Physics P1, P2

Since superscript letters doesn't seem to be supported here (Exception are presets), please take note that the [^] sign translates to 'to the power of'. (e.g. 10^-6 means ten to the power of minus six)

Random General Tips

• DON'T PANIC (Printed in large, friendly, and yellow letters)
  
• Be concise, precise, accurate, clear, complete, and coherent when you are asked to define. (This is also why definitions are the hardest things in physics. Believe me)
• Use multiple short sentences for definitions, one point per sentence. This way, your answer is direct and easily understood. (As your English sucks, you are more likely to obscure the meaning should you put long sentences)
• In diffraction grating, n is always rounded down to the lowest integer. (e.g. n =2.8, so when rounded, it becomes n = 2, not n = 3)
• Read, comprehend, and read again the question. Seriously.
  
• When encountering an unfamiliar sum, the first thing that you must do is to find out which laws to apply to the sum. This is extremely important. (For proof, read question number 2bii, paper 2, physics, summer 2008; was your first thought the principle of conservation of momentum?) Second, attack the sum in a logical manner; never ever ever ever assume anything, and always start from facts given by the sum. Third, a particular sum could involve materials from any chapters, so don't be so narrow-minded and confine your thinking in only one direction; be creative.
• Sometimes, they require you to write the equations in words first (i.e. Work done = area under graph line) before moving on to symbols. (i.e. W = ½Fx) This usually appears when you are asked to prove something.
• When drawing field lines, draw at least (To 6.
  
• May/June 2006's Paper 2 is one of the hardest in the repertoire, (But honestly, I think it is the hardest) so go check that.
  

Chapter 1 - Physical Quantities and Units

All quantities consist of a numerical magnitude and a unit.

The six basic units (In S.I. units) are:

• mass (kg)
• length (m)
• time (s)
• current (A)
• temperature (K)
• amount of substance (mol)

Derived units consists of basic units. (e.g. Velocity is displacement over time [m/s], charge is current times time [As])

Physical equations must be homogeneous. Examples:

• v = u + at (Homogeneous; v is [m/s], u is [m/s], at is [(m/s²)*s] = [m/s])
• v² = u² + 2at (Not homogeneous; v² is [(m/s)²] = [m²/s²], u² is [(m/s)²] = [m²/s²], 2at is [(m/s²)*s] = [m/s])

List of prefixes:

• pico [p] = 10^-12
• nano [n] = 10^-9
• micro [μ] = 10^-6
• mili [m] = 10^-3
• centi [c] = 10^-2
• deci [d] = 10^-1
• kilo [k] = 10^3
• mega [M] = 10^6
• giga [G] = 10^9
• tera [T] = 10^12

Should you ever need to estimate physical quantities, there are 2 things that you must do: 1) Use your common sense, and 2) Common sense.

Scalar quantities have only magnitude and unit, while vector quantities have magnitude, direction, and unit. (e.g. Energy [J], Distance [m] are scalars; force [N], momentum [Ns] are vectors)

In doing equations, the product or division of two vector quantities would result in a scalar quantity, while the product or division of a scalar and a vector quantity would result in a vector quantity. (e.g. Force (Vector) times displacement (Vector) equates to energy (Scalar), acceleration (Vector) times time (Scalar) equates to velocity (Vector))

Vectors can be added or subtracted only if they are coplanar. (i.e. On the same plane; vectors in the x-direction can be added or subtracted, but not vectors in x-direction with vectors in y-direction)

If wishing to add non-coplanar vectors, it is necessary to split the vectors into perpendicular components. (Most of the time, into x-and-y-directions)

Chapter 2 - Measurement Techniques

Formulae in this chapter:

• Δa/a = Δb/b + Δc/c + ... + Δz/z (Fractional uncertainty)
  

Systematic errors,

• causes a set of readings to be either too high or too low from the true value.
• cannot be eliminated or reduced by averaging.
• include: Unaccounted zero error in measuring instrument; using a damaged or poorly calibrated instrument; reaction time not accounted for. (e.g. When using stopwatches)

Random errors,

• causes a set of readings to have generally high scatter. However, the end result could still be accurate.
• can be reduced by averaging.
• include: Parallax error, changes in wind speed and/or direction, changes in temperature, fluctuations in pressure, etc.

Precision refers to the distribution or scatter of values about the true value. A set of readings is said to be precise if it has low scatter.

Accuracy refers to the trueness of values. A set of readings is set to be accurate if the readings are about the true value of the reading.

Therefore, a set of values with very few random errors can be said to be precise, while a set of values with negligible systematic error can be said to be accurate.

Example: (Doing an experiment to find the value of g)

• g = 8.93, 9.34, 9.01, 8.77 (Not precise and not accurate)
• g = 9.34, 9.35, 9.33, 9.34 (Precise, but not accurate)
• g = 9.66, 9.81, 9.99, 9.78 (Not precise, but accurate)
• g = 9.79, 9.82, 9.81, 9.81 (Precise and accurate)

Uncertainty arises because a quantity may never be measured exactly. (e.g. Even if it looks like you've measured 25 cm on a plastic ruler, the actual value could be 25.0012312753 cm)

The uncertainty of a measurement usually is taken from the smallest scale of the measuring instrument (e.g. Suppose you've measured 24.3 mm on the vernier caliper. The smallest reading the vernier caliper can take is 0.1 mm, therefore, the reading is written as: 24.3 ± 0.1 mm. This means that the exact value of the reading is between 24.2 to 24.4 mm)

The fractional uncertainty of a measurement is the ratio of the uncertainty of the measurement over the read value. (Using the above example, its fractional uncertainty is given by [Δx/x] = [0.1/24.3])

The percentage uncertainty of a measurement is its fractional uncertainty expressed in percentage form. (Again, using the above example, the percentage uncertainty is given by [(Δx/x)*100%] = [(0.1/24.3)*100%])

When arithmetically operating on readings with uncertainties, always apply the fractional uncertainty formula. [Δa/a = Δb/b + Δc/c + ... + Δz/z, where Δa represents the uncertainty of quantity a, Δb quantity b, and so on]

For example, you are asked to find the answer of [(5 ± 1)*(7 ± 2)]:

• Δa/a = Δb/b + Δc/c
• (Δa/35) = (1/5) + (2/7)
• Δa = 17
• Answer = 35 ± 17

The 5 and 7 are multiplied as usual, but this is not the case with the uncertainty. This is why it is recommended to stick to the formula.

When measuring distance using a ruler, you are actually measuring two points and taking the difference between those two points. Therefore, the uncertainty should be doubled. (e.g. Smallest scale on ruler is 1 mm. You measured 250 mm. The uncertainty would be 250 ± 2 mm)

Chapter 15 - Waves

The electromagnetic spectrum:

• Gamma-rays (10^-14 < λ m < 10^-11)
• X-rays (10^-11 < λ m < 10^-8)
• Ultraviolet (10^-11 < λ m < 4 * 10^-7)
• Visible light (3.8 * 10^-7 < λ m < 7.5 * 10^-7)
• Infra-red (7.5 * 10^-7 < λ m < 10^-3)
• Microwaves (10^-3 < λ m < 1)
• Radio waves (1 < λ m < 10^9)

Chemistry Paper 2: Blind Spot Part 3

I'll skip nucleophile and electrophile with inorganic chemistry first because of the higher demand of the latter chapter.

For convenience, I will use point form in this section.


Since the Periodic Table tool is malfunctioning, we will use the one on top of this section.

The Basic

• Basically (this word is getting redundant), all of this chapter is about the electrons and protons. That's it!
• This is what I always assume. Imagine an element which possess 7 protons, 7 neutrons and 7 electrons. Then let's say I magically add 1 proton, 1 neutron and 1 electron (nothing less). This changes will affect their chemical and physical characteristics, hence its uses and chemical reactions . But we know that this kind of thing is virtually impossible. Just to give an example on how electrons and protons (and neutrons) is the crucial point in inorganic chemistry.
• In this part, we need to understand first the effect of the electron, especially their position at the outermost shell. As we all know, all chemical reactions depend on the transfer of electron.
  
• Like we all know, as the proton number increases, the number of electron increases.
• As the number of electron increases, hence the number of shell and orbital will also increase, as the electron need a space to move right?
• So as the number of shell increases, wouldn't the size of the atom will increase?
• So as the size of the atom increases, wouldn't the atomic radius increases?

Okay, grasp this concept first before we continue.

The Important Point

• Get back to the important point. So when atomic radius increases, the distance of its valence electrons from its nucleus will increase.
• Hence its attraction will decrease as the distance gets larger, right? (Important point)
  
• When the distance gets larger, the ionization energy will decrease as there is less attraction so less energy is required to remove them, right?
• Like we discussed before, as the proton number increases, so does electron and their shell. Down the group, they all have more proton numbers so their electron will increase too and so their shells and their atomic radius.
• Factors that affects ionization energy is not only their atomic radius. Size of their nucleus (their positive charge) and the shielding effect of electron will also play.
  
• So for group I-III, they will become more reactive down the group because it is easier for them to remove their outermost electrons down the group. However, the non metal which belongs to group V-VII accepts electron rather than donating them. To accept electron, they rely on their attractive force of their nucleus. This is another important point. Down the group, their reactivity will decrease as the distance between the nucleus and the outermost electron will become larger and hence their attraction will decrease.
• Another note for ionic radii. A lot of people often have misconception about this. You may notice in your textbook that, unlike atomic radius, ionic radius of elements have uneven trend. This is the reason. For positive ions, they will remove their outermost electron to form a positive ion. When we remove the electron, we will also remove its shell/orbital hence we will also reduce its size and atomic radius. The opposite things happen for negative ions. As they receive electrons, they will need to create shell/orbital to place their newfound electron. When this happen, the size of the atom will get larger and ionic radius will become larger. (Assuming that all elements will form cation, their ionic radii will decrease throughout the period).
• Electronegativity of elements decrease down the group. The reason is because of their attraction gets weaker as atom gets larger.
• Ionization energy is uneven across a period. This is due the fact that electron occupies different orbital and there is repulsion effect of electron when they occupy the same orbital.
  
• Take period 3 as an example. Sodium electron configuration is [Ne] 3s1, while magnesium's is [Ne] 3s2. Aluminium electronic configuration is [Ne] 3s2 3p1. The p orbital is located farther than the s orbital and it is at higher energy level (less energy required to remove electron at higher energy level). Use this logic to understand why first ionisation energy of aluminium is lower than magnesium, the second ionisation energy of aluminium is higher than magnesium and the third ionisation energy of magnesium is higher than aluminium.
  
• Again we will use period 3 as an example. Phosphorus electronic configuration is [Ne] 3s2 3p3, while sulphur's is [Ne] 3s2 3p4. Refer back to "Chemistry Paper 2: Blind Spot Part 2". The extra electron in sulphur is located at the px orbital. Inside the px orbital in sulphur, there are two electrons whilst in phosphorus there are only one electron. The two electrons will create a repulsion effect and it will decreases the ionisation energy. Hence phosphorus ionisation energy is higher than sulphur.

Group II

• Group II sulphates had its solubility decreases down the group. Oxide, on the other hand, becomes more soluble down the group.
  
• All of Group II nitrates are soluble and all of its carbonate are insoluble.
• Group II thermal stabilities increases down the group which means down the group, the metal carbonates and nitrates will be more stable towards heating down the group. Memorize all the three of them fully.
• Yes I know that Group II oxides are able to dissociate. But only at high temperature. What do I mean by high is 3000 C to 4500C.
  

Group VII

• There is a lot to remember. Firstly, down the group, Group VII reactivity decreases and its colour becomes darker. Their extent and ease of reaction will also decrease means that when it reacts with something, the yield will not be maximum (if it is supposed to form hydrogen halide, the result will be a mixture of hydrogen halide and the halogen)
• Hydrogen halides bond energy will decrease down the group. Hence it will become thermally less stable down the group and their reactivity will increases. Be careful on this part. Hydrides and elements are two different things. Many people (including myself) made mistakes when the questions ask about their states of matter (Iodine is solid, bromine is liquid at r.t.p) result after reaction (when you put a hot iron metal inside compound, example) and their reactivity with acids or salt (NaX).
• Each halides ion can be oxidised by the halogen above it. Example, I-(aq) can be oxidised to Br2 or Cl2.
• About oxidation, down the group it is easier to oxidise halide ions to halide element.
  

Thank you for reading this post. I hope that it will be useful for you guys.

Chemistry Paper 2: Blind Spot Part 2

Next is the ever elusive atomic structure and chemical bonding. The points that we often missed about these topics are the shape of orbital, shape of sigma and pi bond, what is instantaneous dipole or permanent dipole.

Orbital is the region where we can find the electron most of the time. Remember that electron moves randomly at a specific area? Orbital is simply the part of that specific area where we it has the highest probability of having the electron.

For now we only need to know the shape of s orbital and the p orbital.



Picture of s orbital



Picture of p orbital which consists of px, py and pz orbital. Note their orientation.

The difference of 1s orbital with 2s orbital is their size. 1s orbital is smaller than 2s orbital and the same goes for p orbital. It is just about their size.

When we draw the electron in the box diagram, take note of the placement of the arrows and their directions.


Move on to the chemical bonding. Important points here are the shape and how to draw sigma and pi bond. Another important point is the understanding between the difference of temporary dipole-dipole and permanent dipole-dipole.


This is the picture of s orbital overlapping another s orbital forming sigma bond (example H-H)



This is the picture of s orbital overlapping p orbital forming sigma bond (example H-Cl)



This is the picture of p orbital overlapping with another p orbital forming sigma bond (example Cl-Cl)

We can also draw a "sausage" (ellipse) for the px px sigma bond.

Pi bond is formed when two p orbitals overlap sideways to produce regions of electron density above and below the axis joining the two nuclear centers. Pi bond is formed when a species forms double bond or triple bond.



This is an example of py orbital forming a pi bond. Notice the shape of the orbital after the pi bond is formed.

Move on to instantaneous dipole or permanent dipole. This is one of three known intermolecular bonds that we have learned (I have absolutely no idea how many are there).

So basically both of these things are due to the effect of electrostatic attraction of the dipole that is created due to 2 different things:
(-) Random movement of electrons for the temporary/instantaneous dipole.

(-) The effect of polarization caused when an element with higher electronegativity bonded with another element of lower electronegativity for the permanent dipole.

The fact that movement of electron is extremely random will sometimes cause the electron to be positioned in such way:


Here we can notice at the second picture had all of its electron positioned at one side of the atom. Then, a millisecond later it would disperse right away. Then a few milliseconds later it would be positioned as the second picture and so on an on (remember, it is moving at speed of light).

When the electrons are positioned as in the second picture, the ion will be polarized for a while and it will produce a dipole with negative side at the left side of the atom (where the electrons gather). So the left side will be positively charged for a while isn't it?


This is what we called as the temporary dipole-dipole/ instantaneous dipole-dipole/ van der Waals force.

This bond is important only when all other bonds are not present, permanent dipole dipole and hydrogen bond. This is important in alkane and monoatomic element such as noble gases.

Take a look at HCl :



Here, we can see that Cl atom, being much electronegative than H, will polarize H and will cause it to become slightly more positive, whereas Cl will become slightly more negative. Thus when there are two HCl molecules come into contact:


The interactions (the vertical line) is the result of the attraction which is the permanent dipole. Basically, if it is a polar molecule, it will have the permanent dipole bond as the stronger intermolecular force.

Hydrogen bond is a special case. As defined in Wikipedia, hydrogen bond results from a permanent dipole force between hydrogen atom(s) bonded to nitrogen , oxygen or fluorine (thus the name "hydrogen bond", which must not be confused with a covalent bond to hydrogen). It is much stronger than the the permanent dipole force since the polarization here is much stronger (electronegativity of F is greater than Cl).

Consider this. Ethanoic acid (vinegar) is able to form a hydrogen bond where as ether such as methoxymethane (CH3OCH3) can't form any hydrogen bond.

Any question, feel free to ask.
All comments will be deeply regarded.

Thank you for reading this.

Chemistry Paper 2: Blind Spot Part 1

The title is quite aggravating.

So a few more days and we are going to have our P2 chemistry. Though not as difficult as its P1, still it poses a lot of challenge, especially if we want score higher.

Some points to remember is that the questions here are divided into several parts. What I mean is that each questions will usually correspond with one of the major topics in chemistry. So for a better mark, always skip the part that you feel incapable (example: organic chemistry) to do and search through the part that you are confident with (example: thermodynamic), don't waste time doing it linearly. Strange thing is that organic chemistry always seems to appear while the other parts comes out randomly (serious). So a good strategy is that to master your organic chemistry.

I bold "usually" with a reason. The setters may elaborate two topics in one question (periodic table and chemical bonding) or perhaps split one topic in two questions (isomerism and organic chemistry reactions). Study all parts, no matter what.

Some topics that resurges back are the electrolysis, atomic structure and chemical bonding.

Inorganic chemistry had become a frequent topic lately (periodic table, group II, group VII).

The topic that keeps freaking people is the nucleophile and electrophile; especially their reactions and how to differentiate them. Their mechanism, fortunately, is easily memorized once we know which is which.

When asked about electrolysis, the most common format of the question is:
(-) Draw a diagram and describe the electrolysis of bla, bla, blah
(-) Reaction at cathode and anode
(-) The products and their uses

To answer the first part, you can only memorize the diagram that we have learned, namely for the extraction of aluminum from aluminum oxide, electrolysis of brine and purification of copper. Note that we do not need to draw sophisticated diagram, a simple one will do.


This is a simple diagram for the extraction of aluminium.


And this is the picture for the electrolysis of brine. Do not forget to draw the diaphragm!

To determine the reaction at cathode or anode, all we have to do is to determine the cation (positive ion) and anion (negative ion). This is the important point. Cathode is the negative pole whilst anode is the positive pole. Therefore reaction at cathode involves cation (positive ion) while reaction at anode involves anion (negative ion).

The products for each electrolysis are aluminum (extraction of aluminum, duh), sodium hydroxide, chlorine and hydrogen gas (electolysis of brine) and copper (purification of copper). I suppose I don't need to discuss their uses because most textbooks had already discussed them and if you can't find them in textbook, Google is everybody's best friend (or Yahoo maybe).