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Unformatted text preview: Practical Session 2 — Solutions Exercises 1. The machine accuracy is between 10 8 and 10 7 — it is actually 2 24 ≈ 5 . 96 × 10 8 , so to two digits 6 . × 10 8 . Here’s a suitable code. / * machine_acc.c * Program to determine the machine accuracy, i.e. the smallest number * that can be added to one to give a number different from one. * / #include <stdio.h> int main() { float eps; float x; printf("Test value of eps: "); scanf("%f", &eps); x = 1.0 + eps; if (x == 1.0) printf("1.0+%g = 1.0\n", eps); else printf("1.0+%g != 1.0\n", eps); return 0; } 2. (a) You need to solve x 2 + K a x K a = 0 , where K a = 1 . 76 × 10 5 . The solution required is the positive root, which is about 4 . 2 × 10 3 , so the percentage ionisation is about 0.42%. (b) The pH is about 2.4. (c) You need to solve x 2 = K a , which gives x ≈ 4 . 195 × 10 3 M , compared with 4 . 186 × 10 3 M in (a), so they differ by about 0.2%....
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This note was uploaded on 09/29/2009 for the course COSC 1002 taught by Professor Wheatland during the Three '09 term at University of Sydney.
 Three '09
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