# data[].attributes.body
1. find no of mole of each reactants
2. balance equation
3. using stoichiometry
4. determine the limiting reactant

# data[].attributes.body
1. axis mesti ada 3, jgn ubah garisan axis tu.
2. orbital size mesti sama
3. label mesti tulis (USING PEN)
*****lukislah spt yg sy attach di atas.!
****** draw using PENCIL!

# data[].attributes.body
1. matter
2. element
3. compound
4. mixtures
5. homogenous mixture
6. heterogenous mixture

# data[].attributes.body
1. element
2. compund
3. mixtures

# data[].attributes.headline
able to differentiate homogenous and heterogenous mixtures  

# data[].attributes.subject
able to differentiate homogenous and heterogenous mixtures  

# data[].attributes.body
oxygen gas, O₂ ; element
water, H ₂ O; compound
chocholate cookies, salt & water, oil & water ; heterogenous mixture
sugar water, a cup of coffee; homogenous mixture

# data[].attributes.body
1. using stoichiometry to determine how many moles produced by limiting reactant
2. using formula mass = n x Mr

# data[].attributes.body
1. using stoichiometry to determine how many moles of excess reactant used by by limiting reactant.
2. find no of moles excess reactant;
moles excess = moles available - moles used

# data[].attributes.body
****REACTANTS MESTI ADA DUA.
SATU MESTI LIMITING REACTANT
DAN SATU LAGI AKAN JADI EXCESS REACTANT
****MESTI GUNA STOICHIOMETRY UNTUK;
1) TENTUKAN BERAPA MOLES PRODUCT YG TERHASIL.
2) TENTUKAN ANOTHER REACTANT(EXCESS REACTANT) YG DIGUNAKAN (USED)

# data[].attributes.body
1. PASTIKAN BAWA KAD MATRIK (DAH ADA KAD MATRIK TAK PERLU IC)

2. HAFAL NO TEMPAT DUDUK DAN PUSAT DEWAN PEPERIKSAAN. (CTH; A001) (MAKSUDNYA NO TEMPAT DUDUK 001 DAN PUSAT A)

3. PASTIKAN ALAT TULIS, KALKULATOR BAWA & ADA BATERI

4. CARA NAK TULIS NO TEMPAT DUDUK (A001)

# data[].attributes.body
1. MASUK DEWAN BACALAH DOA, LANGKAH KAKI KANAN

2. PASTIKAN MASA MASUK DEWAN,;

3. PAKAI KAD MATRIK

4. ADA ALAT TULIS ; (PEN SECUKUPNYA, PENSIL 2B, PEMADAM, PEMBARIS)

5. BILA DAH MASUK DEWAN, JGN BUAT BISING (TAK PERLU NAK CAKAP, HAI2 KE, GOOD LUCK KE..)

6. TERUS LETAK KAD MATRIK DI SEBELAH SLIP MEJA. HARI PERTAMA TGOK SLIP MEJA BACA DETAILS, JIKA SEMUA BETUL, TURUNKAN TANDATANGAN.

7. BACALAH ZIKIR SEMENTARA TUNGGU ARAHAN KETUA PENGAWAS.

8. DENGAR ARAHAN SEJELASNYA. IKUT ARAHAN SEBAIKNYA.

# data[].attributes.body
OBJECTIVES QUESTIONS (20 MARKS)
20 SOALAN (WAJIB JAWAB SEMUA)

1. JAWAB DLM KERTAS OMR

2. MESTI HITAMKAN DGN PENSIL 2 B

3. JAGA KERTAS OMR, JANGAN TINDIH DGN KERTAS LAIN, JGN TERLIPATKAN KERTAS OMR. (NANTI MESIN TAK BOLEH NAK BACA KERTAS OMR)

4. TERUS HITAMKAN DLM KERTAS OMR SELEPAS TANDA DLM KERTAS SOALAN. (JGN SIMPAN HUJUNG2 WAKTU UTK HITMKAN PD KERTAS OMR)

# data[].attributes.body
5 SOALAN STRUKTUR (WAJIB JAWAB SEMUA)

1. ***JAWAB DALAM BUKU JAWAPAN YG DISEDIAKAN

2. GUNAKAN MUKA SURAT YG BAHARU UTK SOALAN YG SETERUSNYA.

# data[].attributes.body
The elements below exist in Group 17 in Periodic Table EXCEPT

A. F
B. Cl
C. Be
D. I

# data[].attributes.body
(a) Define element (1 mark)
(b) Give three (3) examples of element. (3 marks)
(c) Water can exist in few forms. List all the possible forms and explain the physical and chemical changes that occur. (2 marks)
(d) Differentiate between homogenous and heretogenous mixture. (4 marks)

# data[].attributes.body
The substances loses one or more electrons.

Increase oxidation number

Act as reducing agent

# data[].attributes.body
The substances gains one or more electrons.

Decrease in oxidation number.

Act as oxidising agent.

# data[].attributes.body
IS A REACTION THAT INVOLVES BOTH REDUCTION & OXIDATION

# data[].attributes.body
Step 1:
Divide the equation into two half reactions;
Fe²⁺ + MnO₄⁻ --> Fe3+ + Mn2+
oxidation: Fe2+  --> Fe³⁺
reduction: MnO₄⁻ --> Mn²⁺

# data[].attributes.body
Balance the element other than oxygen and hydrogen
*** Balance oxygen atom by adding H ₂ O
*** Balance hydrogen atom by adding H⁺
oxidation: Fe2+  --> Fe³⁺
reduction: MnO₄- + 8H+  --> Mn²⁺ + 4H₂O

# data[].attributes.body
Balance charge by adding electrons to the side with the greater overall positive charge.
oxidation: Fe2+  --> Fe3+ + e
reduction: MnO₄- + 8H+  + 5e--> Mn²⁺ + 4H₂O

# data[].attributes.body
Balance electrons for both half cell reactions
(oxidation: Fe2+  --> Fe3+ + e) x 5
reduction: MnO₄- + 8H+  + 5e--> Mn²⁺ + 4H₂O
--------------------------------
oxidation: 5Fe2+  --> 5Fe3+ + 5e
reduction: MnO₄- + 8H+  + 5e--> Mn²⁺ + 4H₂O

# data[].attributes.body
Check all the species,
if the same species in the same side --> add it.
if the same species in the opposite side --> cancel it.

oxidation: 5Fe2+  --> 5Fe3+ + 5e
reduction: MnO₄- + 8H+  + 5e--> Mn²⁺ + 4H₂O
-----------------------------------
overall cell reactions: 5Fe2+ + MnO₄- + 8H+ --> 5Fe3+ Mn²⁺ + 4H₂O

# data[].attributes.body
Check the equation to make sure that there are the same number of atoms of each kind and the same total charge on both sides.
5(+2) + (-1) + (+8) --> 5(+3) + (+2) + 4(0)
+17 --> +17

# data[].attributes.body
1. PASTIKAN BAWA KAD MATRIK (DAH ADA KAD MATRIK TAK PERLU IC)

2. HAFAL NO TEMPAT DUDUK DAN PUSAT DEWAN PEPERIKSAAN. (CTH; A001) (MAKSUDNYA NO TEMPAT DUDUK 001 DAN PUSAT A)

3. PASTIKAN ALAT TULIS, KALKULATOR BAWA & ADA BATERI

4. CARA NAK TULIS NO TEMPAT DUDUK (A001)

# data[].attributes.headline
Sedikit panduan utk menulis dlm buku jawapan. 

# data[].attributes.subject
Sedikit panduan utk menulis dlm buku jawapan. 

# data[].attributes.body
1. Pastikan semasa menunggu masa menjawab bermula anda tulis details pd bhgm ats stp helaian buku jawapan.
2. Bila dah mula.. selawat dlm hati byk2, semoga stp jawapan awk dibantu Allah S.W.T
3. Pastikan tulis no soalan cth 1.(a) pd bahagian kiri buku jawapan.
4. Mulakan helaian baru utk no soalan berikutnya.