exam1_WHITE_BCHM461-0102_OFFICIAL_KEY_Fall2022
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BCHM 461-0102
Name:
Official Key (WHITE exams) October 10, 2022
Prof. Julin
Exam I
(100 points)
NOTE:
read
the questions carefully,
think
about your answers, and,
be sure that you
answer all parts of the questions.
Provide specific information
that answers the question being asked.
Always show your work or explain your reasoning.
You may use a calculator, but only
for computation. Any other use is a violation of the
University’s Code of Academic Integrity
.
No other electronic equipment may be used - that is, turn off your cell phone
!
Honor Pledge
Please write
the following sentence in the box, and then sign your name:
“I pledge on my honor that I have not given or received any unauthorized assistance
on this examination.”
Some equations and formulas which might (or might not) be useful:
a) Quadratic formula.
For: ax
2
+ bx + c = 0
x = !
b ± ( b
2
!
4ac )
½
2a
b)
K
w
= [H
+
][OH
!
] = 10
!
14
c)
K
a
= [H
+
][A
!
]/[HA]
pH = pK
a
+ log([A
!
]/[HA])
d) Δ
G = Δ
H !
T
Δ
S
e) Δ
G = Δ
G
EN
+ RT ln{[
products
]/[
reactants
]}
f)
K
eq
= exp(
!
Δ
G
EN
/RT)
g) Δ
G
EN
= !
RTln(K
eq
)
K
eq
= [products] / [reactants]
h)
R = 8.3 J/K/mol = 0.0083 kJ/K/mol
i)
T (K) = T (
E
C) + 273
j)
E
%
Q
1
Q
2 / ε
r
k)
fr dissoc
Ka
H
Ka
H
pH
pKa
pH
pKa
.
.
[
]
[
]
(
)
(
)
1
10
10
1
l)
m
/
z
= (MW + nH)/n H = 1.008
m)
c
= Abs / ε
(
Beer’s Law
)
n) ε
280
= (# Trp) × 5,500 + (# Tyr) × 1,490
1. (20 points)
Given below are the names of five recipients of a Nobel Prize (in various fields).
i) (4 pts)
Consider only the last names
(written in capital letters). Which of these last
names could also be a peptide, written in single-letter code? Explain briefly
and specifically
why
the other last names cannot be a peptide.
1) (Ei-ichi) N E G I S H I
2) (David) B A L T I M O R E
3) (Venkatraman) R A M A K R I S H N A N
4) (Doris) L E S S I N G
5) (Gabriel Garcia) M A R Q U E Z
Answer:
The 6 letters in the alphabet that do not stand for an amino acid are: B, J, O, U, X, Z.
So, David Baltimore (Physiology or Medicine, 1975) and Gabriel Garcia Marquez
(Literature, 1982) are NOT also peptides.
Ei-ichi Negishi (Chemistry, 2010), Venkataraman Ramakrishnan (Chemistry,
2009), and Doris Lessing (Literature, 2007) are also peptides (whether they know it or
not).
ii) (16 pts)
Pick ANY ONE of the last names given above that could be a peptide.
Draw the complete structure
of four consecutive residues from that last name, as a
tetrapeptide.
Your drawing must fit the following conditions:
a)
The amino acid that is the first letter
of the name must be at the amino terminus of
your peptide.
b)
Use the letters (amino acids) in the order (left-to-right) that they appear in the name,
but DO NOT use the same amino acid twice! Skip a repeated letter and use the next one.
c)
Do not use A
or G
(too simple!).
d)
Draw the peptide structure in the form that would predominate in a solution at pH = 7
.
e)
Write the 3-letter abbreviations of the four amino acids that you have drawn (fill in the
blanks, below).
The name that you chose: _____________
3-letter abbreviations: (N-term) - _____ - _____ - _____ - _____
Answer:
See the textbook for amino acid structures and 3 letter abbreviations. In addition
to points deducted for an incorrect structure or incorrect protonation, points were
deducted for: 1) an incorrect 3-letter abbreviation, 2) using A (alanine) and/or G
(glycine) in your structure, 3) the peptide did not start with the amino acid that
corresponds to the first letter in the last name, 4) a letter in the name that is an amino acid
was skipped and omitted from the peptide structure.
2
2. (5 points)
The table given below lists information about four proteins - proteins 1 - 4
. 1)
The last row in the table gives the number of protein chains in the native form of these
proteins.
2)
The second-to-the-last row gives the total number of amino acid residues in one chain
of each protein.
3)
The rest of the table lists the number of each amino acid residue in one chain
of each
protein.
protein 1
protein 2
protein 3
protein 4
Amino acid
#
#
#
#
Alanine
83
28
26
56
Arginine
44
34
12
26
Asparagine
29
22
21
3
Aspartic acid
59
25
22
19
Cysteine
4
11
3
3
Glutamic acid
36
40
25
19
Glutamine
26
21
10
6
Glycine
60
22
27
21
Histidine
14
11
11
12
Isoleucine
41
26
19
10
Leucine
65
40
26
25
Lysine
25
25
19
5
Methionine
19
13
8
7
Phenylalanine
23
20
13
6
Proline
34
23
11
14
Serine
34
25
13
17
Threonine
48
21
17
14
Tryptophan
10
0
4
3
Tyrosine
16
24
15
3
Valine
59
38
28
24
# residues:
729
469
330
293
# chains:
1
2
6
3
Suppose that you analyzed these four proteins by polyacrylamide gel electrophoresis in
the presence of SDS (SDS-PAGE). Which protein would move the farthest
in the gel? Explain
briefly how you decided.
Answer:
The individual chains of a protein separate when a protein is treated with SDS
(detergent) to be run on SDS-PAGE. The mobility of proteins in SDS-PAGE depends on
the molecular weight of their individual chains. The molecular weights of these proteins
is correlated with the number of amino acid residues in the individual chains.
Protein 4
has the fewest amino acids in each chain, so protein 4 runs farthest in
the gel.
3
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3. (10 points)
The table given below lists information about three more proteins. The table
contents are the same as for the previous question.
protein A
protein B
protein C
Amino acid
#
#
#
Alanine
83
74
77
Arginine
44
21
57
Asparagine
29
25
8
Aspartic acid
59
27
23
Cysteine
4
11
6
Glutamic acid
36
41
58
Glutamine
26
14
18
Glycine
60
50
53
Histidine
14
17
15
Isoleucine
41
32
9
Leucine
65
50
76
Lysine
25
28
13
Methionine
19
9
7
Phenylalanine
23
16
33
Proline
34
30
55
Serine
34
15
24
Threonine
48
35
19
Tryptophan
10
9
22
Tyrosine
16
16
27
Valine
59
43
45
# residues:
729
563
645
# chains:
1
1
1
Estimated charges:
A B C
+44
+21
+57 (Arg)
!
59
!
27
!
23 (Asp)
!
0.4
!
1.1
!
0.6 (Cys)
!
36
!
41
!
58 (Glu)
+1.4
+1.7
+1.5 (His)
+25
+28
+13
(Lys)
!
25
!
18.4
!
10.1
:total charge
Suppose that you ran a mixture of these three proteins on an ion exchange
chromatography column in which the column beads are positively-charged
. Which protein
would elute last
from the column? Explain how you decided.
Answer:
The protein with the greatest net negative charge will elute last from the column.
Estimate the net charge on each protein by assuming that all Arg and Lys residues
are positively charged and all Asp and Glu are negatively charged, at pH 7. Each His has
an average charge of about +0.1 at pH 7, and each Cys has charge of !
0.1. The Tyr
residues can be ignored because they are almost completely protonated at pH = 7 and do
not contribute any significant charge. See estimates of the net charges given above.
Protein A
elutes last from the column since it has the greatest net negative
charge.
4
4. (5 points)
Here are the last four proteins. The table contents are the same as for the previous
two questions.
protein I
protein II
protein III
protein IV
Amino acid
#
#
#
#
Alanine
59
28
47
8
Arginine
36
20
16
16
Asparagine
34
13
30
6
Aspartic acid
47
16
24
15
Cysteine
12
1
14
1
Glutamic acid
42
21
35
21
Glutamine
28
16
13
7
Glycine
72
35
34
19
Histidine
24
2
18
8
Isoleucine
56
27
24
11
Leucine
64
43
49
30
Lysine
52
18
38
13
Methionine
10
8
11
9
Phenylalanine
23
8
27
13
Proline
42
11
26
18
Serine
38
14
49
17
Threonine
45
9
24
15
Tryptophan
9
7
3
2
Tyrosine
20
6
21
11
Valine
41
18
45
17
# residues:
754
321
548
257
# chains:
1
1
1
1
Suppose you have four aqueous solutions, each of which contains only one of the four
proteins. The protein concentrations are the same in the four solutions.
You measure the absorbance at 280 nm (A
280
)
of each solution.
Which protein solution would have the greatest
A
280
? Explain briefly how you decided.
Answer:
The relation of protein concentration (c) to A
280
is given by Beer’s law:
c = A
280
/
ε
280
.
The ε
280
of a protein is due to absorbance by the Trp and Tyr residues in the
protein, and Trp contributes greater absorbance than does Tyr.
The solution that contains Protein I would have the greatest A
280
.
5
5. (15 points)
Suppose that you have four aqueous solutions that each contain a different amino
acid:
Solution 1:
Aspartic Acid
pH = 2
Solution 3:
Cysteine pH = 9
Solution 2:
Histidine
pH = 9
Solution 4:
Tyrosine pH = 7
Each solution is buffered to the given pH value. The total concentration of amino acid is the
same in each solution.
Consider ONLY the side chains
of these amino acids in answering the following
questions.
i)
Which amino acid side chain has the largest
value of the fraction dissociated
in these
solutions? Show your work or explain briefly how you decided.
Answer:
The fraction dissociated of an acid is greatest when the pH is above the pKa of the
acid.
The pH of solution 2 (pH = 9) is above the pKa of the acid (His side chain,
pKa = 6) by the greatest amount, of the four solutions, so histidine
(solution 2) has the
largest fraction dissociated.
ii)
For which amino acid is the ratio [A-]/[HA]
the smallest
? Show your work or explain
briefly how you decided.
Answer:
The [A
G
]/[HA] ratio is small when pH is below the pKa, so the [A
G
] is small and
[HA] is large.
The pH of solution 4 (pH 7) is below the pKa of the acid (Tyr side chain,
pKa = 10.07) by the greatest amount, so tyrosine
(solution 4) has the smallest [A-]/[HA]
ratio.
iii)
Consider the total charge
in each solution (due only to the charges on the side
chains). Which solution has greatest overall charge
? Show your work or explain briefly how
you decided.
Answer:
The charge on an acid depends on its fraction dissociated (i.e., pKa relative to
pH), AND on whether the dissociated form (conjugate base) or the undissociated form
(the acidic form) has the charge.
The answer is cysteine
(solution 3): the pH is greater than the pKa of cysteine and
the conjugate base is charged.
You can also calculate exact answers to these questions using equations covered in class:
aa
pH
pK
fr. dissoc
[A-]/[HA]
charge
Asp
2
3.65
0.022
0.022
-0.022
His
9
6
0.999
1000
+0.001
Cys
9
8.18
0.87
6.6
-0.87
Tyr
7
10.07
0.00085
0.00085
-0.00085
6
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6. (12 points)
i) (3 pts)
The figure, below, is a MALDI mass spectrum obtained with a purified
protein. This particular protein sample gave three peaks in the mass spectrum. The m
/
z
values of
the three peaks are shown on the
figure. Calculate the molecular
weight (MW) of the pure protein
that was used in the experiment.
Show your work or explain briefly
how you arrived at your answer.
Answer:
The most likely interpretation of
the MALDI spectrum is that the
protein ions in the three peaks have
charges (
z
) of +1, +2, and +3 (see
the figure).
So, the molecular weight of the protein is MW = 78,000
.
ii)
The figure, below, shows an ESI mass spectrum that was obtained with the same
purified protein
as was used for the mass spectrum shown in part (i), above.
ii.a) (5 pts)
The (m/z) value is given
for one protein ion in the ESI spectrum.
Calculate the value of the charge (
z
) on that
same protein ion. Show your work.
Answer:
Use (
m
/
z
) = [MW + n(H)]/n (H = 1.008)
Solve algebraically for n (= z
for that peak):
n = MW /((
m
/
z
) - H)
n = 78,000/2600 =
30
ii.b) (4 pts)
Calculate the (
m
/
z
) value and the charge (
z
) for the unknown protein ion in
the peak that is indicated by the “?” in the figure. Show your work.
Answer:
The charge on peaks in an ESI mass spectrum increases from right to left. The ion
in the unknown peak has n = 33
.
The (
m
/
z
) of the unknown peak can be calculated from (
m
/
z
) = [MW + n(H)]/n :
(
m
/
z
)
= [78,000 + 33*1.008]/33 = 2364.64
7
7. (8 points)
The figures given below, left, show parts of a peptide, drawn so that the Phi or Psi
angle can be estimated. The other figure is the Ramachandran plot that was shown in class.
i)
Identify one pair of figures (one figure
that shows Phi and one figure that shows Psi) that
would be for an amino acid residue that is in the
forbidden
part of the Ramachandran plot. Explain,
briefly
, how you decided.
Answer:
Estimate the angle values from the figures:
Fig
angle
Fig
angle
1 Phi .
!
60
E
5
Psi .
+150
E
2 Phi .
!
80
E
6
Psi .
!
150
E
3 Phi .
!
100
E
7
Psi .
!
45
E
4 Phi .
+120
E
8
Psi .
+45
E
The angle in Figure 4
(Phi .
+120
E
) is in the forbidden part of the plot,
regardless of the Psi angle.
The angle in Figure 6
(Psi .
!
150
E
) is in the forbidden part of the plot,
regardless of the Phi angle.
ii)
Identify one pair of figures (one Phi, one Psi) for a residue that could be in an alpha-
helix
. Explain, briefly
, how you decided.
Answer:
Figure 1
(Phi .
!
60
E
) and Figure 7
(Psi .
!
45
E
) are closest to the area on the
plot where an alpha-helix is found.
8
8. (8 points)
The amino acid sequence written below is part of the human β
-globin protein. The
arrows indicate four hypothetical mutations that might occur in the protein, where a different
amino acid is substituted for the one that is normally found at the indicated positions in the
sequence.
Î
Arg
Ï
Asn
ƒ
ƒ
Met-Val-His-Leu-Thr-Pro-Glu-Glu-Lys-Ser-Ala-Val-Thr-Ala-Leu-Trp-Gly-Lys-Val-
Ð
Ser
ƒ
Asn-Val-Asp-Glu-Val-Gly-Gly-Glu-Ala-Leu-Gly-Arg-Leu-Leu-Val-Val-Tyr-Pro-Trp-
„
Ñ
Met Amino acid sequence changes (i.e., mutations) can affect the stability of a folded protein,
and can cause a protein to unfold under physiological conditions.
Which one of the four mutations shown above would you predict would have the greatest
effect on the stability of the folded protein? Explain how you decided.
Answer:
A mutation in which the mutated amino acid is very different from the original
amino acid will have the greatest effect on the stability of the folded protein. The four
mutations:
Î
change one positively-charged residue (Lys) to another (Arg),
Ï
change a small, polar and uncharged residue (Thr) to another polar,
uncharged, residue (Asn), Ð
change a hydrophobic residue (Val) to a smaller polar, uncharged, residue
(Ser).
Ñ
change a large non-polar (hydrophobic) residue (Leu) to another non-polar
residue (Met).
Mutation Ð
(Val changed to Ser) would have the greatest effect on the stability of the
folded protein. The Val is most likely in the interior of the folded protein. Ser is more
polar and soluble in water and would want to remain on the surface of the protein where
it could interact favorably with water. This would prevent the protein from folding into
its normal structure. If the protein were to fold with the Ser in the interior, that is likely to
be unstable since it is unlikely that there would be hydrogen-bonding groups in the
vicinity of the Ser, since Val does not form hydrogen bonds.
9
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9. points (17 pts)
i) (5 pts)
Suppose that the peptide shown below were in an alpha-helix
. One
main chain carbonyl group is indicated by an arrow. Draw an arrow to indicate the specific atom
to which the indicated carbonyl group forms a hydrogen bond in the alpha helix. Explain briefly
how you decided.
Answer:
There are 3.6 amino acid residues in one turn of an alpha helix - i.e., more than 3
but less than 4. The carbonyl group of residue i
(i.e., the carbonyl indicated by the arrow)
would form a H-bond to the N
!
H group of residue i
+ 4 - see the arrow on the right side
in the figure.
ii) (6 pts)
Draw a circle around one group of atoms that all lie in the same plane with
each other in the structure drawn below.
Explain briefly why those atoms are co-planar.
Answer:
The atoms in the peptide bond and the two C-alpha bound to the carbonyl-C and
the N
!
H are all constrained to be in the same plane. One set of these six atoms is circled
in the figure, above.
These atoms are co-planar due to the fact that the carbonyl-C, carbonyl-O, and N
are all sp
2
hybridized, and the electrons are delocalized in the p-orbitals on those atoms.
The indicated six atoms must be co-planar to allow that delocalization of electrons.
Question #9 continues on the next page . . .
10
Question #9, continued
iii) (3 pts)
Draw a box around all atoms that constitute one complete amino acid residue
in the figure, below.
iv) (3 pts)
Draw arrows in the figure, below, to label the bonds that you must use to
determine the value of the Phi
angle.
11
BCHM461
‐
0102 (MW)
12
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Use the References to access important values if needed for this question.
The compound cyclohexanol, C6H120, is a good fuel. It is a liquid at ordinary temperatures. When the liquid is burned, the reaction involved is
2 C6H120(t) + 17 02(g)→12 CO2(g) + 12 H20(g)
The standard enthalpy of formation of liquid cyclohexanol at 25 °C is -348.2 kJ mol-1; other relevant enthalpy of formatton values in kJ mol1 are:
C6H120(g) = -286.2 ; CO2(g) = -393.5 ; H20(g) = -241.8
(a) Calculate the enthalpy change in the burning of 3.000 mol liquid cyclohexanol to form gaseous products at 25°C. State explicitly whether the reaction
is endothermic or exothermic.
= oHV
(b) Would more or less heat be evolved if gaseous cyclohexanol were burned…
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(a) The population of a large city grows very quickly for several years. Then the growth begins to slow, and the population remains constant for several
years. After this, the population grows again, but slower than before.
Population
Population
Population
Population
Time
Time
Time
Time
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i think this is physics and chemistry so i hope you can help me to answer this, only TRUE OR FALSE
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ter 9- CHEM A054 520 2 x Question 12 - Chapter 9 part 2 X
с
Chapter 9 part 2 Homework i
12
1
points
ezto.mheducation.com/ext/map/index.html?_con=con&external_browser=0&launchUrl=https%253A%252F%25
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CH6_Chem103 - Kenai Peninsu X +
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Determine the freezing point of an aqueous solution that is 0.584 m sucrose.
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Hawaii Carbon Dioxide Time-Series
425
CO2 Time Series in the North Pacific
8.30
Mauna Loa Atmospheric CO2 (ppm)
ALOHA seawater PCO2 insitu (uatm)
ALOHA seawater pH (insitu)
400-
160°W 158°W 156°W
8.25
23°N
Station ALOHA
22
375-
- 8.20
21°
20°
Station Mauna Loa
8 350-
190
-8.15
325
8.10
300-
8.05
275-
8.00
2021
1958
1967
1976
1985
1994
Year"
Data Mauna Loa (php.ondi noan ovproductsrendaeeee n mlo iN) ALOHA phahana soest hawaieduttdogstexionhl)
ALOHA pi &C0, e calulated a insitu tempere fhom DIC A TA (measured from samples collected on Hawai Ocean Times-series IOT) enuisen)
ing coys (Pelletier, v2so) wih constants Lueker et al. 2000, KSO Dickson, Total boron Lee et al. 2010, & KF secarb
2003
2012
What problem in the oceans does the data collected
from 1985 to 2021 show?
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Kayley Valentine
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AALEKS-Kayley Valentin x
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E Apps
HPrechecks: Test 2.
O MATTER
Calculating mass density
Try again.
Kayley v
A chemist v v
as a safety inspector finds an unmarked bottle in a lab cabinet. A note on the door of the cabinet says the cabinet is used to store bottles of
pentane, dimethyl sulfoxide, acetone, methyl acetate, and chloroform.
The chemist plans to try to identify the unknown liquid by measuring the density and comparing to known densities. First, from his collection of Material Safety
Data Sheets (MSDS), the chemist finds the following information:
liquid
density
0.63
mL
pentane
db
dimethyl sulfoxide
1.1
0.79
acetone
0.93
mL
methyl acetate
1.5
chloroform
3
Next, the chemist measures the volume of the unknown liquid as 1797.…
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因
+ II
P
8
团
w/
S - Cameron Hite Learn
b Chemistry Questions & Answers
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A www-awu.aleks.com/alekscgi/x/Isl.exe/1o_u-IgNslkr7j8P3jH-IVDWKVW_BBZZI6tTytly4Fcfu6zOtOf8oMM9sLjOBhHfJ6ZNhysAi_9KCqM8FRGCtiPxvdUuxrTTETxcEo9QVTr9NhEkLqSLo?1oBw7QYjlbav. E
O CHEMICAL REACTIONS
0/3
Cameron V
Solving for a reactant in solution
One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride
anions in solution will combine with the silver cations to produce bright white silver chloride precipitate.
Suppose an EPA chemist tests a 200. mL sample of groundwater known to be contaminated with tin(II) chloride, which would react with silver nitrate solution
like this:
SnCl,(aq) + 2 AgNO,(aq) → 2 AgCl(s) +
Sn(NO,),(aq)
The chemist adds 23.0 mM silver nitrate solution to the sample until silver chloride stops forming. He then washes, dries, and weighs the precipitate. He finds he
has collected
6.8
of silver…
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Madeline Carlo - CHM 112 HW X
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A docs.google.com/document/d/10m8ZGoSGXKFFcl4KWhwOLh8bFpbtPAmvTWjJPnj15U4/edit
Madeline Carlo - CHM 112 HW DUE Monday April 12, 2021 ☆ D
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I 7 II I|
.1. Refer to the table of standard reduction potentials.
Standard Reduction Potentials at 25 C
Reduction Half-Reaction
(V)
FAg + 2 €
H2O{aq) + 2 H*(aq)+ 2 6
Mno,taq) + 8 H*(aq) + 5 &
Clalg) + 2 e
CrOFMaq)+ 14 H*(aq) + 6 6-
Ozlg + 4 H*(aq) + 4 &
Bra() + 26
Ag*(aq) + e
Fe*(aq) + e
Ozl9) + 2 H'(aq) + 2 €
2(s)+ 26
Ozlg) + 2 HO) + 46
Cu (aq) + 2 €
Sn" (ag)+ 2 6
2 P(aq)
2 H2O()
→ Mn(aq) + 4 H;0()
→2 cr(aq)
2 Cr (aq)+ 7 H0()
→2 HO()
2 Br(aq)
Ag(s)
» Fe*(aq)
→H;Oz(aq)
→2 F(aq)
→4 OH"(aq)
2.87
1.78
1.51
1.36
1.33
3.
1.23
1.09
0.80
0.77…
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STPKsYapxmOBQpqJ9sYHEQf
udwLFoPO2ojmMGw/edit
s Help
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B
U A
G O O- E E E E E E - E - E E
14.5
3 4 5 6
You are asked to perform an experiment to determine the density of
square block below. In order to complete your calculations you will need
to
(AKS 1b, DOK1).
Question options:
1. Use the electronic balance to determine the mass of the block.
2. Use water displacement to determine the volume of the block.
bD
1. Use the electronic balance to determine mass.
2. Use ruler to measure the length, width, and height to determine
volume.
C=
1. Use a pipet to determine the mass of the block.
2. Use water displacement to determine to volume of the block.
Pr
Au
立
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Real-World Chemistry
1. Suppose you wanted to ship bananas from Puerto Rico to New York and the time required for
shipment was 8 days. Will you choose to ship ripe bananas or green bananas? Why?
2. You may have heard the saying "One bad apple will spoil the whole barrel." Based on the results of |
your experiment, do you think the statement may be true? Why?
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MC Kimetsu no Yaiba - Di.
As you work on the questions, have your lab manual and text book handy.
If you have questions go to one of the office hours and remember, do not leave it for the last
minute!
In the titration of 25.00 mL of a water sample, it took 19.090 mL of 3.165x 10-3 M EDTA solution to reach the
endpoint.
Calculate the number of moles of EDTA required to titrate the water sample. (enter your answer with 3 significant
figures)
6.04x10-5
You are correct.
Your receipt no. is 151-9969 ?
Previous Tries
The total hardness is due to one or a combination of Ca2+, Mg2+, and Fe2+ in your sample. It is convenient to express
this hardness as though it was entirely due to Ca2+. Making this assumption, determine the number of moles of Ca2+
present in the bottled water sample titrated. (enter your answer with 3 significant figures)
6.04x10-5
You are…
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.5.
For a reaction in which E° = -0.100 V and n = 1 mol electrons / mol, calculate AG°. State
+
your assumptions, if any, about temperature.
.6.
For a reaction in which E° = 0.100 V andn = 1 mol electrons / mol, calculate AG° at 25°C.
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MnO, (aq) + 8H*(aq) + 5e → Mn²*(aq) + 4 H,O(1)
Al(s) → Alš*(aq) + 3 e
+
.ЗА.
Assign oxidation numbers to all metals whether they be elemental, monatomic ion, or in
a polyatomic ion, in questions (1) and (2),
.3B. Tell what is being oxidized and what is being reduced in each reaction in questions (1) and
(2).
.3C. Identify each half-reaction in question (2) as oxidation half-reaction or reduction
half-reaction.
.4. Do you need to know the temperature to calculate AG° from E°? Explain.
>
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I am not asking for help in writing this report.
Could you solve the calculation parts for the steps on the last page? like 13, 14, 15, 17, 18, and 19.
This experiment is based on this video: https://www.youtube.com/watch?v=SVvGtcFhdls
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