Extra Credit 18
- Page ID
- 82825
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Q17.2.7
Why is a salt bridge necessary in galvanic cells like the one below?
Solutions:
In a galvanic cell, electrons flow from the anode to the cathode. This results in a charge imbalance, as the anode gets more positive while the cathode becomes more negative with the influx of electrons, eventually losing its electrical neutrality and ending the reaction. The salt bridge is needed to maintain electrical neutrality by allowing ions to transfer between the two solutions to balance charges, and completes the circuit by carrying electrical charges.
Q19.1.16
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.)
a. MnCO3(s)+HI(aq)⟶
b. CoO(s)+O2(g)⟶
c. La(s)+O2(g)⟶
d. V(s)+VCl4(s)⟶
e. Co(s)+xsF2(g)⟶
f. CrO3(s)+CsOH(aq)⟶
Solution:
a. Breaking apart this reaction, we first look at the MnCO3 oxidation states. O3 will have a total oxidation state of -6, while C will have an oxidation state of +4. This results in CO3 having an oxidation state of -2, while this results in Mn having an oxidation state of +2. For HI, the oxidation state of H is +1, and I has the oxidation state of -1. This reaction would be a double replacement, pairing Mn with I. Since Mn had an oxidation state of +2, and I has a an oxidation state of -1, there would be 2I with one Mn, forming MnI2. For the other product, H with the oxidation state of +1, and CO3 with an oxidation state of -2, forms H2CO3. However, H2CO3 does not have a big Ksp in water, and it will be decomposed to CO2 and H2O. After balancing the equation by multiplying HI by 2, we get the equation, \[{MnCO_3}_{(s)}+{2HI}_{(aq)}⟶{MnI_2}_{(aq)}+{CO_2}_{(g)}+{H_2O}_{(l)}\]
b. When looking at this reaction, we can see that since there is O on both reactants, this may be a synthesis reaction. Looking at the oxidation state of Co, it will be +2 since the O in the reactant side of the compound will be -2. We have to note that Co becomes oxidized in this reaction and becomes an oxidation state of +3. By combining the two reactants, we get CoO3. When we balance the reaction by multiplying CoO by 2, O2 by 2, and CoO3 by 2, we get the equation \[{2CoO}_{(s)}+{2O_2}_{(g)}⟶ {2CoO_3}_{(s)}\]
c. When looking at this reaction we can see that La(s) and O2 could result in a synthesis reaction. Looking at La(s), we see that it is in the third period, meaning that it has a +3 charge. O2 by itself will have a 0 charge. When predicting the product, by matching the charges on O, which is -2, and La, which is +3, we get the product La2O3. After balancing the reaction, we get the equation \[{2La}_{(s)}+{3O_2}_{(g)}⟶{La_2O_3}_{(s)}\]
d. When looking at this reaction we look at the oxidation states of the reactants. We see that V(s) has an oxidation state of 0 while VCl4 has an oxidation state of +4 for V and -1 for Cl. Looking at the electron configurations for both V and VCl4, we see that V=[Ar]4s23d3 while V4+=[Ar]4s13d3. Taking these states into account, we see that the products should be VCl2 for the most stable configurations. We get that the equations is \[{V}_{(s)}+{VCl_4}_{(s)}⟶{2VCl_2}_{(s)}\]
e. For this reaction, the xs notation in front of F2 means that there is excess F2 in the reactant side which means there will be more F2 used in the reaction than written. The equation of this reaction will be \[{Co}_{(s)}+{xsF_2}_{(g)}⟶{CoF_4}_{(g)}\]
f. When looking at this reaction, we observe the oxidation state of CrO3. Looking at O3 with an oxidation state of -6, we see that Cr must have an oxidation state of +6. When looking at CsOH, we see that O has -2, while H has +1, which means OH has -1 oxidation state. This means that Cs must have a +1 oxidation state. When looking at the products of this reaction, we can see that this will be a double replacement reaction. The products Cr +6 with OH yields Cr(OH)6 while balancing the Cs with O yields Cs2O. After balancing the reaction, we get
\[{CrO_3}_{(s)}+{6CsOH}_{(aq)}⟶{Cr(OH)_6}_{(aq)}+{3Cs_2O}_{(s)}\]
However, this reaction happens in water, and H2O will react with Cs2O. Therefore, the overall reaction will be
\[{CrO_3}_{(s)}+{6CsOH}_{(aq)}+{3H_2O}_{(l)}⟶{Cr(OH)_6}_{(aq)}+{6CsOH}_{(aq)}\]
Q19.3.8
For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the t2g orbitals increases. Which complex in each of the following pairs of complexes is more stable?
a. [Fe(H2O)6]2+ or [Fe(CN)6]4−
b. [Co(NH3)6]3+ or [CoF6]3−
c. [Mn(CN)6]4− or [MnCl6]4−
Solutions:
a. Since the oxidation number is the same for both metal ions, we have to look at the ligands of each complex. By using the electrochemical spectrum, we can see that CN- is a strong field ligand compared to H2O which is a weak field ligand. Since CN is a strong field ligand, it is a low spin ligand, which ensures that more electrons in that complex stay in the t2g orbital, making the [Fe(CN)6]4- more stable than the [Fe(H2O)6]2+ complex.
b. Looking at the ligands for the two complexes, we see that F is a weak field ligand, or high spin ligand, while NH3 is a stronger field ligand, meaning that it would be a low spin ligand. Since NH3 is a low spin ligand, it would result in the complex having more electrons in the t2g orbital, making the [Co(NH3)6]3+ complex more stable than the [CoF6]3- complex.
c. Looking at the ligands for the two complexes, we see that CN- is a strong field ligand (low spin), while Cl- is a weak field ligand (high spin). Since low spin complexes are more stable, the [Mn(CN)6]4- complex is more stable than the [MnCl6]4- complex.
Q12.4.8
What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M? The rate constant for this second-order reaction is 8.0 × 10−8 L/mol/s.
Solutions:
Givens: [NOCl]=0.15, k=8.0 x 10-8 L/mol/s
second-order reaction: \[{t_{1/2}}=\frac{1}{k}\times[NOCl]\]
Calculate by plugging in values:
\[{t_{1/2}}=\frac{1}{0.00000008}\times0.15={83333333}\]
t1/2=\(8.33\times10^7\)seconds
Q21.2.3
For the following isotopes that have missing information, fill in the missing information to complete the notation
\[a. {^{34}_{14}}X\]
\[b. {^{36}_{X}}P\]
\[c. {^{57}_{X}}Mn\]
\[d. {^{121}_{56}}X\]
Solutions:
a. The atomic number of the unknown element is 14, which means that the X element should be Si, Silicon. The notation should be \({^{34}_{14}}Si\).
b. Since the element is Phosphorus, the atomic number X should be 15. The notation should be \({^{36}_{15}}P\).
c. Since the element is Manganese, whose atomic number is 25, then X should be 25. The notation should be \({^{57}_{25}}Mn\).
d.The atomic number of the unknown element X is 56, which means that the element should be Barium (Ba). The notation should be \({^{121}_{56}}Ba\).
Q21.5.6
In usual practice, both a moderator and control rods are necessary to operate a nuclear chain reaction safely for the purpose of energy production. Cite the function of each and explain why both are necessary.
Solutions:
Moderator: A moderator in a nuclear reactor for energy production is a substance that slows down the neutrons in the nuclear reactor. It is necessary for the nuclear chain reaction to produce energy due to the fact that the neutrons produced in the nuclear reactions move too fast to cause nuclear fission, so there must be some substances that slow them down to make energy.
Control Rods: Control rods are rods made out of certain materials that are capable of absorbing neutrons. They are inserted into the nuclear reactor in order to control the fission rate of the reactor by adjusting the number of neutrons in the reactor. They are necessary to operate a nuclear chain reaction safely for the purpose of energy production due to the fact that they help keep the rate of the nuclear chain reaction at a safe level for operators.
Q20.4.4
If the components of a galvanic cell include aluminum and bromine, what is the predicted direction of electron flow? Why?
Solutions:
In the galvanic cell that has aluminum and bromine, the predicted direction of electron flow would be towards bromine because bromine is more electronegative than aluminum, thus making it more likely to attract electrons toward itself.
Q20.7.2
Why does the density of the fluid in lead–acid batteries drop when the battery is discharged?
Solutions:
The electrolyte in the lead-acid battery is made out of a solution of sulfuric acid in water. When the battery is discharged, the sulfuric acid is consumed and more water is produced in the solution, causing the concentration of the ions in the lead-acid batteries to drop.