(s)+O2(g)⟶
- Mn2O3(s)+HCl(aq)⟶
- Ti(s)+xsF2(g)⟶
Answer 19.1.17
In order to predict the products of a reaction. We must look at the reactants given. The given 6 reactions are all double replacement reactions, thus the cation in one molecules is substituted with the cation of the one its reacting with. Same can be said for the anion. After this switch, we must observe whether or not the reactants and products are soluble or incoluble. We do this through observation of solubility rules:
After, we must balance the equation and leave spectator ions (ions appearing on both sides of the reaction).
- Fe(s)+2H3O+(aq)+SO2−4(aq)⟶Fe2+(aq)+SO2−4(aq)+H2(g)+2H2O(l)
- FeCl3(aq)+3Na+(aq)+3OH−(aq)⟶Fe(OH)3(s)+3Na+(aq)+3Cl+(aq)
- Mn(OH)2(s)+2H3O+(aq)+2Br−(aq)⟶Mn2+(aq)+2Br−(aq)+4H2O(l)
- 4Cr(s)+3O2(g)⟶2Cr2O3(s)
- Mn2O3(s)+6H3O+(aq)+6Cl−(aq)⟶2MnCl3(s)+9H2O(l)
- Ti(s)+xsF2(g)⟶TiF4(g)Ti(s)+xsF2(g)⟶TiF4(g)
Question 19.3.9
Trimethylphosphine, P(CH3)3, can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH3)3 can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?
Answer 19.3.9
P(CH3)3 + NiCl2 yields a products that contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH3)3 .
First we need to find out how many moles of Ni, Cl, and P(CH3)3 there are in the blue compound.
Ni: 270g * .215 = 58.05g Ni * 1mol Ni/58.7g = .989mol thus roughly one mole Nickel in the blue compound.
Cl: 270g* .260 = 70.2g Cl * 1mol Cl/35.5g = 1.98mol thus roughly two moles of Cl in the blue compound.
P(CH3)3: 270g * .525 = 141.75 * (1mol P(CH3)3)/72.97 = 1.94mol thus roughly two moles of P(CH3)3 in the blue compound.
The molecular formula of the compound is: [NiP(CH3)3Cl2]
The geometry of the compound is: octahedral
Question 12.4.9
What is the half-life for the decomposition of O3 when the concentration of O3 is 2.35 × 10−6 M? The rate constant for this second-order reaction is 50.4 L/mol/h.
Answer 12.4.9
A half-life is the time it takes for a substance to decay to half of its original concentration. This is a fairly simple calculation using a half-life equation which is specific to the order of the reaction. The table below shows a summary of the kinetic equations by order.
In our case, we are given that the reaction for the decomposition of O3 is second order.
[O3]= 2.35 × 10−6 M
k= 50.4 L/mol/h
t1/2= 1/(k[A]0) => 1/[(2.35 × 10−6 M )(50.4 L/mol/h)]= 8443.1 hours for O3 to decompose by half.
Question 21.2.4
For each of the isotopes in Question 21.2.3, determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope.
Answer 21.2.4
A neutral atom is one where the amount of protons and electrons are the same. To identify the atomic number of the element, we look at the subscript. To identify the atomic mass we look at the superscript. To find the number of neutrons: neutrons= atomic mass - atomic number
a) Si has an atomic number of 14. This is equivalent to 14 protons and because the atom is neutral it also has 14 electrons. The number of neutrons is 20 (34-14).
b) P has an atomic number of 15. This is equivalent to 15 protons and because the atom is neutral it also has 15 electrons. The number of neutrons is 21 (36-15).
c) Mn has an atomic number of 25. This is equivalent to 25 protons and because the atom is neutral it also has 25 electrons. The number of neutrons is 32 (57-25).
d) Ba has an atomic number of 56. This is equivalent to 56 protons and because the atom is neutral it also has 56 electrons. The number of neutrons is 65 (121-56).
Question 21.5.7
Describe how the potential energy of uranium is converted into electrical energy in a nuclear power plant.
Answer 21.5.7
Fission is the process in taking a large nucleus and splitting it into smaller nuclei. This releases energy that can be converted to electrical energy to power, for example, a nuclear power plant. Uranium is very radioactive and contain a lot of potential energy, which is why a process like fission can be so powerful.
The fission of uranium generates heat, which is carried to an external steam generator (boiler). The resulting steam turns a turbine that powers an electrical generator.
Question 20.4.5
Write a cell diagram representing a cell that contains the Ni/Ni2+ couple in one compartment and the SHE in the other compartment. What are the values of E°cathode, E°anode, and E°cell?
Answer 20.4.5
Cell diagrams are composed of the anode on the left and the cathode on the right. According to our cell potential table:
Nickel has a lower reduction potential and thus is more readily oxidized. We put its electrode on the far left, then a vertical bar to signify a phase change as it releases aqueous Ni2+ ions. The double vertical bar signifies a salt bridge needed to maintain the galvanic cell. We put platinum on the very right because the electrode needs to be a solid and hydrogen gas is a gas.
Ni(s)∣Ni2+(aq)∥H+(aq, 1 M)∣H2(g, 1 atm)∣Pt(s)
We calculate E cell by doing: Eo cell = Eo cathode - Eo anode.
|
E∘anode E∘cathode E∘cell |
Ni2+(aq)+2e−→Ni(s);−0.257 V 2H+(aq)+2e−→H2(g); 0.000 V 2H+(aq)+Ni(s)→H2(g)+Ni2+(aq); 0.257 V
|
Question 20.7.3
What type of battery would you use for each application and why?
- powering an electric motor scooter
- a backup battery for a smartphone
- powering an iPod
Answer 20.7.3
Primary batteries cannot be recharged. Examples of primary batteries are: modern, alkaline batteries, lithium iodine battery.
Secondary batteries can be recharged. An example of a secondary battery is a lead acid storage battery.
- Lead storage battery. This battery can be recharged so one would be able to recharge their scooter to use it multiple times.
- Lithium–Iodine battery. Lithium batteries are beneficial for portable electronics because they have a significantly longer life span than many other batteries.
- NiCad, NiMH, or lithium ion battery (rechargeable). A rechargeable lithium ion battery is ideal as they give you maximum performance for their lifespan as well as charge efficiently.