Determining the concentration of Sodium Hypochlorite (NaClO) in commercial bleach solution

Commercial Bleach is compared by the reaction of Cl2 with a base (usually a strong one). For the most part, this base is Sodium Hydroxide (NaOH), which yields NaClO (sometimes written as NaOCl; sodium hypochlorite). When aqueous, NaClO ionizes to Na+ and ClO– (Sodium and hypochlorite ions). When you bleach something, the ClO- ion is used in an oxidation-reduction (change of ionic charge) reaction, and is reduced to Cl-. The reducing agent (being oxidized) is the stain being acted on by the bleach.

Thus, we can easily find the concentration of OCl- in a solution by reacting a known mass of solution (commonly halide ions, for this we will use an Iodine ion) in an acidic solution.

OCl- (aq) + 2I- (aq) + 2H3O+ (aq) → I2 (aq) + Cl- (aq) + H2O (l)

As the iodine is in excess, the reaction of OCl- goes to completion. Now, we must determine the amount of I2 formed by means of titration (with Na2S2O3, sodium thiosulfate)

I2 (aq) + 2S2O3-2 (aq) → 2I- (aq) + S4O6-2 (aq)

S2O3 2- (thiosulfate ion) is reducing the iodine (I2) in order to form the polythionate (tetrathionate) ion S4O6 -2. As the titration reaches the endpoint (point where the reactions are stoichiometrically equal), the I2 concentration decreases as it is converted to the Iodine ion. This is marked by a change in color from a brown to a pale yellow (amber) color, and then finally to a colorless solution. This may be difficult to determine when the colorless solution is reached, so usually the addition of a starch (glucose joined together by glysodic bonds) is done when the color is that of a light yellow (as starch decomposes in the presence of an acid, and the loss of iodine as the attachment of I2 to the starch decomposes would be irreversible, it is imperative that this not be done until the color of the titration is that of a light amber). The addition of this starch will yield a blue color. The additional Thiosulfate ion breaks the I2 + starch complex down, and with the disappearance of the blue color (caused by the combination of I2 and starch) we can observe the endpoint, the amount of thiosulfate ion needed to react with I2, and the concentration of I2.

We can now determine the concentration of ClO- ion in the original bleaching solution by examining the volume of solution titrated, the volume/concentration of titration used, and the total stoichiometry of equations 1 and 2. We can determine the amount of moles of thiosulfate required to titrate x moles of Sodium Thiosulfate.

X moles of Thiosulfate (S2O3 2-) = (volume of Na2S2O3 in liters) * (number of moles of Na2S2O3) *(1mol S2O3 2-/1mol Na2S2O3)

Then we can use our balanced chemical reaction to find moles of I2 produced, then from that moles of ClO-, which would equal moles of NaClO-. From there, we find mass of NaClO, and determine percent solution of NaClO from the mass of the bleaching solution.

Usually, measuring small amounts causes greater error than measuring in large amounts. To alleviate this, we can dilute the commercial bleaching solution by adding more water. We will dilute 10ml of commercial bleach into 100ml of total solution. Then, we will add 100ml distilled water, 25ml of the diluted bleach solution, 10ml of Potassium Iodide (KI; source of Iodide ion) and 4ml of 6M HCl (hydrochloric acid). When we add HCl, there will be a reaction with the air releasing Iodine from the KI; it is imperative to continue titration as quickly as possible to prevent Iodine from disappearing from the system. When then titrate this to find the amount of ClO- ion is in the original concentration. Then, we add starch (1ml) for the excess iodide to bind too (after the titration goes from a brown to a light amber (think light beer color)). After we figure out the amount of S2O3 -2 concentrated, we can then figure out the mass of ClO- ion in the original solution.

A note: sometimes the titration will reach a clear color with black solid at the bottom (the solid is starch). Shaking the solution will return to the blue

Calculations to determine percent solution of NaClO in bleach:

(part A) X moles of Thiosulfate (S2O3 2-) = (volume of Na2S2O3 titrated in liters) * (number of moles used of Na2S2O3) (1 molS2O3 2- = 1 mol Na2S2O3)

(part B) X moles of I2 Produced = (Part A)/2 (2 mol S2O3 = 1 mol I2)

(part C) x moles of ClO- ion = (part B) (1 mol ClO- = 1 mol I2)

(part D) x mass (in grams) of NaOCl = (part c) *74.44g (molecular weight sodium thiosulfate. 1 mole ClO- = 1 mole NaOCl)

(part E) x % NaClO in concentrated solution = (part D)/(1.10g/ml*total ml of bleach in diluted solution (this case is 2.5)) We have .1 parts to 1 part of dilute solution, and we used 25 ml. We have 2.5ml of bleach.

This is usually around 6%. Thus, most commercial bleach contains a concentration of around 6% NaClO by weight. (Density of commercial bleach is usually around 1.10g/ml)

-Dr. Ricard

Questions, comments, concerns? Leave a comment.

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Can 20% (by mass) solution of Sodium Thiosulfate and 6% (by mass) Sodium Hypochlorite cause second degree burns?

Abstract: We were instructed to determine if the combination of 20.0% by mass of Sodium Thiosulfate (Na2S2O3, molar mass 158.11 g/mol or 248.2 g/mol for the pentahydrate 5H2O ) reacting with 6.0% by mass of Sodium Hypochlorite (NaClO, 74.44 g/mol) would be a possible reason for second degree burns on a garbage disposal worker (aforementioned worker was burned due to improper chemical disposal). According to Ripple et. Al, “a time-temperature integral of 1315 degrees C-second above body temperature correlates with heat transfer causing second-degree burns” (1990). Simply put, the antiderivative of the product of time and temperature must be greater than 1315 degrees (plus body temperature, which would yield 1413.6 assuming a body temperature of 98.6) in order to cause second degree burns. From this information, we can already gather that the evidence shows that burn severity is a multi-variable dependent factor (time and temperature). The thiosulfate anion reduces the hypochlorite group (and is in turn oxidized to become Sulfate). While undoubtedly more common, water containing form Sodium Thiosulfate Pentahydrate Na2S2O3·5H2O, the report states only that “sodium thiosulfate” is in the bottle; the pentahydate is not mentioned. Owing to the fact that the pentahydrate is more common (and additionally, because we used it in lab) I am going to speculate that the lack of information is simply a mistake on the part of the authors of the lab manual. Furthermore, I am unaware of the temperature of the solutions (which would affect density, thus molarity, thus percentage by weight). Another factor is the lack of information regarding the volumes of the solutions. Thus, with the lack of information, I am forced to proclaim the speculative nature of all after-mentioned hypotheses. As a ‘second-degree burn’ is a multi-variable factor, and we know that the reaction of Sodium Thiosulfate Pentahydrate and Sodium Hypochlorite is exothermic, we can assume that burns are indeed possible; now we simply figure out the length of time needed to be exposed and reason the probability that this garbage disposal worker spent that time in contact with aforementioned volatile chemicals.
NaClO + (Na2S2O+ 5H2O) = NaCl + NaOH + Na2S4O6
Methods: At the Texas State University Chem lab, we determined using 8 different molar proportions ranging from a ratio of 1:4 (Sodium Thiosulfate:Sodium Hypochlorite) to 4:1 that the reaction between the two took place of a molar ratio of 1:3 (the most heat, or change in heat, was indicated at this ratio which gave off 21 degrees Celsius). In our experiment, we used 0.5M concentrations of both NaClO and Na2S2O3. We can convert the given concentrations (20% Na2S2O3 (x) and 6% NaClO (y)) to molarity by speculating 100g of solution to start. For the Na2S2O3+5H2O, we determine that 20g/100ml (definition). Now we have 200g/1liter. To convert to moles, we divide 200/248.2 = 805801773moles/liter, or a molarity of x= .805801773M. For Y= 6g/100=60g/1000ml= (60/74.44)/liter, or .80601827M.From our scaled experiment using Job’s Method we saw the largest increase of temperature at 25 degrees Celsius at a ratio of 3:1 Sodium Thiosulfate Pentahydrate and NaClO with our .5M concentrations.  We found this using 8 experiments using varying ratios.
Results: Assuming a volume of a liter of each (again, no information given on volume spilled on worker or volume in dumpster), we’d react a total of 0.268600591molesNa2S2O+ 5H2O (limiting reactant) and Sodium Hypochlorite to get a  ΔT 1387.5 degrees Celsius (25degrees was observed with 6ml of Na2S2O+ 5H2O, 25/6*333=1387.5) Going back to our original number calculated in the abstract, we find that 1413.6 must be reached as the integral of time*temperature in order to cause second degree burns.
∫(x)(1387.5) = 1413.6
We get x=1.42695 (seconds), and even taking to account the fact that heat is lost instantaneously (thermodynamics state that systems tend to evolve towards equilibrium; heat would be lost to the surroundings) we can assume that this would probably cause second degree burns.
Conclusion: Assuming a liter of both reactants are used, assuming sodium thiosulfate is in the pentahydrate form, and assuming a realistic initial temperature of 20c, we can see thatthe sanitation worker, if exposed to the chemicals for 1.42965 seconds (a realistic time) would sustain second degree burns.
-Dr. Ricard
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Quantum Theory Part II

Continued from Part I, here. This may be hard to understand if you do not understand the concepts discussed in Part I.

When scientist looked at the emission spectrum (or discrete energy levels) of hydrogen (containing just 4 wavelengths at 410, 434, 486, and 656nm) they were struck by how simple it was (compare to the more complicated ones from part I). In 1885, Johann Balmer related the four wavelengths into a formula, which was later consolidated to the Rydberg Equation, which allows us to accurate calculate the wavelengths of all the spectral lines of Hydrogen. The equation is as follows:

1/λ = (Rh)(1/n1-1/n2)

Where λ is the wavelength, Rh is the Rydberg Constant (1.096776 x 10^7 m^-1), and n1 and n2 are any 2 positive integers where n2>n1 (commonly, n1 is 1 and n2 is 2).

This is astonishingly simple. But why? With Bohr’s theory, the the spectrum of hydrogen is accurately produced; however, other elements are not. Using his three postulates (part I), Bohr discovered that the energies that corresponded to the electron orbits (for hydrogen) fit the following formula;

E= (-hcRh)(1/n^2) = (-2.18 x 10^-18J)(1/n^2)

Where h is Planck’s constant, c is the speed of light, and Rh is Rydberg constant. N, called the principal quantum number is quantized (only can be represented as integers; 1, 2, 3, 4….) Each electron orbital away from the nucleus increases by (1)n. The first allowed orbit is n=1 (n=0 is impossible, as this would place the electron within in the nucleus; The repelling forces of the proton would not allow this). As the electron orbit moves away from the nucleus, we see increasing values of n (again, without decimals). Note at this point all values of n would be negative when input into the above equation. The lower the energy level is, the more stable it is; at n=1, the value of E is lowest. Thus, the atom is most stable at its ground state of n=1, and as n increases (the electron moves away from the nucleus) it becomes less negative, or more unstable. When an electron is at n=2 or above, it becomes more excited (reactive), and is said to be in its excited state. When n=∞, or when the electron is completely separated from the atom, we see that E (energy) = 0. This state is called the reference or zero-energy state.

In Bohr’s third postulate, he proposes the concept that electrons can move from one orbit to another (change of n) by absorbing or emitting photons whose corresponding radient energy (E=hv) equal exactly the difference of energy (ΔE). That is,

ΔE = Ef-Ei = hv

Where ΔE is change of energy, Ef is final energy, Ei is initial, h is planck’s constant, and v is wavelength frequency.

Substituting back into the equation above this one, we get

ΔE = hv = hv/λ = (-2.18 x 10^-18) (1/nf ^2- 1/ni^2)

Where nf is the final n number (state) and ni is the initial n number.

If the final n number is higher than the initial (that is, if nf >ni, or the electron moves away from the nucleus) energy is being absorbed, and if ni is larger than nf (the electron moves towards the nucleus, ni>nf) energy is being released. Example; if n=3 moves to n=1,

ΔE = hv = hv/λ = (-2.18 x 10^-18) (1/nf – 1/ni);

ΔE = hv = hv/λ = (-2.18 x 10^-18) (1/1^2- 1/ni^3)

We get a negative value (-1.94 x 10^-18), meaning energy is being released as the electron moves from n=3 (excited state) to n=1 (ground state). Simply, as an electron moves away from the nucleus, it requires energy (absorbed). If an electron moves towards the nucleus, it needs to release energy.

We still didn’t answer the question we started with. Why is the discrete energy level of hydrogen so simple? It can be attributed to the quantized jump of electrons between energy levels.

As stated in Part I, the limitation of this model is that it only can explain hydrogen atoms; all other elements it fails to predict accurately. What we need to take away from Bohr’s model is that electrons only exist as quantum numbers, and that energy is involved in the transition of one principal quantum number (n) to another, and that n is the size of the electron orbit.

Following Bohr’s model, wave-particle duality became a popular theory. Louis de Broglie asked that if radient energy could behave as a stream of particles (photon), could it also behave as a wave? He suggested that an electron that moved around the nucleus of an atom like a wave (and thus, has a wavelength). This wavelength would depend on the mass and velocity, and so, the following formula was proposed;

λ = h/mv

Where h is Planck’s constant, m is mass, and v is velocity.

Mass times velocity is considered momentum. De Broglie called the wave characteristics of material particles (such as the electron) matter waves. From this an obvious question would arise- would not, then, all material particles (both in the micro- and the macro- worlds) be considered waves? Yes, but if you plug into the equation, you find that the resulting wavelength is arbitrarily small.

When X-Rays pass through a crystal, the interference patten that results is characteristic of the wavelike properties of electromagnetic radiation; this is called X-Ray diffraction. The same pattern is observed when a stream of electrons is passed through the crystal. This proves Louis De Broglie’s theory and reinforces the thought of wave-particle duality.

This notion of wave-particle duality raised curious questions- and answers. Werner Heisenberg speculated in the now famous Heinsberg Uncertainty Principal. This postulate essentially states that the more accurately you know the speed of an electron (or any subatomic particle/wave) the less accurate one is in predicting the location of that object. He gave the following equation to relate uncertainty of position (Δx) and uncertainty of momentum Δ(mv) to Planck’s constant h.

Δx * Δ(mv) ≥ h/4Π

With some calculations, we can observe that the uncertainty really is limited to however small the object in question is. If we were to use a large object, say, a baseball, we can see that uncertainty is so small that it becomes a non-issue. With smaller objects, however, we see the large discrepancy.  With this, we see a revolution in the way we think about atoms. We begin to see the energy of electrons and its location not as concrete numbers, but as probabilities.

More in Part III.

-Dr. Ricard

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Electronic Structure of Atoms, Waves, and Intro to Quantum Theory

“It is often stated that of all the theories proposed in this century, the silliest is quantum theory. In fact, some say that the only thing that quantum theory has going for it is that it is unquestionably correct.” -Dr. Michio Kaku

Quantum theory seems to be a topic with a source of a lot of confusion- and it is. People unfamiliar with the term use Quantum Mechanics as a joke meaning something really difficult to comprehend. Intellectuals think of Quantum Entanglement, String Theory and the like. But what is quantum mechanics? Simply, it’s the process of dealing with things that are really, really, really, reallyreallyreallyreallyreally small. To understand quantum theory we must take a step back into the (relatively) large scale of the atom.

In the early 20th century, Niels Bohr suggested that the electrons of a hydrogen atom moved in a circular orbit around the dense, proton and neutron containing center of the atom. Bohr further suggested that in moving away from the center of the atom (becoming closer to the ‘excited state’ or point with the highest energy) energy was absorbed, while when moving towards the ‘ground state’ (closer to the nucleus) energy was given off. Bohr based his theory on three postulates;

  1. Only orbits with certain radii (corresponding to certain energies) are permitted for electrons in the hydrogen atom.
  2. An electron in orbit is in an ‘allowed’ energy state; it does not radiate energy, or spiral into the nucleus.
  3. Energy is absorbed and emitted by the electron only as the electron changes from an allowed energy state to another, and is in units of photons (more on the photon later in this post).

However, this was found to not be true; for one, physics tells us that if an electron were to be moving around in an orbit around a nucleus, it would continue to lose energy. Furthermore, this theory only held true for the hydrogen atom (which is just a proton and an electron) Let us back track a bit to discover how this was proved false. What occurs when a neon light is activated? The electrons in the neon atoms become excited and go into a higher energy state. The electrons can only remain excited for a short amount of time, and as heat is given off due to the return to the ground state, light is given off. This discovery can be explained by one of the most revolutionary theories of science in the 20th century- the quantum theory.

To truly understand this phenomenon, we must first understand light. The majority of our understanding involving the electronic structure of atoms (the description of arrangement of electrons in atoms) comes from the light absorbed and given off by these substances. Visible light is a type of electromagnetic radiation. Electromagnetic radiation carries energy through space; because of this, it is also known as radient energy.

Electromagnetic Spectrum

Electromagnetic Spectrum. From

Above is the electromagnetic spectrum. Notice the relatively small amount that constitutes the visible light spectrum. Included are sound waves, microwaves, infrared (heat) waves. You probably have heard of the speed of light (commonly 3.00 x 10^8,  estimated to 2.99792458 x 10^8), but in reality, this is the speed of all electromagnetic waves (through a vacuum). Let us take a step even further back; what is a wave?


Waves. From

A wave is a periodic (repeating) interval pattern. Wavelength (denoted by the Greek lamda, λ or Λ) is defined as one full interval. This is usually measured by the length between the peaks (the top of the wave) or the trough (the bottom) but, by definition, this can be measured anywhere on the wave. The function defining the number of complete wavelengths that occur in a given second is the frequency of the wave. Frequency is either denoted as S ^-1 or Hz (which are the same thing). There is an inverse relationship between frequency and wavelength which is defined by the following equation.

 c = λv

c = speed of light, λ = wavelength, and v = Frequency

There are different units for common wavelengths

Å (Angstrom, 10^-10) – X-Ray

nm (Nanometer, 10^-9) – Ultraviolet (UV), Visible

μm (Micrometer, 10^-6) – Infrared (heat)

mm (millimeter, 10^-3) – Microwave

cm (centimeter, 10^-2) – Microwave

m (meter, 1) – Television, Radio

km (kilometer, 1000) – Radio

When given different units, one must convert to meters for wavelength and Hz for frequency.

This existing wave model of light cannot explain, however, the emission of light from objects (blackbody radiation), emission of electrons that occurs when light is shined on metal (photoelectric effect) and the emission of light from electronically excited gas atoms (emission spectra).

When solids are heated, they emit waves of radiation (think of the red color from a hot stove). Many scientists tried to figure out the correlation between temperature and wavelength on these objects, to no avail. Then, in 1900, Max Planck proposed a theory that energy can only be absorbed and released in certain quantized “chunks”. Being quantized means that only certain increments are allowed to be used. The popular analogy is that of a stair and a ramp. You can walk up one stair, two stairs, 5 stairs or 1400 stairs, but you can’t walk up 1.5 stairs, 2.4 stairs, etc. We call a stair quantized. A ramp (inclined plane) can be traversed up any amount; it is non-quantized.

Planck called the smallest amount of energy that could be released or absorbed (through electromagnetic radiation) a quantum. 

According to Planck theory that energy is quantized, energy can only be released by multiples of hv; hv, 2hv, 3hv, 4hv… In 1905 Einstein built on Planck’s theory in the study of the photoelectric effect. A minimum frequency is required by light to be shined on metals in order  for those metals to emit electrons (for example cesium requires a minimum light frequency of 4.60 x 10^14 Hz in order to emit electrons). This amount of energy is called a ‘work function’ is needed in order for the electrons to break the attractive forces that bind them to the metal. Einstein proposed that radient energy  acts like streams of ‘packets’ (or quantize-able amounts) called photons. These photons could be considered ‘particles’ of energy.

He gave an equation that gave the energy E of a single photon:

E = hv

Where E is energy (of one quantum in joules), h= Planck’s constant (6.626 x 10^-34 Joule-seconds) and is wavelength frequency (in S^-1 or Hz)

Imagine if you had a laser (a single wavelength) that you could theoretically turn on and off as fast as possible (thus controlling the amount of total energy emitted). Einstein’s theory states that the smallest amount of energy you could reach would be given by the above equation, or a single photon.

Now we get start getting into the obscure part of quantum mechanics that confuses the layman and scientist alike. Is light a wave, or is it more particle like? We are forced to accept a possibly contradictory position on the matter- that light is both. We call this wave-particle duality. We will expand on this in Part II.

One famous experiment involving wave-particle duality is the infamous double slit experiment. Here’s a video that explains it.

When a source of radient energy is emits a single wavelength (like a laser), it is considered monochromatic. When there are multiple wavelengths it is considered polychromatic. A spectrum is produced when radiation is separated into their components.

What are we looking at here? When we look at light through a prism, we observe the top line spectrum, called the continuous spectrum, named because there are no breaks. We observe this in nature when individual water molecules act as a prism for sunlight; we call it a rainbow. Simple enough. However, as we see with the lower charts, we see that when we take a tube, create a vacuum (meaning that no external elements exist within the tube), and pump it full of X element (shown here are Hydrogen, Helium, Mercury and Uranium) we observe different spectra. A spectrum containing only certain radiation of only specific wavelengths is called is called line spectrum.

Continued in Part II, here. For practice problems involving Wavelength and Energy, go here.

-Dr. Ricard

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Wavelength/Energy Practice Problems

Answers and solutions at the bottom. For theory behind these, go here.

1. The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is the frequency of this radiation?

2. A certain microwave has a wavelength of 0.032 meters. Calculate the frequency of this microwave.

3. A radio station broadcasts at a frequency of 590 KHz. What is the wavelength of the radio waves?

4. Microwave ovens emit microwave energy with a wavelength of 12.9 cm. What is the energy of exactly one photon of this microwave radiation?

5. Calculate the energy of one photon of yellow light that has a wavelength of 589nm.


1.The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is the frequency of this radiation?

We are given wavelength in nanometers and asked to find frequency.

From our equation, we know that c=λv. (speed of light (meters) = wavelength (meters) times frequency (hertz))

c =  3.00 × 10m/s (sometimes this is represented as 2.998 × 10m/s, but the key I pulled these questions from uses the former value. Ask your teacher which value to use.)

 3.00 × 108 m/s = 589nm * v

We must convert nanometers to meters. This is a very commonly missed step! We must have all units when dividing be equal. 589 nanometers=  5.89 × 10-7 meters (review dimensional analysis if you do not know how to get this number).

 3.00 × 10m/s /  5.89 × 10-7 m = v

v = 5.093378608 × 1014 Hz

2.  A certain microwave has a wavelength of 0.032 meters. Calculate the frequency of this microwave.

From our equation, we know that c=λv. c =  3.00 × 10m.

3.00 × 10m/s = .032m * v

This one is nice and easy- we are given the same units. We simply divide here.

3.00 × 10m/s /.032m=v

v=9.375 × 10 Hz

3. A radio station broadcasts at a frequency of 590 KHz. What is the wavelength of the radio waves?

Again, we know our equation c=λv

Here is another simple division problem. First we convert KHz to Hz.

590 KHz = 590× 10 Hz

(3 x 108 m/s)/(590 x 103 Hz) = 500m.

λ = 500m.

4. Microwave ovens emit microwave energy with a wavelength of 12.9 cm. What is the energy of exactly one photon of this microwave radiation?

Here we need to use two equations.

c=λv (from above)

E=hv (Energy (J) = frequency (Hz) times Planck’s constant (Joule-Second or J-S))

Next we define our constants.

c= 2.998 x 108 m/s (this problem wants us to use this number for speed of light), h=6.626 x 10-34J-s

Now we simply plug in, making sure that our units match (convert 12.9cm to meters)

2.998 x 108 m/s = .129m * v

2.998 x 108 m/s / .129m = v

= 2,324,031,008 Hz

Now that we found v, we can solve for E.

E = 2,324,031,008 Hz * 6.626 x 10-34J-s

E= 1.539902946  x 10-24J

E= 1.54 x 10-24J (significant figures)

5. Calculate the energy of one photon of yellow light that has a wavelength of 589nm. (notice how similar this is to number 1!)



Convert nm to meters.

589nm = 5.89 × 10-7 meters.

Plug into equation to find frequency (v).

3.00 × 10m/s / 5.89× 10-7 meters = v

v = 5.09 x 1014 Hz

Now we can plug into our energy equation to find Joules of energy for one photon.

E = 5.09 x 1014 Hz * 6.626 x 10-34J-s

E= 3.37 x 10-19J

The above number is for one photon of energy. What if they wanted 2 photons? Multiply by two.

How about a mole of photons? Multiply by 6.022 x 1023.

– Dr. Ricard

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