Question

determine the location of the last significant place value by placing a bar over the digit. (example: 1.700) 1. 8040 __________ 2. 0.0300 ________ 3. 699.5 ________ 4. 2.000×10² ________ 5. 0.90100 ________ 6. 90,100 ________ 7. 4.7×10⁻⁸ ________ 8. 10,800,000. ________ 9. 3.01×10²¹ ________ 10. 0.000410 __________ emistry if8766 9 ©instructional fair, i

Explanation:

Step1: Recall significant - figure rules

Non - zero digits are always significant. Zeros between non - zero digits are significant. Trailing zeros in a number with a decimal point are significant. Trailing zeros in a whole number without a decimal point may or may not be significant, but here we assume they are not significant unless otherwise indicated.

Step2: Identify last significant digit for each number

1. For 8040, assuming no decimal point, the last significant digit is 4, so it is 80\(\overline{4}\)0.
2. For 0.0300, the last significant digit is the second 0 after the 3, so it is 0.03\(\overline{0}\)0.
3. For 699.5, the last significant digit is 5, so it is 699.\(\overline{5}\).
4. For 2.000×10², the last significant digit is the last 0, so it is 2.00\(\overline{0}\)×10².
5. For 0.90100, the last significant digit is the last 0, so it is 0.901\(\overline{0}\)0.
6. For 90,100, assuming no decimal point, the last significant digit is 1, so it is 90,\(\overline{1}\)00.
7. For 4.7×10⁻⁸, the last significant digit is 7, so it is 4.\(\overline{7}\)×10⁻⁸.
8. For 10,800,000., the last significant digit is the last 0 (because of the decimal point), so it is 10,800,00\(\overline{0}\).
9. For 3.01×10²¹, the last significant digit is 1, so it is 3.0\(\overline{1}\)×10²¹.
10. For 0.000410, the last significant digit is the last 0, so it is 0.00041\(\overline{0}\).

Answer:

1. 80\(\overline{4}\)0
2. 0.03\(\overline{0}\)0
3. 699.\(\overline{5}\)
4. 2.00\(\overline{0}\)×10²
5. 0.901\(\overline{0}\)0
6. 90,\(\overline{1}\)00
7. 4.\(\overline{7}\)×10⁻⁸
8. 10,800,00\(\overline{0}\)
9. 3.0\(\overline{1}\)×10²¹
10. 0.00041\(\overline{0}\)