By Richard Harwood, Ian Lodge
Written by means of hugely skilled authors, Cambridge IGCSE Chemistry Workbook offers whole aid for the IGCSE Chemistry syllabus provided through CIE. This publication comprises routines that supply transparent development to scholars as they cross alongside. a wide selection of questions boost all of the talents had to achieve the IGCSE Chemistry exam. uncomplicated and transparent language makes this booklet available to various skills. This workbook is absolutely recommended via CIE and is meant for use along the Cambridge IGCSE Chemistry Coursebook. A Teacher's source CD-ROM can also be to be had.
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Written through hugely skilled authors, Cambridge IGCSE Chemistry Workbook offers whole help for the IGCSE Chemistry syllabus provided by way of CIE. This e-book comprises workouts that supply transparent development to scholars as they move alongside. a wide selection of questions advance all of the talents had to reach the IGCSE Chemistry exam.
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Additional info for Cambridge IGCSE Chemistry Workbook
Use rules #3 and #5. Multiply the exponents. = (a 2 bc)2 a 2 × 2 b 2c 2 = a6 b 2 c6 × × . Isolate each variable. a4 b2 c2 = a 2 × b 0 × c 4 . Use rule #2. Subtract the exponents for each base. = a2 × 1 × c4 . = a 2c 4 .
3 16. E The radicand in a cube root can be either positive or negative. This differs from the radicand in a square root. If the problem asked for the domain of f ( x) = (4 x2 − 1), then 4x2 − 1 would be have to be greater than or equal to zero. Here, there is no restriction on x, so the domain of the function is the set of all real numbers. 17. E Set up an equation to represent the exponential decay, y = Ce−kt where C is the initial amount of the substance, k is the rate of decay, and t is the time.
C 16. E 26. B 36. C 46. A 7. D 17. E 27. D 37. B 47. E 8. A 18. B 28. B 38. D 48. D 9. C 19. B 29. E 39. B 49. C 10. C 20. A 30. E 40. D 50. E ANSWERS AND SOLUTIONS 1. C 3. D Take either the log or natural log of both sides of the equation to solve for x. ⎛ 1 1⎞ 1 ⎛ 1 1⎞ c−1 ⎝ + ⎠ = ⎝ + ⎠ . a b c a b The LCD of 1 1 and . b a 1⎛ b a⎞ + = c ⎝ ab ab ⎠ 1⎛ b⎞ = a+ ⎝ ab ⎠ c ( a + b) . abc 2. D You don’t need to start multiplying the second trinomial by the first. Instead, group x and y as a single expression.