I also get 9.24 x 10^-5 but I don't think that is the correct number of s.f.
You subtracted 5.57 - 4.9320 = 0.638 but that must be rounded to 0.64 because in addition or subtraction we can't have more places after the decimal than the least precise. Then 0.64/6902 = 9.27 x 10^-5 which would round to 9.3 x 10^-5 g/mL. If you kept the 0.638/6902 = 9.24 x 10^-5 which rounds to 9.2 x 10^-5 g/mL. I have always kept the less precise number in my calculator and do it all at once; i.e., (5.57-4.9320)/6902 =?? instead of subtracting, rounding, then dividing, then rounding again. Therefore, I would go with 9.2 x 10^-5 g/mL instead of the 9.3 but you need to follow the instructions of your prof. That is, do you subtract, round, then divide, and round again or do you do all of the calculations at one time and round only once. I ALWAYS carry one more place in intermediate calculations, then round only one time at the end. My experience is that if I round too many times, I get rounding errors. Again, however, do what you prof has taught you. I would like to know what is being taught these days, just for my own information.
In an experiment 5.57 +/- 0.02 grams of water chemically decomposes into the products hydrogen gas and oxygen gas. The measured mass of oxygen gas produced in the reaction is 4.9320 +/- 0.0001 grams. The hydrogen gas occupies a volume of 6902 mL. Report the density of the hydrogen gas produced to the correct number of significant digits (Do not perform any calculation involving the estimates of uncertainties in the measurements.)
My answer is 9.24 *10^-5....Is it the right num of sig figs?
2 answers
9.24 *10^-5
1 answer