MODUL PDPR PPD KINTA UTARA 2 2020

Kompilasi

DAERAH KINTA UTARA
JILID 2/2020

KERJASAMA ANTARA SISC+,
GURU CEMERLANG &

GURU YANG BERPRESTASI CEMERLANG
DAERAH KINTA UTARA

BIOLOGI KANDUNGAN

Kompilasi

MODUL PDPRDaerah Kinta Utara Jilid 2/2020

1. Form 6: Chapter 6 Photosynthesis– Autotroph 08

§ En. Ong Wei Siang (GC Biology Kolej Tingkatan 6 Seri Putera) 53
84
KIMIA 107
135
1. Tingkatan 5 : Haba Peneutralan
160
§ Pn. Noraini binti Nasikin (GC Kimia SMK Raja Perempuan)
2
2. Form 4: Chapter 7 – Rate of Reaction (Numerical Problem Solving)

§ Pn. Noriza bt Awang (GC Kimia SMK Anderson)

3. Form 4: Chapter 6 – 6.2 Acid, Base & Salt

§ Komathy Veerasinghan (GC Kimia Smk Buntong)

4. Form 4: Chapter 6 – 6.3 Acid, Base & Salt

§ Komathy Veerasinghan (GC Kimia Smk Buntong)

FIZIK

1. Tingkatan 4: Bab 4 – 6.4 Formula Kanta Nipis

§ Pn. Sabariah bt Mat Isa (GC Fizik SMK Bercham)

Kompilasi KANDUNGAN

MODUL PDPRMPV KATERING & PENYAJIAN
Daerah Kinta Utara Jilid 2/2020

1. Tingkatan 4 : Bab 4 – Kek & Hiasan Kek 146

§ Pn Noor Saahida Bt Mohd Isa (GC Katering & Penyajian SMK Ulu Kinta) 184
197
SAINS
204
1. Tingkatan 4 : Bab 11 – Jatuh Bebas 220

§ Pn. Sabariah bt Mat Isa (GC Fizik SMK Bercham) 245

2. Tingkatan 2: Bab 1 - Biodiversiti 3

§ Pn. Norhaini binti Abdul Aziz (SISC+ PPD Kinta Utara)

BAHASA MELAYU
1. Penulisan: Pendahuluan Karangan

§ Encik Rohiman bin Subri (Mantan Guru Bahasa Melayu SMK Raja Permaisuri Bainun)

2. Penulisan: Dulang Pemikiran

§ Pn. Misah binti Sulaiman (SISC+ PPD Kinta Utara)

PENDIDIKAN MORAL
1. Tingkatan 2: Unit 1 - Sumber Moral Asas Pembentukan Akhlak Mulia

§ Pn. Juliana binti Shaharum (SISC+ PPD Kinta Utara)

SENARAI PAUTAN MODUL PDPR

Kompilasi MODUL PDPR

Daerah Kinta UtaraJilid 2/2020

BIL MATA PELAJARAN/ TOPIK PAUTAN

1 Biology - Form 6: Chapter 6 Photosynthesis– Autotroph https://anyflip.com/glmbu/eugi/
2 Kimia Tingkatan 5 : Haba Peneutralan https://anyflip.com/vnga/mecf/
3 Chemistry - Form 4: Chapter 7 – Rate of Reaction (Numerical Problem Solving) https://anyflip.com/qbiqu/gzzq/
4 Chemistry - Form 4: Chapter 6 – 6.2 Acid, Base & Salt http://gg.gg/nddeg
5 Chemistry - Form 4: Chapter 6 – 6.3 Acid, Base & Salt http://gg.gg/nddfz
6 Fizik Tingkatan 4: Bab 4 – 6.4 Formula Kanta Nipis https://anyflip.com/vnga/lxcd/
7 MPV Katering & Penyajian Tingkatan 4 : Bab 4 – Kek & Hiasan Kek https://anyflip.com/vnga/tnuy/
8 Sains Tingkatan 4 : Bab 11 – Jatuh Bebas https://anyflip.com/vnga/rthj/
9 Sains Tingkatan 2: Bab 1 - Biodiversiti https://anyflip.com/vnga/oudm/
10 Bahasa Melayu Penulisan: Pendahuluan Karangan http://gg.gg/neoli
11 Bahasa Melayu Penulisan: Dulang Pemikiran https://anyflip.com/qkpky/qdby/
12 Pendidikan Moral Tingkatan 2: Unit 1 - Sumber Moral Asas Pembentukan Akhlak Mulia https://anyflip.com/vnga/pkec/

4

KATA PENGANTAR

En Mohamad Hazawawi bin Yusof
Pegawai Pendidikan Daerah Kinta Utara

Assalamualaikum dan salam sejahtera kepada semua pendidik dan murid di daerah Kinta Utara.

Kinta Utara Sentiasa Unggul.

Kompilasi Modul Pengajaran dan Pembelajaran di Rumah secara berstrukutur dan terancang mengikut Manual
Pengajaran dan Pembelajaran di Rumah KPM dibangunkan bertujuan untuk membantu guru melaksanakan Pengajaran
dan Pembelajaran di Rumah (PdPR) sebagai alternatif pembelajaran norma baharu. Modul ini sebagai salah satu rujukan
tambahan untuk pentadbir sekolah dan guru-guru jadikan panduan penyediaan modul PdPR.

Penyediaan modul ini memberi penekanan ke atas strategi pembelajaran yang berkesan sesuai dengan
kehendak pendidikan masa kini dan masa depan.

Modul PdPR ini boleh membantu guru mempelbagaikan kaedah pengajaran dan pembelajaran yang
berkesan di samping mewujudkan pembelajaran di rumah yang menggembirakan murid.

PPD Kinta Utara, merakamkan setinggi-tinggi penghargaan dan terima kasih kepada pegawai Sektor
Pembelajaran, Guru Cemerlang, guru-guru dan semua yang terlibat dengan penyediaan modul ini. Di sebalik kekangan
serta cabaran yang dihadapi sepanjang tempoh PKPP dan PKPB ini, usaha berterusan guru-guru demi memastikan
pelajaran anak murid tidak terabai amatlah dipuji. Di samping itu sokongan ibu bapa juga amatlah dihargai. Semoga
bahan kompilasi modul PdPR Daerah Kinta Utara ini dapat dimanfaatkan oleh semua pihak yang terlibat.
Sekian dan selamat maju jaya.

30 November 2020

6

BIOLOGI

7

BIOLOGY MODULE

Chapter

6 PHOTOSYNTHESIS
6.1 AUTOTROPH

PREPARED BY
MR. ONG WEI SIANG

8

LEARNING classify autotroph into photoautotroph and
OUTCOMES chemoautotroph;

describe photosynthetic pigments;

explain the absorption spectrum of
photosynthetic pigments and action spectrum of
photosynthesis.

9

TASK 1 Before you get started with the lesson, you
may watch a prepared video explaining
Classify autotroph photoautotroph and chemoautotroph.
into photoautotroph
Scan the QR code given below to watch
and the video.
chemoautotroph

10

REMEMBERING

Light Simple Complex Autotrophs Nitrate
Carbon dioxide
Chemical reactions Energy

Definition of ___________ synthesize _______ organic
autotroph compounds from ________ inorganic
compounds such as __________, water,
and _______ using energy from ______ or
_____________.

11

Classification of UNDERSTANDING
autotrophs
Photoautotrophs use _______ as their source
of energy to synthesise organic compounds
(such as glucose) from carbon dioxide and water
(hydrogen sulfide or organic compound for
______________) by ____________.

Chemoautotrophs use energy from other
_______ ______ such as oxidation reactions to
synthesise organic compounds from carbon
dioxide and ________ by ___________.

Photoautotroph Chemoautotroph
________ bacteria,
Green plants, ______,
cyanobacteria, purple sulfur iron bacteria
_______, purple _______
bacteria, and brown (red) bacteria 12

TASK 2 Continue to watch the first video given.

Describe You may scan the QR code given below to
photosynthetic watch the video.

pigments

You may refer to next page for the
structures of photosynthetic pigments.

13

Chlorophyll a Chlorophyll b

Bacteriochlorophyll

14

REMEMBERING

Description of Photosynthetic pigments in eukaryotic cells
photosynthetic are found on the _________ ________in
chloroplast.
pigments
There are ____________, ____________,
___________ and xanthophyll.

15

UNDERSTANDING

A chlorophyll molecule has two main parts, a
complex ring structure and a long
_____________ _____.

The ring structure, called a __________ _____,
is composed of carbon, nitrogen atoms and at
the center is a __________ atom.

The porphyrin ring absorbs ________ ______.

The long hydrocarbon side chain that
makes the molecule extremely ________.

_________ the chlorophyll in the thylakoid
membrane.

16

REMEMBERING

__________ consists of carbon rings linked
to a chain of 18 carbon atoms with
alternating single and double bonds.
Two typical carotenoids:

___________
___________

17

TASK 3 Continue to watch the first video given.
You may scan the QR code given below to
Explain the absorption watch the video.
spectrum of
You may refer to next page for the
photosynthetic pigments absorption spectrum and action spectrum.
and action spectrum of
18
photosynthesis

Each photosynthetic pigment has a
characteristic absorption spectrum, the
range and efficiency of photons it is
capable of absorbing from visible light.

Produces a plot of photosynthetic

wavelengths.
The graph is called absorption spectrum
of photosynthetic pigments.

19

An action spectrum of
photosynthesis is a graph of the
relative effectiveness of different
wavelengths of light in promoting
photosynthesis.

A peak shown in an action
spectrum corresponds to a
maximum absorption in the
absorption spectrum of
photosynthetic pigments.

20

UNDERSTANDING
Based on the absorption spectrum, determine the wavelengths of
light that absorbed by chlorophyll a, chlorophyll b and -carotene.

Absorption ____________ absorbs light of 430 nm (blue-violet) and 662 nm
spectrum of (_____).
photosynthetic
____________ absorbs light of 453 nm (blue) and 642 nm
pigments (_____).

____________ absorbs light of 450 nm (blue) and 475 nm (blue-
green).

21

EVALUATING

Explain why most of the leaves shown green colour.
___________________________________________________________
___________________________________________________________
___________________________________________________________

Justify the advantage of having different types of photosynthetic pigments.
___________________________________________________________
___________________________________________________________
___________________________________________________________

22

REFLECTION Define autotroph.
Classify autotroph into photoautotroph
and chemoautotroph.
Describe photosynthetic pigments.
Explain the absorption spectrum of
photosynthetic pigments.
Explain the action spectrum of
photosynthesis.

23

NEXT LESSON Explain the cyclic and non-cyclic
photophosphorylation including
photoactivation of chlorophyll a resulting in
photolysis of water as well as electron
transport system resulting in the production
of ATP and NADPH;

24

SUGGESTED
ANSWERS

25

Autotrophs synthesize complex organic
compounds from simple inorganic
compounds such as carbon dioxide,
water, and nitrate using energy from light
or chemical reactions.

26

Photoautotrophs use light as their source of
energy to synthesise organic compounds (such
as glucose) from carbon dioxide and water
(hydrogen sulfide or organic compound for
photosynthetic bacteria) by photosynthesis.

Chemoautotrophs use energy from other
chemical reactions such as oxidation reactions
to synthesise organic compounds from carbon
dioxide and water by chemosynthesis.

Photoautotroph Chemoautotroph
Nitrifying bacteria,
Green plants, algae,
cyanobacteria, purple sulfur iron bacteria
bacteria, purple nonsulfur
27
bacteria, and brown (red)
bacteria

Description of Photosynthetic pigments in eukaryotic cells
photosynthetic are found on the thylakoid membrane in
chloroplast.
pigments
There are chlorophyll a, chlorophyll b, -
carotene and xanthophyll.

28

A chlorophyll molecule has two main parts, a
complex ring structure and a long
hydrocarbon side chain.

The ring structure, called a porphyrin ring, is
composed of carbon, nitrogen atoms and at the
center is a magnesium atom.

The porphyrin ring absorbs light energy.

The long hydrocarbon side chain that
makes the molecule extremely nonpolar.

Anchors the chlorophyll in the thylakoid
membrane.

29

Carotenoid consists of carbon rings linked
to a chain of 18 carbon atoms with
alternating single and double bonds.
Two typical carotenoids:

-carotene
xanthophyll

30

Based on the absorption spectrum, determine the wavelengths of
light that absorbed by chlorophyll a, chlorophyll b and -carotene.

Absorption
spectrum of
photosynthetic

pigments

Chlorophyll a absorbs light of 430 nm (blue-violet) and 662 nm
(red).

Chlorophyll b absorbs light of 453 nm (blue) and 642 nm (red).

-carotene absorbs light of 450 nm (blue) and 475 nm (blue-
green).

31

Explain why most of the leaves shown green colour.
Different photosynthetic pigments found in photosynthetic cells
absorb most of the wavelengths of light except the green light. Hence,
the green light is reflected from the leaves.

Justify the advantage of having different types of photosynthetic pigments.
Different photosynthetic pigments absorb different wavelengths of
light to maximize the absorption of photon energy for photosynthesis.

32

KIMIA

33

Modul PdPr Kimia
Tingkatan 5
Tajuk :
Haba Peneutralan

Tema Haba
Bidang pembelajaran 11.0 Termokimia
Standard kandungan 11.2 Haba Tindak balas
11.2.3 Menentukan haba peneutralan melalui
Standard pembelajaran melalui aktiviti

Pada akhir modul, a. Menyatakan maksud haba peneutralan.
murid boleh
b. Membuat perbandingan haba peneutralan

c. Menulis persamaan termokimia bagi tindak
balas peneutralan

d. Membina gambarajah aras tenaga

e. Menyelesaikan masalah penghitungan yang
melibatkan haba peneutralan.

Haba yang dibebaskan
apabila 1 mol air, H20 terhasil

daripada tindak balas
peneutralan antara asid dan

alkali.

Eksotermik → Haba dibebaskan ke
persekitaran

Formula : mcØ
m = jisim larutan dalam gram
c = muatan haba tentu dalam Jg-1 oC -1
Ø = perubahan suhu dalam oC

H+ + OH- → H2O

Bahan

- M kJ
Hasil



Kira perubahan haba

Tuliskan persamaan kimia seimbang yang terlibat

Kira bilangan mol bahan tindak balas

Kira haba peneutralan dengan menghubungkait bilangan mol
bahan dengan perubahan haba.
*Langkah 3 dan Langkah 1

Masihkah anda Contoh asid kuat:
ingat maksud asid • Asid hidroklorik, HCl
• Asid nitrik, HNO3
kuat dan alkali • Asid sulfurik, H2SO4
kuat..?
Contoh alkali kuat:
Asid kuat ialah bahan • Natrium hidroksida, NaOH
kimia yang mengion • Kalium hidroksida, KOH
dengan lengkap didalam
air untuk menghasilkan
kepekatan ion hidrogen

yang tinggi

Alkali kuat ialah bahan
kimia yang mengion
dengan lengkap didalam air
untuk menghasilkan
kepekatan ion hidroksida

yang tinggi

Haba peneutralan bagi tindak balas asid kuat dan alkali kuat
adalah -57 kJmol-1.

Haba peneutralan bagi tindak balas asid sulfurik (diprotik asid) dengan
alkali kuat.

2 mol NaOH bertindak balas dengan 1 mol asid sulfurik untuk
menghasilkan 2 mol air. Haba yang dibebaskan = 2 x 57 kJ = 114 kJ.

Walaubagaimanapun, haba peneutralan asid sulfurik dan
alkali kuat tetap 57 kj mol-1 kerana definisi haba peneutralan
adalah haba yang dibebaskan apabila 1 mol air terhasil.

Haba peneutralan bagi tindak balas asid kuat dan alkali kuat
adalah -57 kJmol-1.

Haba peneutralan bagi tindak balas asid sulfurik (diprotik asid) dengan
alkali kuat.

2 mol NaOH bertindak balas dengan 1 mol asid sulfurik untuk
menghasilkan 2 mol air. Haba yang dibebaskan = 2 x 57 kJ = 114 kJ.

Walaubagaimanapun, haba peneutralan asid sulfurik dan
alkali kuat tetap 57 kj mol-1 kerana definisi haba peneutralan
adalah haba yang dibebaskan apabila 1 mol air terhasil.

Magnitud haba peneutralan lebih rendah dari – 57 kJmol-1
Contoh :

ü Asid lemah mengion separa di dalam air.
ü Sebahagian zarah masih kekal dalam bentuk molekul.
ü Tenaga haba diserap untuk mengionkan molekul asid lemah yang

belum mengion.

50 cm3 asid hidroklorik 1.0 mol dm-3 ditambah kepada 50 cm3 larutan
natrium hidroksida 1.0 mol dm-3 . Suhu awal dan akhir direkodkan
seperti di bawah:

50 cm3 asid hidroklorik 1.0 mol dm-3

50 cm3 Larutan natrium hidroksida 1.0 mol dm-3

Suhu awal asid hidroklorik 29.0 0C
Suhu awal larutan natrium hidroksida 29.00C
Suhu tertinggi campuran 35.00C

Kira perubahan haba

Suhu purata awal Perubahan haba,Q
= mcӨ
= !"#!" = 29 = 100g X 4.2 J g-1 oC1 X 6.0 0C
!
= 2520 J / 2.52 kJ
Perubahan suhu, Ө

= 35.0 - 29.0 = 6.0 oC

m = 50+50 = 100 g

c =4.2 J g-1 oC-1

persamaan kimia seimbang

HCl + NaOH → NaCl + H2O

Kira bilangan mol bahan tindak balas

Bilangan mol asid hidroklorik, n= !.# $ %# = 0.005 mol
!###
Asid hidroklorik
V = 50cm³ Daripada persamaan,
M= 1.0 mol dm-3 1 mol asid hidroklorik menghasilkan 1 mol air.
0.005 mol asid hidroklorik menghasilkan 0.005 mol air.

Hubungkait bilangan mol air dengan perubahan haba.
*Langkah 3 dan Langkah 1

0.005 mol air terhasil à membebaskan haba
sebanyak 2520 J

Haba dibebaskan ke ∴ 1 mol air terhasil à membebaskan haba
persekitaran
sebanyak !"!# $ = 504000 J mol-1 / 504 kJ
(kerana suhu meningkat) #.##" &'(

Haba tindak balas, ΔH mol-1

= - X kJ mol-1

Oleh itu, haba peneutralan, ∆# = - 504000 J
mol-1 / - 504 kJ mol-1

Apabila 100 cm3 larutan asid nitrik cair , HNO3 2.0 mol dm-3 ditambahkan ke
dalam 100 cm3 larutan natrium hidroksida , NaOH 2.0 mol dm-3 , suhu tindak

balas meningkat dari 27oC kepada 40.65 oC. Hitungkan haba peneutralannya.

Perubahan haba, Q = mcӨ
= (100+100)g X 4.2 J g-1 oC1 X (40.65-27) 0C
= 11466 J / 11.466 kJ

HNO3 + NaOH → NaNO3 + H2O

Bilangan mol asid nitrik, n = !.# ) *## = 0.2 mol
*###

Daripada persamaan,

1 mol asid nitrik menghasilkan 1 mol air.

0.2 mol asid nitrik menghasilkan 0.2 mol air.

0.2 mol air terhasil à membebaskan haba sebanyak 11466 J

∴ 1 mol air terhasil à membebaskan haba sebanyak **+,, $ =
#.! &'(
57330 J mol-1 / 57.33 kJ mol-1

Oleh itu, haba peneutralan, ∆# = - 57.33 kJ mol-1