Our live are largely built around the rising and setting of the Sun in the sky each day.  Our biology is intimately connected to this via our circadian rhythms:

Have you ever wondered what would happen if the Sun just suddenly blinked out of existence?  When would you know?  Turns out, it takes time for light to travel from the Sun to the Earth.  We’ve just learned that light travels at a constant speed, c, in a vacuum like outer space.  We know that c = 3.0 x 10⁸ m/s.  To calculate how long it takes light to travel from the Sun to Earth, we need to know the distance between the two.  While the orbit of Earth around the Sun is not a perfect circle, on average the Earth is about 93 million miles (mi) from the Sun.  Time for some math!

Have: c = 3.0 x 10⁸ m/s and distance = 93 x 10⁶ mi

Want: Time it takes light to travel from the Sun to Earth

Need: Connection between meters (m) and miles (mi)

Another quick Google search tells us there are 1609.34 meters in 1 mile.  We are in business!

Calculation: (93 x 10⁶ mi) x (1609.34 m / 1 mi) x (1 s / 3.0 x 10⁸ m) = 499 s

Analysis: It takes about 499 seconds for light to travel from the Sun to the Earth!  Divide 499 by 60 and that gives us about 8.3 minutes.  So if the Sun blinked out right now, at this very instant, we wouldn’t know until 8.3 minutes from now.  The Sun is our nearest star, and it still takes 8.3 light-minutes for its light to reach us.

The second closest star to Earth is called Alpha Centauri.  Alpha Centauri is actually a triple star system (three stars in orbit around each other) located approximately 4.37 light-years from Earth.  Traveling at the speed of light, c, it would take 4.37 years to reach Alpha Centauri.  How many miles away is that?

4.37 light-years x (365 days / 1 year) x (24 hours / 1 day) x (60 minutes / 1 hour) x (60 seconds / 1 minute) x (3.0 x 10⁸ meters / second) x (1 mile / 1609.34 meters) = 2.57 x 10¹³ miles, or 25.7 trillion miles away!

That’s a long way!  It also means that if you look at Alpha Centauri in a telescope (or just look in the right part of the night sky), you are actually seeing light that left the star system 4.37 years ago.  You are literally looking back in time!  In fact, every time you look up in the night sky, you are looking back in time.

If traveling 4.37 years at the speed of light seems like a long time, don’t despair.  The distance from Earth to the nearest planet outside our solar system is a bit less.  Discovered in 2016, the planet Proxima Centauri b orbits Alpha Centauri and is “only” about 4.2 light-years from Earth.

Last week, MIT Technology Review announced the likely discovery of an Earth-like planet around a Sun-like star. (To be more accurate, and to give you a sense of how the scientific process works, the exoplanet was observed, the findings were written up into a scientific article and submitted to the scientific journal Astronomy & Astrophysics on October 16, 2019, and after successfully completing the peer-review process, the article was accepted for publication on May 3, 2020 and then published by the journal on June 4, 2020.  Here is the link to the published article.)  The exoplanet is named KOI-456.04 and is 3,140 light-years from Earth.  

While humanity hasn’t yet engineered a solution for how to accelerate a large spacecraft to near the speed of light, the video below introduces some important concepts regarding near light-speed travel.

Finally, if you think a bit more about the idea that looking at the stars is like looking back in time, the same hold true for Earth.  An alien pointing a telescope at Earth would be looking back in time at Earth as it was when the light left Earth.  If the alien is currently 65 million light-years from Earth, then the light they are observing today through their telescope left Earth 65 million years ago.  Is there a sufficiently powerful telescope that would allow the alien to actually see dinosaurs on Earth?  So glad you asked!

Return to Week 38 – Properties of Light and continue working.

When we think about light, we think about what can be seen.  If you’ve ever looked through a prism, you understand that white light is actually a collection of all the colors of the rainbow.  The visible spectrum consists of all of the light that we can see with our eyes.  Let’s go back to the rainbow.  The acronym ROYGBIV is a helpful way of remember the colors of the rainbow in “order” where R=red, followed by Orange, Yellow, Green, Blue, Indigo, and Violet.  It turns out that red light has a wavelength range of 620-750 nanometers (nm), while violet light has a wavelength range of 380-450 nm.  Remember, the shorter the wavelength, the greater the energy.  Therefore, because violet light has a shorter wavelength than red light, violet light is higher energy than red light.  We have specialized photoreceptor cells in our eyes that are excited by specific wavelengths of light.  When white light strikes an object, some wavelengths of light are absorbed by the object while other wavelengths are reflected. The color of an object is actually the wavelength of light that object does not absorb!  When reflected light is detected by our eyes, we see color.

Now for the really interesting part: visible light only comprises a small part of the larger electromagnetic spectrum.  The shortest wavelength of electromagnetic radiation is on the scale of 10^-12 cm (smaller than the diameter of an atom).  Remember, the shorter the wavelength, the greater the energy.  Photons with the shortest wavelength are called gamma rays and they are powerful enough to shred DNA.  We learned about gamma (γ) rays earlier in the school year during our study of nuclear decay (Lesson 15).  Viewing the night sky with gamma ray detectors gives us a very different perspective about the structure of space compared to looking with our eyes.

At the other end of the electromagnetic spectrum are the radio waves, with wavelengths on the scale of 10⁴ cm (the height of the Statue of Liberty).  Radio waves are emitted by stars and planets and can be detected with radio telescopes.  When the night sky is scanned using a radio telescope, we once again see structures in space that are invisible to our eyes.

To learn more about the visible spectrum, gamma rays, radio waves, and all the rest of the electromagnetic spectrum, visit NASA’s Tour of the Electromagnetic Spectrum and prepare to be amazed with the richness of the Universe!

Return to Week 38 – Light and Color and continue working.

Welcome to Week 37!  This week, we will tackle the topic of acids and bases.

Last week was the latest in a long string of really tough weeks for our country.  Rather than try making a light-hearted video introduction, I am simply asking you to visit the Future Voter registration page on the Washington Secretary of State website.  You can register to vote as early as age 16, and then you will be able to vote once you turn 18.  Vote for the country you want and vote in every single election.  While you are waiting to turn 18, remember that every dollar you spend is also a vote in support of wherever your money was spent.  Be intentional with who you choose to give your money to.  Your vote is your voice – scream!

1. Week 37 Attendance Check-In (required by 10am 6/5)
2. [H+] and pH (pH Concepts Google Form assignment)
3. pH Indicators (pH Analysis Gizmo)
4. pH Lab@Home (optional bonus credit lab)

You did it!  Just to make sure, here’s a checklist of items you must complete this week by Sunday, June 7 at 11:59pm:

• Week 37 Attendance Check-In (school district requirement)
• pH Concepts (worth +10 assignment points)
• pH Analysis Gizmo (worth +20 assignment points)
• Optional pH lab (worth +40 bonus lab points)

Remember, you can email me any time.  Office hours for Science are Tuesdays from 11am-12pm and Thursdays from 1pm-2pm.  Check your student Gmail for Zoom instructions.

Don’t forget to complete the Week 37 Bonus Credit Opportunity!  For a complete list of all of the bonus credit opportunities, bonus assignments, and bonus lab reports offered during distance learning, click here.

If you have access to some red cabbage and a few household solutions, you have what it takes to prepare the cabbage juice indicator shown in both Tyler DeWitt’s “magic trick” and Mr. Swart’s Week 36 introduction video.  Either way, earn up to 40 bonus points in the lab portion of your grade by documenting your efforts to:

1. Prepare red cabbage indicator
2. Observe after adding baking soda to the indicator
3. Observe after adding an acid to the indicator
4. Observe after adding a base to the indicator
5. Observe after adding an acid to the indicator, followed by baking soda
6. Observe after adding a base to the indicator, followed by baking soda

To earn the full bonus credit, document all of your work as a lab report and share it with Mr. Swart at david.swart@g.highlineschools.org.  Before you begin, review the Week 34-35 bonus credit lab report post and recall what happens when you mix baking soda and vinegar (an acid) – it gets messy!  That post includes videos, lab report requirements, and lab report templates you are welcome to use as appropriate.  Have fun!

Return to Week 37 – Acids and Bases and continue working.

The words acid, base, and neutral are all familiar.  By now in your schooling you have probably learned that water is a neutral substance, lemon juice is an acid, and bleach is a base.  You may have been introduced to the pH scale where you learned that water has a pH of 7, while acids have a pH of between 0 and less than 7 and bases have a pH of greater than 7 up to 14.  In fact, the pH of lemon juice is around 2.0, while the pH of bleach is around 12.6.  Lemon juice is a fairly strong acid, while bleach is a strong base.  Your blood has a pH of 7.4, making it slightly basic.

So what exactly makes something an acid or base?  Why are some acids and bases stronger than others?  Click the image below (or just click here) and read all about the relationship between hydrogen ion (H⁺) concentration and pH.

Next, watch the video below from Mr. Anderson at Bozeman Science:

Finally, complete the pH Concepts Google Form assignment and show what you know about acids, bases, and pH.  When finished, return to Week 37 – Acids and Bases and continue working.

To determine whether a solution is an acid, neutral, or a base, we need a tool.  In the lab, we can use a pH probe to obtain a quantitative pH value (an actual number).  Back in the day, we used pH strips to estimate the pH of different solutions.  Some strips are more sensitive than others, but the common theme is the strips rely on the user matching the color of the strip to a color chart which then estimates the pH.

At home (or in high school classrooms with limited funding) we can create a colorimetric indicator using cabbage! Colorimetric pH indicators provide us with a semi-quantitative measurement of pH.  By comparing the color of the indicator to a scale showing the color at a known pH, we can estimate the pH visually.  Color alone would be a qualitative data point (a description) while matching the color to a number provides us with a quantitative data point (an actual number).

To be fair, scientists use laboratory-grade colorimetric indicators in the lab all the time, and then use machines called spectrophotometers to quantitatively determine the optical density of the light passing through…remember the ELISA post from last week?

If you watched the Week 36 Intro video, you will soon realize that it also introduced you to this part of our lesson.  Prepare to be dazzled by the wizardry that is red cabbage juice indicator:

What is the actual chemistry behind red cabbage juice indicator?  Click the picture below and find out:

You now have what you need to complete the pH Analysis Gizmo.  The Gizmo was sent as a PDF attachment on Monday morning at around 8:00 am to the Week 37 – Chemistry Lesson email.

Anticipated answers to the question, “How do I turn in the Gizmo?”

• If you have access to a printer, print the Gizmo and then:
  • Scan and email your completed work to Mr. Swart
  • Send pictures of your completed work Mr. Swart
  • Insert pictures of your completed work into a Google Doc and share with Mr. Swart.
• If you do not have access to a printer:
  • Write answers on a piece of paper and then see above.
  • Write answers in a Google Doc and then see above.
  • Add comments to the PDF and share with Mr. Swart
  • This is 2020 – get creative!

Extend your learning!  For more on acid-base indicators, read Lesson 117 in the online textbook.  Note: this is not an assignment and you are not required to turn in any work related to lesson 117.

Return to Week 37 – Acids and Bases and continue working.

Welcome to Week 36!  This week will tackle the concept of Solution Concentration.  Many of the experiments we conduct in chemistry use chemicals in aqueous form.  By the end of this lesson, you will know how to prepare chemical solutions of a specific molarity.  Let’s go!

Checklist for the week:

1. Week 36 Attendance Check-In (required by 10am 5/29)
2. Molarity (Google Form assignment)
3. Dilution (Google Form assignment)
4. ELISA (optional virtual lab with assignment)

You did it!  Just to make sure, here’s a checklist of items you must complete this week by Sunday, May 31 at 11:59pm:

• Weekly Attendance Check-In (school district requirement)
• Molarity (worth +10 assignment points)
• Dilution (worth +10 assignment points)
• Optional Lab  Assignment (worth +30 bonus lab report points)

Remember, you can email me any time.  Office hours for Science are Tuesdays from 11am-12pm and Thursdays from 1pm-2pm.  Check your student Gmail for Zoom instructions.

Don’t forget to complete the Week 36 Bonus Credit Opportunity!

As a biotech researcher for many years, one of lab techniques I used quite often was the Enzyme-Linked Immunosorbent Assay (ELISA for short).  For many years, my research focused on understanding the activities of molecules of the immune system called interleukins (part of the cytokine family of molecules).  Some of my work included focusing on understanding the biology of  a interleukin called interleukin-17A (IL-17A).  I maintained a line of mouse fibroblast cells (NIH-3T3 cells) which had been shown would respond to IL-17A and produce IL-6 (another interleukin), measurable by ELISA (click here for the ELISA procedure).  If we stimulated the cells with IL-17A along with small amounts of other molecules (TNFα or IL-1β), we would see absolutely massive amounts of IL-6 released – well beyond what would be produced by the cells in response to any one of those molecules alone.  At the time, we used this concept to screen blocking antibodies against the IL-17A receptor (IL-17RA) which were then used in a variety of mouse models of different diseases.  Fast forward to today: scientists are accumulating data that the cytokine storm observed in some of the sickest COVID-19 patients may actually the result of those patients releasing too much IL-6, perhaps as a result of the activity of IL-17A released by the body as part of the defense against the virus.  In fact, there is a clinical trial underway for an antibody to IL-6 called tocilizumab.

With all that as background, it’s time to focus our learning around the ELISA.  For starters, if you are using an ELISA kit (like the IL-6 kit I linked to above), most of your assay reagents come packaged up in a tidy little box, and most of the reagents arrive as solids.  To use the reagents, you need to add specific volumes of specific solutions (solution concentration!).  After preparing the ELISA plates to receive samples, the protein standard must be diluted to the proper starting concentration.  Then the protein standard is serially diluted to generate a standard curve.  The samples are also often serially diluted.  When the assay is complete, if all goes according to plan, you can use the standard curve to determine the concentration of protein in your samples.

Your turn!  While not required, you are highly encouraged to work through the HHMI Biointeractive Immunology Virtual Lab.  To guide your learning, complete the Immunology Virtual Lab Worksheet and earn +30 bonus points in the lab report category of your semester grade.

Return to Week 36 – Solution Concentration and continue working.

Often when you are working with chemicals in the lab, the chemicals are already in solution.  For example, imagine you need to use some sodium hydroxide, NaOH, in a chemical reaction.  You have a 1.0 L bottle of 1.0 M NaOH on the shelf, but your reaction calls for a 0.25 M solution.  What to do?  Prepare a dilution by adding solvent (in this case, water) to the solution to lower the concentration of the solute (in this case, NaOH).

For starters, we need to know the volume of 0.25 M NaOH that we need.  Let’s say we need to end up with 1.0 L of the 0.25 M NaOH.  Now we can figure this out.  We know that M = mol/L.  For our 0.25 M solution, M = 0.25 and L = 1.0.  Rearranging the equation to solve for moles and we get mol = M x L = 0.25 x 1.0 = 0.25 mol.  Therefore, we need to end up with 0.25 mol of NaOH in 1.0 L of solution.

Our stock solution of NaOH has a molarity of 1.0 M, or 1.0 mol / L.  We need 0.25 moles of NaOH.  To figure out the volume of stock solution we need to obtain 0.25 moles of NaOH, we can set up a proportion: 1.0 mol / 1.0 L = 0.25 mol / x.  Solving for x, we need 0.25 L of the stock solution.

Finally, now that we know the volume of 1.0 M NaOH stock solution needed to add to prepare our 0.25 M NaOH solution (0.25 L), we need to calculate how much water to add to make the 0.25 M NaOH solution.  We need a total volume of 1.0 L, and 0.25 L is going to come from the 1.0 M NaOH stock solution.  Therefore, we need 1.0 L – 0.25 L = 0.75 L of water.  To prepare the 1.0 L 0.25 M NaOH solution, we need to add 0.75 L of water to our flask, then add 0.25 L of the 1.0 M NaOH stock solution.

Now that you have seen the math and read the reasoning behind it in painstaking detail, let’s try a practice problem.

Question: How would you prepare 2.0 L of a 0.5 M aqueous solution of CuCl₂ from a stock solution of 3.0 M CuCl₂?

Answer: 0.33 L of the 3.0 M stock solution + 1.67 L of water.  Why?  A 3.0 M CuCl₂ solution has 3.0 mol of CuCl₂ per liter of solution.  We want to prepare 2.0 L of a 0.5 M solution, so solving M = mol/L for mol, mol = M x L, so we need 2.0 x 0.5 = 1.0 mol of CuCl₂ in a total volume of 2.0 L.  Our stock solution is 3.0 mol/L so 1.0 mol = 0.33 L.  Therefore, we need to add 0.33 L of the stock solution to 1.67 L of water.

One more question: Vinegar is commonly sold as a 5% acetic acid solution (the other 95% is water).  A 100% acetic acid solution is called glacial acetic acid: glacial because the freezing point is just a few degrees below normal room temperature, so the acetic acid appears like a partially frozen glacier.

The molarity of glacial acetic acid is 17.4 M.  How would you prepare 0.5 L of 1.0 M acetic acid?

Answer: You want to prepare 0.5 L of a 1.0 M acetic acid solution.  First, calculate how many moles you need: mol = M x L so 0.5 x 1.0 = 0.5 mol of acetic acid.  Next, the stock solution of glacial acetic acid has a molarity of 17.4 M, or 17.4 mol/L.  To determine the volume of glacial acetic acid needed to obtain 0.5 mol: 0.5 mol x (1 L / 17.4 mol) = 0.029 L.  Always add acid to water, so first add 0.471 L of water (0.5 L – 0.029 L) to the flask and then add 0.029 L of glacial acetic acid.

Time to show what you know!  Complete the Week 36 – Dilution Google Form assignment and then return to Week 36 – Solution Concentration and continue working.

If you’ve ever made ice tea and decided it needs more sugar, you understand the importance of solution concentration.  Add too little sugar and the sugar concentration is too low to make the ice tea pleasantly sweet.  Add too much sugar and the ice tea just tastes like sugar water because the sugar concentration is too high.  Somewhere in there is the ice tea Goldilocks zone – just the right amount of sweet.  If you have access to water, a measuring cup, a glass, some tea, some sugar, and a teaspoon: fill a glass with one cup (8 ounces) of cold water, add one bag of tea, and let it steep while you continue working.

Back to our lesson: we need to understand the concept of molarity which is defined as the concentration of dissolved substances in a solution, expressed in moles of solute per liter of solution.  We use M as the unit of molarity (M = mol/L).  We already know how to calculate moles from grams, so just take moles and divide by volume (in liters) to calculate molarity.  Here’s an easy example question: glucose has a molar mass of 180 g/mol.  If 90 g of glucose is added to 1 L of water, calculate the molarity of the solution.  Answer: 90 g x 1 mol / 180 g = 0.5 mol, 0.5 mol / 1 L = 0.5 M.  The molarity of the solution is 0.5 M.

Here’s a harder example question: A 20 fluid ounce bottle of Dr. Pepper contains 64 grams of sugar (high fructose corn syrup).  Determine the molarity of sugar Dr. Pepper.

To approach this question, we need to know some additional pieces of information.  Ultimately, we need our answer in units of M, or mol/L.  So we need to convert ounces to liters and we need the molar mass of high fructose corn syrup.

Key information:

• 1 L = 33.814 fluid ounces
• High fructose corn syrup consists of glucose and fructose, both of which have a molar mass of 180 g/mol.

Answer: First, let’s convert 64 grams of sugar to moles: 64 g x (1 mol / 180 g) = 0.36 mol of sugar.  Next, let’s convert 20 ounces to liters: 20 ounce x (1 L / 33.814 ounces) = 0.59 L.  Finally, we just need to divide moles by liters to calculate molarity: 0.36 mol / 0.59 L = 0.61 M.

Let’s turn our attention back to the tea we started making at the beginning of this lesson and wrap this up with a more challenging practice problem.  Question: If we add one teaspoon of table sugar to our 8 ounce glass of tea, what is the molarity of the solution?

Key information:

• 1 L = 33.814 fluid ounces
• molar mass of table sugar (sucrose) = 342 g/mol
• 1 teaspoon of table sugar = 4.2 grams

Answer: Let’s begin by calculating the number of moles of sucrose (table sugar) added to the tea.  One teaspoon = 4.2 grams of sucrose.  4.2 grams x (1 mole / 342 g) = 0.0123 mol of sucrose.  Next, let’s convert 8 ounces to liters: 8 ounces x (1 L / 33.814 ounces) = 0.237 L.  Finally, to calculate molarity (M), divide moles by liters: 0.0123 mol / 0.237 L = 0.052 M.

Time to show what you know!  Complete the Week 36 – Molarity Google Form assignment and then return to Week 36 – Solution Concentration and continue working.