Sunday, June 15, 2008

Function Revision



A. Functions
http://www.fileden.com/files/2008/6/10/1953044/Function_Intro_n_Notation.pdf


B. Composite Function
http://www.fileden.com/files/2008/6/10/1953044/Function_Composite_Function.pdf

More practice for those who are keen to try challenging problems!
Fret not, students . Take a cue from the worked solutions.
C. Functions
http://www.fileden.com/files/2008/6/10/1953044/Function_More_Challenging_Examples.pdf

Thursday, June 12, 2008

:) Here's some revision for basic college algebra for online revision ! :)


All you need to know about Algebra
http://www.fileden.com/files/2008/6/10/1953044/Alg_Solving.pdf


KAL's cartoons (cartoons)
http://www.economist.com/displayStory.cfm?story_id=11525206

Tuesday, June 10, 2008

:) Here are some revision notes on Trigo! :)

ratio of sides of a Right -angles Triangle
sin delta = opp/hyp, cos delta = adj/hyp, tan delta = opp/adj

http://www.fileden.com/files/2008/6/10/1953044/Maths_D_-Trigonometry.pdf
http://www.fileden.com/files/2008/6/10/1953044/Maths_D_-Trigonometry.pdf

Friday, June 6, 2008

Worked Solution for May/June 2006 4024 Mathematics

4024/02/M/J/06 © UCLES 2006 MATHEMATICS (SYLLABUS D) 4024/02
Section A [52 marks]
Answer all questions in this section.
1 (a) Solve the equation 3x2 – 4x – 5 = 0, giving your answers correct to two decimal places. [4]
(b) Remove the brackets and simplify (3a – 4b)2. [2]
(c) Factorise completely 12 + 8t – 3y – 2ty. [2]
______________________________________________________________________________________
2 (a) A solid cuboid measures 7 cm by 5 cm by 3 cm.
(i) Calculate the total surface area of the cuboid. [2]
(ii) A cube has the same volume as the cuboid.
Calculate the length of an edge of this cube. [2]
(b) [The volume of a cone is × base area × height.]
[The area of the curved surface of a cone of radius r and slant height l is πrl.]
A solid cone has a base radius of 8 cm and a height of 15 cm.
Calculate
(i) its volume, [2]
(ii) its slant height, [1]
(iii) its curved surface area, [2]
(iv) its total surface area. [1]
______________________________________________________________________________________
4024/02/M/J/06 © UCLES 2006 [Turn over]
3 (a) In the diagram, the points A, B, C and D lie on a circle, centre O.
DO∧B = 124° and CD∧O = 36°.
Calculate
(i) DC∧B, [1]
(ii) DA∧B, [1]
(iii) OD∧B, [1]
(iv) CB∧O. [1]
(b) The diagram shows a circle, centre O,with the sector POQ shaded.
Given that PO∧Q = 140° and the radius
of the circle is 8 cm, calculate
(i) the area of the shaded region, [2]
(ii) the total perimeter of the unshaded region. [3]
______________________________________________________________________________________
4 (a) These are the prices for a ride in an amusement park.
(i) A family of two adults and three children went on the ride.
They paid with a $20 note.
Calculate the change they received. [1]
(ii) Express $2.25 as a percentage of $3.60. [1]
(b) Diagram I represents part of the framework of the ride.
The points A, B, C, D, E and F are on the framework.
The points H, C, G, E and F lie on a horizontal line.
The lines BH and DG are vertical.
BC = 80 m, HC = 60 m, DG = 40 m, GE = 35 m and DC∧G = 32°.
Diagram I
Calculate
(i) HC∧B, [2]
(ii) CD, [3]
(iii) the angle of depression of E from D. [2]
4024/02/M/J/06 © UCLES 2006 [Turn over]
(c)
Diagram II
Diagram II shows part of the ride.
The carriage that carried the family was 4.6m long.
It was travelling at a constant speed of 15 m/s as it passed the point F.
(i) Calculate, correct to the nearest hundredth of a second, the time taken for the carriage to pass
the point F. [2]
(ii) Express 15 m/s in kilometres per hour. [1]
______________________________________________________________________________________
5
Pattern 1 Pattern 2 Pattern 3
Counters are used to make patterns as shown above.
Pattern 1 contains 6 counters.
The numbers of counters needed to make each pattern form a sequence.
(a) Write down the first four terms of this sequence. [1]
(b) The number of counters needed to make Pattern n is An + 2.
Find the value of A. [1]
(c) Mary has 500 counters.
She uses as many of these counters as she can to make one pattern.
Given that this is Pattern m, find
(i) the value of m, [1]
(ii) how many counters are not used. [1]
______________________________________________________________________________________

4024/02/M/J/06 © UCLES 2006
6 (a) The results of a survey of 31 students are shown in the Venn diagram.
 = {students questioned in the survey}
M= {students who study Mathematics}
P = {students who study Physics}
S = {students who study Spanish}
(i) Write down the value of
(a) x, [1]
(b) n(M ∩ P), [1]
(c) n(M ∪ S), [1]
(d) n(P′). [1]
(ii) Write down a description, in words, of the set that has 16 members. [1]
(b) In the diagram, triangle AQR is similar to triangle ABC.
AQ = 8 cm, QB = 6 cm and AR = 10 cm.
(i) Calculate the length of RC. [2]
(ii) Given that the area of triangle AQR is 32cm2, calculate the area of triangle ABC. [2]
______________________________________________________________________________________
4024/02/M/J/06 © UCLES 2006 [Turn over
Section B [48 marks]
Answer four questions in this section.
Each question in this section carries 12 marks.
7 James and Dan are partners in a small company.
From each year’s profit, James is paid a bonus of $15 000 and the remainder is shared between James and Dan in the ratio 2 : 3.
(a) In 1996 the profit was $20 000.
Show that Dan’s share was $3000. [1]
(b) In 1997 the profit was $21 800.
Calculate
(i) the percentage increase in the profit in 1997 compared to 1996, [2]
(ii) the total amount, including his bonus, that James received in 1997. [2]
(c) In 1998 Dan received $7500.
Calculate the profit in 1998. [3]
(d) In 1999, the profit was $x, where x > 15 000.
(i) Write down an expression, in terms of x, for the amount Dan received. [1]
(ii) Given that Dan received half the profit, write down an equation in x and hence find the
amount that Dan received. [3]
______________________________________________________________________________________
8
4024/02/M/J/06 © UCLES 2006
8 Answer the whole of this question on a sheet of graph paper.
The table below gives some values of x and the corresponding values of y, correct to one decimal place,where y= + – 5.
x 1 1.5 2 2.5 3 4 5 6 7 8
y 13.1 7.3 4.5 3.0 2.1 1.5 1.7 p 3.7 5.3
(a) Find the value of p. [1]
(b) Using a scale of 2 cm to 1 unit, draw a horizontal x-axis for 0  x  8.
Using a scale of 1 cm to 1 unit, draw a vertical y-axis for 0  y  14.
On your axes, plot the points given in the table and join them with a smooth curve. [3]
(c) Use your graph to find
(i) the value of x when y = 8, [1]
(ii) the least value of + for values of x in the range 0  x  8. [1]
(d) By drawing a tangent, find the gradient of the curve at the point where x = 2.5. [2]
(e) On the axes used in part (b), draw the graph of y = 12 – x. [2]
(f) The x coordinates of the points where the two graphs intersect are solutions of the equation
x3 + Ax2 + Bx + 144 = 0.
Find the value of A and the value of B. [2]
______________________________________________________________________________________
9
4024/02/M/J/06 © UCLES 2006 [Turn over
9 In the diagram, A and B are two points on a straight coastline.
B is due east of A and AB = 7 km.
The position of a boat at different times was noted.
(a) At 8 a.m., the boat was at C, where
AC∧B = 66° and AB∧C = 48°.
Calculate
(i) the bearing of B from C, [1]
(ii) the distance AC. [3]
(b) At 9 a.m., the boat was at D, where AD = 6.3 km and DA∧B = 41°.
Calculate
(i) the area of triangle ADB, [2]
(ii) the shortest distance from the boat to the coastline. [2]
(c) At 11 a.m., the boat was at E, where AE = 9 km and BE = 5 km.
Calculate the bearing of E from A. [4]
______________________________________________________________________________________
10
4024/02/M/J/06 © UCLES 2006
10 (a) The lengths of 120 leaves were measured.
The cumulative frequency graph shows the distribution of their lengths.
Use this graph to estimate
(i) the median, [1]
(ii) the interquartile range, [2]
(iii) the number of leaves whose length is more than 31.5 cm. [1]
(b) Each member of a group of 16 children solved a puzzle.
The times they took are summarised in the table below.
(i) Write down an estimate of the number of children who took less than 13 minutes. [1]
(ii) Calculate an estimate of the mean time taken to solve the puzzle. [3]
(iii) Two children are chosen at random.
Calculate, as a fraction in its simplest form, the probability that one of these children took
more than 10 minutes and the other took 10 minutes or less. [2]
(iv) A histogram is drawn to illustrate this information.
The height of the rectangle representing the number of children in the interval 10 < t  12
is 8 cm.
Calculate the height of the rectangle representing the number of children in the interval
5 < t  10. [2]
______________________________________________________________________________________
11
4024/02/M/J/06 © UCLES 2006
11 (a) A = (1 -3)   B=(-2p 3p) C =  (-1 0)
3 -2 -3p p 0 1
(i) Evaluate 4C – 2A. [2]
(ii) Given that B = A–1, find the value of p. [2]
(iii) Find the 2 × 2 matrix X, where AX = C. [2]
(iv) The matrix C represents the single transformation T.
Describe, fully, the transformation T. [2]
(b) PQ→=(3)   PR→=(h)  QU→=(7)  PS→=  (17)
-9 -6 2 k
(i) Given that R lies on PQ, find the value of h. [1]
(ii) Express PU→as a column vector. [1]
(iii) Given that U is the midpoint of QS, find the value of k. [2]
______________________________________________________________________________________

4024/02/M/J/06
BLANK PAGE


1 (a) For numerical p = 4 and r = 6 B1
q = 76 or q = 8.71
x = 2.12 or −0.79
(b) 9 a2 + 16b2− 24ab (c) (4−y)(3+2t)
2 (a) (i) 2(7×5 + 7×3 + 3×5), 142 cm2
(ii) x3=7×5 ×3 soi, 4.7 to 4.72 cm
(b) (i)1/3 π64× 15
1005 to 1010 cm3
(ii) 17 cm (iii) 8 ×17 × π M1
427 to 427.3 cm2 (iv) 628 to 628.6 f.t. cm2 B1 1 f.t. 201 + their 427
3 (a) (i) DĈB = 62 o
(ii) B A D ˆ = 118 o f.t. B1 (f.t. 180 – their 62)
(iii) B D O ˆ = 28 o B1
(iv) B O C ˆ = 26 o B1
(b) (i)140/360soi
78.1 to 78.25 cm2
(ii) 220 o B1 2 ×8 ×π×220/360 M1
46.7 to 46.73 cm

4 (a) (i) $6.05
(ii) 62.5%
(b) (i) cos B C H ˆ =60/80 oe
41.4 o to 41.41 o
(ii) sin 32 = 40/CD
CD=40/ sin 32
75.48 to 75.5 m
(iii) tan d = 40/35
d = 48.8o to 49o
(c) (i) 4.6/15
0.31 s
(ii) 54 km/h
5 (a) 6, 10, 14, 18
(b) 4
(c) (i) 124
(ii) 2
6 (a) (i) (a) 8
(b) 4
(c) 21
(d) 19 f.t.
(b)(i) 8/6 = 10/RC or 8/14= 10/10+RC oe
7.5cm
(ii) (8/14)2 or (14/8)2 oe
98cm2

© University of Cambridge International Examinations 2006
7 (a) 3/5 × 5000
(b)(i) 1800/20000
9%
(ii) ( 2/5) × 21800 -15000
$17 720
(c) (5/3) × 7500
$12 500
$27 500 f.t.
(d)(i) (3/5 ) (x-15000) oe
(ii) their 3/5(x-15000 ) =x/2 f.t
x = 90 000 ⇒ $45000
8 (a) 2.5 B1 1
(c) (i) 1.4 < x < 1.5
(ii) 6.4 to 6.5
(d) Negative value
2.0 to 2.5
(e) Line with negative slope thro’ (0,12)
Also through (6,6)
(f) Attempt to simplify
x2/8 + 18/x -5 = 12-x
Allow M1 for attempt to sub
x = 1.2 and 7.5 and solve
A = 8 AND B = −136
GCE O Level – May/June 2006 4024 02
© University of Cambridge International Examinations 2006
9 (a) (i) 138o
(ii) 7/66 sin=AC/48 sin
AC= 7 sin 48 /sin 66
5.69 to 5.7 km
(b)(i) 1/2× 7×6.3 sin 41
14.46 to 14.5 km2
(ii) 6.3 sin 41 or area/3.5
4.13 to 4.15 km
(c) Cosine Rule involving B ˆ AE
cos A =
= 105/126

33.5 o to 34 o
(0)56o – 56.5o f.t.
10 (a) (i) 31.8 cm
(ii) 32.1 – 31.65 cm
0.42 to 0.48 cm
(iii) 108
(b) (i) 9
(ii) (2 x 7.5) + (4 x 11) + (6 x 13) +
(3 x 15) + (1 x 18) ÷ 16
12.5 min
(iii) 7/30 cao
(iv) 1.6 cm

Note: Please looked through the worked out solutions, I'm sure that you will be understanding
them better and has more confidence in your tests! :)

Thursday, May 22, 2008

Higher 2 Physics Preliminary Exam Papers (2007)- 9745/01

Latest H2 Prelim Physics Examination Papers
(PJC, RJC, TJC, TPJC, VJC, YJC, NJC, CJC, HCI, MI, ACJC)
Syllabus 9745/01, 9745/02, 9745/03
(with Worked Solutions) available at S$49.99 (- worked solutions)
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TPJC H2 Prelim Paper 1,2,3 2007 (9745/01) - Preview
Paper2/Q2:The satellite at a distance of 4.225 x 107 m from the centre of the Earth receives radio waves from an Earth station which is directly beneath it and 0.638 x 107 m from the centre of the Earth. Consider the Earth station as a point source of radio waves of power 8.0 kW. This power spreads out at an angle of 0.500 as shown in Fig. 2.1 below.

Paper1/Q21:In a water heater, water at 27 C flows in through an inlet pipe and gets heated up. The hot water then flows though the outlet pipe at 45 C. The rate of water flow is 4.0 kg per minute and the heat lost to the surrounding is 25 kW. The specific heat capacity of water is 4200 J kg-1 K-1. What is the power of the heater?

Tuesday, May 20, 2008

Advanced Level Calculations Acid-base and other non-redox titration questions

Example : 100 cm3 of a magnesium hydroxide solution required 4.5 cm3 of sulphuric acid (of concentration 0.1 mol dm-3) for complete neutralisation. [atomic masses: Mg = 24.3, O = 16, H = 1)
(a) give the equation for the neutralisation reaction. (b) calculate the moles of sulphuric acid neutralised. (c) calculate the moles of magnesium hydroxide neutralised. (d) calculate the concentration of the magnesium hydroxide in mol dm-3 (molarity). (e) calculate the concentration of the magnesium hydroxide in g cm-3
a) Mg(OH)2(aq) + H2SO4(aq) ==> MgSO4(aq) + 2H2O(l)
(b) moles of sulphuric acid neutralised = 0.1 x 4.5/1000 = 0.00045 mol
(c) moles of magnesium hydroxide neutralised also = 0.00045 (1:1 in equation) in 100 cm3
(d) concentration of the magnesium hydroxide in mol dm-3 = 0.00045 x 1000 ÷ 100 = 0.0045 (scaling up to 1000cm3=1dm3, to get molarity)
(e) molar mass of Mg(OH)2 = 58.3 so concentration of the magnesium hydroxide = 0.0045 x 58.3 = 0.26 g dm-3 (= g per 1000 cm3),so concentration = 0.26 ÷ 1000 = 0.00026 g cm-3

Example : A bulk solution of hydrochloric acid was standardised using pure anhydrous sodium carbonate (Na2CO3, a primary standard). 13.25 g of sodium carbonate was dissolved in about 150 cm3 of deionised water in a beaker. The solution was then transferred, with appropriate washings, into a graduated flask, and the volume of water made up to 250 cm3, and thoroughly shaken (with stopper on!) to ensure complete mixing.
25 cm3 of the sodium carbonate solution was pipetted into a conical flask and screened methyl orange indicator added. The aliquot required 24.65 cm3 of a hydrochloric acid solution, of unknown molarity, to completely neutralise it. [atomic masses: Na = 23, C = 12, O = 16]
(a) Calculate the molarity of the prepared sodium carbonate solution.(b) Write out the equation between sodium carbonate and hydrochloric acid, including state symbols. (c) How many moles of sodium carbonate were titrated? (d) How many moles of hydrochloric acid were used in the titration? (e) What is the molarity of the hydrochloric acid?

(a) moles = mass/f. mass, f. mass Na2CO3 = 106, mol Na2CO3 = 13.25/106 = 0.125
molarity = mol/vol. in dm3, 250 cm3 = 0.25 dm3,molarity Na2CO3 = 0.125/0.25 = 0.50 mol dm-3
(b) Na2CO3(aq) + 2HCl(aq) ==> 2NaCl(aq) + H2O(l) + CO2(g)
(c) mol = molarity x volume
(d) mol Na2CO3 titrated = 0.5 x 25/1000 = 0.0125 mol Na2CO3 (in the 25 cm3 aliquot pipetted)
from equation, mole ratio Na2CO3:HCl is 1:2,
so mol HCl = 2 x mol Na2CO3 = 2 x 0.0125 = 0.025 mol HCl (in the 24.65 cm3 titre)
(e) molarity = mol/vol. in dm3, dm3 = cm3/1000, 24.65/1000 = 0.02465 dm3 therefore molarity HCl = 0.025/0.02465 = 1.014 mol dm-3 (1.01 3sf)


Example : (a) Describe a procedure that can used to determine the molecular mass of an organic acid by titration with standardised sodium hydroxide solution. Indicate any points of the procedure that help obtain an accurate result and explain your choice of indicator.
0.279g of an organic monobasic aromatic carboxylic acid, containing only the elements C, H and O, was dissolved in aqueous ethanol. A few drops of phenolphthalein indicator were added and the mixture titrated with 0.100 mol dm-3 sodium hydroxide solution. It took 20.50 cm3 of the alkali to obtain the first permanent pink. [at. masses: C = 12, H = 1 and O = 16]
(b) How many moles of sodium hydroxide were used in the titration? (c) How many moles of the organic acid were titrated? and explain your reasoning. (d) Calculate the molecular mass of the acid. (e) Suggest possible structures of the acid with your reasoning

(a) An appropriate quantity of the acid is weighed out, preferably on a 4 sf electronic balance. It can be weighed into a conical flask by difference i.e. weight acid added to flask = (weight of boat + acid) - (weight of boat). The acid is dissolved in water, or aqueous-ethanol if not very soluble in water. The solution is titrated with standard sodium hydroxide solution using phenolphthalein indicator until the first permanent pink. The burette should be rinsed with the sodium hydroxide solution first. The flask should be rinsed around the inside to ensure all the acid and alkali react, and drop-wise addition close to the end-point to get it to the nearest drop.
The pKind for phenolphthalein is 9.3, and its effective pH range is 8.3 to 10.0. The pH of a solution of the sodium salt of the acid (from strong base + weak acid) is in this region and so the equivalence point can be detected with this indicator. (page on acid-base equilibria and theory of indicators is in production)
(b) moles = molarity x volume in dm3 (dm3 = cm3/1000) mol NaOH = 0.10 x 20.5/1000 = 0.00205 mol
(c) mol NaOH = mol RCOOH = 0.00205 because 1:1 mole ratio for a monobasic acid: RCOOH + Na+OH- ==> RCOO-Na+ + H2O
(d) moles = mass (g)/Mr, so Mr = mass/mol = 0.279/0.00205 = 136.1
(e) The simplest aromatic carboxylic acid is benzoic acid C6H5COOH, Mr = 122, 136-122 = 14, which suggests an 'extra' CH2 (i.e. -CH3 attached to the benzene ring instead of a H), so, since the COOH is attached to the ring, there are three possible positional/chain isomers of CH3C6H4COOH (Mr = 136) 2-, 3- or 4-methylbenzoic acid.

moles Z = mass of Z gas (g) / atomic or formula mass of gas Z (g/mol)
gas volume of Z = moles of Z x molar volume
Example 9.6: A small teaspoon of sodium hydrogencarbonate (baking soda) weighs 4.2g. Calculate the moles, mass and volume of carbon dioxide formed when it is thermally decomposed in the oven. Assume room temperature for the purpose of the calculation.
2NaHCO3(s) ==> Na2CO3(s) + H2O(g) + CO2(g)
Formula mass of NaHCO3 is 23+1+12+(3x16) = 84 = 84g/mole
Formula mass of CO2 = 12+(2x16) = 44 = 44g/mole (not needed by this method)
or a molar gas volume of 24000 cm3 at RTP (definitely needed for this method)
In the equation 2 moles of NaHCO3 give 1 mole of CO2 (2:1 mole ratio in equation)
Moles NaHCO3 = 4.2/84 = 0.05 moles ==> 0.05/2 = 0.025 mol CO2 on decomposition.
Mass = moles x formula mass, so mass CO2 = 0.025 x 44 = 1.1g CO2
Volume = moles x molar volume = 0.025 x 24000 = 600 cm3 of CO2
Avogadro's Law states that 'equal volumes of gases at the same temperature and pressure contain the same number of molecules' or moles of gas
Example : C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O(l)
(a) What volume of oxygen is required to burn 25cm3 of propane, C3H8.
Theoretical reactant volume ratio is C3H8 : O2 is 1 : 5 for burning the fuel propane.
so actual ratio is 25 : 5x25, so 125cm3 oxygen is needed.
(b) What volume of carbon dioxide is formed if 5dm3 of propane is burned?
Theoretical reactant-product volume ratio is C3H8 : CO2 is 1 : 3
so actual ratio is 5 : 3x5, so 15dm3 carbon dioxide is formed.
(c) What volume of air (1/5th oxygen) is required to burn propane at the rate of 2dm3 per minute in a gas fire?
Theoretical reactant volume ratio is C3H8 : O2 is 1 : 5
so actual ratio is 2 : 5x2, so 10dm3 oxygen per minute is needed,
therefore, since air is only 1/5th O2, 5 x 10 = 50dm3 of air per minute is required
You can use the ideas of relative atomic, molecular or formula mass AND the law of conservation of mass to do quantitative calculations in chemistry.
On analysis, a sample of hard water was found to contain 0.056 mg of calcium hydrogen carbonate per cm3 (0.056 mg/ml). If the water is boiled, calcium hydrogencarbonate Ca(HCO3)2, decomposes to give a precipitate of calcium carbonate CaCO3, water and carbon dioxide.
(a) Give the symbol equation of the decomposition complete with state symbols.
Ca(HCO3)2(aq) ==> CaCO3(s) + H2O(l) + CO2(g)
(b) Calculate the mass of calcium carbonate in grammes deposited if 2 litres (2 dm3, 2000 cm3 or ml) is boiled in a kettle. [ atomic masses: Ca = 40, H = 1, C = 12, O = 16 ]
the relevant ratio is based on: Ca(HCO3)2 ==> CaCO3
The formula masses are 162 (40x1 + 1x2 + 12x2 + 16x6) and 100 (40 + 12 + 16x3) respectively
there the reacting mass ratio is 162 units of Ca(HCO3)2 ==> 100 units of CaCO3
the mass of Ca(HCO3)2 in 2000 cm3 (ml) = 2000 x 0.056 = 112 mg
therefore solving the ratio 162 : 100 and 112 : z mg CaCO3
where z = unknown mass of calcium carbonate
z = 112 x 100/162 = 69.1 mg CaCO3
since 1g = 1000 mg, z = 69.1/1000 = 0.0691 g CaCO3, calcium carbonate
(c) Comment on the result, its consequences and why is it often referred to as 'limescale'?
This precipitate of calcium carbonate will cause a white/grey deposit to be formed on the side of the kettle, especially on the heating element. Although 0.0691 g doesn't seem much, it will build up appreciably after many cups of tea! The precipitate is calcium carbonate, which occurs naturally as the rock limestone, which dissolved in rain water containing carbon dioxide, to give the calcium hydrogen carbonate in the first place. Since the deposit of 'limestone' builds up in layers it is called 'limescale'.

Tuesday, May 13, 2008

Atomic Structure (Combined Science) - Sample Questions

Analysis of the May/June Session 2000 Paper II - 5070/3


Sol to A4 Helium-6 & Helium-7 are isotopes. The nucleon mass number of Helium-6 is 6
and of helium-7 is 7. Compare the number of electrons, neutrons, and protons in one-atom of helium-6 and one-atom of helium-7.
c. Explain why helium does not react with other elements to form compounds
Isotopes have same no. of protons (same element) with differing nos. of neutrons(proton number/atomic number/nucleon numbers/mass number)
bi. Electrons 2 (same) ii. Protons 2(same) iii. Neutron 4 and 5 (one more)
c. Helium has a full outer shell or stable configuration , neither gains or loses electrons



Which electron arrangements is that of a Halogen?
A.electron arrangement Y
Belectron arrangement Z
C.electron arrangement X
D.electron arrangement W
Which electron arrangements correspond to two elements on the same period of the Periodic Table?
A.electron arrangement W and Z
B.electron arrangement Y and Z
C.electron arrangement X and Z
D. electron arrangement W and X

Sol A.The reactive non-metallic Halogens are Group 7, with 7 outer electrons, its the top element in the group, next to the last on Period 2 (2 shells used)
Sol A. both in the same period, period 2, 2 shells used