Extra Credit 4
- Page ID
- 83011
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Q17.1.4
a. Cl^- --> Cl2
(-1) --> (0)
oxidation number : (-1 --> 0), therefore Cl- is oxidized and it's an oxidization process
b. Mn^2+ --> MnO2
(+2) --> (+4)
oxidation number: (+2 --> +4), therefore Mn^2+ is oxidized and it's an oxidation process
c. H2 --> H^+
(0) --> (+1)
oxidation number: (0 --> +1), therefore H2 is oxidized and it's an oxidation process
d. NO3^- --> NO
calculations:
((+5) + 3(-2) = -1) --> ((+2) + (-2) = 0)
oxidation number: ( +5 --> +2), therefore N is reduced and it's an reduction process
Q19.1.2
We use the noble gas [Ar] to represent 1s^2 2s^2 2p^6 3s^2 3p^6
a. Ti : [Ar] 4s^2 3d^2
Ti^2+: [Ar] 3d^2
Ti^3+: [Ar] 3d^1
Ti^4+: [Ar]
As you can see the s orbitals are removed before the d orbitals because atoms try to keep electrons at their lowest stable energy possible also known as the Aufbau Process
Q19.2.4
a. [Pt(H2O)2Br2] (square planar)
The first one is cis because the H2O's and Br's are 90 degrees apart. The second one is trans because the H2O's and Br's are 180 degrees apart. This is drawn as square planar because the molecules are all 90 degrees spaced apart.
b. [Pt(NH3)(py)(Cl)(Br)] (square planar, py = pyridine, C5H5N)
The molecule is drawn as square planar because they are all equally spread apart at 90 degrees and there are no cis and trans isomers because there are no all the molecules are different and there are no optical isomers because there is a plane where the molecule can be cut so that each side mirrors the other.
c. [Zn(NH3)3Cl]+(tetrahedral)
The molecule is drawn as tetrahedral because the molecules are spaced apart at 107.9 degrees. There is no cis or trans isomer because the molecules are spaced out at 107.9 degrees and there are no optical isomers because there is a plan where the molecule can be cut so that each side mirrors the other.
d. [Pt(NH3)3Cl]+ (square planar)
The molecule is drawn as a square planar because the molecules are separated at equal distances of 90 degrees apart. There are no cis and trans isomers because the molecule can only be drawn one way and there is only one Cl vs 3 NH3, therefore are no other ways for the molecule to be oriented.
e. [Ni(H2O)4Cl2]
This is drawn as a tetrahedral because there are 6 ligands that are evenly spaced apart 90 degrees. The trans molecule is the first one where the Cl's are 180 degrees apart. The cis molecule is the second one where the Cl's are 90 degrees apart.
f. [Co(C2O4)2Cl2]3−
This is drawn as a octahedral because there are 6 ligands and there are 3 optical isomers.
Q12.3.17
([0.7]^x/[0.35]^x)= (2.268x10^-2)/(1.34x10^-2)
--> 2^x=2 --> x=1 --> [H2]^1 (first order)
([0.6]^x/[0.3]^x)=(1.134x10^-2)/(2.83x10^-3)
--> 2^x=4 --> x=2 --> [NO]^2 (second order)
rate=k[H2]^1[NO]^2
plug in: rate= 2.83x10^-3 , [H2]= 0.35, [NO]=0.3
solve for k: (2.83x10^-3)/((0.3)^2(0.35)^2)
k= 0.09 m^-2 s^-1
Q12.6.8
K[NO][H2]
This is the solution because it is the slowest reaction and does not contain any intermediates
Q12.4.20
Half life of Carbon-14= 5730
ln ([A/A0]) = -kt
ln ([0.5A0]/[A0]) = -k(5730)
solve for k= -ln(0.5)/5730
k= 1.21x10^-4
plug k back into the equation: ln ([0.0825Ao]/[Ao])= -(1.21x10^-4)(t)
solve for t: ln(0.0825)/(-1.21x10^-4)= t
t= 20614.5
Q12.3.8
a) NO3^- + H^+ --> HNO2 E0= 0.98 V
SO3^2- --> SO4^2- E0= -0.93 V
b) NO3^- + H^+ --> HNO2 (cathode)
SO3^2- --> SO4^2- (anode)
c) positively charged electrode: (anode) SO4^2-
negatively charged electrode: (cathode) SO3^2-
Q12.5.9
a) Fe^2+ (aq) + Ce^4+ (aq) --> Ce^3+ (aq) + Fe^3+ (aq)
b)
c) Since the solution of Ce^+4 was 0.1 M originally and it took 9mL of the 0.1 M solution to reach the end point, there is 0.9 millimoles of Ce^+4 used used in the reaction and because the oxidation of Fe^2+ by Ce^+4 has a 1 : 1 ratio then there is 0.9 millimoles of Fe^2+ in the original solution.