# Questions On Chemistry

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Pages: 23

1

CHAPTER 18
18.1 (a)

f1 (10) = 0.90309 +

1.0791812 − 0.90309 (10 − 8) = 0.991136 12 − 8

εt =
(b)

1 − 0.991136 × 100% = 0.886% 1

f1 (10) = 0.9542425 +

1.0413927 − 0.9542425 (10 − 9) = 0.997818 11 − 9

εt =

1 − 0.997818 × 100% = 0.218% 1

18.2 First, order the points x0 = 9 x1 = 11 x2 = 8 f(x0) = 0.9542425 f(x1) = 1.0413927 f(x2) = 0.9030900

Applying Eq. (18.4) b0 = 0.9542425

Equation (18.5) yields b1 =

1.0413927 − 0.9542425 = 0.0435751 11 − 9

Equation (18.6) gives
0.9030900 − 1.0413927 − 0.0435751 0.0461009 − 0.0435751 8 − 11 b2 = = = −0.0025258 8−9 8−9

Substituting these values into Eq. (18.3) yields the quadratic formula f 2 ( x) = 0.9542425 + 0.0435751( x − 9) − 0.0025258( x − 9)( x − 11)

which can be evaluated at x = 10 for f 2 (10) = 0.9542425 + 0.0435751(10 − 9) − 0.0025258(10 − 9)(10 − 11) = 1.0003434

18.3 First, order the points x0 = 9 x1 = 11 x2 = 8 x3 = 12 f(x0) = 0.9542425 f(x1) = 1.0413927 f(x2) = 0.9030900 f(x3) = 1.0791812

The first divided differences can be computed as
PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

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1.0413927 − 0.9542425 = 0.0435751 11 − 9 0.9030900 − 1.0413927 = 0.0461009 f [ x2 , x1 ] = 8 − 11 1.0791812 − 0.9030900 = 0.0440228 f [ x3 , x2 ] = 12 − 8 f [ x1 , x0 ] =

The second divided differences are 0.0461009 − 0.0435751 = −0.0025258 8−9 0.0440228 − 0.0461009 = −0.0020781 f [ x3 , x2 , x1 ] = 12 − 11 f [ x2 , x1 , x0 ] =

The third divided difference is f [ x3 , x2 , x1 , x0 ] =

−0.0020781 − (−0.0025258) = 0.00014924 12 − 9

Substituting the appropriate values into Eq. (18.7) gives f3 ( x) = 0.9542425 + 0.0435751( x − 9) − 0.0025258( x − 9)( x − 11) + 0.00014924( x − 9)( x − 11)( x − 8) which can be evaluated at x = 10 for f3 ( x) = 0.9542425 + 0.0435751(10 − 9) − 0.0025258(10 − 9)(10 − 11) + 0.00014924(10 − 9)(10 − 11)(10 − 8) = 1.0000449

18.4

18.1 (a): x0 = 8 x1 = 12 f1 (10) =

f(x0) = 0.9030900 f(x1) = 1.0791812
10 − 12 10 − 8 0.9030900 + 1.0791812 = 0.991136 8 − 12 12 − 8

18.1 (b): x0 = 9 x1 = 11 f1 (10) = 18.2: x0 = 8 x1 = 9 x2 = 11

f(x0) = 0.9542425 f(x1) = 1.0413927 10 − 11 10 − 9 0.9542425 + 1.0413927 = 0.997818 9 − 11 11 − 9

f(x0) = 0.9030900 f(x1) = 0.9542425 f(x2) = 1.0413927

PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

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f 2 (10) =

(10 − 9)(10 − 11) (10 − 8)(10 − 11) 0.9030900 + 0.9542425 (8 − 9)(8 − 11) (9 − 8)(9 − 11) (10 − 8)(10 − 9) + 1.0413927 = 1.0003434 (11 − 8)(11 − 9)

18.3: x0 = 8 x1 = 9 x2 = 11 x3 = 12 f3 (10) =

f(x0) = 0.9030900 f(x1) = 0.9542425 f(x2) = 1.0413927 f(x3) = 1.0791812
(10 − 9)(10 − 11)(10 − 12) (10 − 8)(10 − 11)(10 − 12) 0.9030900 + 0.9542425 (8 − 9)(8 − 11)(8 − 12) (9 − 8)(9 − 11)(9 − 12) (10 − 8)(10 − 9)(10 − 12) (10 − 8)(10 − 9)(10 − 11) + 1.0413927 + 1.0791812 = 1.0000449 (11 − 8)(11 − 9)(11 − 12) (12 − 8)(12 − 9)(12 − 11)

18.5 First, order the points so that they are as close to and as centered about the unknown as possible

x0 = 2.5 x1 = 3.2 x2 = 2 x3 = 4 x4 = 1.6

f(x0) = 14 f(x1) = 15 f(x2) = 8 f(x3) = 8 f(x4) = 2

Next, the divided differences can be computed and displayed in the format of Fig. 18.5, i 0 1 2 3 4 xi 2.5 3.2 2 4 1.6 f(xi) 14 15 8 8 2