Extra Credit 2

Q17.1.2

For the scenario in the previous question, how many electrons moved through the circuit?

S17.1.2

Going off of the previous question, you found the answer to be 5.25 × 10³ C. So after this, to find the amount of electrons moving through the circuit, you must divide this number by e = 1.6 x 10^-19 C. This number is the charge of one electron. Thus, the solution becomes...

\(1 A= 1 { C\over s}\)

\(2.5 {A} \times {60 s \over 1 min} \times {35 min} = 5250 {C}\)

\(5250 C \over {1.6 \times 10^{-19} C}\)

\(= {3.28 \times 10^{22} electrons}\)

=3.28 × 10²² number of electrons moved through the circuit.

Watch out for significant figures!

(All in Latex)

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%.

S17.7.5

1. First off you need to write a balanced chemical equation:

\(Zn^{2+} (aq) +2e^{-} \rightarrow Zn (s)\)

2. Secondly, you must calculate the moles of electrons passing through the cell.

So you take the current and convert it from amps into seconds. Then, you divide that value by the charge of one electron which is 1.6 x 10^-19 C

=\( 1 hr \times {60 min \over 1 hr} \times {60 s\over 1 min} \times {2.599 {C \over s}} = 9356.4 C\)

3. Now you divide 9356.4C by the charge of one electron (1.6 x 10^-19)

=\({9356.4 C \over {1.6 \times 10^{-19} {C \over electron}}} = {5.85 \times 10^{22} electrons}\)

4. Now you divide that number by Avogadro's number (6.02× 10²³) to find the moles of electrons.

= 0.0971 moles of electrons

Since two moles of zinc are used up to product one mole of zinc, you divide the moles by 2.

Zn = 0.0486 mol

In all,

\({5.85 \times 10^{22} electrons} \times {1 mol \over {6.022 \times 10^{23} electrons}} \times {1 mol Zn \over 2 mol e^-} = 0.0486 mol Zn\)

5. Then, you use the molar mass of Zinc to find the grams of it. You multiply the moles by the molar mass found on the periodic table (65.39 g/mol).

Mass of Zn = 0.0486 mol × 65.39 g/mol

= 3.18 grams of Zinc

6. Lastly, you divide the grams of zinc by the density given 7.14 g/cm³

Volume of zinc = (3.18 g) / (7.14 g/cm³ )

= .445 cm³

In all,

\(0.0486 mol Zn \times {65.38 g Zn \over 1 mol Zn} \times {1 cm^3 \over 7.140 gr Zn} = 0.445 cm^3\)

7. And divide that value by the 0.01123mm layer of zinc converted into centimeters.

\(0.01123 mm \times {1 cm \over 10 mm} = 0.001123 cm\) layer of zinc

8. Surface area = \({0.445 cm^3 \over 0.001123 cm} = 395.4 cm^2\)

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

To write in a formula, one must write the metal atom or ion before the ligands in the chemical equation. Also, [metal then neutral ligand then anionic ligand], [cation][anion]. Furthermore, one must be familiar with different prefixes to be able to determine the amount of the certain ligand attached. For example, diamine means 2 amine ligands.

Coordination Number is determined by the amount of ligands bonded to the metal.

1. Formula: [Zn(OH)₄]^2- Coordination Number: 4

In this case, the metal is zinc with four OH ligands attached to it. OH is called hydroxo and has a negative one (-1) charge multiplied by four so in total have a negative four charge (-4). Zinc is called zincate and has a positive two (+2) charge. So, the overall charge of the complex is negative two. The coordination number is four because it has four hydrxo ligands attached to the zinc. Tetra means four.

2. Formula: [Pd(CN)₆]^2- Coordination Number: 6

In this case, the metal is palladium with six cyano ligands attached to it. CN is called cyano and it has a negative one charge multiplied by six to in total have a negative six charge (-6). Pd is palladium with a positive four charge to make the overall charge of the complex negative two. Hexa means six. The coordination number is six because it is octahedral with six cyano ligands attached to the palladium.

3. Formula: [AuCl₂]^- Coordination Number: 2

In this case, the metal is gold with two chloro ligands attached to it. Cl is called chloro and have a negative one charge multiplied by two in total to have a negative two charge. Au is gold and is called aurate in this complex. It has a positive one charge to make the overall charge of the complex negative one. Di means two. The coordination number is two due to the two chloro ligands attached to the gold.

4. Formula:[Pt(NH3)₂Cl₂] Coordnation Number: 4

In this case, the metal is platinum with two different ligands attached to it. NH3 is called amine and has a neutral charge. Cl is called chloro with a negative one charge multiplied by two in total to have a negative two charge. Pt is platinum with a positive two charge to make the overall charge of the complex neutral. Di means two. The coordination number is four due to two amine and two cholo ligands being attached to the platinum.

5. Formula: K[Cr(NH3)₂Cl₄] Coordination Number: 6

In this case, the metal is chromium with different ligands attached to it. NH3 is called amine and has a neutral charge. Cl is called chloro with a negative one charge multiplied by four in total to have a negative two charge. K is potassium with a positive one charge. Cr is chromium with a positive three charge to in total give the complex a neutral charge. The coordination number is six due to the two amine and four chloro ligands attached to the chromium.

6. Formula:[Co(NH3)₆][Cr(CN)₆] Coordination Number: 6

In this case, there are two complexes, one being the cation and the other being the anion. First, the cation, is composed of the metal cobalt with six neutral amine ligands attached to it. The cobalt has a positive three charge because it is cobalt (III). The anion is chromium with six cyano ligands attached to it. The cyano ligand (CN) has an overall negative six charge, meanwhile the chromium has a positive three charge. Overall both formulas combined equals a neutral complex. The coordination number is six because there are six amine ligands bonded to cobalt and six cyano ligands bonded to chromium.

7. Formula: [Co(en)₂Br₂]NO₃ Coordination Number: 6

In this case, the metal is cobalt with different ligands attached to it. en is ethylenediamine, a neutral ligand, however it is also a bidentate ligand meaning that one of the ligand is actually two and two of the ligand is actually four, etc. Br is called bromo and has a negative one charge multiplied by two to equal a negative two charge. NO₃ is called nitrate with a negative one charge. Since it is written after the metal in it's name then it is not included within the brackets of the complex. Co is cobalt and has a positive three charge thus making the overall charge neutral. The coordination number is six because of the two bromo ligands and the two ethylenediamine ligands (total four because bidentate).

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

S12.3.14

For determining the rate law, you take the difference between the second and the first [A] values given as well as that for the rates given. You solve for the power of "x" attached to [A]

[A2/A1]^x=[Rate2/Rate1]

1. (2.66/1.33)^x=(1.52 x 10^-6)/(3.8 x 10^-7)

2^x=4

x=2

2. Then, writing the rate law..

rate=k[]

so, rate=k[A]²

3. Now to solve for "k" we can plug in values from the table..

\({3.80 \times 10^{-7}} {M \over h} = k [1.33 \times 10^{-2} M]^2\)

\(k = {2.15 \times 10^{-3}} {1\over {M \times h}}\)

k=.002148 M^-1h^-1

The units of k is M^-1h^-1 because the overall reaction is 2 and h is set at a power of -1 and thus M is -(overall-1)=-1.

Lastly, the order is second order because of the power value of x being 2. Meaning that there are two "A" particles being multiplied by each other.

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?

(a) Cl₂+CO⟶Cl₂CO

rate=k[Cl2]^3/2[CO]

(b) PCl₃+Cl₂⟶PCl₅P

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]

S12.6.6

In an elementary reaction, the rate law will be the same as the balanced reaction because the rate law is not broken down into anymore reaction steps. Thus, only the rate laws that match the balanced equation will have the same elementary and overall reactions. The solution for this would be b, d, and e.

Q21.4.18

The isotope Sr3890 Sr3890 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.

S21.4.18

For this problem, you would use the equation: ln[A]=ln[Ao]-kt

[A] is the newer value given of the sample, meaning the sample was originally .5g then diminished to .393g.

[A] is 0.393g

[A_o] is 0.500g

t is 10 years

1. To find the half life you must plug in these values, so the equation becomes..

\(ln[0.393 g] = ln[0.500 g]-k(10.0 years)\)

2. then you divide .393 by .5 and take the natural log of both sides

-0.2408=-k(10)

k=.0241

3. Then, you plug in k to the equation: t1/2=ln(2)/k

\(t_{1 \over 2} = {ln (2) \over k}\)

\(t_{1 \over 2} = {ln (2) \over 0.0241 {1 \over year}}\)

\(t_{1 \over 2} = 28.76 years\)

Q20.3.6

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

The voltage produced by a redox reaction can be measured more accurately using two electrodes immersed in a single beaker, containing an electrolyte that completes the circuit. With having two different beakers, there is a greater chance that they will have different ion concentrations and diffusion rates, which in turn can affect the rate flow. The single beaker reduces errors due to the junction potential, which is a boundary that resists the flow of charge. Also, differences in the cation and anion diffusion rates increases the resistance in the salt bridge and limits the electron flow through the circuit. Resistances like these decreases the amount of energy to do work which creates a larger difference between what the actual potential is and the potential measured.

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?

S20.5.17

First off, you must write a chemical equation. You are given Ca2+ and Na(s) as reactants and Ca(s) and Na+ as products. You must balance the charges on both sides of the equation. To do this we but the coefficient 2 in front of Na+, but then we need to balance out the number of Na s on the other side so we add another coefficient 2. Now, it is positive 2 on both sides and there are equal amounts of each element.

Ca²+ + 2Na(s) ⟶ Ca(s) + 2Na⁺

Now, we use the reduction half reaction chart to determine the E° values of each.

Cathode: \(Ca^{2+} (aq) + 2e^- → Ca(s)\) E° = -2.84 V

Anode: \(Na^{+} (aq) + e^- → Na(s)\) E° = -2.71 V

Then, we must determine which one is the anode and which one is the cathode. We do this by telling which one is being reduced or oxidized. We find that Calcium is being reduced because it is going from positive to more negative, thus gaining electrons. Meanwhile, sodium is being oxidized because it is becoming more positive and losing elections. We then plug in the values to this equation..

\(E°_{cell} = E°_{cathode} - E°_{anode}\)

\(E°_{cell} = -2.84 V - (-2.71 V)\) = - 0.127 V

Now we must find ΔG° through this equation.... n is the amount of electrons used. n=2 because it requires 2 electrons to balance the reduction and oxidation reaction for the overall reaction. F is Farraday's constant: 96485 C/mole.

\(ΔG°= - nFE°\)

\(ΔG°= - {2 mol} \times {96485 C } {- 0.127 V}\)

=24507.19 J

Divide 24507.9J by 1000 to get the value in kJ

\(ΔG°= 24507.9 J = 24.5 kJ\)

Based off of this, you would expect the elements to be close to each other on the periodic table because of such a small magnitude of ΔG°. Also, the positive ΔG° means that the reaction is nonspontaneous. It is expected since the elements are so close to each other that it is more likely for the reaction to be nonspontaneous.