Consider the expression. 7.90 × 10–5 x(0.100 + x)? You can solve for x using a technique called successive approximations. Step 1: If you assume that x is very small compared to 0.100, such that 0.100 + x x 0.100, th (call it x1) can be calculated as 7.90 x 10-5 = x1 (0.100)² Calculate the first approximation of x. Express all answers to three or more significant figures. = Ix Step 2: Now, take your first approximation of x and plug it into the full equation. 7.90 × 10-5 = x2 (0.100 + x1)² Calculate the second approximation of x. X2 = Step 3: Each successive approximation uses the value from the previous approximation. 7.90 x 10-5 = x3(0.100 + x2)² Calculate the third approximation of x. X3 =

Consider the expression. 7.90 × 10–5 x(0.100 + x)? You can solve for x using a technique called successive approximations. Step 1: If you assume that x is very small compared to 0.100, such that 0.100 + x x 0.100, th (call it x1) can be calculated as 7.90 x 10-5 = x1 (0.100)² Calculate the first approximation of x. Express all answers to three or more significant figures. = Ix Step 2: Now, take your first approximation of x and plug it into the full equation. 7.90 × 10-5 = x2 (0.100 + x1)² Calculate the second approximation of x. X2 = Step 3: Each successive approximation uses the value from the previous approximation. 7.90 x 10-5 = x3(0.100 + x2)² Calculate the third approximation of x. X3 =

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter4: Energy And Chemical Reactions

Section: Chapter Questions

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