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No, of course not. They are my freehand drawings.

I want to draw attention to what appears to be an error or drafting mistake in the NSW Engineering Studies examination. Question 21 was the first of the extended response questions, and it was examining Aeronautical Engineering. The question was worth 11 marks. It had four parts (a) worth 2...

EMF is not a force. EMF is a field potential or field potential difference. The units of EMF are joules per coulomb. The EMF of a point in an electric field is defined as the work done to bring 1 unit of positive test charge from infinity to that point within the electric field. The electric...

Thanks for the [!!!]WARNING[!!!]. Better stop swimming in the ocean - too much chloride

Ignore all the conversation about commutators and radial magnetic fields. Just consider a single current loop in a uniform B-field as shown below: The vertical segments are experiencing a constant force shown by the red vectors, and the two forces constitute a couple or moment. The horizontal...

The mass loses kinetic energy and gains GPE as it moves from A to B. Let the velocity at the bottom be v1 and the velocity at the top be v2 The change of gpe is U = 2mgr. The change of KE is U = -1/2 m (v12 - v22) Equate the two expressions and eliminate the common factor "m" you get: 2gr =...

The mass of solution is M= 14 g + 100 g = 114 g The density rho ρ of the solution is ρ= 1.060 g/mL so divide M by ρ and you get V=107 mL of solution. The 14 grams of the hydrous MgSO4.7H2O contains n = 0.0568 moles. Therefore divide n by V and you get 0.53 mol/L

Ava-Gardner

You are using the Arrhenius theory of a strong base. The Brönsted-Lowry theory is not restricted to aqueous solutions. If Ca(OH)2 is a good proton acceptor (which it is), then it is a strong base according to the latter theory.

I think any further debate will likely result in arcane hair-splitting. However, for students sitting their HSC Chemistry exam, there will be just three states of matter, they are solid, liquid and gas and the only other mixture they are likely to be examined on is the aqueous mixture, and we...

Thanks for that contribution, you got me thinking that can't be right because IUPAC has reserved the term "sol" for colloidal suspensions, so I went to the references. In the Journal of Pure Appl. Chem., Vol. 80, No. 2, pp. 233–276, 2008. doi:10.1351/pac200880020233 © 2008 IUPAC they give the...

Obviously this student was in a bit of a hurry and missed the equivalence point by a substantial margin. :( If you add the last drops of NaOH slowly you should be able to achieve a precision of 0.1 mL in the aliquot. Look at the results at 20.0 mL and 24.0 mL. This is where the inflection...

Ahh, no. We call them liquids because that is their standard state at 25oC. If the ester coincidentally happens to be dissolved in a reaction mixture like an alcohol, then you still call it a liquid. The adjective "aqueous" (aq) should be reserved for substances that are dissolved in water, or...

Yes, it should be tan θ. I disagree with that equation given in the sample answer. The result is correct, but the equation is wrong. Here is the free body diagram of the car. The car is subjected to just two forces, they are W the force of gravity and FN the normal force from the road. The net...

Most esters are liquids at room temperature.

You state that gases are (g) and liquids are (l). Students are not required to memorise the physical states of organic compounds for the HSC Chemistry exam. You should be able to work these facts out from the context of the question. If a question requires you to write a chemical equation...

No you're not. You're doing just fine. Just make sure you read all parts of the question, and underline the important facts.

In the original post, it says Io is the intensity after leaving polarising filter #1. Therefore Malus' Law only applies to the interaction between polarizing filter #2 and filter #3. Therefore the intensity after leaving filter #3 varies from zero to 0.5 Io.

This is a common misconception, but the influence of the first polarizing filter over the third polarizing filter (or the second analyser if you want to refer to it that) is irrelevant once you introduce the middle polarising filter. I have done this experiment many times in class with three...

Yes, the intensity of the light will halve after passing through the first filter, but it will not halve after passing through the second filter. The law about halving the intensity only applies to unpolarised light that has passed through an ideal polarising filter. Once the light has become...