i have a question on my chem homework that i cant figure out. i really dont understand significant figures. the 0s confuse me because inever know when they are supposed to count. anyone who knows about sigfigs please help! lol
heres the question:
4. How many significant figures are found in each of the following:
i also forget how to figure out percent error when given an observed answer and actual answer.
i could use help on tis one too! thanks so much
5. In lab, the molar mass of a substance was found to be 78.9g/mole, if the real molar mass is 82.6 g/mole, what is the experimental error?
ok .. one more that i dont get hah sorry guys.
7. Complete the operation below and give the final answer in proper significant figures and scientific notation:
7.77 x10^5 x 6.98 x10^7
4.5 x 10^8
- 1 decade agoFavorite Answer
It is different than some of the others, but I am certain that I am right. (For the purposes of doing math or science homework.)
Significant figures tell you which digits you actually measured. Any non-zero number is significant. The zeros are significant if they are actually standing for the number '0', not if they are just 'placeholders'.
Ending zeros right of the decimal are significant. (There is no reason to write them if they are not. 2.5 means you averaged to the nearest tenth. 2.50 means you averaged to the nearest hundreth and that number is '0'. It is significant because it is 2.50, not 2.51 or 2.49, but 2.50)
Ending zeros on an integer (no decimal) are not significant. (They are only there as 'placeholders' for the different columns, 1's, 10's, 100's, etc.)
A tricky one is your letter c, that has leading zeros. 0.00678 The leading zeros are place holders for the one's column, the tenths column, and the hundreths column.
If you have an integer with ending zeros that you want to be significant, the correct notation is putting a line over the right-most zero that is significant. For example, 6,000,000 means six million, rounded to the nearest million. The number is closer to six million than to 5 million or 7 million. But if you have 6,000,000 with a line over the zero in the tens column, you have six million rounded to the nearest ten. The number is closer to 6,000,000 than to 5,999,990 and 6,000,010.
Double check at http://en.wikipedia.org/wiki/Significant_figures
a. 6.0080 ....... 5 .....all of the digits are significant including the zero behind the 8 because they are all right of the decimal.
b. 5.000 x 10 to the 34 ......... 4 ......all of the digits are significant because once again they are right of the decimal
c. 0.00678 ........ just three. The other zeros are 'place holders'
d. 56.90 ....... 4 ........ all of the digits
e. 6,000,000 ........ just one. The other zeros are 'place holders'
Another example, you measure the length of an object with a ruler. If you say it is 3.2 cm, you are saying you measured it and you are certain of the three and that the end of the object was closer to the .2 than the .3 or .1. If you are using calipers and say that the object is 3.24 cm, you are saying that you are certain of the three cm and the .2 cm and that the end of the object was closer to the .24 than the .25 or .23. It is a more precise measurement, thus has more significant figures.
You already have the right answers to your percent error. The percent error is the difference between the measured value and the correct value divided by the correct value.
(82.6 - 78.9)/82.6 = 4.48% By the way, the answer has three significant figures because the most precise measurement in the calculation has three significant figures (in this case all of them). If you were doing a different calculation, for example, calculating your speed if you travel 47 miles in 1.25 hours ...
47 mi/ 1.25 hr = 38 mph (only two significant figures because your most precise measurement has only two significant figures.)
Last question... first do 7.77 x 6.98 / 4.5 (ignoring the exponentials) = 12 (only two significant figures because of the 4.5 having only two significant figures)
Then do the exponentials : 10^5 x 10^7 / 10^8
5+7-8 = 4 ........ so 10^4
Combine for 12 x 10^4, but that is not in the correct scientific notation form. You need to make it 1.2 x 10^5 (Scientific notation has the first significant digit in the ones column and all other significant digits to the right of the decimal.)Source(s): Physics major and taught high school physics.
- Barry CLv 71 decade ago
It depends, see below, but for hw I would go with:
IMHO the zeroes don't really matter IF the method of measurement can have measured as precisely as the right-most zero.
Let's use e: as an example.
If I say there are "about" 6 million (6,000,000) people in my city, then there is 1 significant figure.
But if I say "I counted every single person in my city ands there was exactly 6,000,000" then there are 7 significant figures.
Just as if I said "I counted al the people here and there are 6,000,001 people" - 7 significant figures.
The purpose of "significant figures" is as an estimate of how precise the counting or measuring is.
Have you done how to write a number in exponential form yet?
That helps with sig figs because the zeroes come out of it, or at least is it more clear as to why they are there.
E.g. 6 million = 6 x 10^6 --> 1 sig fig
6,000,001 = 6,000,001 x 10^1 = 7 sig figs
and maybe, as above, 6,000,000 = 6,000,000 x 10^1 = 7 sig figs.
Hope this helps!Source(s): math and physics degree from top university, then spent a while as a rocket scientist with NASA where this stuff mattered a lot!
- zanthusLv 51 decade ago
well in the number 6,000,000
there is only 1 significant digit, because there is NO decimal
for more info, look up significant digits on wikipedia
for th other question, figure out the difference between the actual and measured. then divide the difference by the actual
eg if something weighs 100 pounds, but your scale is wrong and it weighs the item at 90. Then there is a difference of 10 pounds right...10/100 is a 10% measuring error.
- lemmonsLv 44 years ago
a million sig fig= 6 2 sig figs= 6.a million 3 sig figs= 6.13 substantial figures are in basic terms numbers that certainly arent 0 if u had 3.140567 to around to 4 sig figs it could be 3.1406 through fact 0 doesnt count selection as one and there replaced right into a 6 after the 5 so u might around that to the subsequent selection desire this helped :) x
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- Anonymous1 decade ago
I believe this is correct but I haven't done sigfi, in so long that it might be wrong BTW,
I have no clue on Number 5
- lenpol7Lv 71 decade ago
a. 6.008 (4 sf)
b. 5.000x 1034 = 5170 (3sf)
c. 0.00678 (3sf)
d. 56.9 (3sf)
e. 6,000,000 (1sf)
NB Ignoring the decimal point, count up the number of 'non-zero' digits.
- 1 decade ago
% error= (act. value-obs. value)/actual
- Anonymous1 decade ago