Extra Credit 2
- Page ID
- 83250
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Q17.1.2
For the scenario in the previous question, how many electrons moved through the circuit?
S17.1.2
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Given: Total Charge: 5.3 x 103 C |
Step 1. From the previous question, there were 5.3 x 103 C in a 2.5 A current that ran for 35 minutes. |
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Know: Electrons charge: 1.6 x 10-19 C |
Step 2. We know that the charge of an electron is 1.6 x 10-19 C. |
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Solve # of electrons: \[\frac{5.3\times {10^3} C}{1.6 \times 10^{-19}C}=3.3\times 10^{22} electrons \ moving \ through \ circuit\] |
Step 3. Divide the total charge by charge per electron to find the number of electrons. |
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
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Given: Current: 2.599 A Time: \[1 \ hour = \frac{60 \ min}{1 \ hour}=\frac{60 \ sec}{60 \ min}\]
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Know: Electrons charge: 1.6 x 10-19 C Density of zinc: 7.140 g/cm3 Mass of zinc: 65.39 g/mol |
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Solve: 1 coulomb = 1 ampere x 1 second \{2.599 A\times 60 minutes\times 60 seconds} = 9,356.4 C ne- = total charge/electron charge = \[\frac{9,356.4 C}{1.6 \times 10^{-19}C}=5.85\times 10^{22} electrons] = \[\frac{5.85\times 10^{22}{6.02 \times 10^{22}mol^{-1}}=0.971 mole of electrons] |
1. Convert ampere to coulombs.
2. Calculate number of electrons.
3. Calculate moles of electrons. |
This question is referring to the process of electroplating.
step (1) calculate the moles of electrons passing through the cell.
no. electrons= total charge/ charge on electron
=(2.599 A × 60 min× 60 seconds) / 1.6× 10-19 C
=(5.85× 1022 number of electrons)/ 6.02× 1022
= 0.971 mol of electrons
step (2) write the balanced equation
Zn2+ +2e- →Zn which tells us the stoichiometry of the reaction.
therefore moles of Zn = 0.486 mol
mass of Zn = 0.486 mol × 65.39 g/mol
= 31.8 grams of Zinc
therefore volume of zinc = (31.8 g) / (7.14 g/cm3 )
= 4.45 cm3
therefore the surface area = (4.45 cm3 ) / 0.0001123 cm = 396 m2
Q19.2.2--- (Done by Jenny)
Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:
- tetrahydroxozincate(II) ion (tetrahedral): [Zn(OH)4]2-
- hexacyanopalladate(IV) ion: [Pd(CN)6]2-
- dichloroaurate ion (note that aurum is Latin for "gold"): [AuCl2]
- diaminedichloroplatinum(II): [PtCl2(NH3)2] geometric, cis/trans
- potassium diaminetetrachlorochromate(III): K[CrCl4(NH3)2] geometric, cis/trans
- hexaaminecobalt(III) hexacyanochromate(III): [Co(NH3)6][Cr(CN)6]
- dibromobis(ethylenediamine)cobalt(III) nitrate: [CoBr2(en)2]NO3
Q12.3.14 --- (Done by Jenny)
From the following data, determine the rate equation, the rate constant, and the order with respect to A for the reaction A⟶2C.
| [A] (M) | 1.33 × 10−2 | 2.66 × 10−2 | 3.99 × 10−2 |
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| Rate (mol/L/h) | 3.80 × 10−7 | 1.52 × 10−6 | 3.42 × 10−6 |
We can find this by compare proportions of [A] and the rates of the reaction.
1.52 × 10−6/3.80 × 10−7= (2.66 × 10−2/1.33 × 10−2)x
4=2x
x=2, indicating that the order is 2.
3.42 × 10−6/1.52 × 10−6=(3.99 × 10−2/2.66 × 10−2)x
2.25=1.5x
x = 2, also indicating that the order is 2.
From the two comparisons, we can confirm the order with respect to A is 2 and that the rate equation is:
rate = k[A]2
Q12.6.6 --- (Done by Jenny)
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]32[CO]
We first look at all the reactants make sure the rate laws have the correct reactants. We next look at the coefficients of each equation to check if they match the exponents of the corresponding reactant molarity. If they match, it indicates that the elementary reaction and overall reaction is the same. In this case, the coefficient for Cl2 does not match the exponent for [Cl2], thus indicating that the elementary reaction and the overall reaction are not the same.
(b) PCl3+Cl2⟶PCl5
rate=k[PCl3][Cl2]
We follow the same steps as above and find that the coefficients match the exponents for the reactant molarity in the rate law. The elementary reaction and the overall reaction are the same.
(c) 2NO+H2⟶N2+H2O
rate=k[NO][H2]
The coefficient for NO in the reaction and the exponent for the molarity for [NO] in the rate law do not match, so the elementary reaction and the overall reaction are not the same.
(d) 2NO+O2⟶2NO2
rate=k[NO]2[O2]
The coefficients for the reaction equation match the exponents for each of the reactant molarity in the rate law. The elementary reaction and the overall reaction are the same.
(e) NO+O3⟶NO2+O2
rate=k[NO][O3]
Coefficients match exponents. The elementary and overall reaction are the same.
Q21.4.18--- (Done by Jenny)
The isotope 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.
We know that radioactive decay follows the equation: Nt=Noekt where k is a constant. We first plug in what we are given to find k.
0.393 = 0.500ek(10)
Now, using algebra, we can find k.
1. 0.393/0.500=ek(10)
2. ln(0.393/0.500)=ln(ek(10) )=k(10)
3. k=-0.02408
Now we can plug k back in to find half life:
0.5No=Noekt => 0.5(0.500)=0.500e-0.02408t
We isolate t using algebra:
1. 0.5=e-0.02408t
2. ln(0.5)=ln(e-0.02408t) = -0.02408t
t=28.785 yrs
Q20.3.6 --- (Done by Jenny)
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?
Using two beakers requires a salt bridge which is a factor that can change the potential due to the various transfers of ions in the solutions.
Q20.5.17 ---(Done by Jenny)
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?
We first find the standard reduction potentials to find the overall change of the reaction.
| Ca2+ + 2e− ⇌ Ca(s) | –2.84 |
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Na++e−⇌Na(s) |
–2.713 |
We observe oxidation states to determine what was reduced and what was oxidized. Ca2+ (cathode) was reduced to Ca(s) and Na(s) (anode) was oxidized to Na+.
Using Ecell=Ecathode-Eanode, we find the standard potential of the reaction. Ecell= -2.84-(-2.713) = -0.11V
Next, we use ΔG°=-nFEcell where n is the number of electrons transferred and F is Faraday's constant (96,485 C/mol)
We plug everything in:
ΔG°=-(2)(96,485)(-0.11)=21225.7J = 21.2257kJ
ΔG° is positive indicating a non-spontaneous reaction, as expected since Ecell was negative.