Hooray!!! It’s SPM trial! (2010) – Johor

July 30, 2010

Attention: The Johor state SPM trial will start on 12/8/2010! Every Form 5 students will need to start and get busy preparing for the trial! For the other states in Malaysia, the date should not deviate too far from it either. (The “hooray” was just to loosen up the stressful environment)

SPM Trial 2010 – Johor State : 12th August 2010

Please note that the SPM trial examinations is to be taken seriously. This is due to the trials being:

1. used as forecast results for your SPM in event you wish to start applying for colleges earlier and have not yet obtained your results.
2. used as forecast for early scholarship application. (although most likely that the final decision is pending your actual SPM results)
3. a chance for you to gauge your ability prior to the actual SPM. This allows you to identify your weaker points and give you a warning to work on it.

So, all Berry Readers would have to sharpen their mind and 2B pencil! Students need to get prepare simple notes. For some general tips for SPM trials:

• Chemistry: Get ready with Periodic Table and Electrochemical series.
• Biology: Get ready with all the mechanisms, structures (ruminant and non-ruminant), blood circulation and processes!

Good luck to all taking the trials examination.

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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 4)

July 27, 2010

Welcome back to Berry Berry Easy. Today, the Berry Berry Teacher will like to share with all Berry Readers, Part 4 of the SPM Form 5 Chemistry notes on “Oxidation and Reduction”. In the previous part, oxidation and reduction were viewed in terms of changes made to the oxidation numbers. However, we also mentioned that it can also be understood through electron transfer. So yes, you guessed it! This part will focus on redox reaction in terms of electron transfer. Similar to very easy “change in oxidation number”, understanding redox reactions in terms of electron change are just as easy.

Do also try to memorise the table presented at the end of this post. Most of the time, objective questions will ask about the colours of the precipitation when iron (II) or iron (III) ions are present in the various reagents. So don’t forget this tips. Do pop quiz with your friends to memorise the table below.

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 4)

Oxidation and Reduction in Terms of Electron Transfer

2I^- (aq) –> I₂ (aq) + 2e
Oxidation: Iodide ion, I^- is a reducing agent because it donates/loses electrons to become I₂.

Br₂ + 2e –> 2Br^- (aq)
Reduction: Bromine water, Br₂ is an oxidising agent because it receives/accepts electrons from I^- to form bromide ions, Br^-.

–> Overall reaction: 2I^- (aq) + Br₂ –> I₂ (aq) + 2Br^- (aq)

Conversion of Fe²⁺ Ions to Fe³⁺ Ions and Vice Versa

A) Common oxidising agent (change Fe²⁺ ions to Fe³⁺ ions):

• bromine, Br₂
• chlorine, Cl₂
• hydrogen peroxide, H₂O₂
• concentrated nitric acid, HNO₃
• acidified potassium manganate(VII), KMnO₄ solution
• acidified potassium dichromate(VI), K₂Cr₂O₇ solution

Fe²⁺ (aq) –> Fe³⁺ (aq) + e
Oxidation: Iron(II) ion, Fe²⁺ is a reducing agent because it donates/loses one electron to become Fe³⁺.

Br₂ (aq) + 2e –> 2Br^- (aq)
Reduction: Bromine water, Br₂ is an oxidising agent because it receives/accepts electrons from Fe²⁺ to form bromide ions, Br^-.

–> Observation: iron(II) sulphate solution changes colour from pale green to yellowish-brown.
–> Overall reaction: 2Fe²⁺ (aq) + Br₂ (aq) –> 2Fe³⁺ (aq) +2Br^- (aq)

B) Common reducing agent (change Fe³⁺ ions to Fe²⁺ions):

• zinc powder, Zn
• aluminium, Al
• Magnesium, Mg
• Calcium, Ca
• Sulphur dioxide, SO₂
• Hydrogen sulphide, H₂S
• Sodium sulphide solution, Na₂SO₃
• Tin(II) chloride solution, SnCl₂

Zn (s) –> Zn²⁺ (aq) + 2e
Oxidation: Zinc powder, Zn is a reducing agent because it donates/loses two electrons to form zinc ions, Zn²⁺.

Fe³⁺ (aq) + e –> Fe²⁺ (aq)
Reduction: Iron(III) ion, Fe³⁺ is an oxidising agent because it receives/accepts one electron to become Fe²⁺.

–> Observation: iron(III) sulphate solution changes colour from yellowish-brown to pale green.
–> Overall reaction: 2Fe³⁺ (aq) + Zn (aq) –> 2Fe²⁺ (aq) + Zn²⁺ (aq)

C) Investigate the presence of iron(II) and iron(III) ions

The next Part in this series contains arguably the most important and memorable “series” in your SPM Chemistry studies, namely the Electrochemical Series. It’ll be something that you’ll memorise even after you leave school. So stay tune and log in frequently to Berry Berry Easy.

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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 3)

July 24, 2010

As promised in the previous post in this series, let Berry Berry Easy present all Berry Readers with Part 3 of the SPM Form 5 Chemistry notes on “Oxidation and Reduction”. The previous two parts were focused on the concepts of redox reactions and oxidation numbers. (You ought to understand how to write a redox reaction from the two half reaction components by now, else do revise from the beginning before starting this part) Now, we are going to show all of you how oxidation and reduction can be understood in terms of changes in oxidation numbers. (Do remember that you can also view it through the gain and loss of electrons, but that’s for another time).

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 3)

Oxidation and Reduction in Terms of Changes in Oxidation Numbers

Redox reactions – oxidation number of all elements change.

Rusting of iron, combustion, displacement of metal from its salt solution, displacement of halogen from its halide solution and electrolysis are redox reaction.

-10 …. -3  -2  -1  0  +1  +2  +3  …  +10

<———-  Reduction || Oxidation ———->

• H₂ (g) + CuO (s) –> H₂O (l) + Cu (s)
  Hydrogen: 0 –> +1 (Oxidised to water & Hydrogen is a reducing agent)
  Copper oxide: +2 –> 0 (Reduced to copper & Copper oxide is a oxidising agent)

• 2Zn (s) + O₂ (g) –> 2ZnO (s)
  Zinc: 0 –> +2 (Oxidised to zinc ion & Zinc is a reducing agent)
  Oxygen: 0 –> -2 (Reduced to oxide ion & Oxygen is an oxidising agent)

• 2Mg (s) + CO₂ (g) –> 2MgO (s) + C (s)
  Magnesium: 0 –> +2 (Oxidised to magnesium ion & Magnesium is a reducing agent)
  Carbon dioxide: +4 –> 0 (Reduced to carbon & Carbon dioxide is an oxidising agent)

• Br₂ (l) + 2HI (aq) –> 2HBr (aq) + I₂ (s)
  Hydroiodic acid / Hydrogen iodide: -1 –> 0 (Oxidised to iodine & Hydroiodic acid is a reducing agent)
  Bromine: 0 –> -1 (Reduced to hydrobromic acid & Bromine is a oxidising agent)

Non-redox reactions – oxidation number of all elements remain unchanged.

Precipitation, decomposition and neutralisation are not redox reaction (non-redox reaction)

Precipitation:

• AgNO₃ (aq) + NaCl (aq) –> AgCl (s) + NaNO₃ (aq)
  +1 +5 3(-2)      +1  -1              +1  -1        +1 +5  3(-2)

No change in the oxidation numbers.

Decomposition:

• ZnCO₃ (s) –> ZnO (s) + CO₂ (g)
  +2 +4  3(-2)    +2 -2       +4  2(-2)

No change in the oxidation numbers.

Neutralisation:

• NaOH (aq) + HCl (aq) –> NaCl (aq) + H₂O (l)
  +1 -2 +1          +1 -1             +1  -1             2(+1)  -2

No change in the oxidation numbers.

Revision time:

Questions to ask yourself at this point:

1. Do you understand what is a redox reaction?
2. Can you write the two half reactions out of a redox reaction?
3. Do you understand the concept of oxidation number?
4. Give three examples of an oxidising agent and the example of the reaction involved.
5. Give three examples of a reducing agent and the example of the reaction involved.
6. Can you differentiate a redox reaction with a non-redox reaction?
7. What is the characteristics of a non-redox reaction?

If you can answer all these questions, it shows that you understand the basics of oxidation and reduction.

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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 2)

July 22, 2010

Berry Berry Easy would like to present all with Part 2 of SPM Form 5 Chemistry “Oxidation and Reduction”. In the previous part, the concept of redox reaction was introduced to all students. By now, all of you should be comfortable in identifying the “two reactions” (reduction and oxidation) that make up a redox reaction. In this part, the intention is to convey the concept of “oxidation number“. Ample of examples on it and sample calculations are also given. However, do take note also that some metals might break convention and show different oxidation numbers (as shown in the tables towards the end).

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction(Part 2)

Oxidation Number – is the charge that the atom of the element would have if complete transfer of electron takes place.

Oxidation number

(i) Free elements have an oxidation number of zero.

Na = 0
Mg = 0
C = 0
H₂ = 0
Br₂ = 0

(ii) Oxidation number of a simple monoatomic ions is the same as its charge.

Na⁺ ion = +1
Mg²⁺ ion = +2
O^2- ion = -2
Cl^- ion = -1
H⁺ ion = +1

(iii) Sum of the oxidation number for a neutral compound is zero.

CaH₂
(+2) + 2(-1)
= 0
Sum of oxidation number is 0

Al₂O₃
2(+3) + 3(-2)
= 0
Sum of oxidation number is 0

Iodine, Bromine, Chlorine, Nitrogen, Oxygen, Fluorine

—> Electronegativity increase

Cl₂O
2(+1) + (-2)
= 0
Sum of oxidation number is 0.

(Chlorine, bromine and iodine usually have the oxidation number of -1 except when combine with a more electronegative element.)

HClO
(+1) + (+1) + (-2)
= 0
Sum of oxidation number is 0.

(Chlorine, bromine and iodine usually have the oxidation number of -1 except when combine with a more electronegative element.)

(iv) Polyatomic ion, the sum of the oxidation numbers of all the atoms equals the charge on the ion.

SO₄ ^2-
(+6) + 4 (-2)
= +6 + (-8)
= -2
Sum of oxidation number is -2

Cr₂O₇^2-
2(+6) + 7(-2)
= -2
Sum of oxidation number is -2

(v) Calculating the oxidation numbers of elements in compounds or ions.

K₂Cr₂O₇
2 (+1) + 2x + 7 (-2) = 0
x = +6
Oxidation number of chromium in K₂Cr₂O₇ is +6

NO^3-
x + 3(-2) = -1
x = +5
Oxidation number of nitrogen in NO^3- is +5

Hydrogen peroxide, H₂O₂
2(+1) + 2x = 0
x = -1
Oxidation number of oxygen in H₂O₂ is -1 (and not -2)
(Usually oxidation number for combined oxygen usually is -2 except in peroxides)

F₂O
2(-1) + x = 0
x = +2
Oxidation number of oxygen in F₂O is +2 (and not -2)
(Usually oxidation number for combined oxygen usually is -2 except in fluorine compounds)

NaH
(+1) + x = 0
x = -1
Oxidation number of hydrogen in NaH is -1 (and not +1)
(Usually oxidation number for combined hydrogen usually is +1 except in metal hydrides)

AlH₃
(+3) + 3x = 0
x = -1
Oxidation number of hydrogen in AlH₃ is -1 (and not +1)
(Usually oxidation number for combined hydrogen usually is +1 except in metal hydrides)

MgH₂
(+2) + 2x = 0
x = -1
Oxidation number of hydrogen in MgH₂ is -1 (and not +1)
(Usually oxidation number for combined hydrogen usually is +1 except in metal hydrides)

(vi) Some metals show different oxidation numbers.

(vii) Usually non-metals have negative oxidation numbers but Cl, Br & I can have positive or negative oxidation number.

Stay tune for the next installment in the “Oxidation and Reduction” series with focus on the difference between redox reaction and non-redox reactions.

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Berry Role Model No.5 – Aspiring Designer – Cheryl Chu

July 16, 2010

Welcome to Part 5 of the Berry Role Model Series. The featured Role Model this issue is Ms Cheryl Chu, an aspiring designer. For most people, when we see the word “designer” we would immediately think of art stream students. While this might be true most of the time, but it is predominantly due to societal pressure forcing conformity. Academically bright students are always ushered into science stream, rather than being encourage to take up what they excel in and where their passion lies. However, we have with us today a young lass by the name of Cheryl Chu who dares to follow her dreams and break with convention. She thinks it is okay to be unconventional and live life with no “what ifs”. And to walk the talk, she switched from science stream in SPM to pursue her dream in an arts based course (Diploma in Design and Media majoring in Advertising). Not only is she offered a place at the Nanyang Academy of Fine Arts (NAFA) in Singapore, she is also awarded a tuition grant from Singapore MOE and a bursary award from the Lee Foundation.

We are sure without a doubt that our Berry Readers would love to know a bit more about this go-getter and dare devil in life. You can read about her life and thoughts at <http://cherylchu.com>. In her blog, you would find her portfolio. (You won’t believe what she has amassed in her portfolio at the tender age of 17. But considering that she started learning Photoshop at 13, we can now fathom the phenomenon) We thought of not wanting to spoil the surprise of what you can find when you view her portfolio, but Berry Berry Easy cannot resist spilling the beans on the fact that she and her friends was The Star’s NiE Mag Inc. 2009 National Competition winner. She has also a design site (Cherying Design) producing fresh designs at <http://www.cherying.org>. Be sure to check them out.

Despite her grandiose portfolio and achievements, we were most impress by her bravery and maturity in being daring enough to go against convention to chase her dreams. We hope Berry Readers can take a leaf from her and chase your dreams too. Let the interview with Cheryl Chu begins.

Role Model #6 – Aspiring Designer – Cheryl Chu

1) Who are you?

> Coming from the city of Kuching, Sarawak, I’m a soon to be 18 year old girl who just finished her SPM examination last year and will be pursuing tertiary education in the field of Design and Media in due time.

2) So what do you do daily as an aspiring designer?

> Frankly, being a high school graduate who is anticipating for college to commence, I do not have much of a fixed daily routine. As a freelance designer now, I usually check my emails on a daily basis because I communicate through email or instant messaging with my clients based outside of Kuching. For clients based in Kuching, I would mostly meet them out to discuss as I find it easier to communicate with a person face to face. I would usually then work on projects from websites to print publishing and finally do a lot of reading on the web. The Internet is a place where one can find vast information and resources and it’s why I like to take time off to get inspired while I still can. I believe that when I start college, life will be much different and time for my own stuff would lessen considerably.

3) What do you aspire to achieve in the future, considering that you have already achieve relatively much for your age?

> I feel that every young mind have ideas and dreams of their own and it’s the same for me. Besides focusing on my studies, I would love to work on achieving a name for myself so that more people would be exposed to my work. If possible, I’d like to be more than just a designer in the future, maybe enter entrepreneurship or more so, contribute back to society.

4) Tell us about your passion in design and how it started? Who/What inspired you to take up designing seriously?

> The story of how it all started has always amused me till now because never once then, would I think that these things I did would have changed my life so much. It initiated purely as entertainment when I created a “fansite” for my idol, Hilary Duff. Funny it may seem, when one sees a 13 year old doing something like this may find it much of a pointless and time-consuming thing to do, but as I look back at that 13 year old me, a shy girl who did not dare share to the world what she was doing behind the computer, I’ve learnt a lot of myself over the years. On top of that, I used my spare time as a teenager to learn about creating websites and reading online tutorials on techniques in designing graphics and sites with Adobe Photoshop.

Back in high school, being granted the opportunity of designing for my school’s magazine, prefectorial board, the class and for clubs had also contributed to me wanting to take up designing. I feel that the self-satisfaction and also appreciation I get from people who like my work had been one of the co-factors that has been pushing me all these while to pursue this passion. Besides, being able to make my family proud and believe in the things I do makes me trust my instincts and decisions I make.

Basically, my inspiration comes from my environment, and more specifically from the World Wide Web. Any creative piece of art/design I see that is worth giving a thumbs-up to, drives me to improve myself everyday for the better.

5) What are the traits that you have that has brought you success so far? (Our Berry Readers who would like to follow your footsteps would know what to aim for)

> Have lots of faith. I have always been a firm believer in putting a lot of faith in the things I aim for and to always, always strive hard to reach those goals. I am adamant that working hard pays off in the end.

6) So, share with us your other secret of how you (and your friends) achieved national success in the Star NiE Mag Inc 2009 competition? How would you advise our Berry Readers if they want to replicate your success?

> I don’t think there was any real secret behind winning this competition. We all worked hard together as a team, did some brainstorming together, focused on every single detail we could think of in those pages and just pulled ourselves through while having the confidence that we did our best.

7) Before we part, can you tell us what made you go against the convention to pursue your dreams? Were there any external pressure stopping you from pursuing your dreams (you dont have to mention the names though)? How did you overcome them?

> It’s very simple, I followed my heart. =)

There are certain people that tend to look down on those pursuing art related courses because they feel that these are careers that cannot bring one far enough to succeed. They do so sometimes not through speech but in the tone of their voices or facial expressions when I am asked of my plans for tertiary education. In the beginning, I avoided such questions or gave the typical “I’m not sure yet” answer because I was uncomfortable with being judged based on the opinions of others. I do understand that a lot of people care and feel that way because they fear I may be making the wrong choices which I might regret in the future. However, I realized though I cannot prove to people with such mindsets wrong at this very point of time but I can do so in the future, as long as I continue to stick with what I feel is right and not be easily shaken by others.

8 ) Oh, before we really end this interview you, can you share with us some “words of wisdom” which have brought you success over the years.

> Though I doubt I have experienced or achieved enough to actually impart “words of wisdom” to others, I have one thing to add however cliché or cheesy it may be, that is for anything one dreams in life, dare to pursue them no matter how crazy they are because it’s one’s determination and willpower that brings that person far in life.

- End -

Here are some of Cheryl’s other fantastic achievements:

• Historical figure portrait drawing competition 2005 (School) – 1st place
• “History of Sarawak” album making competition (State) 2008 – Consolation
• North Star Art Project (National) 2008 – Shortlisted to finals
• Bag the Awareness bag design competition 2009 & Breast Cancer exhibition at Bangsar Village II (National) – Consolation
• Star-NiE: Mag Inc 2009 (National) – Grand prize

Now reward yourself with a sub-sample or teaser of her work (better still hop over to her website for a view of her portfolio):

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STPM Chemistry Form 6 Notes – Terminology and Concepts: Liquid and Solid States (Part 5 – Final)

July 15, 2010

Berry Berry Easy would like to present the final part on long-running topic of “Liquid and Solid States” for STPM Form 6 Chemistry. Despite being the final part in the series, this is certainly the most important part in the whole chapter due to the berry important phase diagram, triple point (uniqueness of three phases coexisting), the uniqueness of water (be thankful that water had the properties it had or else we would be living today). The aforementioned subtopics are rather popular in topical tests and major examinations. So make sure you understand them fully. Do also make use of the super summary table at the of the post. (Remember to draw out the phase diagrams)

STPM Chemistry Form 6 – Terminology and Concepts: Liquid and Solid States (Part 5)

Summary: Definition of the states of matter

Phase – refers to a single homogeneous physical state of a heterogeneous system. There are three phases with the same composition solid, liquid and gas.

Triple point – the point of a condition of temperature and pressure at which the solid, liquid and vapour phases exist simultaneously at equilibrium.

Critical point – is the highest temperature and highest pressure at which there is a difference between liquid and vapour states. At either a temperature or a pressure over the critical point, only a single fluid state exists, and there is a smooth transition from a dense, liquid-like fluid to a tenuous, gas-like fluid/or pressure that is required to liquefy a gas at its critical temperature.

Supercooling – metastable condition where a liquid can exist below its freezing point.

Phase Diagrams

In laboratory, experiments are being carried out on two environmental factors which is temperature and pressure (referred to as independent variables).

A) The Phase Diagram of Water
- ice (solid), water (liquid) and water vapour / steam (gas)

• Vapour Pressure Curve
  - critical point = critical temperature (374˚C) and critical pressure (200 atmospheres)
  - temperature above 374˚C and critical pressure 200 atmospheres, the vapour and liquid are indistinguishable (no longer two separate phases) because the densities of the gas and liquid are equal (meniscus separating a liquid from its vapour disappears).

• Melting Temperature Curve
  - melting temperature point decrease with pressure- supercooling is the cooling of a liquid to below its freezing point without a change taking place from the liquid to the solid state. A phenomenon (metastable condition) shows the vapour pressure of water below its freezing point.

• Triple Point
  - Water triple point is at temperature 0.01˚C and pressure 0.006 atm (610 N m^-2). All the three phases (ice, water and water vapour) coexist at equilibrium.

• Normal Melting Temperature Point
  - the temperature at which both the solid and the liquid states of the substance exist in equilibrium at a pressure of 1 atm (101 kNm^-2)

• Normal Freezing Temperature Point
  - the temperature at which both the liquid and the solid states of the substance exist in equilibrium at a pressure of 1 atm (101 kNm^-2)

Unsual Behaviour of Water

• i) Why ice can float?
  - the volume of water increase when the change of phase from liquid to solid.
  Reasons: Ice (solid) has an open structure (hydrogen bond).

• ii) Why the melting temperature curve slopes to the left (melting point decreases with pressure)?
  - (In most of substances (except water), an increase in pressure will push the molecules even closer / Increase in pressure favours the physical state which is higher density)
  Reasons: Increasing the pressure favours the formation of liquid water due to the latent heat of fusion is absorbed from the surroundings during melting.

B) The Phase Diagram of Carbon Dioxide
- solid carbon dioxide (dry ice), liquid carbon dioxide and gas carbon dioxide

• Vapour Pressure Curve
  critical point = critical temperature (374˚C) and critical pressure (217 atmospheres)
  - temperature above 374˚C and critical pressure 217 atmospheres, the vapour and liquid are indistinguishable because the densities of the gas and liquid are equal. At this point, carbon dioxide gas can be liquefied.

• Melting Temperature Curve
  - melting temperature point increase with pressure
  - melting temperature curve slopes to the right
  - density of dry ice (solid carbon dioxide) is higher than the density of liquid carbon dioxide. It is because the carbon dioxide molecules are held closer together (smaller volume).
  - Increasing the pressure favours the formation of solid carbon dioxide due to the latent heat of fusion is liberated (given out) to the surroundings.

• Triple Point
  - Carbon dioxide triple point is at temperature -57˚C and pressure 5.1 atm. All the three phases (solid, liquid and gas) coexist at equilibrium.

• Normal Sublime Temperature Point
  - the temperature at which both the solid and the gas states of the substance exist in equilibrium at a pressure of 1 atm (101 kNm^-2)
  - at atmospheric pressure, solid carbon dioxide (dry ice) sublimes to form carbon dioxide gas at -78˚C.

Berry Important Points of Water Phase Diagram and Carbon Dioxide Phase Diagram

Now we have reached the conclusion of this topic. Be sure to understand this topic in full if you want to score in chemistry. Do also learn the Berry Important Points table of the water and carbon dioxide phase diagram by heart.

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Berry Event No.1 – National Robotics Competition (NRC) 2010 – Zone H : Johor State Level

July 12, 2010

[This is the inaugural post of the new Berry Berry Easy category called Berry Event. Posts from this category is dedicated to the student-related events which team members from Berry Berry Easy participated or were invited to cover the event. Yeah, you heard us right! You may invite us to cover your student-related event if you want to raise the profile of your event. All in all, post from this category would be mainly to highlight the activities of students under the Malaysian education system, so that students who were unaware of the events may take part in the future, while parents who were worried about what their children is up to will be able to see the event first hand.]

In this very first Berry Event, the Berry Berry Teacher would like to bring attention to all Berry Readers and parents about the National Robotics Competition (NRC) 2010 for the Johor Zone (State-level) competition. For those who are unfamiliar with this nationwide robotics competition for all school going children in Malaysia, the 2010 event is the 6th edition of the NRC and second for the Berry Berry Teacher (and her students in SMK Dato Jaafar).

Figure 1: The gladiators from SMK Dato Jaafar Contingent for NRC 2010 (Back row (L-R): Toh Ian Kim, Darwin, Luqman, Kelvin, (slightly front) Aaron, Founder Jin Heng, Founder Wai Loon, and Founder Zebedee. Front row (L-R): Aniq, Fikri, Ian, Clerence, Tahir, Yee, Wen Wen and Berry Berry Teacher SDJ Mentor Isabelle Wong.

Berry Event #1 – National Robotics Competition 2010 – Zone H (Johor State Level)

• Event: National Robotics Competition 2010 Zone H (Johor State)
• Date: 26th June 2010 (Primary School), 27th June 2010 (Secondary School)
• Venue: Persada Johor International Convention Centre , Johor Bahru
• Event Covered by: Miss Isabelle Wong – The Berry Berry Teacher – SDJ Robotics Mentor
• Dedicated to: The founding batch of the SDJ Robotics Club and this year’s competing robotics team from SDJ

Unlike in 2009, this year’s NRC competition for Johor district/local level competition were scrapped and teams were fast-tracked into the state-level copetition. So competition for this year were stiffer than in the previous year and teams only had one chance to show their mettle against their friendly opponents.

1) Treasure Hunt – Regular Category for Lower Secondary

This category was a keenly contested category with 69 teams from all over the state of Johor competing for the sole slot to advance to the national finals. The great SMK Bandar Putra (with strong history) teams once again dominated the field with a 1-2 finish in the category (their other teams finished 12th and 13th, which are very commendable too). The winning team B49 of SMK Bandar Putra blazed the trail with a perfect 100 score and a time of 24.94 seconds. A total of 5 teams scored perfect 100 score, and they are:

1. SMK Bandar Putra (Kulai) – 100 points – 24.94 seconds
2. SMK Bandar Putra (Kulai) – 100 points – 30.07 seconds
3. SMK Dato Menteri (Batu Pahat) – 100 points – 30.38 seconds
4. SMK Dato Jaafar (Johor Bahru) – 100 points – 32.06 seconds (Yup, this is my school team)
5. SM Sains Kota Tinggi (Kota Tinggi) – 100 points – 45.44 seconds

Congrats to the medalists in this category. All of you thoroughly deserve your victories as it must have been tough training for the competition. It is also heartbreaking on a personal note for myself and my students who missed out on a bronze by 1.2 seconds. Nonetheless, great effort to be no.4 in Johor and highest scorer of Johor Bahru. However, it must be said that everybody is a winner, as we all learnt some new stuffs and struck out new friendships.

Other statistics (at the moment you can find the analysed data only from Berry Berry Easy):

• The average score (mean) for all teams are 31.449 points. So all teams that scored above 30 points can be said to have done very well.
• The most common score (mode) for teams in this category is 0 points. A total of 25 teams (~36% of all teams) did not register any score. This showed the difficulty of this category.
• The median score for teams in this category is 20 points. So any teams who scored more than 20 points can declare themselves to have won more than half of the teams on the field.

2) Vertical Limit – Regular Category for Upper Secondary

Teams in this category were even more evenly matched with more than 10% of all the 56 teams scoring perfect 100 scores. However, the percentage is misleading as 4 of the 6 teams who scored perfect 100 scores were from the dominating SMK Tinggi Batu Pahat. It is no wonder that they also got 1st, 2nd, 4th and 5th place for the category. The only team to break the stranglehold is SMK Taman Daya 2 and SMK Teknik Johor Bahru with a 3rd and 6th place, respectively. The perfect score teams are:

1. SMK Tinggi Batu Pahat (Batu Pahat) – 100 points – 11.53 seconds
2. SMK Tinggi Batu Pahat (Batu Pahat) – 100 points – 11.81 seconds
3. SMK Taman Daya 2 (Johor Bahru) – 100 points – 11.97 seconds
4. SMK Tinggi Batu Pahat (Batu Pahat) – 100 points – 12.38 seconds
5. SMK Tinggi Batu Pahat (Batu Pahat) – 100 points – 15.12 seconds
6. SMk Teknik Johor Bahru (Johor Bahru) – 100 points – 28.47 seconds

Congrats to the SMK Tinggi Batu Pahat for dominating this category. A special congratulation also to SMK Taman Daya 2 for garnering the best team in Johor Bahru district. On a personal note, I’m pleased that my school team from SMK Dato Jaafar did a commendable job of scoring 60 marks with a 16th placing.

Other statistics (at the moment you can find the analysed data only from Berry Berry Easy):

• The average score (mean) for all teams are 44.82 points. So all teams that scored above 40 points can be said to have done very well.
• The most common score (mode) for teams in this category is 60 points. A total of 25 teams (~44.6% of all teams) scored 60 points, which is equivalent of the successful climbing of the first tower. This showed the difficulty of this category lies in transferring the robot from Tower A to Tower B.
• The median score for teams in this category is 60 points. So any teams who scored more than 60 points can declare themselves to have won more than half of the teams on the field.

3) Open Category for Lower Secondary

Scores in this category were not published publicly. However, it must be said that all teams performed well in this category. Countless of photos has been published on Facebook and other photo sharing sites which showed the excellence achieved by the lower secondary school students. (At the moment, do not have the list of the winners yet)

Figure 2: The open category team from SMK Dato Jaafar with their creation of Wan the Robotic Dragon. The spirit of 1Malaysia is displayed when Malaysian irrespective of race come together to promote Dragon dance to the rest of the world.

I’m delighted with the performance of the Form 1 students from my school, SMK Dato Jaafar who battled against the elder boys from the other schools valiantly. Although a bronze is all we can show, but this should be something that we can build upon in the future, in order to launch an assault on the gold medal.

Figure 3: The open category team from SMK Dato Jaafar with their bronze medal in the Lower Secondary Open Category in the NRC 2010 Johor State level.

Awards given to the SMK Dato Jaafar team:

• Bronze medal in Johor State level
• Best Spirit Award in Johor State level – to the entire school team
• Best Mentor in Johor State level – to the guiding mentor

Figure 4: The spoils of victory from the NRC 2010 Johor State level competition. Good job done from all the competitors of the SMK Dato Jaafar Robotics Club.

I must say that I’m very happy to have won the best mentor award for my category in my second year in charge of the SDJ Robotics Club and first venture into the open category. I would like to say that all the mentor for all teams are equal winners in my eye. All our sacrifice are worth it in the end when we see the product of our student’s ability.

Figure 5: Berry Berry Teacher collecting her NRC 2010 Best Mentor (Johor State Level) award.

I’m also happy with the undying spirit shown by my student, not only in the open category, but all from Form 1 to Form 5 who have embodied the Jaafarian spirit. So we are humbled to have won the Best Spirit award. In all honesty, we would like to share this award with all teams especially those who travelled from the other district just for this competition. They should too be given best spirit award.

Finally, a hearty congratualtions to the winners in this category, best innovation and best presentation. (Will update this when I get the full list) Hope to see you all again in next years competition. Hopefully, our Johor state winner will do us proud and win the national and world championship.

Other categories that were not covered in this report include:

• Open category for upper secondary
• Primary school competition

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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 1)

July 9, 2010

Berry Berry Easy would like to present all Berry Readers with one of the most interesting topic (as voted by most students) in SPM Form 5 Chemistry. Without prolonging the suspense, the interesting topic as mentioned by most chemistry takers is “Oxidation and Reduction“. So why is this interesting? Based on a straw poll, it is simply because of the coolness of the word Redox, which is a shorthand for REDuction/OXidation = REDOX!!! Teachers only need to mention this once, and students can remember this virtually forever. Its concept are also rather simple, but at the hardest end (not covered by the syllabus), it can be hard to identify a redox reaction.

One obvious tip to understand redox reactions is to think of them as components of a bigger reaction. So a redox reactions is best imagined as two smallers parts of oxidation and reduction. (If you would like to impress your friends, you can also tell them that, non-redox reactions are called methathesis reactions.)

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 1)

Redox reaction – chemical reactions in which both oxidation and reduction occur simultaneously.

1) Oxidation

• gain of oxygen, O₂ by a substance
• loss of hydrogen, H₂ from a substance
• a loss of electrons
• occurs when there is an increase in oxidation number

2) Reduction

• loss of oxygen, O₂ by a substance
• gain of hydrogen, H₂ from a substance
• a gain of electrons
• occurs when there is an decrease in oxidation number

Oxidation Number – is the charge that the atom of the element would have if complete transfer of electron takes place.

IUPAC nomenclature – name inorganic compounds in order to avoid confusion due to elements have variable oxidation numbers.

Oxidation and Reduction in Terms of Gain and Loss of Oxygen

2CuO (s) + C (s) –> 2Cu (s) + CO₂ (g)

• Reduction:
  CuO loses its oxygen to form copper. CuO (oxidising agent) is being reduced.
• Oxidation:
  Carbon gains oxygen to form CO₂. Carbon (reducing agent) is said to be oxidised.

PbO (s) + CO (g) –> Pb (s) + CO₂ (g)

• Reduction:
  PbO loses its oxygen to form lead. PbO (oxidising agent) is being reduced.
• Oxidation:
  Carbon monoxide (CO) gains oxygen to form CO₂. Carbon monoxide (reducing agent) is said to be oxidised.

H₂ (g) + CuO (s) –> H₂O (l) + Cu (s)

• Reduction:
  CuO loses its oxygen to form copper. CuO (oxidising agent) is being reduced.
• Oxidation:
  Hydrogen (H₂) gains oxygen to form H₂O. Hydrogen (reducing agent) is said to be oxidised.

Oxidation and Reduction in Terms of Gain and Loss of Hydrogen

H₂S (g) + Cl₂ (g) –> S (s) + 2HCl (g)

• Reduction:
  Cl₂ gains hydrogen to form hydrogen chloride. Cl₂ (oxidising agent) is being reduced.
• Oxidation:
  Hydrogen sulphide loses hydrogen to form sulphur. Hydrogen sulphide (reducing agent) is said to be oxidised.

2NH₃ (g) + 3Br₂ (g) –> N₂ (g) + 6HBr (g)

• Reduction:
  Bromine gains hydrogen to form hydrogen bromide. Br₂ (oxidising agent) is being reduced.
• Oxidation:
  Ammonia loses hydrogen to form nitrogen. Ammonia (reducing agent) is said to be oxidised.

Do drop by Berry Berry Easy soon again for the continuation of this interesting chapter.

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