6th edition, 11.23

Moderators: Chem_Mod, Chem_Admin

Jonathan Christie 1I
Posts: 32
Joined: Mon Apr 09, 2018 1:38 pm

6th edition, 11.23

Postby Jonathan Christie 1I » Wed Jan 09, 2019 7:32 pm

the question used values of 500.0, 0.031, 0.495, 0.145 but the answer in the solutions manual gave the answer as 1.4.

I just wanted to make sure I brushed up on my sig figs but shouldn't the answer have 3 sig figs instead of two?

whitney_2C
Posts: 74
Joined: Fri Sep 28, 2018 12:28 am

Re: 6th edition, 11.23

Postby whitney_2C » Wed Jan 09, 2019 7:35 pm

Two sig figs is actually right because the value 0.031 only has two sig figs since neither of the zeros are significant. If the number were .0310 then it would have three sig figs though because 0s that come after non-zeros in decimals are significant.

Grace Kim 1J
Posts: 60
Joined: Fri Sep 28, 2018 12:18 am
Been upvoted: 1 time

Re: 6th edition, 11.23

Postby Grace Kim 1J » Wed Jan 09, 2019 7:37 pm

I believe the answer should have two sig figs because the Kc given in the problem is 0.031, which is 2 sig figs. Even though the other given values have three sig figs, you always want to use the one with the least amount of sig figs, so the final answer for Br2 would have two sig figs.

Hope this helped!

Kim Tran 1J
Posts: 62
Joined: Fri Sep 28, 2018 12:24 am

Re: 6th edition, 11.23

Postby Kim Tran 1J » Wed Jan 09, 2019 7:38 pm

One of the sig fig rules is your answer should have no more significant figures than the least accurately known number. So, 0.031 which has 2 sig figs is the "least accurately known number" and is the number of sig figs you should base your answer on.

Jonathan Christie 1I
Posts: 32
Joined: Mon Apr 09, 2018 1:38 pm

Re: 6th edition, 11.23

Postby Jonathan Christie 1I » Wed Jan 09, 2019 7:45 pm

Thanks guys, I had a feeling I was missing something


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 8 guests