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O Level Chemistry Tuition Singapore/Chemistry O Level Tuition/Tutor

Introduction to Organic Chemistry – Important Definitions

1. Organic compounds are compounds that contain the element carbon.

2. A hydroarbon is an organic compound that contains only carbon and hydrogen.

3. A homologous series is a family of organic compounds with the same general formula and similar chemical properties.

4. A functional group is an atom or group of atoms that gives characteristic properties to an organic compound.

5. Petrochemicals are chemical made from petroleum or natural gas.

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O-Level Singapore/O-Level/Pure Physics Tuition/Physics Tutor

Key Points – DC Circuits

1. Total emf Et of n identical cells arranged in series
Et = E1 + E2 + E3 …… + En

2. Total emf Et of n identical cells arranged in parallel
Et = E1 = E2 = E3 ……+ En

3. At any point in the circuit, total incoming current is equal total outgoing current

4. The resistance of thermistors decreases as temperature increases.

5. The resistance of light-dependent resistors (LDRs) decreases as light intensity increases.

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A-Level Biology Tuition Singapore/H2 Biology Tuition/JC Biology Tutor

TOPIC 1: CELLULAR FUNCTIONS – Part 3

LEARNING OUTCOME

(c)Describe the formation and breakage of a glycosidic bond.

ESSAY ANSWER

Formation of an
-1,4 glycosidic bond

A condensation reaction between the hydroxyl group at carbon 1 of 1
-
glucose and the hydroxyl group at carbon 4 of another
-glucose results in the formation of an
-1,4 glycosidic bond. 1 water molecule is lost in this reaction.

Formation of an
-1,6 glycosidic bond

A condensation reaction between the hydroxyl group at carbon 1 of 1
-
glucose and the hydroxyl group at carbon 6 of another
-glucose results in the formation of an
-1,6 glycosidic bond. 1 water molecule is lost in this reaction.

Formation of an -1,4 glycosidic bond

A condensation reaction between the hydroxyl group at carbon 1 of 1 -
glucose and the hydroxyl group at carbon 4 of another -glucose results in the formation of -1,4 glycosidic bond. In order the hydroxyl group on carbon 1 to line up alongside with the hydroxyl group on carbon atom 4, every alternate - glucose must be rotated 180o. 1 water molecule is lost in this reaction.

Breakage of
-1,4-glycosidic bond

A hydrolysis reaction breaks the glycosidic bond between the
-glucose
residues. One water molecule is used in this reaction. This restores the
hydroxyl group in carbon positions 1 and 4.

*Note: This answering technique may be applied to the breakage of
-1,6
glycosidic bond and -1,4 glycosidic bond.

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A-Level Economics Tuition Singapore/H2/H1 Economics Tuition

Hi H2/H1 Economics Tuition Students

A. Explain the concept of barriers to entry. [10]

B. Assess the validity of the view that the profits and efficiency of firms in a market is determined by the level of competition. [15]

Suggested Answer:

Synopsis: Essay should define barriers of entry and explain with examples the natural and artificial barriers to entry. Explanations should show within each category the differences in strengths of the different features facilitating/allowing the barriers to be maintained.

a. 1. Define: Barriers to entry
2. Classified into 2 categories: Natural barriers and Artificial barriers
3. Explanation of 2 or more examples of barriers to entry for EACH category.

Intro: Define BTE: Barriers to entry are obstacles that prevent new competitors from competing on an equal basis with established firms in an industry. The obstacles may be naturally existing or artificially created by firms.

Impact on firms: The level of barriers to entry determines the profit equilibrium or what type of profits a firm could earn in the long run.
Type of BTE: Natural and Artificial BTE
(A) Natural Barriers to Entry:

– Existing firm experiencing Economies of Scale = Firm achieves economies of scale –> lowers average cost per unit  potential firms wanting to enter unable to compete since not able to enjoy cost advantage –> barrier to entry
– High Sunk Cost = Poses a type of barrier to entry since high sunk cost would become a big risk since the sunk cost cannot be recovered if firm decides to leave the industry
– Natural Monopoly due to small market = a situation where long- run average costs would be lower if an industry were under monopoly than if it were shared between two or more competitors.
– High Fixed Costs = nature of production requires high fixed cost –> the average cost would also be high since average fixed cost is high –> requires a large amount of output in order for average cost to eventually fall –> discourages the less financially able firms from entering

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A Level GP/General Paper Tuition Singapore

How has the regulation of arts and the media evolved in Singapore? How has new media affected this development?

A. INTRODUCTION

Why regulate?

It is important for democratic societies to have a wide range of independent and autonomous means of communication, in order to be able to reflect a diversity of ideas and opinions…subject to such conditions and restrictions as are prescribed by law and necessary in a democratic society.

The exclusions cover: the prevention of disorder or crime, the protection of health or morals, the protection of the reputation and rights of others (including the right to privacy), preventing the disclosure of information received in confidence, and maintaining the authority and impartiality of the judiciary.

B. BACKGROUND ON THE SINGAPORE CONTEXT
The Key Players in Singapore
1) MICA (Ministry of Information, Communication and the Arts)
2) MDA (Media Development Authority of Singapore)
3) The Individual

Principles of Censorship in Singapore
The guiding principles behind MDA’s regulatory work are:
 To protect the young while providing more choice for adults;
 To uphold community values and support racial and religious harmony;
 To safeguard national and public interest;
In deliberating on and responding to the CRC (Censorship Review Committee) recommendations, we have relied on two key principles. Firstly, we should move with, rather than move ahead of society. While we want to increase content choices for adults, we also have to first ensure that society is generally comfortable with the direction and pace of the changes. Secondly, within the broad parameters defined by our norms and values, we want to make more choices available while we enable parents to exercise more effective control over these choices on behalf of their children.

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A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

Atoms, Molecules and Stoichiometry Part 4

Concept: Back titration and calculation of Percentage Purity

The chemical used for detecting proteins, biuret reagent, H2NCONHCONH2, can be formed by heating urea, (NH2)2CO.

2(NH2)2CO → H2NCONHCONH2 + NH3 — (1)

3.88 g of impure sample of urea (NH2)2CO was heated strongly above its melting point. The ammonia liberated was absorbed in 32.0 cm3 of 2.00 mol dm-3 sulfuric acid. The resulting solution was made up to 500 cm3 with distilled water. 25.0 cm3 of the solution required 25.50 cm3 of 0.20 mol dm-3 sodium hydroxide for neutralization, using methyl orange as an indicator.
(a) Calculate the percentage purity of urea in the sample. [4]

2NH3 + H2SO4 (excess) → (NH4)2SO4 — (2)

H2SO4 (remaining) + 2NaOH → Na2SO4 + 2H2O — (3)

No. of moles of NaOH in 25.50 cm3 = 25.50 x (0.20/)1000 = 5.10 x 10-3

From (3) 2NaOH Ξ H2SO4 (remaining)

No. of moles of H2SO4 in 25.0 cm3 = (5.10 x 10-3)/2 = 2.55 x 10-3

No. of moles of H2SO4 in 500 cm3 = 2.55 x 10-3 x (500/25) = 0.0510

Initial no. of moles of H2SO4 in 32.0 cm3 = 2.00 x (32/1000) = 0.0640

No. of moles of H2SO4 reacted with NH3 = 0.0640 – 0.0510 = 0.0130

From (2) and (1) 2NH3 Ξ H2SO4 (reacted) Ξ 4(NH2)2CO

Mass of urea, (NH2)2CO produced = 0.0130 x 4 x 60 = 3.12 g

Percentage purity of urea, (NH2)2CO = (3.12/3.88) × 100 = 80.4 %

(b) Determine the maximum number of hydrogen atoms present in the biuret reagent after the impure sample of urea was heated. [2]

From (2) and (1) 2NH3 Ξ H2SO4 (reacted) Ξ 2H2NCONHCONH2

No. of moles of H2NCONHCONH2 = 0.0130 x 2 = 0.0260

H2NCONHCONH2 Ξ 5H

No. of H atoms in biuret reagent = 0.0260 x 5 x 6.02 x 10^23
= 7.83 x 10^22

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A-Level Mathematics Tuition Singapore/JC Maths/H2 Math Tuition and Tutor

Hi A-Level/H2 Math Students

Integration by Substitution

Many expressions may be simplified by means of a suitable substitution i.e. change of variable to recognisable forms ie. one of the standard forms.

The following example illustrates the steps involved in the substitution method.

Explanation of Steps:

Step 1: Using the substitution provided, differentiate with respect to x.

Step 2: Replace terms in the integrand involving x with terms involving new variable,u. The operator dx has to be changed to du accordingly. Simplify if necessary.

Step 3: Integrate the integrand with respect to u.

Note that the new integrand should be easier to integrate (if not, it defeats the purpose of carrying out the substitution). Otherwise, check steps 1 and/or 2 for mistakes.

Step 4: Replace all the terms in u with that in x.

Note: Under the syllabus, the substitution to be used will be given to you in the question if you are expected to integrate using a substitution.

However, even if the substitution is not given, you may still use the substitution method if you know the appropriate substitution to use. (But there will usually be another method of doing it.)

Example
By using the substitution u = ln x , find

∫[(ln x )^6 + 1]/x dx

Step 1:
Let u = ln x
du/dx = 1/x
dx/x = du

Step 2: ∫[(ln x )^6 + 1]/x dx
= ∫ (u^6 + 1)du

Step 3: = u^7/7 + u + c

Step 4: = (ln x)^7/7 = ln x +c

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A-Level Physics Tuition Singapore/H2 Physics Tuition/JC Physics Tutor

Hi A-level/H2/JC Physics Tuition students

Mastering Qualitative Questions

Chapter 6: Thermal Physics Part 1

1. A student says that “when two objects are in thermal equilibrium, there will not be heat flow between the two objects and that they have the same internal energy.” Give reasons to support how far you agree with the statement above.[2]

2. State two reasons why temperature of a body is not a measure of the quantity of the thermal energy in the body. [2]

3. Compare the pattern of movement and speed of molecules in water and water vapour at the same temperature.[4]

4. State what is meant by saying that a temperature is on absolute scale. [1]

5 Explain the concept of absolute zero in the thermodynamic temperature scale? [1]

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O-Level Singapore/O-Level/Physics and Chemistry Tuition/Physics Tutor

Key Points – Current Electricity Part 2

7. Resistance R of an electrical component is the ratio of the potential difference V across it to the electric current I going through it.

R = V/I

8. SI unit of resistance is ohm

9. Ohm’s law states that the current I passing through a conductor is directly proportional to the potential difference V across it when the physical conditions are unchanged.

I is proportional to V

10. Conductors that obeys Ohm’s Law are said to be ohmic. Their resistance remains constant under constant physical conditions an their I-V graphs are linear.

11. Non-ohmic conductors do not obey Ohm’s Law. Their resistance varies.

12. Resistivity p is a measure of how much a material opposes the flow of electric current.

13. The resistance R of a wire is directly proportional to its length, inversely proportional to its cross-sectional area A, and directly proportional to the reisitivity p of the material.

R = (pl)/A

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O-Level Singapore/O-Level/Physics and Chemistry Tuition/Physics Tutor

The Atmosphere and Environment – Important Definitions

1. Air pollution is the condition in which air contains substances that are harmful to living things or the environment. These substances are known as air pollutants.

2. A photochemical reaction is a chemical reaction caused by light or ultraviolet radiation.

3. Acid rain is rainwater with pH 4 or less than 4

4. Flue gases are waste gases produced when fossil fuels undergo commutation.

5. Chlorofluorocarbons CFCs are compounds containing the elements carbon, fluorine and chlorine.

6. The carbon cycle is the movement of carbon from carbon dioxide in the in the air through living things and back to the air again.

7. Greenhouse effect is the absorption of infrared radiation by some gases in the air which leads to atmospheric warming.

8. Global warming is the gradual warming of the Earth’s atmosphere due to the increased concentration of greenhouse gases in the atmosphere.

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O Level Chemistry Tuition Singapore/Chemistry O Level Tuition/Tutor

The Atmosphere and Environment – Important Definitions

1. Air pollution is the condition in which air contains substances that are harmful to living things or the environment. These substances are known as air pollutants.

2. A photochemical reaction is a chemical reaction caused by light or ultraviolet radiation.

3. Acid rain is rainwater with pH 4 or less than 4

4. Flue gases are waste gases produced when fossil fuels undergo commutation.

5. Chlorofluorocarbons CFCs are compounds containing the elements carbon, fluorine and chlorine.

6. The carbon cycle is the movement of carbon from carbon dioxide in the in the air through living things and back to the air again.

7. Greenhouse effect is the absorption of infrared radiation by some gases in the air which leads to atmospheric warming.

8. Global warming is the gradual warming of the Earth’s atmosphere due to the increased concentration of greenhouse gases in the atmosphere.

If you need help in the above topic, please contact Angie @96790479 or Mr Ong 98639633

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O-Level Singapore/O-Level/Pure Physics Tuition/Physics Tutor

Key Points – Current Electricity Part 2

7. Resistance R of an electrical component is the ratio of the potential difference V across it to the electric current I going through it.

R = V/I

8. SI unit of resistance is ohm

9. Ohm’s law states that the current I passing through a conductor is directly proportional to the potential difference V across it when the physical conditions are unchanged.

I is proportional to V

10. Conductors that obeys Ohm’s Law are said to be ohmic. Their resistance remains constant under constant physical conditions an their I-V graphs are linear.

11. Non-ohmic conductors do not obey Ohm’s Law. Their resistance varies.

12. Resistivity p is a measure of how much a material opposes the flow of electric current.

13. The resistance R of a wire is directly proportional to its length, inversely proportional to its cross-sectional area A, and directly proportional to the reisitivity p of the material.

R = (pl)/A

If you need help in the above topics, please contact Angie @96790479 or Mr Ong @98639633

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O-Level Additional Mathematics Tuition Singapore

Objective
Identify non-linear equations and use substitution method to solve the equations.

Any equation not in the form of ay + bx = c is considered non-linear equation.

Note:
When solving 2 simultaneous equations, the solutions are actually the intersection points ( x and y coordinates) of the 2 intersecting curves (or straight lines) represented by the simultaneous equations.

Stop and Think
3x +4y = 2
x^2 + 8xy +12 = 0

Ans: x = -6/5, y = 7/5 or x = 2, y = -1

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O Level E Maths Tuition Singapore/Tuition O Level E Maths/Tutor

Reciprocal

1. The reciprocal of a number n is 1/n

Example: The reciprocal of 2 is 1/2

2. The reciprocal of a fraction a/b is b/a

Example: The reciprocal of 2/3 is 3/2

3. The product of an number and its reciprocal is 1

Example: 2 x 1/2 = 1

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A-Level Biology Tuition Singapore/H2 Biology Tuition/JC Biology Tutor

TOPIC 1: CELLULAR FUNCTIONS – Part 2

LEARNING OUTCOME

(b)Outline the functions of the membrane systems and organelles listed in Cellular Functions in Part 1.

ESSAY ANSWER

Cell surface membrane
The cell surface membrane acts as a boundary to separate the cell contents
from the external environment. It serves to control exchange of substance
between the cell and its external environment. The cell surface membrane is also a site for chemical reactions, recognition of external stimuli and
attachment of other cells, the extracellular matrix and the cytoskeleton.

Rough endoplasmic reticulum
The rough endoplasmic reticulum (RER) is associated with ribosomes that
synthesis proteins via translation. It is also part of the endomembrane
system, functioning as an intracellular transport network. Proteins that are synthesised by ribosomes bound to it are packaged for transport to other parts of the cell. Some chemical modifications are also made to proteins. The RER provides a large surface area for biochemical reactions to occur.

Smooth endoplasmic reticulum
The smooth endoplasmic reticulum (SER) is the site of lipid and steroid
synthesis. It is also part of the endomembrane system, being involved in the packing of certain proteins from the RER. Some chemical modifications are also made of proteins. The SER is also involved in detoxification reactions and is a site of Cl- and Ca2+ storage.

Golgi body
The golgi body is part of the endomembrane system. It chemically modifies
proteins and lipids such as by attachment of carbohydrate chains to proteins and lipids to form glycoproteins and glycolipids. It also packs these substances into vesicles for storage, for transport to other parts of the cell or for secretion. Lysosomes are formed from the golgi body.

Lysosome
Lysosomes contain hydrolytic enzymes within an acidic environment. They
are involved in digestion of materials taken up by endocytosis and
phagocytosis, autolysis, autophagy and remodelling of the extracellular
matrix.

Mitochondria
Mitochondria are the sites of ATP synthesis via aerobic respiration. ATP is produced via substrate-level phosphorylation in the matrix via the link
reaction and Krebs cycle and at the inner mitochondrial membrane via
oxidative phosphorylation. Lipid metabolism and synthesis also occur in
mitochondria.

Chloroplast
Chloroplasts are the sites of photosynthesis. The light dependent reactions occur at the thylakoid membranes. Light energy is harvested and used to synthesis ATP via photophosphorylation. The light independent reactions occur in the stroma. Carbon dioxide is fixed and reduced to form
carbohydrates. Starch is also temporarily stored in the stroma.

Nucleus
The nucleus contains genetic information in the form of DNA. It controls
cellular activities including cell division and protein synthesis.

Nuclear envelope
The nuclear envelope is perforated by channels known as nuclear pores which regulate the exchange of substance between the nucleus and the rest of the cell.

Nucleolus
The nucleolus is the site of transcription of rRNA genes to form rRNA. rRNA is then combined with ribosomal proteins to form ribosomal subunits.

Ribosomes
Ribosomes are the site of protein synthesis via translation. Ribosomes read the codons on mRNA and allow base-pairing with complementary tRNA anticodons. The large ribosomal subunit carries peptidyl transferase activity allowing the formation of peptide bonds between amino acids joined to tRNAs.

Centrioles
Centrioles are involved in the organisation of spindle fibres during cell
division, and serve as an anchorage for cilia and flagella.

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