McClicker
McClicker
McClicker Questions for the Semester - if you don't know the answer, go to the class notes for that day and find the McClicker question.
Order is now in REVERSE. Most recent at the top.
Refer to Question on classroom screen.
A.
B.
C.
D.
E.
explanation:
Dr. McCord described a "fix" for his 1977 Camaro for jump starting it. He duct taped 8 D-cell batteries together so the voltage was 1.5 × 8 = 12 Volts. He then jump started his Camaro. Did this work?
A. Not at all - an epic fail.
B. Worked like a charm - car started right up.
explanation:
This will not work do to the fact that alkaline cells in a single series can NOT supply enough current to start a car.
What is the potential in millivolts (mV) of a concentration cell made with Cu2+/Cu half cells where the cathodic compartment has a concetration of 1.2 M and the anodic compartment has 0.060 M?
A. –77.0 mV
B. –154 mV
C. +38.5 mV
D. +379 mV
E. +77.0 mV
F. –38.5 mV
G. +154 mV
H. +340 mV
explanation:
Use Nernst equation. E = 0 - (0.05916/2)log(0.06/1.2) = 0.0385 V = +38.5 mV
How many grams of copper is plated after we pass 4.85 A current for 4 hours in a copper(II) sulfate solution?
explanation:
4.85(4·3600)/(2·96485) = 0.362 mol Cu
0.362(63.55) = 23 g Cu
What is the E° for this cell?
Bi(s) | Bi3+(aq) || Fe2+(aq) | Fe(s)
A. +0.22 V
B. +0.64 V
C. –0.22 V
D. –0.97 V
E. –0.64 V
F. +0.97 V
explanation:
Ecell = –0.44 V – +0.20 V = –0.64 V
Balance the reaction... or at least enough to answer the question.
CrO42– + M+ → Cr3+ + M3+
After fully balancing this equation, how many total electrons are being transferred in the over all equation?
A. 5
B. 8
C. 10
D. 12
E. 15
F. 3
G. 4
H. 6
I. 2
explanation:
The fully balanced reaction is:
16H+ + 2CrO42– + 3M+ → 2Cr3+ + 3M3+ + 8H2O
The chromium changes from +6 to +3, a 3 electron transfer. The M changes from +1 to +3 which is a 2 electron transfer. The only way to match them up is to have a total electron transfer of 6 electrons.
Consider the five changes in oxidation state for five different elements. Which change in state corresponds to a reduction?
A. Fe to Fe2+
B. Cu+ to Cu2+
C. Cl– to Cl
D. Zn2+ to Zn
E. O2– to O–
explanation:
The oxidation number increases for all of these except the Zn one. It starts at +2 and DROPS to 0 (zero). Oxidation number dropping (or decrease) is a reduction.
A rare acid what is the oxidation number for the element 😎 in the anion H😎O3–?
A. –5
B. –4
C. –3
D. –2
E. –1
F. 0
G. +1
H. +2
I. +3
J. +4
K. +5
L. +6
explanation:
Consider the rather simple reaction that occurs in one simple step:
A → 2B
The rate constant is equal to 0.0383 min–1. If the initial concentration of A is 0.15 M, what is the concentration of product B after 42 minutes?
A. 0.30 M
B. 0.030 M
C. 0.12 M
D. 0.060 M
E. 0.18 M
F. 0.24 M
G. 0.15 M
H. 0.32 M
explanation:
Use 1st order integrated rate law and solve for final concentration of A. ln(0.15/[A]) = 0.0383(42)
[A] = 0.03 M
This means that 0.12 M of A completely reacted and turned into B. But, there is a doubling due to the coefficient of 2 for B. That means that 2(0.12) or 0.24 M B is formed at 42 minutes.
Which statement is NOT true about a catalyst?
A. A catalyst alters the reaction mechanism.
B. A catalyst will change the free energies of the reactants and products so as to speed up the reaction.
C. A catalyst will enter the mechanism on a given step and exit the mechanism on a later step.
D. A catalyst lowers the activation energy for a reaction.
E. A catalyst can be heterogenous or homogeneous with the reaction mixture.
explanation:
All are true except B. Nothing will change the fundamental (standard) energies of the reactants and products.
A sample containing biologically derived C-14 was analyzed. The amount of C-14 was shown to be 42.0% as the amount in current bio samples containing C-14. What is the approximate age of this sample? The half-life of C-14 is 5730 years.
A. 6780 yr
B. 7170 yr
C. 9120 yr
D. 8120 yr
E. 5830 yr
explanation:
k = ln2/thalf = 1.21×10-4 yr-1
ln(100/42)/1.21×10-4 = 7169.4 yr
Consider the rxn: 2A + 3B → C + 4D
If the rate of B is -0.12 M/s, what is the correct rate for D?
A. –0.16 M/s
B. +0.090 M/s
C. –0.090 M/s
D. +0.040 M/s
E. +0.16 M/s
F. –0.040 M/s
G. +0.10 M/s
H. +0.24 M/s
I. –0.24 M/s
J. –0.10 M/s
explanation:
stoichiometry shows that you get 4 D's for every 3 B's reacted. Therefore you multiply the B rate by 4/3 to get +0.16 M/s. The signs are obvious, or should be... you are losing B as it reacts (negative rate of change) and D is appearing as the reaction proceeds which is a positive rate of change.25 mL of 0.0035 M Sr(NO3)2 is mixed with 75 mL of 0.036 M Na2SO4. What happens? Ksp for SrSO4 is 3.2×10–7.
A. A precipitate of SrSO4 will form.
B. A precipitate of NaNO3 will form.
C. Nothing will happen.
explanation:
Get Qsp and compare to Ksp.
25(.0035)/100 = 0.000875 M Sc2+
75(0.036)/100 = 0.027 M SO42-
Q = (0.000875)(0.027) = 2.36×10–5.
So Qsp >> Ksp which means the reaction goes in reverse which in this case is a precipitation of SrSO4.
A slightly soluble salt (ZrX2) was allowed to saturate an aqueous solution according to the following reaction:
ZrX2(s) ⇌ Zr2+(aq) + 2X–(aq)
After saturation, the concentration of Zr2+ was found to be 182 ppm. What is the value of Ksp for ZrX2?
Here’s a link to our Periodic Table.
A. 1.6 × 10-11
B. 3.2 × 10-8
C. 7.9 × 10-9
D. 4.0 × 10-6
E. 2.1 × 10-7
explanation:
Convert to molarity: 0.182 g/L (1 mol Zr / 91.22 g Zr) = 2.0×10-3 or 0.0020 M
A 1:2 salt will have Ksp = 4x3 = 4(0.002)3 = 3.2×10–8
A. 4.0 × 10–3 M
B. 4.3 × 10–4 M
C. 8.2 × 10–4 M
D. 3.0 × 10–3 M
E. 4.7 × 10–5 M
F. 5.2 × 10–3 M
explanation:
MX3 ⇌ M3+ + 3X–
Ksp = [M][X]3, from RICE table Ksp = (x)(3x)3 = 27x4
x = (Ksp/27)1/4 = (2.2 × 10–9/27)0.25 = 3.0 × 10-3 M
If I mix a solution of an acid (HX) at a concentration of 0.0042 M and the pH is 3.44, what is the percent ionization?
A. 8.6 %
B. 5.8 %
C. 10 %
D. 6.5 %
E. 2.7 %
explanation:
The pH means that [H+] = 10-3.44 = 3.63 × 10-4. Now find the percentage of that vs the original concentration:
%ion = 3.63 × 10-4/0.0042 × 100% = 8.6 %
A short protein chain has two side chains on it that are affected by pH. One is an acid with a pKa = 3.90, and the other is a base with pKb = 3.46. If this protein is in a solution with a pH of 7.60, what is the over all charge on this protein?
A. –2
B. –1
C. 0
D. +1
E. +2
explanation:
The pH of 7.60 is well above the pKa for the acid and it is therefore DEprotonated and has a -1 charge. Converting the pKb to the pKa of the conjugate we get 10.54. Because pH 7.60 is well below that, the base will be protonated and have a charge of +1. Combining those two charges results in ZERO charge.
0.80 M propylamine
(pH = 12.24)
0.50 M HCl
(pH = 0.30)
Equal volumes of the two solutions shown in the diagram are completely mixed together. What is the final pH of the solution after the mixing?
Propylamine has a Kb equal to 3.7 × 10–4 (pKb = 3.43).
explanation:
Easiest to pick an easy volume like 10 mL. So if both are mixed at 10 mL each, that means that 8 mmol of B and 5 mmol of HCl are mixed. The reaction is
B + HCl -> BH+ + Cl-
This is like a titration of a weak base with a strong acid and the acid is the limiting reactant - so the titration/neutralization is not yet complete. The 5 mmol of acid converts 5 mmol of B to BH+. Do the stoichiometry and you get 5 mmol of BH+ (an acid) and 3 mmol of B (a base) left in solution. We are parked in the buffer zone of this weak base. So the Henderson-Hasselbalch equation is our best bet.
\({\rm pH = pKa + \log\left({[base]\over [acid]}\right)}\)
pKa for BH+ is 14-3.43 = 10.57
Subbing into the H-H equation:
pH = \(10.57 +\log\left({3\over 5}\right)\)
pH = \(10.57 - 0.22 = 10.35\)
So the pH is equal to 10.35 - Thank you, Tate McRae!
What is the value of Kb for cyanide ion?
Use the Table of Ka Values on gchem site.
explanation:
Use Kb = Kw/Ka = 10e-14/6.2e-10 = 1.6e-5
(Part 3) 240 mL of 0.063 M HCl is added to 60.0 mL of the buffer from Part 1 (Buffer = 0.25 M trimethylamine, (CH3)3N, and 0.25 M trimethylammonium chloride, (CH3)3NHCl). What is the pH of the buffer solution now?
Kb trimethylamine = 6.5 × 10–5
A. 2.88
B. 3.40
C. 4.18
D. 7.00
E. 9.02
F. 9.54
G. 9.81
H. 10.08
I. 10.60
explanation:
240mL(0.063M) = 15.12 mmol of H+ added. This completely neutralizes ALL of the base, B, to BH+. And there are still 0.12 mmol of H+ left over in a total of 300 mL solution (the 60 mL + 240 mL). Now just calc the [H+]:
0.12mmol/300mL = 0.00040 M [H+], take -LOG and get the answer of 3.40 for pH.
(Part 2) The buffer from Part 1 is 200 mL of 0.25 M trimethylamine, (CH3)3N, and 0.25 M trimethylammonium chloride, (CH3)3NHCl. 50 mL of 0.30 M HNO3 is added to this soluion. What is the pH of the buffer solution now?
Kb trimethylamine = 6.5 × 10–5
A. 2.88
B. 3.40
C. 4.18
D. 7.00
E. 9.02
F. 9.54
G. 9.81
H. 10.08
I. 10.60
explanation:
Do the stoichiometry with mmol. 200 mL of 0.50M each of B and BH+ is 50 mmol each. The added strong acid is 50mL(0.30M) = 15 mmol H+. If you do the RICE table and put in 15 for x... you'll lose 15 from the base (B) and gain 15 on the acid (BH+). So that means you'll have 35 mmol of B and 65 mmol of BH+. Now plug into the H-H equation:
pH = 9.81 + log(35/65) = 9.81 – 0.27 = 9.54 (the answer)
(Part 1) A "perfect" buffer is made by mixing 100 mL of 0.50 M trimethylamine, (CH3)3N, and 100 mL of 0.50 M trimethylammonium chloride, (CH3)3NHCl. What is the pH of this perfect buffer?
Kb trimethylamine = 6.5 × 10–5
A. 2.88
B. 3.40
C. 4.18
D. 7.00
E. 9.02
F. 9.54
G. 9.81
H. 10.08
I. 10.60
explanation:
Get mmol of each species:
0.5M(100mL) = 50 mmol B (and same for BH+)
So the ratio is 50/50 or 1. Therefore the pH = pKa for the acid species.
pKb = -log(6.5 × 10–5) = 4.19
pKa = 14 - pKb = 14-4.19 = 9.81 which is the pH here
(Part 2 of 2) After nailing the end point on the titration, Steve then just added an additional arbitrary amount of the acid solution. He added 2.30 mL more of the HCl solution. What is the new pH after this addition?
Note: Remember, in part 1 he had already added 25 mL of the acid to 100 mL of the 0.10 M NH3 ( Kb = 1.8 × 10‑5 ) solution.
explanation:
This is known as overshoot. He overshot the endpoint by 2.30 mL which amount to 2.3(.4) = 0.92 mmol extra H+. There is NO base to react with, so that IS the amount of H+ in 125+2.3mL = 127.3 mL of solution. Do the math:
0.92/127.3 = 0.007227 M [H+]
pH = -log(0.007227) = 2.14
❤️ Happy Valentine's Day! 2/14 ❤️
(Part 1 of 2) Steve decided to titrate 100 mL of 0.10 M ammonia ( NH3, Kb = 1.8 × 10–5 ). He used 0.40 M HCl as the titrant. He perfectly nailed the end point. What is the pH at the end point?
explanation:
Conjugate acid (NH4+) has \(K_{\rm a} = K_{\rm w}/K_{\rm b} = 5.55\times 10^{-10}\)
The amount of base is 10 mmol and the needed amount of acid solution is 25 mL (25(.4) = 10 mmol HCl). So at the end point (equivalence point because he nailed it) there are 10 mmol of conjugate acid (NH4+, Ka = 5.55×10–10) in 125 mL of solution. That is a concentration of 10/125 which is 0.080 M weak acid, BH+. You CAN use the assumption here to solve for concetration of protons:
[H+] = \(\sqrt{5.55\times 10^{-10}(0.08)}\) = 6.67×10–6
pH = -log(6.67×10–6) = 5.18
Consider 2 beakers with the following contents:
Beaker 1: 600 mL of 0.025 M HCl
Beaker 2: 400 mL of 0.038 M NaOH
Now beaker 2 contents are poured into beaker 1 and mixed. What is the resulting pH?
explanation:
600mL(0.025 M) = 15.0 mmol of H+
400mL(0.038 M) = 15.2 mmol OH–
After neutralization (acid is limiting reactant) there are still 0.20 mmol of OH– left over in 1000 mL
0.2 mmol/1000mL = 0.00020 M OH–. Take -LOG and get 3.70 for pOH. Subtract from 14 and get 10.30 for pH.
What is the pH of a 0.035 M solution of potassium benzoate?
A. 5.63
B. 8.37
C. 9.04
D. 7.96
E. 10.51
explanation:
First, realize you need the Ka for the weak acid called benzoic acid which is the parent acid of the conj ion, benzoate. Ka for benzoic acid is 6.4×10–5.
Now convert that Ka into a Kb:
10–14/6.4×10–5 = 1.5625×10–10 = Kb
The shortcut assumption will work here which is
[OH-] = (Kb · CA-)1/2 = (1.5625×10–10 · 0.035)1/2
[OH-] = 2.339×10–6
pOH = 5.63
pH = 8.37
Which of the following is not a strong acid?
A. hydrochloric acid
B. chloric acid
C. hypochlorous acid
D. nitric acid
E. hydrobromic acid
explanation:
hypochlorous acid (HClO) is not a strong acid
An exothermic reaction that is currently at equilibrium is heated up by around 15 °C. Which of the following is true?
A. now K has decreased and the reaction shifts to the LEFT
B. now K has increased and the reaction shifts to the RIGHT
C. Q = K and you are still at equilibrium
D. now K has decreased and the reaction shifts to the RIGHT
E. now K has increased and the reaction shifts to the LEFT
explanation:
Being exothermic, HEAT is written on the product side of the reaction. This means that a temperature increase will push the reaction to the LEFT. Going left is an increase in reactants and a decrease in products which means that K will decrease.
Consider a simple endothermic reaction, A ⇌ B , that is currently at equilibrium at 25 °C. I now heat the reaction vessel up a bit. What is the response of the reaction, if any?
A. it will proceed forward
B. it will go in reverse
C. it will remain unchanged
explanation:
heating an endothermic reaction will always drive it forward
I balanced an equation like this
A ⇌ 2B
and then I determined the value of K to be 0.0625. My friend balanced the same reaction like this
B ⇌ ½ A
What value did he get for K for his version of this reaction?
A. 4
B. 16
C. 0.25
D. 32
E. 0.125
explanation:
His reaction is reverse and 1/2 of mine. Therefore his K is equal to (1/K)1/2 of my K.
friends K = (1/0.0625)1/2 = 4
You want your pasta to cook faster. You decide to add salt to increase the boiling point. You figure raising the boiling point up to 105 °C would do the trick. You will be cooking one pound (453.6 g) of pasta in just over a gallon of water (say 4.00 L of water).
How much salt (NaCl, 58.44 g/mol) will it take to raise the boiling point of water up to 105 °C ?
kb for H2O = 0.512 K/m
ρwater = 1.00 g/mL or 1 kg/L
A. about 1 pound
B. about 2.5 pounds
C. about 1/2 a pound
D. slightly less than a 1/4 pound
E. just over 5 pounds
explanation:
ΔT = i · m · kb
m = ΔT / i · kf
NaCl is a 1:1 electrolyte and therefore i=2
m = 5 / 0.512(2) = 4.88 m solution
You are using 4 L of water which is the same as 4 kg of water
4(4.88) = 19.53 mol of NaCl needed
19.53(58.44) = 1141 g of NaCl
1141/453.6 = 2.5 lbs of NaCl
A liquid has a vapor pressure of 100 Torr at 250 K and 500 Torr at 325 K. What is this liquid's heat of vaporization (ΔHvap)?
A. 1.74 kJ/mol
B. 40.7 kJ/mol
C. 71.8 kJ/mol
D. 14.5 kJ/mol
E. 22.4 kJ/mol
F. 34.5 kJ/mol
explanation:
Use Clausius-Clapeyron Equation.
\(\ln{P_2\over P_1}={\Delta H_{\rm vap}\over R} \left( {1\over T_1} - {1\over T_2} \right) \)
\(\ln{500\over 100}={\Delta H_{\rm vap}\over 8.314} \left( {1\over 250} - {1\over 325} \right) \)
\(14496 \;{\rm J} = \Delta H_{\rm vap} \)
Consider the following reaction which really more of a change of state:
CO2(g) → CO2(aq)
Which of the following is true?
A. ΔH > 0
B. ΔH = 0
C. ΔH < 0
explanation:
Gases have effectively zero IMFs. Once a gas is solvated (aqueous) there are now attractive IMFs in play and therefore the enthalpy is lowered. This means the process has a negative ∆H and is exothermic.
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