 "> | 1. Distance - time graph 2. Speed - time graph 3. How to find acceleration and area under graph 4. Formulae of all forces 5. How to increase the stability of an object 6. Changing of states in terms of kinetic theory 7. Relationship between pressure and temperature/volume 8. Why metals conduct heat the fastest 9. Convection current 10. What makes a good radiation absorber/emitter 11. Wave terms [ Amplitude,wave length] 12. Formulae of waves and light 13. Characteristics of image formed by thin converging lens 14. Properties of electromangetic spectrum rest are application -_- lol |
| "> | Heat capacity will be up tmr night! This is strongly recommended as a must to memorize in chemistry 1. How to form formulae 2. How to do balance + ionic equation 3. Chapter 1 -_- 4. Explaining the changing of states in terms of kinetic theory 5. Drawing of atomic structure [ionic compound] and [covalent compound] 6. Properties of covalent and ionic compound and gaint molecular substance 7. Explanation of properties of covalent and ionic compound except solubility 8. The 6formulae of mole concept 9. The identification of Acid Base and Salt [Includes Acidic, Basic, Amphoteric, Neutral Oxide] 10. The solubility of Base and Salt 11. Indicators and its colour changes 12. Acid Base Salt and Metal equations 13. Reactivity series 14. Properties of Acid and Bases 15. The 3 method in preparation of salt 16. Properties of alkali metals, halogens and noble gases [Its trend] 17. Metals reaction with water and steam 18. Blast Furnace 19. Uses of metals [prevent rusting] |
| "> | I choose to upload my geography notes first. for those we do not know how to use host-web file to download, please inform me to send it over to your mail-box my apology being unable to do the notes on chapter 10 due to insufficient time http://rapidshare.com/files/60779870/Geography_Notes.doc http://www.megaupload.com/?d=RP41OB0T http://download.yousendit.com/EEE945595F12F478 http://files-upload.com/files/543802/Geography%20Notes.doc http://www.yourfilehost.com/media.php?cat=other&file=Geography_Notes.doc yousendit will expire in a week megaupload will expire in a month rapidshare will expire in a month if im not wrong |
| "> | For heat capacity and latent heat, there is no need to use formula, it will make you even confuse. In fact, there is only 1 formula needed to tackle a difficult question. First, you have to understand wat is the unit is talking. Whatelse, the units are the one that give you the defination of heat cacpty, speceific heat capacity and specific latent heat. Whenever you see the word 'specific', it involves with mass. Therefore, Heat capacity = J / K Specific heat capacity = J/ Kg K Specific latent heat = J/Kg these units are simply telling the amount of energy PER 1Kg and 1K if it is not 'specific' it will only have 1k The first problem is tackle questions that require u to find J,K or Kg Note: the C below means degree celcius Skill 1 : Calculating Energy A 2kg copper object is heated from 30 C to 35 C. How much increase in thermal energy is required, given that the specific heat capacity of copper is 400J/Kg K first step, use the heat capacity as your guide. write it as 400J/ 1Kg 1Know the qns asked for 2kg, so simply multiply both side by 2 400J / 1Kg 1K[when 1side has be multiplied by 2, another side will be the same]800J / 2Kg 1K Note: please do not multiply by 2 such that it became 2Kg 2K, this means you are actually multiplying by 4! it works like this, 2 x 2 = 4 see the link? now the next time is the find the temperature difference, 35 - 30 = 5 so do the same thing, multiply both sides by 5 4000J / 2Kg 5K noticed that the 2kg remains unchange. so thats your ans, you need 4000J ! Skill 2 : Calculating mass 4000J of energy is used to heat up a copper block from 50 C to 70 C. What is the mass of the copper block, given that the specific heat capacity of copper is 400J / Kg K? this part is a little bit different, its about finding the mass but just do the same thing, write it down first 400J / 1Kg 1K it mention 20 C so lets just multiply by 20 8000J / 1Kg 20k now you have 8000J, its way of course of wat the question is asking which is only 4000J so now jus bring it down, by dividing it by 2, however, u cannot divide temperature change as the qns said that it increases by 20 C. so simple, jus divide Kg by 2 4000J/0.5Kg 20K hence, the answer is 0.5kg. easy? Skill 3 : Calculating temperature change 4000J of energy is used to heat up a 5Kg copper block from an initial temperature of 20 C. What is the final temperature of the copper block, given that the specific heat capacity of copper is 400J / Kg K 400J / 1Kg 1K copper is 5Kg, so multiply by 5 2000J / 5Kg 1K u need 4000J so to change from 2000J to 4000J you have to muliply 2 again, Kg should not be change 4000J / 5Kg 2K Hence, it raises 2 K, final temperature is 22 C thats half of capacity covered. Simple eh? time for latent heat no need formula, how calculate latent heat etc. its the same method as heat capacity wats more, there isn't any temperature to calculate! well but i think i better go through 1 last time Eg: what is the amount of energy needed to melt 10kg of ice when the specific latent heat of fusion is 336000J/Kg To get 10kg, multiply by both sides, 3 360 000J/10Kg so your ans will be 3360000 Another type of question that you have to learn how to tackle what is the total energy required to heat 2kg of pure ice at -5C to 10C of water? Specific heat capacity of ice : 2100J/Kg K Specific heat capacity of water : 4200J/Kg K Specific latent heat of fusion of ice : 336 000J/Kg This question may seems a complicated question but it is still only by one simple concept. This process has three stages. As everyone know, every substnaces and states have different specific heat capacity hence, we cannot directly take the temperature change as 15. In this question, there are 3 things you have to find. 1. Amount of energy to raise the temperature of ice 2. Amount of energy to melt 3. Amount of energy to raise the temperature of water why do i have to split it up? this is concern about different amount energy in different states, thats is y, total of 3 different heat capacity/latent heat has to be used 1st, you have to calculate the amount of energy needed in the first stage, raising temperature of the ice. The maximum temperature of ice can reach is of coz 0C as it has to undergo the change of state from solid to liquid now 1st stage: 2100J/Kg K --> x2 each side for its mass 4200J/2Kg K --> x5 for each every K change 21000J/2Kg 5K --> the ice has now reached 0C Therefore, energy needed = 21000 in the 1st stage 2nd stage: since its changing state, you have to use latent heat now 336 000J/Kg ---> x2 for mass 672000J/2Kg Therefore, energy needed to melt the ice = 672 000J 3rd stage: Now it has changed into water, bear this in mind, the water is still at 0C 4200J/Kg K ---> x2 for mass again 8400J/2Kg K ---> x10 for temperature 84000J/2Kg 10K Therefore, energy needed in this stage is 84000J now final step, ADD EVERYTHING TOGETHER FOR TOTAL ENERGY! 84000J + 672000J + 21000J = 777 000J (Answer) you hafta practise more urself :D FAQ: why do we still have to multiply the mass again when we had alr multiply it in the first place? the mass never change wat A: thats y i say, heat capacity is only base on 1 concept, when 2kg of ice melted, there is still 2kg of water remains! the same thing has to be repeated FAQ: What about Watts leh? A: take the last qns for eg: how long does it take for a heater wif 200W to complete this process? 200W can jus simply covert it to 200J/s so you can do proportion easier back to the qns, since it is 200J/s ask primary sch qus to urself, how many '200' do i need to reach 777 000J? so jus simply 777 000 divide it by 200 and tada! ans is 777 000 / 200 = 3885seconds! last part of heat capacity this is recommanded not to know wats the formula is talking about. understand it when ur heat capacity is very stable. so at the moment, jus study it w/o knowing wats happening the formula is : mc(temp change) = mc(temp change) m = mass, c = specific heat capacity this formula is used when 2 substances are mixed together at different temperature and will give u a final temperature Example 2kg of copper ball at 120C is thrown in 10Kg of water at 40C. Assuming that no heat is lost to the surrounding, what is the final temperature of the mixture? Specific heat capacity of copper = 400J/Kg K Specific heat capacity of water = 4000J/Kg K lets take copper for the left side and water at the right side X is the final temperature (2)(400)(X-120) = (10)(4000)(40-X) NOTE!!!: the temperature change part, cannot be in the same sequence means when x is on the left side of the bracket, the another side of the eqn CAN NOT hav X on the left side of its bracket too 800(x-120) = 40000(40-x) 800x-96000 = 1600000 - 40000x x = 41.6C thats the ans, if you were given final temp and find mass of specific heat capacity, jus put the final temperature into X and find the other unknown anything gotta do wif finaly temperature can use this formula to solve THE END! PRACTISE AND WORK HARD! |
| "> | Mole concept is actually not that difficult to learn. The first thing you have to do is the memorise these formula and make sure its at ur fingertip! 1. Mol = Mass / Ar or Mr 2. No. of Particles = Mol x 6 x 10^23 3. Concentration = Mol / dm^3 or Mass / dm^3 4. Volume of gas = Mol x 24dm^3 5. 1dm^3 = 1000cm^3 Notice that ALL the formulae has something got to do with mol? Hence, when you are doing mol qns, 1st step: Find Mol by using the 4formulae 2nd step: Use known mol to find the unknown you are required to find try this problem example 1 What is the ratio of the volume of 2g of hydrogen to the volume of 16g of methane, both volumes at r.t.p? when you give up / check answer pls highlight the space below :) Follow the step, find mole first. mol of hydrogen : 2g/2 = 1mol=1mol Note: hydrogen is exist as H2 when left alone, otherwise in ion form]
now, apply the volume of gas formula : mol x 24dm^3 so 1 x 24dm^3 = 24dm^3 Vol. of Hydrogen : Vol. of Methane ..........24.............:...........24 ...........1..............:............1 ( answer ) easy? jus keep applying that 4 basic formulae and you have mastered 40% of mol concept. Fast? LOL now next i will teach you how to tackle another type of question which is about pencentage mass heres the basic concept between the relationship between Ar / Mr and mass Mr = total number of neutrons + protons Since neutrons and protons are weigh same, it does not matter if neutrons are more than protons. imagine Mr is the same as mass but DO NOT use it as mass! [Eg: a 56Ar of Fe does not means Fe has 56g, i give u 1 bar of iron and it weigh 56g? LOL not logic right?] The Mr only provide us the UNITS confused? don worry... now the method im going to teach is PROPORTION METHOD! this method in fact works with concentration, heat capacity, air pressure and waves problem example 2: Calculate the mass of water in 80g of MgSO4.7H2O The 'dot' means the number of water molecules surrounding the compound. Hence, in this case, there are 7 water molecules surround 1 MgSO4 molecules. Clear? The crystal weigh 80g, and since Mr is the CAUSE of weight so... Find the Mr of the compound : 24 + 32 + 4(16) + 7(18) = 246 now, take it as units 246units ---> 80g your finding mass of water so its Mr is : 7(18) = 126 so basic maths, you are finding 126 units, so you have to find 1unit first 1unit ---> 80g / 246 126units ---> 80g/246 x 126 = 41g simple? Now try this: Calculate the mass of zinc in 48g of zinc carbonate Mr of ZnCO3 = 125 125 = 48g Ar of Zn: 65 Mass of Zn: 65 / 125 x 48 = 25g Using proportion method on concentration: 0.1mol in 50cm^3 1st, put it together as 0.1mol/50cm^3 change the 50cm^3 to 1dm^3 which is, you have to times by 200 when one side times by 200, the other side must also times by the same no. that is, 200 hence, 0.1 x 20/ 50 x 20 = 2mol/dm^3 now done, lets proceed to, in tackling equations question that everyone hates ^^ heres the steps for eqn problem solving 1st step: find the mol ratio 2nd step: use the mol ratio to find the mol using any information given and the 4formulae 3rd step: using proportion method, mol ratio provide UNITS, use it to find the mol of another substance 4th step: after getting the mol of the substance you want, use the 4basic formulae again to find whatever the question asked However, there are difficult eqn qns that will involves limiting reagent, percentage yield and molecular formula. now, i will demostrate 1 simple equation but you will have to REALLY DO MORE on 10yrs series or other assesment books Limiting reagent is the substances that is LACKING in the reaction which means, this substance will use up first Percentage yield is nothing to be terrified, all you have to do is to find mol, then find mass. its formula is Actual Yield / Theoretical yield x 100%. blur? not a problem, just remember, the 2 mass you have gotten, the smaller will ALWAYS divided by the larger number. forgot? use common sense, can u get more than 100%? :D no~ right? so its simple. the actual yield is given in the question most of the time.
problem example 3: 2H2 + O2 ---> 2H2O Given that 72dm^3 of hydrogen react with 48dm^3 of oxygen (a) Which is the limiting reagent? (b) Find the volume of oxygen reacted? (c) Find the mass of water produced? (d) What is the total number of water molecules formed? (e) What is the percentage mass of oxygen in water? (f) Find the percentage yield when only 60g of water produced (g) Find the concentration of the solution if there is 21g of salt[NaF] dissoved into 500cm^3 of water Solution: a) Hydrogen , when 72dm^3 of hydrogen is reacted b) 36dm^3 mol of hydrogen = 72/24 = 3 using mol ratio Hydrogen : Oxygen = 2:1 so mol of oxygen is 3/2 = 1.5 Volume of oxygen : 24dm^3 x 1.5 = 36dm^3 remember...mol ratio are only units c)since hydrogen and water have the same mol ratio, they have the same mol which is 3 using the formula: 3 x 18 = 54g d) once again using the formula: 3 x 6 x 10^23 = 1.8 x 10^24 e) proportion method again: 16/18 x 100% = 88.9% f) for concentration, you have to find the amount of substances dissolved hence since concentration is mol/dm^3 you have to find the mol of NaF mol of NaF: 21/42 = 0.5 now, put everything together again: 0.5mol/0.5dm^3 = both sides x2 = 1mol/1dm^3g) always the smaller divided by larger...so.. percentage yield : 54/60 x 100% = 90% ok last question on molecular formula In a reaction, 4g of carbon reacts with 0.6g of hydrogen. The compound formed has a molecular mass 28g (a) Determine the emptical formula for the compound (b) Determine the molecular formula for the compound Solution: (a) due to presentation expectation from koh yilin =.= we have to draw table zzz as usual, always find mol first ....................C.........H Mass given.. 4......... 0.6 Ar is ...........12......... 1 Mol........... 0.3........ 0.6 hence, through mol ratio that you have jus find out, it is the no. of atom with different element in the formula [ in this case since carbon to hydrogen is 1:2 so the formula is CH2] simple? b) molecular formula is the easiest, just divide the Mr of the molcular formula compound (most of the time is given in the question) by the Mr of the empirical formula compound. Mr of CH2 = 14 Mr of molecular=28 divide it, 28 / 14 = 2 this number that u get, MULTIPLE it into the formula (CH2)2 == C2H4 NOTE: PLS DO NOT WRITE 2CH2 ITS WRONG! this is reEDITed version, i've found out why every1 is confused at concentration le~ concentration is the amount of SOLUTE dissolved in the solution, hence you can 'split' the concentration into parts for you to get the informations its like dissolving 50g of solute [eg: NaCl] into 100cm^3 of water so your concentration can be written as 50g/100cm^3 however, most of the time we leave it in mol/dm^3 form. the method is simple by proportion method, you have to change the 100cm^3 into 1dm^3, therefore, u simply jus multply by 10 to get 1000cm^3=1dm^3 now ur current concentration is written as 500g/dm^3 so how to convert it into mol/dm^3? this will be jus another basic, simply view the 500g separately. using 1 of the 4basic formula, convert mass into mol, 500/58.5 = 8.54mol so now the concentration is 8.54mol/dm^3 ! Note: you can always view the mol or mass part separately from the concentration BUT BUT BUT! ALWAYS, take note of the volume, if its 50cm^3 you have to divide it by 200 from 1dm^3 Titration formula: n1/n2 = m1v1/m2v2 n= mol ratio, m=mol,v=volume this is another method that can use to find the unknown but ONLY in titration thats all for mol...u have to do quite a lot urself or else its not easy to master |
| "> | This chapter MAY be much easier if you were to follow my method. The first thing you always have to learn in acid base and salt is to IDENTIFY them. I know its tough to memorise and master them, but its the best way to identify acid, its basically very simple all u hav to rmb theres an H or H2 in front of the formulae. these are different acids u have to learn to identify Identification: any front can go with any back in the formula ignore acid Front - Back H Cl H NO3 H2 CO3 H2 SO4 Organic Acid [Time Taken to master: 10seconds] Front Back CH3 COOH Heres bases, the key here is jus simply having O or HO behind the formula. The following below are only soluble base. Other bases that are not mention here are all insoluble Soluble Bases [Time Taken to master: 5mins] Front - Back K...........O Na.........OH Li NH4 NH3 Note: For NH3, it is alone, and it is not bonded with O or OH, NH3 itself is a soluble base. For NH4, it is uses only wif O or OH eg: NH4OH (aqueous ammonia) Form when ammonia is dissolved into water Furthur identification of bases [Time Taken to master: 5mins] Acidic Oxide Front - Back Non-Metal + O2 Basic Oxide Front - Back Metal + O Amphoteric Oxide/Hydroxide Front - Back Pb........O Al........ OH Zn There are only 3 amphoteric oxide, other metal + oxygen are basic oxde Neutral oxide Front - Back Non-metal + O [only 1 oxygen plus a non-metal will be neutral oxide] {Eg: CO, SO, H2O} Solublilty in Salts [Time Taken to master: 10mins] Whenever u see nitrate, sulphate, halogen and carbonate in the formulae, its a salt. Type of Salt - Soluble ......................- Insoluble Nitrate.........- All metals.................... - None Sulphate.....- Almost all metals......... - Ca, Pb, Ba Chloride..... - Almost all metals......... - Ag, Pb Carbonate.. - Group I metals/NH4...... - Mostly all .............................................................Metals [Eg: CaSO4 is a sulphate salt, and its insoluble] Memory Guide in General: Whenever you see NITRATE ,Group I metal and Ammonium in the formula, its soluble Memory Guide in Salt: If you DO NOT see nitrate and there is only Pb, it is insoluble the rest are up to ur hardworkingness :) When you have master the idenfication of ABS, you can proceed to learn preparation of salt 1st step: IDENTIFY the ABS 2nd step: IDENTIFY SOLUBILITY 3rd step: Decide the methods to use in preparation of salts For TITRATION IMPORTANT: this is ONLY for preparing group I and ammonium salt because both during titration, both reactants are in aqueous solution hence, excess acid or alkaline have to be removed 1. Use a pipette to draw an accurate fixed volume of acid to a conical flask 2. Add 2 to 3 drops of methyl orange into the flask 3. Titrate drop by drop of alkali from the burette into the acid. Swirl the solution each time a drop of alkali is added 4. When the first appearance of orange colour observed in the solution, stop the titration and note the volume of alkali used 5. Repeat the experiment using the exact volume of acid and alkali from the first experiment without the indicator (so that the salt is free from contamination of indicator) 6. The salt solution is purified by crystallisation process Note: Universal Indictor is not suitable because the colour change cannot be observed distinctly For Filtration IMPORTANT: This is ALSO to produce soluble salts but not Group I metal or ammonium salt. In filtration, one of the REACTANT IS SOLID 1. Excess 'solid reactant' is added to a fixed volune of dilute acid to make sure all acid is used up in this reaction ( important ) 2. Filter the solution to remove the excess 'solid reactant' from the 'salt formed' solution 3. Evaporate the filtrate until it becomes saturated so that the salt crystals can be formed easily 4. Let the solution cool so salt crystal can be formed 5. Wash the salt crystals with little bit of water and let it dry For Precipitation method This is easy however, yet confusing. This method is for preparing INSOLUBLE salt. how do u know which + which to get you insoluble salt? Hence, mastering the solublity salt is a MUST. there are two ways, acid + soluble salt or soluble salt+ soluble salt. Depending on what salt the question asked u to prepare, you have to know how to split the insoluble salt into its ion form in order to see their reactants. Eg: AgCl an insoluble salt split into its ion form, Ag+ and Cl- now, this gives you a clue of the two soluble salts you shld use. Look, the 1st salt should contains Ag and the 2nd salt should contains Cl Since you are looking for soluble, nitrate salt is the best since all nitrate salt are soluble. Hence: The 1st salt can be AgNO3 Now, u can use ANY POSITIVE element for Cl salt EXCEPT for Pb and Ag, remember? and since group I metal or ammonium salt are always soluble, lets take Na Henece: The 2nd salt can be NaCl So now, contruct the equation: AgNO3 + NaCl ----> AgCl + NaNO3(just form the 2nd salt in the product with whatever u are remaining. In this case, you are left with NO3 and Na after getting AgCl) Another example with Acid + Soluble Salt AgNO3 + HCl ----> AgCl + HNO3 finally, 3easy steps to memorise 1. Add the salt solution/acid to a beaker containing 'another salt solution you have chosen' 2. Stir the solution. A whte precipitate of the 'Insoluble Salt' is formed 3. Filter off the precipitate, wash with a little cold water and leave the precipitate to dry Take Note for ABS: IONIC EQUATION is very likely to be tested, hence if the substance is soluble, be sure the state is in AQ state! now lets proceed to reaction with metals,carbonates,ammonium salt, acid and bases. These are the list of equation you have to memorise in order to score well and construct its equation. 1. Acid + Metal Carbonate ----> Salt + Carbon Dioxide + Water 2. Acid + Base/Alkali ----> Salt + Water 3. Base/Alkali + Ammonium salts ----> Ammonia + Salt + Water 4. Acid + Metal ----> Salt + Hydrogen gas 5. Acid + Metal oxide/hydroxide ----> Salt + Water 6. Metal + Water ----> Metal Hydroxide + Hydrogen gas 7. Metal + Steam ----> Metal Oxide + Hydrogen gas 8. Metal Carbonate + Heating ----> Metal Oxide + Carbon Dioxide These are the total number of equation you have to memorise. It is combined with another chapter, metals The last part is just properties of the acid and bases you have to memorise, also indicators and its respective colour change Acidic Oxide - Reacts with bases to form salt and water - Reacts with water to form acid Basic Oxide - Reacts with acids to form salt and water - Reacts with water to form alkali Amphoteric Oxide - Can behaves like basic oxide or acidic oxide Neutral Oxide - They behave neither like Basic oxide nor Acidic oxide For emergency cases where you forgot the product of equations: - When there is just a metal in the reactant, there will be hydrogen gas in product, the other product will be whatever left behind - When you see the element CARBON(exclude acid), in any circumstances, there will be carbon dioxide in the product - When you see ammonium, there will always be ammonia in the products - If you manage to get the salt first, the remaining elements left, are hydrogen and oxygen, it will be water. Long? yeah, to master ABS u need hours of 'training' LOL you may copy post if you wan to. the main problem in ABS is only the identification part and solublity after that, its easy to learn ABS jus endure for 1-2hr memorising it =\ kekeke |
