Extra Credit 2

Q17.1.2

For the scenario in the previous question, "if a 2.5 A current is run through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?" how many electrons moved through the circuit?

S17.1.2

Before we calculate numbers of electrons in the circuit, we need to calculate Coulomb moved through circuit

\[\mathrm{2.5\;A=2.5\frac{C}{seconds}\times\frac{60\;seconds}{1\;min}\times35\;minutes=5250\;C}\]

Using the Faradays constant and Avogardro's number we can derive the charge per electron.

\[\mathrm{\frac{96485\;Coulombs}{mole\;e^{-}}\times\frac{1\;mole\;e^{-}}{6.022\times10^{23}\;electrons}}=1.6\times10^{-19}\frac{C}{electron}\]

Now we have to components we need to calculate total numbers of electrons

\[\mathrm{Total\;number\;of\;electonrs=\frac{5250\;C}{1.6\times10^{-19}\frac{C}{electron}}=3.28\times10^{22}}\]

Q17.7.5

An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a Zn(NO₃)₂ solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123-mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is 7.140 g/cm³. Assume the efficiency is 100%.

Q17.7.5

This question is referring to the process of electroplating.

First we need to calculate the moles of electrons passing through the cell.

\[\mathrm{Q=ne}\]

Q is total charge

n is number of electrons

e us charge of 1 electrons

Rearranging formula gives us

\[\mathrm{n=\frac{Q}{e}}\]

Which is Number of electrons= Total charge/ Charge on electron

\[\mathrm{A\;(Ampere)=\frac{C}{seconds}}\]

Therefore, we need to convert an hour into seconds

\[\mathrm{Total\;Charge=2.599\frac{C}{seconds}\times\frac{60\;seconds}{1\;min}\times\frac{60\;mins}{1hr}\times1hr=9356.4\;C}\]

\[\mathrm{Charge\;on\;Electron=\frac{96485\;Coulombs}{mole\;e^{-}}\times\frac{1\;mole\;e^{-}}{6.022\times10^{23}\;electrons}}=1.6\times10^{-19}\frac{C}{e^{-}}\]

\[\mathrm{Number\;of\;Electrons=\frac{Total\;Charge}{Charge\;on\;electron}=\frac{9356.4\;C}{1.6\times10^{-19}\frac{C}{electron}}=5.85\times10^{22}\;electrons}\]

Now we have to get moles of electron by dividing by Avogadro's number

\[\mathrm{moles\;of\;electrons=\frac{5.85\times10^{22}\;electrons}{6.02\times10^{23}\frac{electrons}{1\;mol}}=0.0971\;moles}\]

Now that we have moles of electrons, we need to tep write the balanced equation

\[Zn^{2+} +2e^- →Zn (s)\]

which tells us the stoichiometry of the reaction. It tells us that for every Zinc 2 moles of electrons are tranferred.

Therefore,moles of \(Zn\) = 0.0486mol

\[\mathrm{Mass\;of\;Zn=0.0486\;mol\times65.39\frac{g}{mol}=3.18\;g\;of\;Zinc}\]

We can find volume of Zinc by mass of Zinc by density of Zinc

\[\mathrm{Volume\;of\;Zinc=\frac{3.18\;g}{7.14\frac{g}{cm^{3}}}=0.445\;cm^{3}}\]

\[\mathrm{Surface\;Area=\frac{0.445\;cm^{3}}{0.01123\;mm\frac{1\;cm}{10\;mm}}=396.3\;cm^{2}}\]

Original answer had Avogadro's number wrong which messed up entire following calculations.

Q19.2.2

Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:

1. tetrahydroxozincate(II) ion (tetrahedral)
2. hexacyanopalladate(IV) ion
3. dichloroaurate ion (note that aurum is Latin for "gold")
4. diaminedichloroplatinum(II)
5. potassium diaminetetrachlorochromate(III)
6. hexaaminecobalt(III) hexacyanochromate(III)
7. dibromobis(ethylenediamine) cobalt(III) nitrate

S19.2.2

The coordination numbers are simply how many bonds the central metals have with ligands. Also the general order for nomenclature is 1.metal 2. neutral ligand 3.anionic ligand.

A.tetra- suggests that number of hydroxo is and zincate part suggest that central metal is zinc lastly oxidation number of zinc is two which means +2 -4(-1 for one hydroxo)=2- for overall charge. With these parts, we have [Zn(OH)₄]^2-. Since hydroxo is monodentate ligand, the coordination is 4

B.Hexa- suggests that there are 6 cyanide and palladate suggests central metal is palladium. For charge, we have -6(six cyanide)+4(palldium)=-2.[Pd(CN)_6]^2-. Cyanide is also monodentate so the coordination number is 6.

C. di-means there are two chloride and aurate means gold is central metal. For the overall charge,-2 from chloride. [AuCl₂]^2-. Chloride is also monodentate thus the coordination number is 2.

D. Like we saw before we have 2 amine and 2 chloride. Also, we have platinum as central atom. For the overall charge, 0(anime is neutral ligand)+2-2=0. So we have [Pt(NH3)₂Cl₂]. Amine and chloride are monodentate. Thus have 4 as coordination number.

E. We see that potassium is separated by space so we know that potassium is counter ion. Diamine means two amine and tetrachloride means 4 chloride so it has 6 coordination number. K[Cr(NH₃)₂Cl₄]

F. We have complexes now. One is cation and the other one is anion. First we have cation complex, hexaaminecobalt means 6 amine attached to cobalt. So we have [Co(NH₃)₆]³⁺. Now for anion, hexacyanochromate, we have six cyanide attached to chromate and it has -6+3=-3 overall charge. So [Cr(CN)₃]^3-. Putting these together, we have [Co(NH₃)₆][Cr(CN)₃]. Coordination is 6

G. In this case we have, cation complex with anion counter ion. dibromo means 2 bromine and bis(ethylenediamine) means two ethylenediamine which is abbreviated by en in a chemical formula adding.For bidentate ligans we use bis- to indicate number. For charge,-2(bromine)+3(cobalt)=+1. Adding all these we have [Co(en)₂Br₂]⁺, to this we add NO₃^- so overall we have [Co(en)₂Br₂]NO_3.

Q19.2.1

Indicate the coordination number for the central metal atom in each of the following coordination compounds:

1. \([Pt(H_2O)_2Br_2]\)
2. [Pt(NH₃)(py)(Cl)(Br)] (py = pyridine, C₅H₅N)
3. [Zn(NH₃)₂Cl₂]
4. [Zn(NH₃)(py)(Cl)(Br)]
5. [Ni(H₂O)₄Cl₂]
6. [Fe(en)₂(CN)₂]⁺ (en = ethylenediamine, C₂H₈N₂)

S19.2.1

1. The 2 aqua and the 2 bromo ligands form a total of 4 coordinate covalent bonds and as a result cordibation number is 4.
2. The ammine, pyridine, chloro and bromo each form one coordinate covalent bond that gives a total of 4 and hence CN=4
3. two ammine and two chloro ligands give a total of 4 coordinate covalent bonds and a CN = 4.
4. one ammine, a pyrimidine, a chloro and a bromo ligand give a total of 4 covalent bonds, resulting in CN = 4.
5. 4 aqua ligands and 2 chloro ligands form a total of 6 coordinate covalent bonds and a CN =6
6. ethylenediamine is a bidentate ligand that forms two coordinate covalent bonds and along with two cyano ligands it forma a total of 6 bonds. hence has a CN=6

Q12.6.6

Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?
$/*<![CDATA[*/\ce{2NO + H2 ⟶ N2 + H2O}/*]]>*/$

(a) Cl₂+CO⟶Cl₂CO

rate=k[Cl₂]^3/2[CO]

(b) PCl₃+Cl₂⟶PCl₅

rate=k[PCl₃][Cl₂]

(c) 2NO+H₂⟶N₂+H₂O

rate=k[NO][H₂]

(d) 2NO+O₂⟶2NO₂

rate=k[NO]²[O₂]

(e) NO+O₃⟶NO₂+O₂

rate=k[NO][O₃]

Q12.6.6

Elementary reaction is a chemical reaction in which reactants react directly to form products in a single step. In another word, rate law for overall reaction is same as experimentally found rate law. Out of 5 options, option (b),(d), and (e) are such reactions

Q20.5.17

What is the standard change in free energy for the reaction between Ca²⁺ and Na(s) to give Ca(s) and Na⁺? Do the sign and magnitude of ΔG° agree with what you would expect based on the positions of these elements in the periodic table? Why or why not?

Q20.5.17

The standard reduction potential values for Ca²⁺ is -2.84 V and the standard reduction potential for Na⁺ , Na⁺(aq)+e^- → Na(s) is equal to -2.71 V. Balancing the two equations gives 2 electrons being transferred.

\[\mathrm{E^{\circ}_{cell}=E_{cathode}-E_{anode}}\]

Since Calcium ion has the more negative reduction potential it is the cathode.

E^o_cell = -2.84-(-2.71) =-0.13 V

Therefore use the the equation to calculate Gibbs free energy.

Change in G= -n F E^o _cell

change in G= (-2) . (96,485 C). (-0.13V)

= 25086.1 J =2.5 KJ

The change is positive which indicates that the reaction is nonspontaneous and since standard reduction potential of Calcium and Sodium are close in value they should be close on the periodic table.

Original answer had value of cathode and anode switched when calculating for E^o_cell.

S20.3.6

It is often more accurate to measure the potential of a redox reaction by immersing two electrodes in a single beaker rather than in two beakers. Why?

S20.3.6

Using one beaker with two electrodes in a single beaker is better than in two beakers. If we use two different beakers, different ion concentrations and diffusion rates affect rate flow. Also, difference in cation and anion diffusion rates causes resistance in the salt bridge which can make the potential measure inaccurate. Using a single beaker has greater accuracy due to liquid junction potential.

Q21.4.18

The isotope $/*<![CDATA[*/\ce{^{90}_{38}Sr}/*]]>*/$ is one of the extremely hazardous species in the residues from nuclear power generation. The strontium in a 0.500-g sample diminishes to 0.393 g in 10.0 y. Calculate the half-life.

Q21.4.18

Radioactive decay is a first order reaction mechanism.

Therefore half life is independent of the initial concentration.

Half life= (ln2)/ k

First calculate the rate constant

\[\mathrm{A=A_{o}e^{-kt}}\]

\[\mathrm{0.393=0.500e^{-10k}}\]

\[\mathrm{0.786=e^{-10k}}\]

\[\mathrm{ln(0.786)=ln(e^{-10k})}\]

\[\mathrm{k=\frac{ln(0.786)}{-10}=0.024}\]

\[\mathrm{Half\;life=\frac{ln(2)}{0.024}=28.8\;years}\]

Q12.3.14

From the following data, determine the rate equation, the rate constant, and the order with respect to A for the reaction A⟶2C. (Check question for the data information)

Doubling the concentration causes the rate to increase by a factor of 4 and increasing the concentration by a factor of 3 causes the rate to go up by a factor of 9. This suggests that Reaction is second order with respect to [A] and because A is the only reactant it is second order with respect to the overall reaction.

\[\mathrm{Rate=k[A]^{2}}\]

Rate constant can be found by plugging in concentration of A and Rate into rate law

\[\mathrm{3.8\times10^{-7}=k[1.33\times10^{-2}]^{2}}\]

\[\mathrm{k=2.15\times10^{-3}\frac{1}{M\;Hour}}\]