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5. determine the number of significant figures in the following numbers…

Question

  1. determine the number of significant figures in the following numbers:

1.5060 x 10⁻⁷ cm
______
b) 0.00400630 l
______

Explanation:

Response
Part a: \( 1.5060 \times 10^{-7} \) cm

Step 1: Recall significant figure rules

For numbers in scientific notation \( a \times 10^n \), we count the significant figures in \( a \). Non - zero digits are significant, zeros between non - zero digits are significant, and trailing zeros after a decimal in \( a \) are significant.

Step 2: Analyze \( 1.5060 \)

The digits in \( 1.5060 \) are 1, 5, 0, 6, 0. 1 and 5 and 6 are non - zero (significant), the zero between 5 and 6 is significant (between non - zeros), and the trailing zero after 6 (after the decimal) is significant. So we count them: 1, 5, 0, 6, 0. That's 5 significant figures.

Part b: \( 0.00400630 \) L

Step 1: Recall significant figure rules

Leading zeros (zeros before the first non - zero digit) are not significant. Zeros between non - zero digits are significant, and trailing zeros after a non - zero digit and after the decimal are significant.

Step 2: Analyze \( 0.00400630 \)

The first non - zero digit is 4. Then we have zeros between 4 and 6? No, the digits after 4 are 0, 0, 6, 3, 0. The zeros after 4 (but before 6) are between non - zero? Wait, 4 is non - zero, then 0 (significant, between 4 and 6? Wait, 4, then 0, 0, 6, 3, 0. Let's list the significant digits: 4, 0, 0, 6, 3, 0. Leading zeros (the three zeros before 4) are not significant. The zeros between 4 and 6 (the two zeros) are significant (because they are between non - zero? Wait, 4 is non - zero, then 0, 0, then 6 (non - zero), so the zeros between 4 and 6 are significant. Then 6, 3 (non - zero), and the trailing zero after 3 (after the decimal) is significant. So counting: 4, 0, 0, 6, 3, 0. That's 6 significant figures.

Answer:

a) The number of significant figures in \( 1.5060\times10^{-7}\) cm is 5.
b) The number of significant figures in \( 0.00400630\) L is 6.