Presentation is loading. Please wait. #  Sig figs consist of all the digits known with certainty plus one final digit, which is somewhat uncertain or is estimated.  Say a nail is between 6.3cm.

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 Sig figs consist of all the digits known with certainty plus one final digit, which is somewhat uncertain or is estimated.  Say a nail is between 6.3cm and 6.4cm. And a final answer of around 6.36 cm. All digits including the uncertain one are significant.  Significant does not mean certain.

 Process is easy if there are no zeros. It is just how many numbers there are.  Ex:  13.2 has 3 sig figs  9857 has 4 sig figs  3 has 1 sig figs

 Zeros appearing between nonzero digits are significant. Ex. 40.7 has 3 sig figs. 87009 has 5 sig figs.  Zeros appearing in front of all nonzero digits are not significant. Ex. 0.095897 has 5 sig figs..00009 has 1 sig fig.  Zeros at the end of a number to the right of a decimal point are significant. Ex. 85.00 has 4 sig figs. 9.000000000 has 10 sig figs.

 Zeros at the end of a number but to the left of a decimal point may or may not be significant. If a zero has not been measured or estimated but is just a placeholder, it is not significant. A decimal point placed after zeros indicates that they are significant.  Ex. 2000 may contain from 1 to 4 sig figs, depending on how many zeros are placeholders. For measures in text assume that 2000 has 1 sig fig. 2000. contains 4 sig figs indicated by the presence of the decimal point.

1. 28.6 g 2. 3440. cm 3. 910 m 4..04604 L 5..0067000 kg

1. 804.05 g 2. 0.0144030 km 3. 1002 m 4. 400 mL 5. 30000. cm 6..00062500 kg

 When adding or subtracting decimals, the answer must have the same number of digits to the right of the decimal point as there are in the measurement having the fewest digits to the right of the decimal.  Ex. 25.1 g + 2.03 g = 27.13 g = 27.1 g  When using whole numbers, the answer should be rounded so that the final sig digit is in the same place as the left most uncertain digit.  Ex. 5400 + 365 = 5765 = 5800

 150.0 +.507 =  7.56 +.375 + 14.0223 =  98 – 50 =  80. – 16 =  76.90 + 1007.1 =

 The answer can have no more sig figs than are in the measurement with the fewest number of sig figs.  Ex. 3.05 g ÷ 8.47 mL = 0.36009 g/mL = 0.360  Ex. 2.4 g/mL X 15.82 mL = 37.968 = 38 g

 10.1 X 12.07 =  70 x 2.4 =  3.98 ÷ 2.3 =  12.4 X 34 =  320. ÷ 0.076 =

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