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Sig Figs Significant figure Sig Figs Significant figure

Sig Figs Significant figure - PowerPoint Presentation

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Uploaded On 2018-03-10

Sig Figs Significant figure - PPT Presentation

a digit that is reliably known Ex 62 has 2 sig figs The hundredth place is not reliably known and thus not reported The easiest way to determine sig figs in a number is to write the number in scientific notation ID: 645666

figs number significant sig number figs sig significant answer decimal reliably mass places accurate match figures trailing notation volume

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Slide1

Sig FigsSlide2

Significant figure

 a digit that is reliably known.

Ex: 6.2 has 2 sig figs. The hundredth place is not reliably known and, thus, not reported.Slide3

The easiest way to determine

sig figs in a number

is to write the number in scientific notation.

0.00620 = 6.20 x 10

-3

The trailing zero is reliably known, so it is SIGNIFICANT.

The number of significant figures is the number of digits when written in

sci

notation.Slide4

Some Rules for Sig Figs

The number of sig figs

does not equal

the number of decimal places.

In whole numbers, trailing zeroes

are not

sig figs.

Ex: 320,000 is 3.2 x 10

5

It has 2 sig figs.

Changing units may shift the decimal points but it doesn’t change the number of sig figs. Slide5

When multiplying/dividing (or taking roots), the number of sig figs in the answer should match the number of sig figs of the

least

precise number in the calculation.

When adding/subtracting, the number of sig figs in the answer should match the

smallest

amount of decimal places of any number in the calculation.Slide6

It’s fine to keep extra numbers around in the intermediate steps, but your final answer

MUST

be accurate in terms of significant figures.

Otherwise, you’ve performed magic and have something more accurate that we can verify.Slide7

Example

The volume of a piece of Al is 4.44 x 10

-4

m

3

. Given a density of 2.7 x10

3

kg/m

3

, what is the total mass of the piece of Al? (mass = density x volume)Slide8

m=

4.44 x 10

-4

m

3

(

2.7 x10

3

kg/m

3

)

m= 1.199 kg

m= 1.2 kgSlide9

If we have a second mass of Al, measured to be 6.47 kg, how much total Al do we have?Slide10

`

6.47 kg + 1.2 kg

=

7.7 kg