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anonymous

  • one year ago

Anyone on? I have a few questions on significant figures and I was wondering if you could check my answers?

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  1. anonymous
    • one year ago
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    @ganeshie8

  2. anonymous
    • one year ago
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    @Nnesha

  3. taramgrant0543664
    • one year ago
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    Rules: 1. All non-zero digits are significant. For example: 123. (3 sig figs) 2. Zeros between non-zero digits are significant. For example: 12.507 (5 sig figs) 3. Zeros to the left of the first non-zero digit are not significant. For example: 1.02 (3 sig figs) 0.12 (2 sig figs) 0.012 (2 sig figs) 4. If a number ends in zeros to the right of the decimal point, those zeros are significant. For example: 2.0 (2 sig figs) 2.00 (3 sig figs) {This signifies greater accuracy.} 5. If a number ends in zeros to the left of the decimal point, those zeros may or may not be significant. For example: If we make a statement that the weight of an object is 120 g, how do we convey our knowledge of whether the balance was accurate to ± 1 g or ± 10 g? Answer: The ambiguity can be removed by using exponential notation. The weight can be expressed as 12. x 101 g or 1.2 x 102 g if we wish to quote unambiguously to 2 sig figs, and 12.0 x 101 g or 1.20 x 102 g if we have a confidence level extending to 3 sig figs. Note: We cannot write 120.0 g since this requires known accuracy of ± 0.1 g.

  4. anonymous
    • one year ago
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    Wow this is an amazing description, thank you so much!

  5. arindameducationusc
    • one year ago
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    yes Tara is awesome....@twistnflip

  6. anonymous
    • one year ago
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    I usually help people, but I just started AP chemistry and I completely blanked on sig figs. My teacher assigned homework without doing notes or reviewing (which is ok except for the fact that I don't have my textbook yet lol)

  7. taramgrant0543664
    • one year ago
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    Calculations with numbers having different accuracies Multiplication or Division: the result can have no more sig figs than the least accurate number. For example: If an object has mass of 29.1143 g and a volume of 25.0 cm3, then its density is given by Density = 29.1143 g / 25.0cm-3 = 1.164572 g cm-3 = 1.16 g cm-3 Addition or Subtraction: the result must be reported to the same number of decimal places as the number with the fewest decimal places. For example: 19.2g+ 0.4745g+127. g=146.6745g= 147. g because one weight is known only to the nearest 1g NOTE: Round off numbers only at the END of calculations; otherwise, errors may be inadvertently carried through.

  8. anonymous
    • one year ago
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    ok, got it! Thanks again! you the bomb

  9. taramgrant0543664
    • one year ago
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    No problem!! It was good for me to review them a little too since school is starting back up soon so it helped me too!!

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