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Archive – Q4 week 2 – 21-22

Week of 4/19 – 4/23 – PLEASE REFRESH EVERY TIME!

Please Refresh every time you open– this page is changing often!
The 4 day – A, B, C, D cycle looks like this:
                                                       Day                               Period
                                                                           2                       3                       4       
                
                                   In class:                          Lab                   Lab               Single Class
                                   Remote:                          Lab                   Lab               Single Class
 
                                    In class:         B          Single Class        LAB                   LAB
                                                          Remote:                     Single Class        LAB                   LAB
 
                                          In class:              C          Single Class              LAB                    LAB
                                   Remote:                     Single Class        LAB                    LAB
 
Monday                    In class:          D               Lab                   Lab              Single Class
                                   Remote:                            Lab                   Lab              Single Class
                 
This weeks 5 day Schedule:
 
4/19  –  Monday –  “D” Day    – period 2,3 (Lab)  –  I   2(B,D) 3(D) AP CHEMISTRY
                                                        – period 2,3 (Lab) –  R  2(B,D) 3(D) REMOTE INSTRUCTION
  
                                                             – period 4 – I  3(B) 4(B,D) AP CHEMISTRY
                                                             – period 4 – R 3(B) 4(B,D) REMOTE INSTRUCTION
 
4/20  – Tuesday –  “A” Day         – period 2,3 (Lab) –     2(A,C) 3(A)  AP CHEMISTRY
                                                             – period 2,3 (Lab) –  R   2(A,C) 3(A) REMOTE INSTRUCTION
                                                             – period 4 – I   3(C) 4(A,C) AP CHEMISTRY 
                                                             – period 4 – R  3(C) 4(A,C) REMOTE INSTRUCTION
4/21  – Wednesday –  “B” Day     period 2,  –  I   2(B,D) 3(D) AP CHEMISTRY
                                                              -period 2, –  R   2(B,D) 3(D) REMOTE INSTRUCTION
 
                                                            -period 3,4 (LAB) – I   3(B) 4(B,D) AP CHEMISTRY
                                                            -period 3,4 (LAB) – R  3(B) 4(B,D) REMOTE INSTRUCTION
4/22  –  Thursday-  “C” Day      – period 2, –  I    2(A,C) 3(A)  AP CHEMISTRY
                                                           – period 2, –  R   2(A,C) 3(A) REMOTE INSTRUCTION
                
                                                           -period 3,4 (LAB) – I.  3(C) 4(A,C) AP CHEMISTRY 
                                                           -period 3,4 (LAB) – R  3(C) 4(A,C) REMOTE INSTRUCTION
 
 
4/23  –  Friday –  “D” Day          – period 2,3 (Lab)  –  I   2(B,D) 3(D) AP CHEMISTRY
                                                           – period 2,3 (Lab) –  R  2(B,D) 3(D) REMOTE INSTRUCTION
  
                                                             – period 4 – I  3(B) 4(B,D) AP CHEMISTRY
                                                             – period 4 – R 3(B) 4(B,D) REMOTE INSTRUCTION
 

4/19  –  Monday –  “D” Day       – period 2,3 (Lab)  –  I   2(B,D) 3(D) AP CHEMISTRY
                                                          – period 2,3 (Lab) –  R  2(B,D) 3(D) REMOTE INSTRUCTION
  
                                                          – period 4 – I  3(B) 4(B,D) AP CHEMISTRY
                                                          – period 4 – R 3(B) 4(B,D) REMOTE INSTRUCTION
Zoom Link:
Topic: AP Chemistry – 04.19 – Periods 2,3,4
Time: Apr 19, 2021 08:00 AM Eastern Time (US and Canada)
 
Join Zoom Meeting
https://us02web.zoom.us/j/81279244863?pwd=YmpCaDNPcURwWFhPWkY3OGo5cUJqQT09
 
Meeting ID: 812 7924 4863
Passcode: wjgs0f
One tap mobile
+16465588656,,81279244863#,,,,*404416# US (New York)
 
1. Review this weekends Forms:
This weekends forms keys:
Rate Law Multiple Choice Form:
 
Half-Life Form:
 
2.  Review of the weekends Free Response Homework – 
Kinetics (rate law),Ksp Free Response, 2004B, 2009 FRQ .pdf
View Download
 
 
3. Iodine clock reaction – 
 
    a) rate law constant (k) dependent upon Temperature and Ea
A catalyst lowers the activation energy (Ea) by moving the reaction in a different pathway or MECHANISM that normally the reactants would not travel through.  A lowered activation energy will increase the rate of reaction as more energy is available for effective collisions!
 
I am using an iodine ion catalyst that creates the mechanism listed in last nights homework for the hydrogen peroxide decomposition.
                                                             
4.  Acid Base Basics Lesson (again!!!) 
      Acid Base Basics* 
*. WE DID MOST OF THIS UNIT IN NOVEMBER so all we have left is the equilibrium (ICE tables) part of the unit.
 

        A) Auto Ionization of water:              H2O  <->  H+  +   OH
                                                                    H2O + H2O  <-> H3O+  +  OH
 
        B) Acid Base Definitions (Arrhenius,  Bronsted Lowry,  Lewis)
        C) pH and pOH determination, Kw = [H+/H3O+] [OH] = 1 x 10-14
        D) pKw =pH  + pOH
        E) Kw = Ka x Kb  or pKw = pKa  +  pKb
       F) Strength of Acids = Ka – acid dissociation constant (equilibrium constant for acid conjugate base equilibria)
       G) Strong Bases
5.  Strong Acid / Base Titration:
 
a) Strong Acid ,  Strong Base – No equilibrium here
25.0 ml of HCl is titrated with 0.10 M NaOH. 2.8 second = 1 ml of base added.
                HCl (aq)   +   NaOH(aq)  —> NaCl (aq)  +   HOH (l)
 
                 H+(aq)  +   OH (aq)   —-> HOH (l)
Strong Acid Strong Base Titration Graph.pdf
View Download
 
6.  Weak Acid / Base Titration:
 
 b) Weak Acid, Strong Base – Equilibrium is need to complete the points we did not complete in                                                                 November.
25.0 ml of acetic acid is titrated with 0.10 M NaOH. 2.8 seconds = 1 ml of base added.
           
         HC2H3O2      +       OH      —–>    _________   +   H2O
             ?M, 25 ml              0.10M

 

Weak Acid (Acetic- 25ml) Strong Base (0.12M) .pdf
 
4/19  –  Monday –  “D” Homework:
 
1.   You can use any piece of graph paper or a blank piece of paper if you estimate carefully and will be asked to sketch the titration curve with limited data.  You will be calculating 5 points in a weak acid and strong base titration.
 
These points are:
 
1: initial ph of acid in a beaker
2. max buffering position (and volume).
3. a point between max buffering position and equivalence point.
4. Equivalence point.
5. final ph of solution (after a known amount of base has been added).
 
These 5 points need to be placed on the graph and you will sketch the graph with the supporting calculations for the 5 positions above.
 
Here is an example 1:
Blank:
Acid Base Blank graph for Titration Prediction 1617 p.pdf
View Download
 
 
Please watch the video with me to complete all points!  If you hand this completed 
you will get a 100 on your first test this quarter!  Due tomorrow in class!
Lecture on how to complete this example 1 :
 
In these next 2 Free Response Questions Please THINK ABOUT WHAT POINTS IN THE GRAPH ARE WE WORKING ON!!!
 
 
2.  Please complete the Free Response Question below:
 
Acid Base 3 – 2007 .pdf
Acid Base 3 – 2007 Key grodski & college board keys p.pdf
View Download
 
Lecture that reviews Acid Base 3 -2007.pdf  – specifically the last question using the Henderson Hasselbalch Equation (buffer equation).
3.  Please complete the following FRQ and review with the key
Acid Base 6 – 2003 B.pdf
View Download
Acid Base 6 2003A – key AP Central .pdf
View Download
End of Monday!

4/20  – Tuesday –  “A” Day         – period 2,3 (Lab) –     2(A,C) 3(A)  AP CHEMISTRY
                                                             – period 2,3 (Lab) –  R   2(A,C) 3(A) REMOTE INSTRUCTION
                                                             – period 4 – I   3(C) 4(A,C) AP CHEMISTRY 
                                                             – period 4 – R  3(C) 4(A,C) REMOTE INSTRUCTION
 

Zoom Link:

Topic: AP Chemistry – 04.20 Periods 2,3,4
Time: Apr 20, 2021 07:00 AM Eastern Time (US and Canada)
 
Join Zoom Meeting
https://us02web.zoom.us/j/86385170974?pwd=aWRzUW14QjhTV1ZDN0NaZTFXb05UUT09
 
Meeting ID: 863 8517 0974
Passcode: Cx1Cun
One tap mobile
+16465588656,,86385170974#,,,,*864513# US (New York)
 
 
1. Collect Homework – Titration Graph
 
2. Review Free Response Homework
Acid Base 3 – 2007 Key grodski & college board keys p.pdf
View Download
 
Acid Base 6 2003A – key AP Central .pdf
View Download
3. Buffered solutions again!
 
Buffer discussion –   What are buffer solutions?
 
A buffer solution is one where a conjugate acid and a conjugate base ARE both present in a solution AND they resist pH changes in the solution.
 
*Remember that conjugate acid/base pairs are the SAME chemical plus or minus a proton (H+).  
 
Having an acid (conjugate acid) and a base (conjugate base) will “buffer” OR maintain the pH of a solution BECAUSE the addition of an acid to “buffered solution” will neutralize the addition of an acid as the conjugate base will react with the added acid AND prevent the pH from changing.  If a base is added to a “buffered” solution the conjugate acid will neutralize added base AND prevent the pH from changing.
 
Lets look at ammonia NH3 (weak base) in water:
 
reaction 1:                              NH3      +       H2O    <—–>       NH4+      +      OH
                                             conjugate base                          conjugate acid
Because NH3 is  weak base it will reach equilibrium and both conjugate base  and  conjugate acid  wil be present in solution.  A conjugate acid /base pair will never neutralize each other because their ability to donate a proton (acid) and ability to accept a proton (base) are inversely related to each other in terms of Kw (1 x 10-14).   Lets revisit a Derivation that we have done:
 
Lets start with the NH3 in aqueous solution and view its corresponding equilibrium expression
 
Now lets view the the conjugate acid,  NH4in aqueous solution and its equilibrium expression
 
These 2 chemical reaction are competing when they are in water, remember that water is the chemical that is in largest quantity and that these 2 reaction are really the forward and reverse reaction of reaction 1 above.
 
So lets add them together to get a perspective of how the forward reaction and the reverse are related:

                          Wait?  Isn’t this the auto ionization of water that has a Kw = 1 x 10-14  ???

 
             YES!!! And if we are adding the reactions we are also multiplying the Ka and Kb:
 

   So what this means that as the conjugate acid strength increases its conjugate base strength decreases!! The ability of conjugate acid to act like an acid is inversely related to the ability of the conjugate base to act like a base. 

 
“Stronger the conjugate acid (higher Ka) the weaker its conjugate base (lower Kb)!”
“Weaker the conjugate acid the stronger the conjugate base!”
 
Given the following Ka’s for 2 weak acids, which conjugate base is the strongest?
 
                                                HNO2   —–>    H+    +    NO2–        Ka = 4.6 x 10-4
 
                                                     HF     —–>    H+    +       F           Ka = 3.5 x 10-4
 
answer: F    – This means that F–   ionize water (make into OH) better than NO2– because it is a stronger base than NO2– because its conjugate acid (HF) is weaker than the conjugate acid (HNO2) of NO2 .
 

So now it should be clear why A conjugate acid /base pair will never neutralize each other because their ability to donate a proton (acid) and ability to accept a proton (base) are inversely related to each other in terms of Kw (1 x 10-14)!

 
Because the conjugate acid and base do not react with each other they are available to react with both acid or base that are added to a buffer solution. As long as there is enough of the conjugate acid or the conjugate base the buffer solution will not change it pH if an acid or a base is added.  That is what buffers do!!!!!
 
So if I have the conjugate acid base pair from reaction 1 above the conjugate base will neutralize the acid that is added:  
 
NH3   +   Acid added to buffered solution     —–>    NH4 +   +    conjugate base of acid added.
 
And if I add a base to the buffered solution from reaction 1 above, the the conjugate acid will neutralize the base added :
 
 NH4 +  Base added to buffered solution   —–>   NH3    +  conjugate acid of base added.
 
                  So a base or an acid added to this buffered solution will maintain this pH!
 
We calibrate our pH probes with a buffered solution at a pH of 7.  I use the following 

  The conjugate acid is delivered by the salt: 

           potassium phosphate monobasic: 

          KH2PO —->  K+  +  H2PO4-1

   The conjugate base is delivered by the salt: 

          sodium phosphate Dibasic: 

         Na2HPO —–>   2Na+  +  HPO4-2
 
This product contains a mixture of these 2 salts in a plastic capsule that is to be dissolved in 100 ml of water.

Lets look at some titration curves of weak acids that we have titrated or will titrate:

 

-Notice that the points labeled the “Buffered Region” is where a Buffer solution is being made as this is the place in the graphs that both the conjugate acid AND conjugate base are both present.

Notice that the half equivalent points determine the pKa of the acid WHICH also tells us the maximum buffer pH or the median pH range of that particular buffer.  We determine appropriate based on the pKa’s or Ka’s of the conjugate acid!  This means if I need a buffer to maintain a pH at 9 I would use a buffer solution with approximately equal amount of  NH4+ and NH3 .

If we titrate “partially” meaning we add enough base to drive the reaction forward enough to produce some conjugate base but do not add enough to reach the equivalence point we create a buffer and we are in the “Buffer Region”!


The best Buffer solution has 

                                    1. Equal amounts of Conjugate Pairs – This would allow for a greater range                                                           buffering ability. Think about the half- equivalence point.  At that position the                                                                   solution can equally buffer an acidic or basic additions.

                                           2.  Has the largest concentration of Conjugate Pairs – This would allow     
                                                the buffer to RESIST pH changes the most. The greater the moles of the conjugate  
                                                pairs,  the greater amount of acid or base the buffer solution can neutralize before it  
                                                runs out of one the conjugate pairs and pH will drastically change.
 
                                           3.  The pKa of the conjugate acid approximates the pH region you  
                                          want to buffer. – The half equivalence point will be the that region!!!

So we make a buffered solution 3 ways:

                1:  Making an aqueous solution of a conjugate acid or conjugate base .
 
                    It the reaction will go to equilibrium quick making the the other conjugate partner.
 
Step 1 is the Not a Preferred method because the reaction will not make much of the other conjugate ion since these acid and base are weak.
 
              2:  Partial titration – Stoichiometrically drive the reaction to a point that is below the equivalence point so that there is a significant amount of both the conjugate base and conjugate acid.
 
            *3:  Deliver the conjugate base or conjugate acid via a salt!! (students often do not recognize  
                    this way as they forget to look for the ion in the formula that is the conjugate partner).
 
3. TITRATION DEMO’s:
 
H3PO4 Titration Virtual Lab – Demo 1:
 
40.0 ml of an unknown concentration of phosphoric acid was titration with 30.0 ml of 0.10 M sodium hydroxide using a pH meter and a drop-counter. The drop-counter was calibrated at 20.3 drops per milliliter. 3 drops of Phenolphthalein was also added.
Titrating unknown concentration of 40.0 ml of H3PO4 with 100 ml of 0.10 M NaOH
In this titration we are using a polyprotic acid that has multiple Ka’s.
Logger Pro File for the Today’s Titration:
Table 6 PolyP Acid SB Titration good example.cmbl
 
Phosphoric Acid Titration Curve.pdf
View Download

 

Na2CO3 TitrationVirtual Lab – Demo 2:
 
40.0 ml of an unknown concentration of sodium carbonate is titrated with 30.0 ml of 1.00 M hydrochloric acid using a drop counter. The drop counter was calibrated at 20.3 drops per milliliter.
Titrating unknown concentration of 40.0 ml of Na2CO3 with a total 50 ml of 1.0M HCl
In this titration we are using a salt (ionic compound) Na2CO3 that is basic due to the anion CO3-2 having the ability to accept 2 protons (H+) thus the salt is dibasic.
 
Sodium Carbonate Titration Curve.pdf
View Download
 

 

 
4. Voltaic cells – Review with thermodynamics     
 
2010 AP voltaic cell question with scoring guide.pdf – I went over this one!
View Download
 
I introduced the last formula that connected volts from electrochemistry to delta G!
 
If you want to see how it might appear in electrolytic cells: 
Electrolytic cells
2007 – Electrolytic cell question with scoring guide.pdf – 
View Download
 
Electrolytic cells
            1) electrolysis of fused salts
            2) electroplating – calculating with amperage
                  Faraday’s constant, New delta G Formula that ties voltage with Delta G.
 
From your Reference Tables:

 

 Electrolytic – (electroplating/electrolysis)
Non-spontaneous
volts = negative

∆G = POSITIVE


Endothermic:  Electricity  +  Reactants –> Products
Anode = Place of oxidation
Cathode = Place of reduction
Electron flow from Anode to Cathode

 Voltaic/Galvanic Cell – 
Spontaneous
volts = positive

∆G = NEGATIVE


Exothermic:    Reactants –> Products   +   Electricity
Anode = place of oxidation
Cathode = Place of reduction
Electron flow from Anode to Cathode

   
4/20  – Tuesday –  “A” Day   Homework:
Period 4 will only complete the following for the homework:
1. Please complete the FRQ below and review with the key and or the video below:
 
Acid Base 4 – 2005.pdf
Acid Base 4 – 2005 Key grodski and college board keys p.pdf
View Download
Lecture on Acid Base 4 worksheet: 
 
2. The Acid Base Form.
Period 2,3 will complete the following below:
 
1. Complete the following Free Response questions and review with the keys.
 
Voltaic cells
2010 AP voltaic cell question with scoring guide.pdf – 
View Download
 
 
Electrolytic cells
2007 – Electrolytic cell question with scoring guide.pdf – 
View Download
 
Additional Electrochemistry practice:
 
2013 – Electrochemistry Question with scoring guide.pdf
View Download
 
2002 – Acid Base Part 2 question with Grodski and College B keys.pdf
View Download
2. Complete the Acid/Base Form: (Previously released MC questions)
 
Acid/Base MC Form: (Its on Auto-Reply and you have 2 submissions)

Acid Base MC Form 1920

End of Tuesday and the end of the course for period 2,3!

4/21  – Wednesday –  “B” Day     period 2,  –  I   2(B,D) 3(D) AP CHEMISTRY
                                                              -period 2, –  R   2(B,D) 3(D) REMOTE INSTRUCTION
 
                                                            -period 3,4 (LAB) – I   3(B) 4(B,D) AP CHEMISTRY
                                                            -period 3,4 (LAB) – R  3(B) 4(B,D) REMOTE INSTRUCTION
Zoom Link:
Topic: AP Chemistry – 04.21 Periods 2,3,4
Time: Apr 21, 2021 08:00 AM Eastern Time (US and Canada)
Join Zoom Meeting
https://us02web.zoom.us/j/84490352646?pwd=cXhENHk4NGxCN01vdzJ1WDFGelJtdz09
Meeting ID: 844 9035 2646
Passcode: VV8vB7
One tap mobile
+16465588656,,84490352646#,,,,*012860# US (New York)
 
 
Period 2,4:
 
 
 
1. Practice test AP Exam 2 – – 4th quarter Exam Grade – completing a full test for AP Grade of 1 – 5 .
 
Part 1: Multiple Choice (normally 60 questions) – 90 minutes
 
90 minutes for 60 questions = 90 minutes/60 questions = 1.5 minutes per question is YOUR Pace!
 
 Practice test 2 has 50 questions = 50 x 1.5 = 75 minutes needed to complete part 1.
 
Today we do 30 – 35 minutes of 75 minutes.
 
We will now use the newer AP Chemistry Reference table!
2013 AP Chemistry reference tables.pdf
View Download
 
 
 – period 3
 
Complete the course – 
 
Review of 2005 question – Buffer question
Phosphoric acid titration-
strength of acids- 
Electrochemistry- 
 

Spectrophotometry

 

                                
4/21  – Wednesday –  “B” Day – homework:
 
Period 3,4 will complete the following below:
 
1. Complete the following Free Response questions and review with the keys.
 
Voltaic cells
2010 AP voltaic cell question with scoring guide.pdf – 
View Download
 
 
Electrolytic cells
2007 – Electrolytic cell question with scoring guide.pdf – 
View Download
 
Additional Electrochemistry practice:
 
2013 – Electrochemistry Question with scoring guide.pdf
View Download
 
2002 – Acid Base Part 2 question with Grodski and College B keys.pdf
View Download
 
 
 
Period 2 class will complete the following below:
 
Please complete the older part 2 questions that will review spectrophotometry labs.  There is no need to time yourself here.  I would like you to think about these questions and how they relate to the 2 spectrophotometry labs that we completed.  Please read through my notes on how we completed 2 labs using the spectrophotometer.
 
In your reference tables:   A = abc
 
        A = absorbance (- log of Transmittance)  
                 Transmittance is the ratio of light reaching the detector to the light being emitted from the source.
                 The more that light is absorbed, the lower the light hitting the detector and thus the lower the          
                 Transmittance.  This number will be less than one so we perform a .-log on its value to make it positive                      and a larger that 1 value and when we do this it becomes Absorbance.
        a = absorptivity constant (proportionality constant like R for gas laws or k for rate law)
         b = path length (distance that the source light travels to the detector)
        c = Concentration of colored chemical (Molarity chromophore or crystal field)
 
So based on A = abc the Absorbance is dependent upon the concentration of the colored chemical and the distance that the light has to travel through the cuvette.
 
% mass of copper in brass: 
Brass shot (alloy of Cu & Zinc) was dissolved with nitric acid. The solution was blue due to the crystal field splitting of Cu+2 with water. The Zinc in the alloy (Brass) also dissolved but it has no color.  We performed a serial dilution of a known concentration of [ Cu+2 ] and we put theses known concentrations in cuvettes and determined the Absorbance of each known concentration (after we set the spectrophotometer to absorb a certain wavelength of the Cu+2  solution called the peak wavelength).We plotted a Beer’s Law graph (linear relationship of diluted Cu+2 ] with lowered Absorbance. We (logger pro) plotted the graph of measured absorbance with Known concentration (we used the keep button to add the concentrations for each measures absorbance of the known concentrations).
We made a line of best fit and we used the slope and y intercept data to help with the next part
 
We then took the dissolved Cu+2  from the brass shot and made a 100.0ml solution.  We put that in a cuvette and measured the absorbance. Using this absorbance, and the Beer’s Law graph (utilizing linear algebra, y = mx +b from the regression line of the graph) we determined the concentration of the dissolved Cu+2 ].  We used that to calculate moles and then percent by mass.
 
Love this lab!
Determining the order of  reaction (with respect to Crystal Violet) of reaction (CV) with OH_
In this lab used placed a known solution of [CV] with is a chemical indicator (or a chromophore!) with a known concentration of OH into a cuvette in a spectrophotometer that plotted the Absorbance decreases in real time for 30 seconds. We identified the order of the CV while keeping the [OH] constant.  (The concentration of the OH was so much larger than the dilute CV that we also assumed that its concentration did not change significantly.) From the absorbance data we collected, we assumed that the concentration of the CV was proportionate to the changes of the Absorbance. (A = abc and because a and b are fixed values for the experiment we used the concept that A is proportionate to c which is concentration) Remember that over time the cuvette changes from a purple color to clear. How much it turned clear was an indication of the RATE of the reaction!  Using the absorbance values collected for each time interval as the proportionate concentration we changed the absorbance values to ln [A] and 1/[A) respectively and then made a new graph of each.  The graph that was the most linear determined the rate order with respect for CV.
 
Ok I love this lab too!
2004 Free Response question 2 spec.pdf
View Download
 
In the above question (e) the Arrhenius equation used to be part of the course and it was on the old reference tables:
when you look at this equation you get a linear function.
                                                                                             y   =        m       x      +     b

We used this equation to understand the relationship of activation energy (Ea) and temperature with the rate law constant but the equation is not part of course anymore so question (e) no longer is applicable.

 
2004 free response scoring guidelines ques 2 spec.pdf
View Download
 
2006 Free response question 5 – spec.pdf
View Download
 
2006 Free response scoring guidlelines ques 5 spec.pdf
View Download
 
End of Wednesday and the end of the course for Period 4!

4/22  –  Thursday-  “C” Day      – period 2, –  I    2(A,C) 3(A)  AP CHEMISTRY
                                                           – period 2, –  R   2(A,C) 3(A) REMOTE INSTRUCTION
                
                                                           -period 3,4 (LAB) – I.  3(C) 4(A,C) AP CHEMISTRY 
                                                           -period 3,4 (LAB) – R  3(C) 4(A,C) REMOTE INSTRUCTION
 
Zoom Link:
Topic: AP Chemistry – 04.22 – Period 2,3,4
Time: Apr 22, 2021 08:00 AM Eastern Time (US and Canada)
Join Zoom Meeting
https://us02web.zoom.us/j/87538687093?pwd=YUFFTTA0cmR1cVJXZU1pU1N3OHJ1UT09
Meeting ID: 875 3868 7093
Passcode: 8i9eks
One tap mobile
+16465588656,,87538687093#,,,,*486100# US (New York)
 
Period 2:
 
25 minutes of time was used to start Practice Test 2 – 
 
25 minutes x 1 question/ 1.50 minutes = about 17 questions should of been completed
 
75 minutes – 25 = 50 minutes left
 
35 minutes class- time will be used.
 
 
15 minutes left.
 
 
Period 3/4
 
19 minutes of time was used to start Practice Test 2 – 
 
19 minutes x 1 question / 1.50 minutes = about 13 questions should of been completed
 
75 – 19 = 56 minutes left
 
We will finish part 1 in class today..
When everyone has completed the Part 1 of Practice Test 2 –  please enter your answers into the form which is on auto-grade.
 

Practice Exam 2 Multiple Choice Test Complete 2021

 
LINK to MC Review videohttps://youtu.be/Mn0yzstQDpc
 
4/22  –  Thursday-  “C” Day  Homework:
Period 2:
 
1.  Period 2 will complete the Part 1 of practice test on their own (15 minutes).
 
Please enter your answers into the form posted above.
 
When you receive your email on your score please START reviewing with me using my video that reviews every question.  I link everyone a private link to the video through your private gmail account that you have been using with me. YOU WILL HAVE TO BE LOGGED INTO GOOGLE WITH THAT EMAIL TO HAVE ACCESS.
 
All you have to do is click on the timecode posted in the description to move directly to the question you want reviewed.
 
LINK to MC Review videohttps://youtu.be/Mn0yzstQDpc
Period 3/4
 
1. You should have completed the test in class and entered your scores into the Form posted  
    above.
 
Please review your incorrect answers with the privately linked video.
 
 
2. Please complete the older part 2 questions that will review spectrophotometry labs.  There is no need to time yourself here.  I would like you to think about these questions and how they relate to the 2 spectrophotometry labs that we completed.  Please read through my notes on how we completed 2 labs using the spectrophotometer.
 
In your reference tables:   A = abc
 
        A = absorbance (- log of Transmittance)  
                 Transmittance is the ratio of light reaching the detector to the light being emitted from the source.
                 The more that light is absorbed, the lower the light hitting the detector and thus the lower the          
                 Transmittance.  This number will be less than one so we perform a .-log on its value to make it positive                      and a larger that 1 value and when we do this it becomes Absorbance.
        a = absorptivity constant (proportionality constant like R for gas laws or k for rate law)
         b = path length (distance that the source light travels to the detector)
        c = Concentration of colored chemical (Molarity chromophore or crystal field)
 
So based on A = abc the Absorbance is dependent upon the concentration of the colored chemical and the distance that the light has to travel through the cuvette.
 
% mass of copper in brass: 
Brass shot (alloy of Cu & Zinc) was dissolved with nitric acid. The solution was blue due to the crystal field splitting of Cu+2 with water. The Zinc in the alloy (Brass) also dissolved but it has no color.  We performed a serial dilution of a known concentration of [ Cu+2 ] and we put theses known concentrations in cuvettes and determined the Absorbance of each known concentration (after we set the spectrophotometer to absorb a certain wavelength of the Cu+2  solution called the peak wavelength).We plotted a Beer’s Law graph (linear relationship of diluted Cu+2 ] with lowered Absorbance. We (logger pro) plotted the graph of measured absorbance with Known concentration (we used the keep button to add the concentrations for each measures absorbance of the known concentrations).
We made a line of best fit and we used the slope and y intercept data to help with the next part
 
We then took the dissolved Cu+2  from the brass shot and made a 100.0ml solution.  We put that in a cuvette and measured the absorbance. Using this absorbance, and the Beer’s Law graph (utilizing linear algebra, y = mx +b from the regression line of the graph) we determined the concentration of the dissolved Cu+2 ].  We used that to calculate moles and then percent by mass.
 
Love this lab!
Determining the order of  reaction (with respect to Crystal Violet) of reaction (CV) with OH_
In this lab used placed a known solution of [CV] with is a chemical indicator (or a chromophore!) with a known concentration of OH into a cuvette in a spectrophotometer that plotted the Absorbance decreases in real time for 30 seconds. We identified the order of the CV while keeping the [OH] constant.  (The concentration of the OH was so much larger than the dilute CV that we also assumed that its concentration did not change significantly.) From the absorbance data we collected, we assumed that the concentration of the CV was proportionate to the changes of the Absorbance. (A = abc and because a and b are fixed values for the experiment we used the concept that A is proportionate to c which is concentration) Remember that over time the cuvette changes from a purple color to clear. How much it turned clear was an indication of the RATE of the reaction!  Using the absorbance values collected for each time interval as the proportionate concentration we changed the absorbance values to ln [A] and 1/[A) respectively and then made a new graph of each.  The graph that was the most linear determined the rate order with respect for CV.
 
Ok I love this lab too!
2004 Free Response question 2 spec.pdf
View Download
 
In the above question (e) the Arrhenius equation used to be part of the course and it was on the old reference tables:
when you look at this equation you get a linear function.
                                                                               y   =        m       x      +     b

We used this equation to understand the relationship of activation energy (Ea) and temperature with the rate law constant but the equation is not part of course anymore so question (e) no longer is applicable.

 
2004 free response scoring guidelines ques 2 spec.pdf
View Download
 
2006 Free response question 5 – spec.pdf
View Download
 
2006 Free response scoring guidlelines ques 5 spec.pdf
View Download
End of Thursday..

4/23  –  Friday –  “D” Day          – period 2,3 (Lab)  –  I   2(B,D) 3(D) AP CHEMISTRY
                                                           – period 2,3 (Lab) –  R  2(B,D) 3(D) REMOTE INSTRUCTION
  
                                                             – period 4 – I  3(B) 4(B,D) AP CHEMISTRY
                                                             – period 4 – R 3(B) 4(B,D) REMOTE INSTRUCTION
 
Zoom Link:
Topic: AP Chemistry – 04.23 Periods 2,3,4
Time: Apr 23, 2021 08:00 AM Eastern Time (US and Canada)
Join Zoom Meeting
https://us02web.zoom.us/j/82797112959?pwd=THIrbTVVWDA5d2ZRbGFxbVlLRjF3UT09
Meeting ID: 827 9711 2959
Passcode: 7hKYYp
One tap mobile
+16465588656,,82797112959#,,,,*371720# US (New York)
Period 2/3:
 
You should have completed the MC at home using the 15 minutes left.  You should have started the review of your Practice Test 2 Multiple Choice with the video posted in Thursday.
 
We will begin the Part 2 – Free Response section: 105 minutes total.
 
            Question 1:     10 points
            Question 2:     10 points
            Question 3:     10 points
            Question 4:      4 points
            Question 5:      4 points
            Question 6:      4 points
            Question 7:      4 points
 
                                          46 points 
 
 
            105 minutes / 46 points  = 2.28 minutes per point
 
 
Today we completed about 60 minutes of the Free Response.
 
60 minutes x 1 point /2.28 minutes = about 26 points
 
If you go in order you should have completed questions 1,2 and about half of 3.
 
 
Period 4:
 
Free Response Practice Test 2 :
 
Today we completed 25 minutes of the Free Response.
 
25 minutes x 1 point/2.28 minutes = about 11 points should have been completed..
 
 
4/23  –  Friday –  “D” Day   Homework:
 
1.View Statistics on the MC section of Practice Test 2:
 
Practice 2 Test Data: In Class test from this week – 23 students -(50 MC questions)
mean =  24.04 – raw (48.08% = 24.04 /50 raw score) correct answers,  SD =  (9.23 raw)
Did you get the same questions wrong as others in your class?? 
Did you get questions wrong that others got correct?
               
 1.   65%  16.  26%  31.  61%  46.   30%
 2.   57%  17.  78%  32.  17%  47.   30%
 3.    57%  18.  43%  33.  30%  48.   39%
 4.    83%  19.  57%  34.  17%  49.   43%
 5.    91%  20.  39%  35.  39%  50.   17%
 6.   61%  21.  22%  36.  35%  
 7.    43%  22.  52%  37.  44%    
 8.   61%  23.  65%  38.  44%    
 9.   52%  24.  65%  39.  17%    
 10.   48 %  25.  22%  40.  39%    
 11.   70%  26.  43%  41.  43%  
 12.  48%  27.  57%  42.  48%    
 13.  43%  28.  74%  43.  57%    
 14.  48%  29.  48%  44.  65%    
 15.  61%  30.  57%  45.  52%  

 

2.  Complete your review of Practice Test 2 Multiple Choice with my review Video if you did not complete this Thursday evening:
 
LINK to MC Review video for Practice Test 2https://youtu.be/Mn0yzstQDpc
2. a)
 
Only after you have completely reviewed the Practice Test 2 Multiple Choice you will begin Practice Test 3.  To be able to complete as many tests as possible you will complete the Multiple Choice this weekend for Practice Test 4. If you reviewed your guessed questions and incorrect questions from Practice Test 2 then you are ready to complete another multiple choice section.  You will complete the Free Response for Practice Test 3 in class later in the week so it will count!
 
To alleviate problems with students “finding” a block of time to complete this MC section I have broken up this section in half using 2 forms.  The first form will consist of questions 1 – 25 and the second will consist of questions 26 – 50.  Each form IS TIMED and you will have 38 minutes to complete each form. If you do not make a guess for those questions that you could not do or did not have time for these questions will be graded incorrectly just as it would be on the AP.  Pacing is important!
 
*Remember you have 1.5 minutes per each multiple choice.
 
Please use these pacing guidelines to monitor your pace.
1o minutes x 1 question/1.5 minutes =  7 questions completed
20 minutes x 1 question/1.5 minutes = 13 questions completed
30 minutes x 1 question/1.5 minutes = 20 questions completed
38 minutes”                                                   “25 questions completed
 
Practice Test 4 Multiple Choice 1 – 25 (38 minutes)– timed form
 
Click on link to begin questions 1 – 25:
 
Practice Test 4 Multiple Choice 26 – 50 (38 minutes)– timed form
 
Click on link to begin questions 26 – 50:
 
 
2b). Please Review your Multiple Choice of Practice Test 3 with my video below. Please use the timecode under the description to move directly to the question you want to review.
 
 
This will be posted Sunday, maybe.. It will be raining…
 
 
End of week 2…