SPM 2010 Chemistry – Common Mistakes in Paper 2 Section B
November 21, 2010
Continuing from the previous post on Section A, Berry Berry Easy would like to present all with “Common Mistakes for SPM 2010 Chemistry Paper 2 Section B“. Make sure you identify possible weaknesses in your own answers based on the common mistakes found for students in the SPM 2010 Trial – Johor State papers. (If you are a Johorean, make doubly sure that you check your answers to see if you have commited any of the below mentioned common errors). The 20 marks in this section is yours, so don’t drop too many marks in this easy section (considered as easy once you know the techniques). For the full paper, Berry Reader can download it from http://edu.joshuatly.com) So do check out the 3 examples of common errors commited by students.
Common Mistakes for SPM 2010 Chemistry Paper 2 Section B (based on Johor State Trial SPM 2010 Paper)
@ Kesalahan Lazim/Biasa SPM 2010 Kimia Kertas 2 Bahagian B
SPM question Paper 2 Section B – essay questions (Question 7 & 8 ) (Choose 1 to answer) consist of 20 marks in total. In this section, students need to answer any 1 question from this section. There are some important rules or answering techniques that students need to follow. We will be discussing these points by using 2010 Johor state trial paper 2 (Section B – Question number 7 and 8).
1) Can students use point form technique in answering essay question?
Yes, in SPM Chemistry Paper 2 (essay) students are allow to use point formed but it must be a complete sentence. (I don’t encourage students to do so unless students have language barrier or not enough time to answer the question.)
Describe the formation of the following chemical bonds and draw the electron arrangement of compounds formed. (Huraikan pembentukan bagi ikatan kimia berikut dan lukis susunan electron bagi sebatian yang terbentuk.)
* Ionic bond between magnesium atom and chlorine atom. (Ikatan ionic antara atom magnesium dengan atom klorin.) Q7 (b) (i) (5 marks)
* Magnesium atom donates.
* Releases 2 electrons to form magnesium.
* Chlorine atom accepts.
* Receives 1 electron to form chloride.
* Achieve electron arrangement.
* Magnesium and chlorine are attracted by electrostatic force.
No marks will be given if students did not provide a complete sentence and clear answer.
* Magnesium atom donates two electrons to form magnesium ion which achieve stable electron arrangement.
* Chlorine atom accepts one electron to form chlorine ion which achieve stable electron arrangement.
* Magnesium ion and chloride ion are attracted to one another by electrostatic force.
2) Can students use table for comparison or explanation?
Yes, it is highly recommended by most of the Chemistry teachers. Students would not miss out any important key-points.
Explain each of the following statements. (Terangkan setiap pernyataan berikut.)
* Compound formed in (b) (i) can conduct electricity in the molten state while compound formed in (b) (ii) cannot conduct electricity in any state. (Sebatian yang terbentuk dalam (b) (i) boleh mengkonduksikan elektrik dalam keadaan leburan manakala sebatian yang terbentuk dalam (b) (ii) tidak dapat mengkonduksikan elektrik dalam semua keadaan.) Q7 (c) (i) (4 marks)
Ions in MgCl2 are held together by strong electrostatic force.
MgCl2 need a lot of heat energy to overcome the strong electrostatic force/fixed crystal lattice structure.
Molecules in CCl4 are held together by weak intermolecular forces / van der Waals.
CCl4 need less energy to overcome the forces.
3) How to get full marks for the graph?
Students need to make sure that your graph must consist of 3 important things
(SAL / SAP)
S = Shape of the graph
A = Axis (must correct and labeled with units)
L / P = Label / Points plotted
And do remember to write the title of the graph.
4) Is the unit important for the calculation?
Yes. Students need to include the unit in the final answer.
Diagram 8 shows the energy level diagram of the heat of combustion of ethanol. Calculate the energy released if 6.9 grams of ethanol is burnt completely in air.
(Rajah 8 menunjukkan gambarajah aras tenaga bagi haba pembakaran ethanol. Kira haba yang dibebaskan jika 6.9 gram ethanol di bakar dengan lengkap di dalam udara.) Q8 (d) (ii) (4 marks)
Number of mole of ethanol = 6.9 / 46 = 0.15
0.15 mole will form 0.15 x 715 = 107.25 kJ mol-1
Number of mole of ethanol = 6.9 / 46 = 0.15
0.15 mole will release 0.15 x 715 = 107.25 kJ of heat
Credit to: http://berryberryeasy.com/2010/11/spm-2010-chemistry-common-mistakes-paper2-section-b/
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