Problem G5

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haleyervin7
Posts: 61
Joined: Fri Sep 28, 2018 12:15 am

Problem G5

Postby haleyervin7 » Wed Oct 03, 2018 10:51 pm

On problem G5, how do you solve part a and b where it is just asking about part of the compound?

Ania Chavez Dis 4D
Posts: 14
Joined: Fri Sep 28, 2018 12:27 am

Re: Problem G5

Postby Ania Chavez Dis 4D » Thu Oct 04, 2018 12:26 am

Start out by finding the molarity of NaCO3:
-find out the moles of NaCO3: (2.111 g NaCO3) / (105.99 g/mol) = 0.01992 mol
-find out the molarity: (0.01992 mol) / (.250 mL) =.0796
For Part A, because for every mole of Na2CO3 you have 2 moles of Na+, you have to multiply the molarity of NaCO3 (0.0796) to the number of moles of Na+ (2)
Then you take 2.15 mmol Na+, which is equal to 2.15 X10-3 or 0.00215 mol, and divide it by the answer you got above (0.0796 X 2) and that should give you your final answer.
Hope this helps!

maldonadojs
Posts: 57
Joined: Fri Sep 28, 2018 12:27 am

Re: Problem G5

Postby maldonadojs » Thu Oct 04, 2018 1:11 pm

When finding the answer for 'a', would that be considered using the M1V1=M2V2 equation? I am just trying to conceptualize this process.

Jasmin Argueta 1K
Posts: 69
Joined: Fri Sep 28, 2018 12:16 am

Re: Problem G5

Postby Jasmin Argueta 1K » Thu Oct 04, 2018 1:22 pm

From my understanding you are using the M1V1=M2V2 equation. Just keep in mind that they give you 2.15 mmoles NA+ which is M2*V2. 2.15 mmoles is what you get when you multiples Moles/liter(M1)* liter(V2). The liters cancel out and you're left with M1V2=2.15mmoles. I hope this helps!

Emma Randolph 1J
Posts: 65
Joined: Fri Sep 28, 2018 12:29 am

Re: Problem G5

Postby Emma Randolph 1J » Thu Oct 04, 2018 3:55 pm

I'm still confused as to how the MinitialVinitial=mfinalvfinal formula applies to this problem, which values correspond to each part exactly?

anthony_trieu2L
Posts: 60
Joined: Fri Sep 28, 2018 12:29 am

Re: Problem G5

Postby anthony_trieu2L » Thu Oct 04, 2018 4:08 pm

Yes, in this case we are using the M1V1=M2V2 formula. M1V1 is equal to the number of moles of Na2CO3, which is 1.08 x 10^-3 mol. You then set this equal to M2V2. M2 can be calculated using the given number of grams of Na2CO3 (2.111g) and volume (250.0mL). With this information, you set the equations equal to one another and calculate V2, which is your final answer. Hope this wasn't too confusing!

Ashley Odibo Dis3E
Posts: 31
Joined: Wed Nov 15, 2017 3:03 am

Re: Problem G5

Postby Ashley Odibo Dis3E » Fri Oct 05, 2018 12:08 am

If we have another way to do the problem, but come sup with the same answer, will the professor deduct points? I completed the problem by finding the molarity, but I didn't use the m(1)v(1)=m(2)v(2) formula.

Erin Nash - 4G
Posts: 29
Joined: Fri Sep 28, 2018 12:27 am

Re: Problem G5

Postby Erin Nash - 4G » Fri Oct 05, 2018 3:42 pm

So I understand how to use the M1V1 = M2V2 equation but for this specific problem I wasn't sure how to tell that we needed to use it. Could someone explain?

Rebecca Park
Posts: 30
Joined: Fri Sep 28, 2018 12:17 am

Re: Problem G5

Postby Rebecca Park » Fri Oct 05, 2018 4:18 pm

Molarity=moles of solute (n)/volume of solution (v) is essentially the same thing as Mi*Vi=Mf*Vf.
If you multiply the volume of solution over to Molarity, you get M*V=n and n initial= n final, which gives you m initial* v initial= mfinal* vfinal.
You just see what information is given to you and then plug in the numbers


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