Extra Credit 37
- Page ID
- 83433
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
S19.15A
Use the equation ΔG° = -nFE°cell
Write the balanced reduction and oxidation reactions for the chemical reaction. The electrons should cancel out.
N2H4(aq) + 4e- ----> N2(g) + 4H+
O2(g) + 4H+ ----> 2H2O(l) + 4e-
Therefore, n = 4 mol e-
And F = 96485 C/mol e- , and E°cell = 1.559 V
So G° = -(4 mol e-)(96485 C/mol e-)(1.559 V) = -6.0168E5 J
(-6.0168E5 J)(1 kJ/1000 J) = -601.7 kJ
-601.7 kJ = ΔG[N2(g)] + 2ΔG[H2O(l)] - ΔG[N2H4(aq)] - ΔG[O2(g)]
-601.7 kJ = 0.00 kJ + 2(-237.2 kJ) + 6.017 kJ - ΔG[N2H4(aq)]
Then ΔG[N2H4(aq)] = 133.31 kJ
S20.1B
a. Co2+: [Ar]3d7
b. Ni2+: [Ar] 3d8
c. Zn2+: [Ar] 3d10
d. Cu2+: [Ar] 3d9
e. Ag+: [Kr]4d10
When ionizing transition metals, they will always lose the electrons in the s shell before the d shell, because the energy of s shell electrons are higher than in the d shell. Transition metals are most stable when they have a full shell--d10 configurations.
S21.11A
a. Tetrahedral have no geometric isomers. Geometric isomers are defined as metal complexes that differ in which ligands are adjacent to one another (cis) or directly across from one another (trans). Since tetrahedral molecules have only 109.5 bond angles between the four ligands, there is only one orientation that the ligands can bond to the metal.
b. Square planar complexes can be either cis or trans, because they form bond angles at 180 and 90 degrees. Because of this, two pairs of ligands can be adjacent (cis) to each other or directly across from one another (trans). A square planar molecule can never be cis and trans at the same time, because the coordination number is 4.
c. Since linear complexes only have a 180 degree bond angle, it cannot have cis or trans isomers. There is only one possible way to bond the two ligands to the metal in the coordination complex.
S24.17C
a. Use equation: ln([A]/[A]o) = -kt for a first order reaction.
Let x = percent remaining and [A] = x[A]o where [A] = remaining after time t
ln(x[A]o/[A]o) = -kt
Where k = ln2/t1/2 for a first order reaction and t1/2 = 250 s
so k = ln2/250s = 0.002773 1/s
ln(x) = -(0.002773)(1500s)
ln(x) = -4.1595
x = 0.015615
Percent remaining = (0.015615)(100) = 1.56%
b. If rate = k[A],
Then rate = (0.002773 1/2)(0.5M) = 0.00139 M/s
S25.1D
a. 201Tl ----> 4He + 193Au
In alpha decay, the Tl nucleus loses an alpha particle, reducing the atomic number by 2 and the atomic weight by 4. The resulting nucleus is Au, atomic mass 193 and atomic number 79.
b. 238Np ----> 0β + 238U
In beta emission, the nucleus loses one beta particle, but it does not effect the atomic number. Np emits one beta particle, so the resulting nucleus has an atomic number that is one less than Np. The resulting nucleus is U, with atomic number 92.
S25.37A
Use only the mass of the nucleus, so subtract the mass of electrons from the total mass of the atom to get the mass of the nucleus. Use the change in mass, which is the mass of products minus the mass of reactants.
Δmass= [(mass O - 8 e-) + (mass H - 1 e-)] - [(mass N - mass 7 e-) + (mass He - mass 2 e-)
= [17.99916 u + 1.00783 u] - [15.00011 u + 4.00260 u] - 0 e-
= 0.00428 u
Convert to megaV: (0.00428 u)(931.5 MeV/u) = 3.9868 MeV
S21.2.5
The conversion of lead to gold is not energetically favorable. In order to make lead out of gold, chemists would need to remove 3 neutrons from the nucleus of lead. Not only are you removing neutrons from an atom that is already stable, but chemists would be removing neutrons and attempting to synthesize an element that would be above the belt of stability--the atomic number would decrease, but the mass number would stay the same. The synthesis of gold is energetically unfavorable, because chemists would produce a very unstable element.
S20.3.3
A Galvanic or voltaic cell is an electrochemical cell that gets its energy from the transfer of electrons during spontaneous redox reactions. A salt bridge is used in electrochemical cells to control the flow electrons. Galvanic cells should always use salt bridges in order to control the transfer of electrons from the anode to the cathode in redox reactions.