F.15

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Jamie Hsu
Posts: 31
Joined: Fri Sep 28, 2018 12:19 am

F.15

Postby Jamie Hsu » Tue Oct 02, 2018 6:07 pm

For question F.15 from the problem sets, I noticed that when calculating the number of mols of H, the setup is 4.60gH/1.008g mol-1
According to the sig fig rules 4.60 has 3 sig figs and 1.008 has 4 sig figs, and when dividing we should take the lowest number of sig figs, however the answer is 4.563 mols H. Has anyone else come across this?

804991762_4A
Posts: 47
Joined: Fri Apr 06, 2018 11:04 am

Re: F.15

Postby 804991762_4A » Tue Oct 02, 2018 6:22 pm

In this case, because the answer is 4.563 I think it safe to make it 4.56 since like you said the lowest sig fig is 3.

Nina Do 4L
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

Re: F.15

Postby Nina Do 4L » Tue Oct 02, 2018 9:12 pm

When you are adding or dividing or multiplying or subtracting with significant figures, your answer will have the least amount of sig figs because your answer can not be more precise than the least precise measurement. In this case, yes, I also agree that the answer would be 4.56 because 3 is less than 5 so we would round down.

gillianozawa4I
Posts: 31
Joined: Fri Sep 28, 2018 12:27 am

Re: F.15

Postby gillianozawa4I » Wed Oct 03, 2018 12:12 am

When multiplying and dividing values with different numbers of sig figs, round down. This is because the answer is no more precise than the value with the least number of sig figs.

Stevin1H
Posts: 89
Joined: Fri Sep 28, 2018 12:17 am

Re: F.15

Postby Stevin1H » Wed Oct 03, 2018 11:54 am

Since the rule for multiplying and dividing significant figures is to have as many sig figs in your answer relative to the lowest amount of sig figs in your calculations, I would agree that the answer 4.56 is sufficient because the lowest number of sig figs in the calculations, as you said, was 3.


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