Density Problem - Please lookover my answer

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The density of the unknown liquid was calculated as 0.8839 g/mL based on the mass difference between the flask with the liquid and the empty flask. The calculation used a volume of 10.00 mL and a mass of 8.839 g for the liquid. There is a discussion about significant figures, with one participant suggesting rounding to 0.88 g/mL due to the significant figures in the volume measurement. The consensus is that the density should be rounded to 0.884 g/mL, as the significant figures in the volume dictate the rounding. The final density value is confirmed to be 0.884 g/mL.
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Homework Statement



A 10.00 mL volume of unknown liquid was added to stoppered erlemeyer flask, which weighed 48.217 g. If weigh of the unknown liquid plus the stoppered flask was 57.056 g. Calculate density of unknown to correct number of signifigent figure.


Homework Equations



D = m /v


3. The Attempt at a Solution [/b

D = m / v

v = 10.00 ml

m = 57.056 - 48.217 = 8.839 g

d = 8.839 g / 10.00 ml = 0.8839 g/ml

Answer is 0.8839 g/ml

is this correct ? or should I round it to 0.884 g/ml ]
 
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benworld said:

A 10.00 mL volume of unknown liquid was added to stoppered erlenmeyer flask, which weighed 48.217 g. If weigh of the unknown liquid plus the stoppered flask was 57.056 g. Calculate density of unknown to correct number of signifigent figure.

Seems correct, but what is that about a "signifigent figure"?
 
I actually believe that it would be .88. With sig figs you round to the lowest amount of sig figs found in the given information. In this problem it would be the 10.00 mL, the 0 after the decimal are just place holders, insignificant.

Joe
 

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