Edexcel June 2019 Pure mathematics 2 worked solutions

Question 1

Edexcel June 2019 Pure mathematics 2 worked solutions

Question 2 (a)

Edexcel June 2019 Pure mathematics 2 worked solutions

Question 2 (b)

Question 3 (a & b)

Question 4

Question 5

Question 6 (b)

Question 6 (a)

Question 6(c) 

Question 6 (d)

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Edexcel IGCSE Biology (2017)

2.45B Investigate the effect of light on net gas exchange from a leaf, using hydrogen-carbonate indicator

(1) Experimental setup 

1. Setup four tests tubes each containing equal volume of hydrogen-carbonate indicator solution.
2. Place three large leaves in three test tubes sealing it with a bung.
3. The four test tubes are exposed to bright light for 24 hours.  

Figure 1 – Experimental setup to investigate the effect of light on net gas exchange in leaves, using hydrogen-carbonate indicator solution.

 

 

1 – Control (no leaf)

2 – Tube with leaf

3-  Tube with leaf covered with aluminium foil (blocks out light)

4-  Tube with leaf covered with guaze (partially lets light through)

 

 

 

(2) Results

Control tube vs. tube with leaf

Figure 2 – Colour change observed in test tube exposed to light (tube 2).

Result 

Control tube showed no change in colour. The hydrogen-carbonate indicator solution remained orange. The tube exposed to light changed colour changed colour from orange to purple.

Explanation

In bright light, the rate of photosynthesis is greater than rate of respiration. So, leaf takes in CO₂ from the air in the test tube. This reduces the CO₂ inside the test tube and  the colour of hydrogen-carbonate indicator solution changes from orange to purple as CO₂ is taken away form the solution.

Test tube covered in aluminium foil

Figure 3 – Colour change in test tube covered with aluminium foil.

Result

The test tube covered with aluminium foil changed colour from orange to yellow.

Explanation

Aluminium foil prevents light entering the test tube. The rate of respiration is greater than rate of photosynthesis (as light is needed for photosynthesis). So, leaf gives out CO₂, which dissolves in the hydrogen-carbonate indicator solution to change its colour from orange to yellow.

 

Test tube covered with guaze

Figure 4 – Colour change in test tube covered with guaze.

Result 

The test tube covered with guaze didn’t show any colour change. The colour of the hydrogen-carbonate indicator solution remained orange.

Explanation

Guaze partially allows the light to enter the test tube. Therefore, the rate of respiration equals the rate of photosynthesis. So, there is no change in CO₂ in the test tube, which keeps the hydrogen-carbonate indicator solution orange in colour.

Edexcel IGCSE Biology Past exam questions

Ans (a) Light intensity affects photosynthesis.  

Ans (b): 1) Size of the leaf  2) Volume of indicator

Ans (c) Tube D acts as a control. This tube was setup to see whether the indicator will change colour with no leaf.

Ans (d) i The rate of photosynthesis is more than the rate respiration so less CO₂ is present in the test tube. This causes CO₂ to leave the indicator solution and a purple colour is observed.  

Ans (d) ii : In tube B, no photosynthesis takes place. CO₂ is released from respiration, which dissolves in the indicator solution to produce a colour change from orange to yellow. 

Ans (e) i: In tube C, the rate of respiration equals the rate of photosynthesis. Hence, there is no net change of CO2 in the test tube resulting in no change of colour of the hydrogen-carbonate indicator. 

Ans (e) ii : In tube D, there was no leaf. Hence, no change observed. 

Ans (f) : Limewater can only show an increase in CO₂ concentration. It will not be able to show a decrease in CO₂ concentration. 

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IB Biology Essential Idea

The evolution of multicellular organisms allowed cell specialisation and cell replacement

The cell theory

According to cell theory, living organisms are composed of cells

The topic 1 of International Baccalaureate (IB) SL & HL Biology starts with Cell Theory. Here is a brief introduction to ‘Understanding point 1’ of the Topic 1: Cell Biology which is as follows: 

• Students should know that according to Cell Theory all cells are composed of cells. 

 By 17th century scientists knew that the human body is composed of organs, which in turn was made up of tissues. With the advent of microscope, scientists started probing the plant and animal tissues and developed cell theory, which states that the cells are the fundamental buildings blocks of all living organisms. The smallest organism is made up of single cell and therefore known as unicellular whereas as larger organisms are multi-cellular as they are composed of many cells. Therefore, the cells are the fundamental building blocks of all living organisms.

Features of cells

1. Cells are separated from their surroundings by membrane – cell surface membrane.
2. Cells contain genetic material, which control the functions of the cell – DNA.
3. Chemical reactions in cells are are catalysed by enzymes.
4. Cells have metabolic processes in place to produces energy in order to power the cellular activities – Mitochondria.

Theory, predictions and discrepancies

Scientists look for trends that can found in variety of different organisms. This forms the early stages of scientific investigation. Robert Hooke was the first to use the word cell for structures in living organisms in 1665 after examining cork cells. His drawing of cells from cork tissues in shown in fig. 1.

Figure 1  

Robert Hooke looked at many different types of plant tissues and found similar structures across all the tissues. He had discovered a general trend (things that appear to be found generally rather than just in specific cases) across many of the tissues. These trends lead to the development of cell theory. Since 17th century many more scientists have looked into variety of different tissues in different organisms. Many of the tissues are shown to consist of cells although some discrepancies have been found. In future more discrepancies are likely to be discovered but cell theory is unlikely to be discarded in the face of such anomalies because cell theory has proved to be true is large number of tissues.

• Scientific investigation into trends lead to the development of cell theory. Theories allow making predictions and explain the natural world. If discrepancies become common then the theory is discarded.
  
  

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A-level Chemistry Specifications

This article will cover the following topics from A-level chemistry specifications:

OCR AS Chemistry

• The changing models of atomic structure over time and the use of evidence to accept or reject particular models.
• Atomic structure in terms of the numbers of protons, neutrons and electrons for atoms and ions, given the atomic number, mass number and any ionic charge.

AQA AS Chemistry

• Appreciate how the understanding of atom has evolved over time.
• An atomic structure in terms of a nucleus containing protons and neutrons surrounded by electrons.

Edexcel AS Chemistry

• Students should consider how models of atomic structure have developed over time, as new evidence has become available.
• Know the structure of an atom in terms of electrons, protons and neutrons.

CIE AS Chemistry

• Identify and describe protons, neutrons and electrons in terms of their relative charges and relative masses.
• Describe the distribution of mass and charges within an atom.
• Deduce the numbers of protons, neutrons and electrons present in both atoms and ions given proton and nucleon numbers.
• Describe the contribution of protons and neutrons to atomic nuclei in terms of the proton number and nucleon number (ii) distinguish between isotopes on the basis of different numbers of neutrons present (iii) recognise and use the symbolism  _y^xA for isotopes, where x is the nucleon number and y is the proton number.

Topic 1: The changing models of atomic structure

This topic will highlight the changing models of atomic structure and how sequence of experiments, observations, analysis and conclusions led to the modern day model of atom. The content of the article will be restricted to AS/A-level students.

Democritus model of atom ~ 400 BC

Fig. 1. The Democritus concept of atom. If you keep cutting a metal eventually a point is reached where cutting the metal any further would not be possible. The smallest unit of metal is known as atom.

The concept of atom evolved from the understanding of Greek philosopher Democritus. The understanding was merely a philosophical & theological reasoning rather than based on hard evidence. He understood matter as being made up of tiny particles that were indivisible and called such particles atoms. In his view, there were many different types of atoms, which differ in shapes, sizes and nature of connectivities between them. For example, iron was made up of solid atoms with hooks locking atoms to each other, salt was made up of pointed and sharp atoms because of their taste and the atoms present in air were thought to be lightweight and swirling around in space.

John Dalton’s atom ~1800

Fig. 2. John Dalton used his own symbols to represent atoms and also developed the first table of atomic masses in New system of Chemical Philosophy (1808).

John Dalton was the founder of the atomic theory. The main points of his atomic theory were:

1. Elements are made up of tiny particles called atoms.
2. Atoms cannot be divided, altered or destroyed by chemical means.
3. Atoms of an element are all the same in terms of shape and size and atoms of different elements differ in size and shape.
4. Elements are characterised by mass of their atoms so all the atoms of an element have identical weight. Different elements have atoms of different atomic weight.
5. Atoms of different elements can combine together to form chemical compounds.
6. When elements undergo chemical reactions, their atoms combine in simple, whole number ratios or more than one simple whole number ratios. Dalton noted that tin can combine with either one or two oxygen atoms which was in line with the percentages of masses observed for tin oxides. Similarly he used the same principle to explain that the different ratios of nitrogen to oxygen in various nitrogen oxides were simple multiples of each other.

However, John Dalton wrongly postulated that when atoms combine only in one ratio, it must be assumed to be binary (in two) one. He got formula of water as HO and ammonia as NH that were both incorrect as he got the atomic weights of oxygen and nitrogen incorrect. A lot of his own experimental evidence didn’t coincide with his theory of atomic structure and was at odds with the scientific community.

Joseph John Thompson’s model of atom ~ 1897

 Until 1897 atoms were thought to be the smallest possible division of matter. However, J.J. Thompson discovered ‘electron’ through series of experiments on cathode rays, which disproved the notion that atoms were indivisible as proposed by ancient Greeks and Dalton. This led to the discovery of sub-atomic particles, which are considered to be the fundamental units of the modern day atomic structure.

Fig. 3. Design of cathode ray tube (CRT) used by J.J Thompson to discover the nature of cathode rays.

Before we look into how J.J. Thompson discovered electron, we need to understand how cathode ray tube works. A cathode ray tube (CRT) is a sealed glass container in which a cathode (negatively charged) and anode (positively charged) are separated by a vacuum as shown in fig. 3. When a voltage is applied across the electrodes, cathode rays are generated from the cathode and are attracted towards the anode. A small hole in the anode allows some electrons to pass through creating a beam of cathode rays producing a glowing patch when they strike the phosphorous-coated glass at the opposite end of the tube or register an electric current when the end is attached to an electrometer. Earlier physicists had discovered cathode rays as yet unknown entity that registered an electric charge on an electrometer. Scientists debated whether cathode rays were immaterial like light or particles of matter. Thompson carried out a series of experiments using cathode rays generated in CRT and was able to demonstrate the following properties of the cathode rays:

• Cathode rays were negatively charged particles. Thompson introduced metal plates on either side of the CRT to bend the cathode rays emerging from the hole present in the anode. He observed that cathode rays were deflected towards the positive electrode of the electric field indicating that cathode rays were negatively charged (Fig.4).

Fig. 4. The deflection of cathode rays by electric field.

• The particles were deflected by magnetic as well as electrical field.
• The mass of the particles was very, very small. The particles were about 2000 times lighter than the hydrogen ions.

J.J.Thompson named the particle as ‘corpuscle’ but the name was later changed to ‘electron’. Since the electrons originated from within the electrodes, they led Thompson to conclude that atoms were divisible and that electrons (corpuscles) were the building blocks of the atom. He proposed that electrons were moving randomly in a sea of uniformly distributed positive charge to account for the neutral charge of the atom as plums in a plum pudding. This model is called the ‘plum pudding model’ of atomic structure (fig.5).

Fig.5. The plum pudding model of the atomic structure.

Ernest Rutherford model of atomic structure ~ 1909

J.J. Thompsons plum-pudding model was disproved by the experiment carried out by Hans Geiger and Ernest Marsden (Geiger-Marsden experiment) working at the behest of Ernest Rutherford. This led to the development of the nuclear model of atomic structure. In Geiger-Marsden experiment (a.k.a gold-leaf experiment), alpha particles were fired at thin sheet of gold foil and  the deflection of the particles was measured once they crossed the gold foil. Based on the Thompson’s plum pudding model of atom (small mass of electrons & low concentration of positive charge), the scientists expected that the alpha particles to pass through the gold foil without any deflection. They made following observations:

Expected results from the plum pudding model of Thompson 

alpha particles passing through the atom with negligible deflection

Actual results from the gold-leaf experiment: 

A small proportion of particles were deflected by the dense positive charge of the nucleus.

1.     Majority of the particles were not deflected at all.

2.     A small percentage of particles were deflected through large angles.

3.     A very small fraction of particles were deflected back to the source.

Rutherford concluded that the positive charge of the atom must be concentrated in a very small volume to produce electric field of sufficient strength to repel the alpha particles.  Rutherford proposed the following model of nuclear atom in 1911 (fig.6).

1.     The positive charge of an atom is concentrated in the nucleus, which was assumed to be present in the center.

2.     Clouds of electrons orbit the nucleus.

3.     Most of the atom’s volume lies between tiny nucleus and cloud of electrons.

4.     There is no net charge on the atom as positive and negative charges cancel out each other.

Fig. 6. The proposed structure of the nuclear atom by Ernest Rutherford.

Henry Moseley’s contribution to the atomic structure ~ 1913

Before Mosley, the atomic number of elements was an arbitrary number assigned by scientists to order the elements in the periodic table. Moseley used a technique called X-ray spectroscopy to establish a mathematical relationship between the atomic number of the elements and the wavelength of the X-rays. Moseley’s work redefined atomic numbers into a value that can be used to order the elements in the periodic table. His work supported the concept of atomic structure put forward by Ernest Rutherford, which had compact positively charged nucleus surrounded by orbiting cloud of electrons.  Moseley understood atomic number to be the exact number of positive charges (later found by Ernest Rutherford to be protons) in the nucleus of an atom.

Bohr’s model of atomic structure ~ 1913

In 1913 Neils Bohr proposed the Bohr model of atom that incorporated the idea of quantum physics postulated by Albert Einstein and Max Planck that light energy is emitted or absorbed in discrete amounts known as quanta. His model allows electrons to follow certain paths (shells) around the nucleus with discrete energy. Under this model the electrons in shells could not loose energy and can only jump from one energy level to another. When this happened light was emitted or absorbed at a frequency proportional to the energy difference between shells (Fig.7). Bohr’s model of atomic structure helped explain some periodic properties such as spectral lines seen in emission spectra and the energy of electrons at different distances from the nucleus.

Fig. 7. The planetary model of Neils Bohr’s atomic structure.

Quantum model of atomic structure ~ 1923

In 1924, Louis de Broglie proposed that the behaviour of electrons could be explained both in terms of wave and particle. In 1926 Erwin Schrödinger described electron as a wave particle in the form of Schrödinger equation. The equation was difficult to visualise and Max Born reconciled the opposing ideas of describing electron as wave and particle by introducing the theory of particle-wave duality. He proposed that Schrödinger equation describes the position of electrons in an atom in terms of probabilities. This invalidated the concept of Bohr’s Model of atom, which had clear defined paths of electrons in circular orbits. Rather the position of electrons in an atom is now defined in terms of probabilities. Based on this, the region with the highest probability of finding an electron around the nucleus is called the atomic orbital.

Quantum mechanics model

Electrons are found in regions of space (orbitals) whose shape is described by mathematical equations

Discovery of nuclear particles ~ 1917 – 1935

In 1918, Ernest Rutherford discovered protons from his work on bombarding nitrogen gas with alpha particles and observing the hydrogen nuclei being emitted from the gas. Further experimentation led him to conclude that the mass of the nucleus was greater than the protons it contained. This was attributed to the unknown neutral particles present in the nucleus, which he called neutrons. This was validated by the experiments of James Chadwick in 1932, who bombarded high-energy α-particles to beryllium and observed the mysterious highly penetrating and electrically neutral radiation. He further demonstrated that the neutral radiation had mass similar to proton and hence cannot be weight-less gamma rays. James Chadwick had discovered Rutherford’s neutrons.

Summary: Timeline of the atomic structure

Topic 2: Atomic Structure

Current model of atomic structure

The current model of the structure of an atom is as follows:

1. The nucleus contains protons and neutrons, which are located at the center of the atom.
2. Electrons orbit the nucleus in shells, which contain the set of orbitals.
3. The nucleus is small, dense mass at the centre of the atom.
4. Most of the atom is empty space between the nucleus and the electron shells.

 Fig. 1 The current model of the atomic structure.

 Table 1 shows the relative mass and charge of sub-atomic particles.

• Protons are positively charged and electrons are negatively charged while neutrons have no charge.
• An atom has same number of protons and electrons so the atom is electrically neutral.
• A proton has same mass as a neutron.

Table 1 shows charges and masses of sub-atomic particles.

Isotopes

• Isotopes are atoms of the same element with the same number of protons, but different numbers of neutrons. 
• Isotopes have similar chemical properties (react in a similar way) because they have the same electronic structure (number of electrons in the outer shells).
• Isotopes may have slightly varying physical properties because of different masses (e.g. different melting point and boiling points).
• Elements do not usually contain all identical atoms but are made up of a mixture of isotopes. The natural abundances may vary in different sources.

Symbol notation of elements in the periodic table

Carbon and hydrogen isotopes

Table 2 – Atomic structures of hydrogen and carbon isotopes.

Atomic structure of ions

 Atoms form ions by gaining and loosing electrons. The ions are charged because the positively charged protons and negatively charged electrons do not cancel each other. The electrons are lost to form positively charged ions while electrons are gained to form negatively charged ions.

Table 4 show examples of the atomic structure of ions.

This completes the AS level topic on the changing models of atom and atomic structure. 

Atomic Structure Worksheet

DOWNLOAD ATOMIC STRUCTURE WORKSHEET

SCROLL DOWN FOR VIDEO SOLUTION TO THE ATOMIC STRUCTURE WORKSHEET

Section 1: Multiple Choice Questions

Q1. The isotope cobalt-60 (⁶⁰Co) is used to destroy cancer cells in the human body.

Which statements about an atom of cobalt-60 are correct?

1. It contains 33 neutrons.

2. Its nucleus has a relative charge of 27+.

3. It has a different number of neutrons from the atoms of other isotopes of cobalt.

(A) 1, 2 & 3 are correct.

(B) 1 & 2 are correct.

(C) 2 & 3 are correct.

(D) 1 only is correct.

Q2. In which species are the numbers of electrons and neutrons equal?

A ⁹Be

B ¹⁹F

C ²³Na⁺

D ¹⁸O^2–

Q3. A radioactive isotope of thallium, ²⁰¹₈₁Tl, is used to assess damage in heart muscles after a heart attack.

Which statement about ²⁰¹ ₈₁Tl is correct?

A This isotope has a nucleon number of 120.

B The number of electrons in one atom of this isotope is 81.

C The number of neutrons in one atom of this isotope is 201.

D ²⁰¹ ₈₂X is an isotope of ²⁰¹ ₈₁Tl.

Q4. John Dalton’s atomic structure theory, published in 1808, contained four predictions about atoms. Which of his predictions is still considered to be correct?

A   Atoms are very small in size.

B   No atom can be split into simpler parts.

C   All the atoms of a particular element have the same mass.

D All the atoms of one element are different in mass from all the atoms of other elements.

Q5. The technetium–99 isotope (⁹⁹Tc) is radioactive and has been found in lobsters and seaweed adjacent to nuclear fuel reprocessing plants.

Which statements are correct about an atom of ⁹⁹Tc?

1 It has 13 more neutrons than protons.

2 It has 43 protons.

3 It has 99 nucleons.

(A) 1, 2 & 3 are correct.

(B) 1 & 2 are correct.

(C) 2 & 3 are correct.

(D) 1 only is correct.

Q6. Skin cancer can be treated using a radioactive isotope of phosphorus, ³²P. A compound containing the phosphide ion ³²₁₅P³⁻, wrapped in a plastic sheet, is strapped to the affected area.

 What is the composition of the phosphide ion, ³²₁₅P³⁻ ?

Q7. One of the most important chemical species responsible for the removal of ozone from the stratosphere is a free radical of chlorine, ³⁵Cl^•.

What is the atomic structure of ³⁵Cl^• ?

Q8. Unnilpentium is an artificial element. One of its isotopes is ²⁶²₁₀₅Unp. Which of the following statements is correct?

A ²⁶²₁₀₅Unp has a nucleon number of 105.

B The atom ²⁶⁰X is an isotope of ²⁶²₁₀₅Unp.

C There are 262 neutrons in ²⁶²₁₀₅Unp.

D The proton number of ²⁶²₁₀₅Unp is 262.

Q9. Use of the Data Booklet is relevant to this question.

 It is now thought that where an element exists as several isotopes, the stable ones usually contain a ‘magic number’ of neutrons. One of these magic numbers is 126.

Which isotope is unstable?

A ²⁰⁹Bi                        B ²⁰⁸ Pb                      C ²¹⁰Po                     D ²⁰⁸ Tl

Q10. Which ion has more electrons than protons and more protons than neutrons?

[H = ¹₁H; D = ² ₁H; O = ¹⁶₈ O]

A    D^–                          B  H₃O⁺                          C  OD^–                          D  OH^–

Q11. Use of the Data Booklet is relevant to this question.

Hard water contains calcium ions and hydrogencarbonate ions arising from dissolved calcium hydrogencarbonate, Ca(HCO₃)₂.

How many electrons are present in the hydrogencarbonate anion?

A 30                                  B 31                                C 32                              D 33

Q12. The nucleus of a radon atom contains

           A           222 protons and 86 neutrons               B   86 protons and 136 neutrons

           C            86 protons and 222 neutrons               D 86 protons, 136 neutrons and 86 electrons

Q13. In which pair do the species contain the same number of neutrons?

Q14. Complete the table below by giving the numbers of protons, neutrons and electron [4]

Q15. Consider the composition of the species W, X, Y and Z below. Which species is a positive ion?

A               W                                    B               X                                  C               Y                                         D               Z

Q16. Chlorine occurs as two natural isotopes, Chlorine-35 and Chlorine-37. Which one of the following statements concerning chlorine atoms is FALSE.

A           All chlorine atoms have the same nuclear charge.

B           The nuclei have diameters which are approximately 1/10 000 of the diameter of chlorine atoms.

C            Some naturally occurring chlorine nuclei contain 18 neutrons.

D           Some naturally occurring chlorine nuclei contain 20 electrons.

Q17.  Isotopes are elements with

A           the same atomic number and the same number of neutrons

B           the same mass number but a different number of neutrons

C            the same atomic number but a different number of neutrons

D           different atomic and mass numbers but the same number of neutrons

Q18. Which of the following particles contains more electrons than neutrons

A           I only                B               II only                     C               I and II only             D   II and III only

 Q.19 Which species has 54 electrons and 52 protons?

^{(A) 128}₅₂Te²⁻

^{(B) 132} ₅₄Xe²⁺

^{(C) 132} ₅₄Xe²⁻

^{(D) 128} ₅₂Te²⁺

Q20. Which of the following represents a pair of isotopes?

A   ¹⁴₆C and ¹⁴₇N

B   ³²₁₆S and ³²₁₆S^2–

C     O₂ and O₃

D     ²⁰⁶₈₂Pb and ²⁰⁸₈₂Pb

Q21. An isotope of an element, atomic number z, has mass number 2z + 4. How many neutrons are in the nucleus of the element?

A      z+ 4

B      z+ 2

C     z

D     4

Q22. The nucleus of a ²³ ₁₁Na atom contains

A 11 protons and 12 neutrons.

B 11 protons and 12 electrons.

C 23 protons and 11 neutrons.

D 23 protons and 11 electrons.

Q23. The isotopes of magnesium, ²⁴₁₂Mg and ²⁵₁₂Mg, both form ions with charge 2+. Which of the following statements about these ions is true?

A       Both ions have electronic configuration 1s² 2s² 2p⁶ 3s².

B     ²⁵Mg²⁺ has more protons than ²⁴Mg²⁺.

C      The ions have the same number of electrons but different numbers of neutrons.

D      The ions have the same number of neutrons but different numbers of protons.

Q24. Which of the following does not have exactly 10 electrons?

A  An ion of fluorine, F^–

B  A molecule of methane, CH₄

C  A molecule of nitrogen, N₂

D An ion of sodium, Na⁺

Q25. Which of the following statements about ²⁵₁₂Mg is true?

(A) It has the same number of neutrons and protons.

(B) It can be oxidized to form an Mg²⁺

(C) It is the most common isotope of magnesium.

(D) It has higher ionization energy than ²⁴₁₂Mg.

Q26. 

Section 2: Atomic structure & the changing models of atom (Short Answer Questions)

Q1. Sir James Jeans, who was a great populariser of science, once described atomic structure of carbon as being like six bees buzzing around a space the size of a football stadium.

(a) (i) Suggest what were represented by the six bees in this description.

……………………………………………………………………………………………………………………………….[1]

(ii) Explain (in terms of an atom of carbon) what stopped the bees from flying away from the space of the football stadium.

…………………………………………………………………………………………………………………………………..

……………………………………………………………………………………………………………………………….[1]

(iii) What is missing from Jeans’ description when applied to an atom of carbon?

…………………………………………………………………………………………………………………………………..

……………………………………………………………………………………………………………………………….[1]

Q2. (a) Define an isotope in terms of its sub-atomic particles.

………………………………………………………………………………………………………………………………….

………………………………………………………………………………………………………………………………..[2]

Q3. Iron and cobalt are adjacent elements in the Periodic Table. Iron has three main naturally occurring isotopes, cobalt has one.

(a) Explain the meaning of the term isotope ? 
…………………………………………………………………………………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………..[2]

(b)  The most common isotope of iron is ⁵⁶Fe; the only naturally occurring isotope of cobalt is ⁵⁹Co. 
Use the Data Booklet to complete the table below to show the atomic structure of ⁵⁶Fe and of ⁵⁹Co.

Q4. In the 19th and 20th centuries, scientists established the atomic theory and showed that three sub-atomic particles, electron, neutron and proton, exist. The masses and charges of these three particles were subsequently determined.

(i) Define the term proton number.

…………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………[1]

(ii) Why is the proton number of an atom of an element usually different from the nucleon number of an atom of the element?

………………………………………………………………………………………………………………….

…………………………………………………………………………………………………………………[1]

(iii) Protons and neutrons have been used in nuclear reactions which result in the formation of artificial elements. In such processes, protons or neutrons are accelerated to high speeds and then fired like ‘bullets’ at the nucleus of an atom of an element.

Suggest why neutrons are more effective than protons as ‘nuclear bullets’.

………………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………..[2]

(iv) In some cases, when neutrons are fired at atoms of an element, the neutrons become part of the nucleus of those atoms.

What effect does the presence of an extra neutron have on the chemical properties of the new atoms formed? Explain your answer.

………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………..[2]

Q5. The relative atomic masses of the elements are based upon the ¹²C scale.

(a)     What is the composition of the ¹²C nucleus?

……………………………………………………………………………………..[1]

(b)   Carbon also exists as the ¹⁴C isotope. How does the composition of the ¹⁴C isotope differ from that of the ¹²C isotope?

…………………………………………………………………………………………………………………… [1]

Q6. Hydrogen has three isotopes, ¹H, known as protium, ²H, deuterium, and ³H, tritium.

(a) In terms of sub-atomic particles, give the similarities and differences between atoms of these three isotopes of hydrogen.

………………………………………………………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………………………………………………….[3]

Q7.    (a)     Give the relative charge and relative mass of an electron.

Relative charge

.…………………………………………………………………………………….[1]

Relative mass

…………………………………………………………………………………….[1].

(b)     Isotopes of chromium include ⁵⁴Cr and ⁵²Cr

(i)      Give the number of protons present in an atom of ⁵⁴Cr

……………………………………………………………………………………..[1]

(ii)     Deduce the number of neutrons present in an atom of ⁵²Cr

……………………………………………………………………………………..[1]

Q8.    (a)     Complete the following table. [2]

(b)     An atom of element Q contains the same number of neutrons as are found in an atom of ²⁷Al. An atom of Q also contains 14 protons.

(i)      Give the number of protons in an atom of ²⁷Al.

………..………………………………………………………………[1]

(ii)     Deduce the symbol, including mass number and atomic number, for this atom of element Q.

………..………………………………………………………………………………………………………………………[1]

Q10. Geiger and Marsden studied the structure of atoms. The diagram shows part of the equipment, which they used.

(i)  The block of lead helped to shield the scientists from radiation. State another purpose of the block of lead. ………………………………………………………………………………………………………………………………………………………………………………………………………………………..
(1)

(ii)  Most of the alpha particles went straight through the gold foil. State another purpose of the block of lead. 
………………………………………………………………………………………………………………………………………………………………………………………………………………………..(1)

(iii)  Some of the alpha particles were deflected. What explanation did the scientists suggest for the deflection? 
…………………………………………………………………………………………………………………………………………………………………………..
……………………………………………
(2)

(iv) Only a small proportion of the alpha particles deflected through a large angle. What explanation did the scientists suggest for the proportion being small? 
…………………………………………………………………………………………………………………………………………………………………………………………………………………………(1)

(v)  Some alpha particles were deflected less than others. What two reasons did the 
scientists suggest for this?
 …………………………………………………………. 
………………………………………………………………………………………………………………………………………………………………………………………………………………………… 
(2)

(vi) The alpha particles were detected when they hit a zinc sulphide screen. How did the scientists know that an alpha particle had hit the screen?

……………………………………………………………………………………………………………….. …………………………………………………………………………………………………………[1]

Q11. In Geiger and Marsden’s α-particle scattering experiment, α-particles were directed at a very thin gold foil. Fig. 11 shows five of the nuclei of the atoms in one layer in the gold foil. Also shown are the paths of three α-particles directed at the foil.

(a) On Fig.11.1, complete the paths of the three α-particles. [3]

(b) (i) State the result of the experiment that shows that an atom consists of a very tiny, charged core, containing almost all the mass of the atom. [1]

(ii) State the sign of the charge on this core. ……………………………………………………………………………………………………………………………………………….. [1]

(iii) State what occupies the space between these charged cores. [1]

(c) The nuclide notation for an α-particle is ⁴₂ α.
State the number of protons and neutrons in an α-particle [2]

protons = ——————————————————————————————————————-

Neutrons= —————————————————————————————————————–

Q12. Use the following information to help you answer the questions.

The gold foil experiment

Scientists used to think that electrons were spread out through a positively charged atom. They called this the ‘plum pudding’ model of atomic structure.

To test this idea, scientists aimed alpha particles at thin gold foil. They expected the alpha particles to pass straight through.

The results showed that almost all the alpha particles did pass straight through, but a few did not. About 1 in every 8000 was deflected away at a very large angle. It was these ‘anomalous’ results that led to a new understanding of the atom.

(a) What was the prediction in this experiment? [1]

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

(b) (i) What do scientists mean by anomalous results? [1]

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

(ii) How should scientists deal with anomalous results? [1]

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

8 c) Explain how these anomalous results led to the idea of a positive charge at the center of an atom. [2]

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

This marks the end of the worksheet on Atomic Structure and The Changing Models of atom. 

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Structure Determination using IR Spectroscopy

The major use of IR spectroscopy is in determining the structures of organic compounds.

In an IR spectrometer IR radiation in the range 400-4000 cm^-1 is passed through a sample. The printout of the spectrum then shows which frequencies (wavenumbers) are absorbed.

IR spectra are always looked at with the baseline (representing 100% transmittance, i.e. zero absorbance) at the top. The troughs (usually called “bands”) thus represent wavenumbers at which radiation is absorbed.

Infra-red Spectrum of propanone

The infra red spectrum can be used to determine the bonds present in a molecule.

Thus, in the region above 1500 cm^-1 in the IR spectrum of propanone there are two bands corresponding to the C-H stretch and the C=O stretch.

The region below 1500 cm^-1 is called the “fingerprint region” and is characteristic of the molecule as a whole.  Comparison of the spectrum in the fingerprint region with spectra in databases of IR spectra can be used to identify the molecule. e.g. the infra red spectra of butanone and propanone can be distinguished using the fingerprint region. They will both show very similar bands in the region above 1500 cm^-1 because they have the same functional group but they will have different fingerprint regions.

We are usually interested in identifying the bonds/functional groups in an organic molecule. To a good approximation the various bonds in a molecule can be considered to vibrate independently of each other. The wavenumbers at which some bonds vibrate are shown in Table A.2.

Table A.2

We can use an IR spectrum to identify the bonds present in a molecule but cannot always distinguish between functional groups. For example, using Table A2 we could identify the presence of C=O in a molecule but would not be able to distinguish between an aldehydes and a ketone.

Consider the IR spectrum of butanoic acid shown.

In order to identify the bonds present in the molecule we first of all look at the region above 1500 cm^-1.

This contains a C-O bond and therefore we should now look in the fingerprint region to confirm the presence of an absorption in the region 1000-1300 cm^-1, which is, indeed, the case. If there were no band in this region we would have to review our hypothesis that the molecule is a carboxylic acid.

The IR spectrum of propan-1-ol is shown above:

Again, by comparison of the bands with the values in Table A.2 we can identify O-H (about 3350 cm^-1) and C-H bonds (about 2900-3000 cm^-1). A C-O bond should also be present and we can see that there is a band in the region 1000-1300 cm^-1.

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Biotechnology and Gene Technologies

5.2.3 Genomes and Gene Technologies

I find it astonishing that so many students complain about module 2 of F215 Biotechnology and Gene Technologies. I have taken the challenge of explaining Module 2 of F215 through my blog.

I have divided Module 2 of F215: Module 2 Biotechnology and Gene Technologies into three sections:

1. Genome sequencing and the techniques involved in the sequencing of genome.
2. Genetic engineering e.g. making recombinant human insulin.
3. The ethics of genetic manipulation.

                                                                                _

 Section 1: Genome sequencing and the techniques used to sequence the genome.

  What is a genome ?

(A) Chromosome structure

(B)  DNA Structure

• Genome is the genetic material of an organism. It is a complete set of deoxyribonucleic acid (DNA), a chemical compound that contains the genetic instructions needed to develop and direct the activities of every organism. Human genome consists of DNA sequences in 23 chromosome pairs and also a small DNA molecule in the mitochondria. Human genome contains approximately 3.2 billion base pairs which includes genes that code for proteins and non-coding DNA.

• Every human cell (apart from gametes) contains 23 chromosomes in the nucleus. The chromosomes are visible under light microscope as X-shaped structures during the mitosis (cell division). The arms of chromosome are known as ‘sister chromatids’, which are joined at centromere.  The chromosomes are highly organised and compact structures  of DNA wrapped around proteins called histones and other associated proteins such as transcription factors.

• DNA forms a double helix consisting of two polynucleotide strands running anti-parallel to each each other. The  exposed nitrogenous bases form hydrogen bonds with complementary bases on the opposite strand. Adenine bases pairs with Thymine and Guanine base pairs with Cytosine according to complementary base pairing rules. The individual nucleotides are joined via condensation reaction between the phosphate group  of one nucleotide and 3-prime OH group on the neighbouring nucleotide to form a negatively charged sugar-phosphate backbone. The two polynucleotide strands twist around to form double helical form of the structure.
• Human genome project was a 13 year-long public funded initiative setup in 1990 with objective  of sequencing all 3.2 billion base pairs in human genome. It was successfully completed on 14th April 2003.

                       ! Students should know:-

1. Genome is the complete set of genetic material of an organism.
2. Genome sequencing involves determining the ordered sequence of bases in the genome.

! Next

– Genome sequencing step 1: Chromosome mapping.

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I set my iGCSE Biology student homework on the ‘Gross Structure Of The Heart’ – She came up with this amazing diagram showing the vertical cross-section of heart. Though simplistic I really liked her effort to draw it as neatly as possible and also labelling it clearly. At iGCSE level it has all the important information needed. Alongside her biology, her artwork also impresses me a lot.

                             Student’s Artwork                                                                                         Vertical View of the Heart

Edexcel International GCSE in Biology

Section    h) Tranport

2.63 Describe the structure of the heart ?

Ans: Student’s should learn to label the vertical and external view of the heart (B) and also the brief outline of the flow of blood in the heart (A).

(A)

Flow of blood in the vertical Cross Section of Heart (IGCSE biology specific) 

Deoxygenated blood enters the heart via vena cava (inferior and superior) and drains into the right atrium of the heart.The contraction of the right atrium pushes the blood into the right ventricle through tricuspid valves. The contraction of ventricles pushes the blood into the pulmonary artery through the semi-lunar valves where the blood is oxygenated (a process known as gas exchange). The oxygenated blood returns to the heart via pulmonary vein and drains into the left atrium of the heart. The blood is then pushed into the left ventricles by the contraction of left atrium. The contraction of left ventricle pushes the blood through the semi-lunar valve in Aorta to the rest of the body.

(B)

Vertical Cross Section Of Heart                                                                            External View of Heart

Key labelings –

1. Superior and Inferior vena cava.

2. Semi-lunar Valves.

3. Atrioventricular Valves (Bicuspid & Tricuspid valves).

4. Aorta

5. Pulmonary artery

6. Pulmonary vein

7. Left and right atrium

8. Left and right ventricle

9. Coronary artries.

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Edexcel GCE Maths C3 Jan 2007 Q8 (ii) (a)

Edexcel GCE C3 JAN 2007 Q8 (ii)

Examiner comments on Edexcel GCE C3 Jan 2007 Q8 (ii)

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