Physics, chemistry: closed system vs isolated system

Tutoring physics or chemistry, definitions are always important. The tutor compares closed system with isolated system.

A closed system cannot lose or gain matter; it cannot exchange matter with the surrounding environment.

An isolated system is closed as above, but also with regards to energy. That is to say, an isolated system cannot exchange matter or energy with the surrounding environment.


Giancoli, Douglas C. Physics, 5th ed. New Jersey: Prentice Hall, 1998.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Biology, chemistry: the difference between cellulose and lignin

Tutoring biology, you deal with organic molecules. The tutor mentions the difference between lignin and cellulose.

Cellulose and lignin are both found in cell walls.

Cellulose is composed of many glucose units bonded together.

Lignin, on the other hand, consists of phenyl propane units bonded together.

Therefore, cellulose is a polysaccaride, or starch, whereas lignin is aromatic.


Mader, Sylvia S. Inquiry into Life, 9th ed. Toronto: McGraw-Hill, 2000.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Chemistry: why nitric acid is more acidic than nitrous

Tutoring chemistry, the concept of acid strength is important. The tutor explains why nitric acid is more acidic that nitrous.

Nitric acid (HNO3) and nitrous acid (HNO2) both have N as the central atom, and share the structure H-O-N….However, nitric acid has an extra oxygen on the other side of N, which causes more electron drain from the N atom. The N atom, in turn, pulls electron density from the O between it and the H. With less electron availability, the O adjacent to H in HNO3 can less effectively attract H+, so loses it more readily than does HNO2.


Mortimer, Charles E. Chemistry, sixth ed. Belmont: Wadsworth, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Calculator usage: linear regression on the Casio fx-991ES PLUS C

Tutoring statistics, you cover linear regression. The tutor shows how to get a best-fit line on the Casio fx-991ES PLUS C.

Let’s imagine you have the following data

x y
10.1 14.2
17.3 19.5
25.4 22.9
40.0 31.8

Furthermore, you’d like to find a line of the form y=A+Bx that fits the data. Here’s how you might do it using the Casio fx-991ES PLUS C:

  1. Press Mode then 3 for Stat mode.
  2. Press 2 for y=A+Bx
  3. In the table that appears, enter the x and y values.
  4. After all the x and y values have been entered, press AC.
  5. Now, press Shift then 1.
  6. Press 5
  7. You’ll see choices for A, B, and other stats. Select the one you want, then press Enter.
  8. If, for example, you select A first and get its value, press Stat then 1 then 5 to return to the other choices. You can then choose B.


Casio fx-991ES PLUS C User’s Guide.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Chemistry: molecular weight of fats and oils

Tutoring organic chemistry, the composition of a fat or oil might be of interest. The tutor shares a resource.

Today, my reading sent me after the molecular mass of a typical fat or oil. I found the triglyceride molecular weight calculator (which I will call TMWC). It definitely answered some of my questions.

Fats and oils are collectively called lipids, but also triglycerides. The TMWC allows you to prescribe various parts of the fat or oil molecule (“build your own”), or else just select a typical one from a dropdown list. Then, it calculates the molecular mass of the fat or oil you’ve described.

I learned from the TMWC that soybean oil has an average molecular mass (some would say molecular weight) of 872.33. Its average molar mass would then be 872.33g. Beef tallow, interestingly enough, has a lower average MM: 850.92g.

I’ll be talking more about fats and oils in coming posts:)


triglyceride molecular weight calculator

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Chemistry: Graham’s Law of Effusion

Tutoring chemistry, you may mention kinetic energy of particles, diffusion, and effusion. The tutor gives a brief explanation.

In my post from Jan 18, 2017, I define effuse: it means to escape from a container through a porous boundary.

Graham’s Law of Effusion compares the rate at which two different gases will effuse, based on their comparative molecular masses. If eA is the effusion rate of gas A, and eB that of gas B, then

eA/eB = (MMB/MMA)1/2

where MMA is molecular mass of gas A, and MMB, that of gas B.

The reasoning behind the formula is that effusion depends on molecular motion, which is quantified by kinetic energy, KE:

KE = 0.5MMv2

where, once again, MM means molecular mass, while v means velocity.

Two gases of the same temperature have the same kinetic energy:


Dividing both sides by 0.5MMAvB2 gives

vA2/vB2 = MMB/MMA

square rooting both sides gives

vA/vB = (MMB/MMA)1/2

Since effusion is motion through pores, the ratio of velocities is the ratio of effusion:

eA/eB=vA/vB = (MMB/MMA)1/2

Example: Compare the effusion rate of methane, CH4, with that of propane, C3H8.

Solution: the ratio of effusion should be


Note that methane, CH4, has MM=16, while propane, C3H8, has MM=44. Therefore,


Methane should escape from a porous container 1.66 times the rate that propane escapes.

Note that, in a more general sense, this is a law of diffusion. Therefore, if propane and methane are released at one end of a room, methane should reach the other end 1.66 times as quickly as propane.



Mortimer, Charles E. Chemistry, sixth ed. Belmont: Wadsworth, 1986.

White, J. Edmund. Physical Chemistry, College Outline Series. New York: Harcourt Brace Jovanovich, 1987.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Battery output vs discharge rate: Peukert’s Law

Tutoring physics or chemistry, you might encounter Peukert’s Law, although it’s probably used more by industry.

In my March 2 post I mentioned reserve capacity and Amp*hours as two ways to measure a battery’s potential output. Numerically they are convertible backwards and forwards, but the reality is not necessarily so simple, because the speed of discharge affects the total output a battery can manage. Specifically, a higher discharge rate lessens the battery’s efficiency, so that the total output will decrease the faster the battery is discharged.

Peukert’s Law gives the equation

t=T(C/(I*T))k, where

t is the actual time the battery will deliver arbitrary current I,

T is the discharge time corresponding to the given amp*hour rating,

C is the given amp*hour rating,

k is a physical constant that depends on the type of battery (around 1.4 for lead-acid).

Because of Peukert’s Law, an amp*hour rating must be given with a specific time for which it’s valid. (My reading suggests that 20 hours might be a typical time.) Therefore, an amp*hour rating might read “120Ah over 20 hours”. Such a rating implies a discharge rate of 6A for 20 hours. How long the battery can deliver a different amperage can be calculated by Peukert’s Law.

Example: Imagine a battery rated 120Ah over 20 hours. How long can it deliver 120A?

Solution: Theoretically, a 120Ah battery can deliver 120A for 1 hour, although we already know not to expect it. Peukert’s Law gives



t=0.30 hours, or about 18 minutes.

I’ll be talking more about lead acid batteries in future posts:)


Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Chemistry: anhydrides, acid or base

Tutoring chemistry, the term anhydride may pop up.

An anhydride is a molecule that results from the removal of water.

An acid anhydride will assume its acidic tendency if water is added. Nonmetal oxides typically are acid anhydrides – for example, SO3:

SO3 + H2O → H2SO4

Metal oxides are typically base anhydrides:

CaO + H2O → Ca2+ + 2OH


Mortimer, Charles E. Chemistry, sixth ed. Belmont: Wadsworth, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Chemistry: zone refining

Tutoring chemistry remains interesting because it’s full of neat concepts and applications.

Zone refining is a technique that can produce highly pure samples; the semiconductors germanium and silicon can be zone-refined.

For zone refining to work, the substance must reject impurities as it cools. Mathematically, the ratio between the % impurity it will tolerate in solid form versus liquid form is k, the distribution coefficient. When k<1, the molten substance, as it solidifies, will exclude impurities; the impurities will be pushed to the boundary between solid and liquid. Zone refining is carried out as follows:

  1. A long rod of the impure substance, desired to be refined, is obtained.
  2. A cylindrical heater, much shorter than the rod, is fitted around one end.
  3. The heater melts the portion of the rod it surrounds, then is moved along to the adjacent section, which it also heats and melts.
  4. The portion of the rod that has been left behind now cools and solidifies.
  5. Because k<1, the solidifying part of the rod excludes the impurities, which are drawn to the adjacent, molten section of the rod.
  6. As the process continues, the impurities are drawn along the rod, eventually to its far end, which can be discarded.
  7. With each melting and re-hardening, perhaps not all impurities are removed. However, the process can be repeated to gain increased purity.


Mortimer, Charles E. Chemistry, sixth ed. Belmont: Wadsworth, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Chemistry: what is back bonding?

Tutoring chemistry, back bonding might come up.

Back bonding is a concept from molecular orbitals. In a molecule with a central 3rd-period nonmetal (such as sulfur or phosphorus), double bonding is possible so that the central atom exceeds an octet. The phenomenon is back bonding.

The reason back bonding is possible is that the central atom, being a third-period nonmetal, has (vacant) d-orbitals. It is thought that a peripheral atom bonding with it can form a second bond by merging one of its p-orbitals with one of the central atom’s vacant d-orbitals. The peripheral atom then contributes two electrons to this p-d hybrid; the result is a pπ-dπ bond, aka back bonding.

Back bonding is thought to explain shorter-than-expected P-O bonds in phosphoric acid and S-O bonds in sulfuric acid.


Mortimer, Charles E. Chemistry, sixth edition. Belmont: Wadsworth, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.