✍️ Free Response Questions
AP Chemistry Free Response Questions
⚛️ Unit 1 - Atomic Structure and Properties
1.1Moles and Molar Mass
1.2Mass Spectroscopy of Elements
1.3Elemental Composition of Pure Substances
1.4Composition of Mixtures
1.5Atomic Structure and Electron Configurations
1.6Photoelectron Spectroscopy & Graph Interp.
🤓 Unit 2 - Molecular and Ionic Compound Structures and Properties
2.0Unit 2 Overview: Molecular and Ionic Bonding
2.1Types of Chemical Bonds
2.2Intramolecular Force and Potential Energy
2.3Ionic Bonding and Ionic Solids
2.4Metallic Bonding and Alloys
2.5Lewis Dot Diagrams
2.6Resonance and Formal Charge
🌀 Unit 3 - Intermolecular Forces and Properties
3.0Unit 3 Overview: Intermolecular Forces and Properties
3.2Properties of Solids
3.3Solids, Liquids, and Gases
3.4The Ideal Gas Law
3.5The Kinetic Molecular Theory of Gases
3.6Deviations from the Ideal Gas Law
3.7Mixtures and Solutions
3.8Representations of Solutions
3.9Separation of Solids/Mixtures
3.10Solubility and Solubility Rules
3.11Spectroscopy and the Electromagnetic Spectrum
3.12Quantum Mechanics and the Photoelectric Effect
🧪 Unit 4 - Chemical Reactions
4.0Unit 4 Overview: Chemical Reactions
4.1Recognizing Chemical Reactions
4.2Net Ionic Equations
4.4Physical vs. Chemical Changes
4.5Stoichiometry & Calculations
4.6Titrations - Intro and Calculations
4.8Intro to Acid-Base Neutralization Reactions
👟 Unit 5 - Kinetics
5.0Unit 5 Overview: Kinetics
5.1Defining Rate of Reaction
5.2Introduction to Rate Laws
5.3Rate and Concentration Change
5.4Writing Rate Laws
5.5Collision Model of Kinetics
5.6Reaction Energy and Graphs w/ Energy
5.7Reaction Mechanisms and Elementary Steps
5.8Writing Rate Laws Using Mechanisms
🔥 Unit 6 - Thermodynamics
6.0 Unit 6 Overview: Thermochemistry and Reaction Thermodynamics
6.1Endothermic Processes vs. Exothermic Processes
6.2Energy Diagrams of Reactions
6.3Kinetic Energy, Heat Transfer, and Thermal Equilibrium
6.4Heat Capacity and Coffee-Cup Calorimetry
6.5Phase Changes and Energy
6.6Introduction to Enthalpy of Reaction
6.7Bond Enthalpy and Bond Dissociation Energy
6.8Enthalpies of Formation
⚖️ Unit 7 - Equilibrium
🍊 Unit 8 - Acids and Bases
8.0Unit 8 Overview: Acids and Bases
8.1Introduction to Acids and Bases
Unit 9 - Applications of Thermodynamics
🤺 AP Chemistry Essentials
🧐 Multiple Choice Questions
AP Chemistry Self-Study and Homeschool
⏱️ 4 min read
August 6, 2020
In chemistry, many times we want to calculate how light reacts to a certain colored solution with solute in it. Well, spectrophotometry can do just that!
To get a better understanding of what spectrophotometry, let's take a look at a spectrophotometer itself:
Image Courtesy of ResearchGate
There are 3 main parts to this machine: the monochromator, the sample, and the detector.
The monochromator has three main parts: the entrance slit, the dispersion device, and the exit slit. The entrance and exit slips are simple, they're where a beam of light enters and exits the monochromator. However, the dispersion device is the important part to the monochromator. Essentially, the dispersion device takes a beam of white light and splits it into the full spectrum of colors🌈. The exit slit is placed based on what wavelength of light the experimenter needs for the solution at hand, typically the opposite color (red solution --> green light, etc.)
The sample is just that - the sample! It can be any colored solution you wish, from a solution of Copper (II) Sulfate, to red Gatorade, which we'll take a look at in a minute. Light from the monochromator is run through the sample and light is absorbed in the sample (That's why we want to use the opposite color, that's the most absorbed light!).
Image Courtesy of Research Gate
On the right is the Incident Light (I0) and on the left is the transmitted light (I).
Finally, after all of this has occurred, the remaining light is detected and the amount absorbed is spit out on a screen.
Let's say a chemist is interested in calculating the molar concentration of red food dye🔴 in Gatorade. Well, this is a perfect time to talk about spectrophotometry in action! A chemist would fill a small test tube🧪 (though spectrophotometer test tubes are rectangular) with red gatorade and run it through a spectrophotometer with green light at around 560–520 nm. Then, he would get results as to the absorption, but how would he find the concentration? Well, luckily scientists have a formula for that:
The formula for the Beer-Lambert Law is actually quite simple. The Beer-Lambert law (also known as Beer's law) is a linear relationship between the absorption of light and the concentration of the absorbing species.
A = abc where: A = absorbance, a = molar absorptivity in L/[(mol)(cm)], b = path length in cm, and c is the concentration of the solution.
Once all written out, it may seem a little confusing, but the two variables a and b have are both quite simple to understand and are most often constant.
a = The molar absorptivity of a solute is a measurement of how strongly a chemical species reduces light (reduce here meaning absorbs, NOT takes electrons like in most other contexts) at a given wavelength.
b = The path length is simply the length of that rectangular test tube used by a chemist in a spectrophotometer. In fact, we often measure absorption on a linear scale by assuming a and b to be constant, leading to the equation A = mc, where m = ab:
Back to our chemist friend. Let's say he looked up the molar absorptivity of his food dye, Red-40 and found it to be 2.13 * 10^4 L/(mol)(cm). He runs his solution through the spectrophotometer and finds an absorption of 0.500. Assuming a path length of 1cm, we can use Beer's law to calculate the molarity!
A = abc
0.5 = (2.13 * 10^4)(1)(c)
c = 0.5/2.13 * 10^4 = 2.3 * 10^-5 mol/L
And boom. We've found the concentration of red-40 in gatorade with an absorption of 0.5.
🎥 Watch: AP Chemistry - The Beer-Lambert Law
To enact the lab you would do for this key topic, you can use PhET simulations:
👉Engage: AP Chemistry - Beer's Law Lab
In one of our previous guides, we went over part a of #5 on the 2019 AP Chemistry Exam. Let's go over part b since we now know the formulas🤓.
To look over the in depth explanation of part a, read 1.6!
Part b is asking us for the wavelength of something. First things first, there are two equations that involve wavelength of light:
E=hv, where h = 6.626 x 10^-34
c=λv, where c = 2.998 x 10^8
That something is the energy needed to remove an electron from the valence shell, or the ionization energy/binding energy⚡. So let's solve for the wavelength of this energy!
The only piece of information they give us is the actual energy itself, but there are 6 different types of energy listed😕.
To recall, where the binding energy is the highest is where the nucleus is located. Thinking about valence electrons, we should be using the lowest binding energy. Now we know that E = 0.980 x 10^-18 and we can solve for frequency!
E = hv --> 0.980 x 10^-18 = (6.626 x 10^-34) (v)
v = 1.48 x 10^15
Now that we have frequency, we can solve for wavelength using c=λv.
2.998 x 10^8 = (λ) (1.48 x 10^15)
λ = 2.03 x 10^-7 m
This is an example of using both equations to solve an FRQ!🧠
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