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re: Chemistry question for the OT

Posted on 12/7/15 at 9:11 pm to
Posted by FutureMikeVIII
Houston
Member since Sep 2011
1082 posts
Posted on 12/7/15 at 9:11 pm to
I think it's ~63 gal, too. I think everyone that is saying 1.5 gal is assuming the fresh water has s.g. of 0 instead of ~1.

I think the correct equation is:

C1V1 + C2V2 = C3V3
Posted by tigerbacon
Arkansas
Member since Aug 2010
3704 posts
Posted on 12/7/15 at 9:15 pm to
Get a salinity meter and add fresh water until it reads what you want.
Posted by Bmath
LA
Member since Aug 2010
18681 posts
Posted on 12/7/15 at 9:34 pm to
quote:

I think it's ~63 gal, too. I think everyone that is saying 1.5 gal is assuming the fresh water has s.g. of 0 instead of ~1.

I think the correct equation is:

C1V1 + C2V2 = C3V3




You really think it would take 63 gallons , or 20% of the current volume, to change the water SG by 0.005 kg/l?
Posted by FutureMikeVIII
Houston
Member since Sep 2011
1082 posts
Posted on 12/7/15 at 9:39 pm to
When the water you're adding only has a difference of 0.03 kg/L...yes.

Explain my mistake, I'm all ears.
This post was edited on 12/7/15 at 9:41 pm
Posted by Ingeniero
Baton Rouge
Member since Dec 2013
18323 posts
Posted on 12/7/15 at 9:54 pm to
Let x be the amount of water you need to add.

[315/(315+x)]*(1.03)+[x/(315+x)]*(1.00)=1.025
Multiply both sides by (315+x)
315*1.03+x*1.00=1.025(315+x)
324.45+x=322.875+1.025x
.025x=1.575
x=63 gallons
Posted by Bmath
LA
Member since Aug 2010
18681 posts
Posted on 12/7/15 at 9:56 pm to
Because C1V1 + C2V2 = C3V3 is used when combining two solutions to make a third. C1V1 = C2V2 is the correct dilution equation. That comes out to be ~1.5 gal.
Posted by FutureMikeVIII
Houston
Member since Sep 2011
1082 posts
Posted on 12/7/15 at 10:04 pm to
quote:

Because C1V1 + C2V2 = C3V3 is used when combining two solutions to make a third.


That's exactly what we're doing here, Einstein.

C1 = 1.03 g/mL
V1 = 315 gal
C2 = 1.00 g/mL
V2 = ?
Posted by Bmath
LA
Member since Aug 2010
18681 posts
Posted on 12/7/15 at 10:16 pm to
quote:

That's exactly what we're doing here, Einstein.



No need for the ad hominem attack. Besides, didn't Einstein fail a math course growing up?

quote:

C1 = 1.03 g/mL
V1 = 315 gal
C2 = 1.00 g/mL
V2 = ?


You're not diluting with, you are diluting to. So, my argument is this:
C1 = 1.03 g/mL
V1 = 315 gal
C2 =1.025 g/mL
V2 = 315.54 gal
Posted by X123F45
Member since Apr 2015
27480 posts
Posted on 12/7/15 at 10:20 pm to
quote:

Shut the frick up, rock nerd. He asked about chemistry, not what a vagina doesn't feel like.




...one of the greatest posts in tiger droppings history was a better version of that joke.
Posted by FutureMikeVIII
Houston
Member since Sep 2011
1082 posts
Posted on 12/7/15 at 10:29 pm to
quote:

No need for the ad hominem attack


Your laughing at my correct answer made me salty...couldn't help myself.

Anyway, you're diluting with pure water with a density of 1.0 g/mL to get to water with a density of 1.025 g/mL.

So,

C1 = 1.03 g/mL
V1 = 315 gal
C2 = 1.00 g/mL
V2 = ?
C3 = 1.025 g/mL
V3 = V1 + V2

C1V1 + C2V2 = C3V3

Since we're actually talking about density, not concentration, the dilution water has a non-zero mass that must be taken into account.
Posted by LSUsuperfresh
Member since Oct 2010
8336 posts
Posted on 12/7/15 at 10:36 pm to
63 gal brah. 315 gal * 1.03 + x = 1.025 * (315 gal + x)


Solve for x.
Posted by jimbeam
University of LSU
Member since Oct 2011
75703 posts
Posted on 12/7/15 at 10:41 pm to
Yeah I'm going with 63. You're talking a pretty significant change honestly.
Posted by Bmath
LA
Member since Aug 2010
18681 posts
Posted on 12/7/15 at 10:49 pm to
quote:

Your laughing at my correct answer made me salty...couldn't help myself.


Fair enough

quote:

Anyway, you're diluting with pure water with a density of 1.0 g/mL


Right, which is what I always assume with making dilutions in the lab. Perhaps what I'm not looking at is I generally am looking to dilute based on percentage. Density is a form of concentration, but it has specific units. Therefore, I could see how the other equation would be more correct. However, it seems odd to me that it would take a full order of magnitude difference in water to make a 0.005 g/mL change.
Posted by Lionnation1993
Member since Nov 2013
6103 posts
Posted on 12/7/15 at 10:50 pm to
This should work

This post was edited on 12/7/15 at 11:11 pm
Posted by CuseTiger
On the road
Member since Jul 2013
8230 posts
Posted on 12/7/15 at 11:34 pm to
quote:

C1 = 1.03 g/mL
V1 = 315 gal
C2 = 1.00 g/mL
V2 = ?
C3 = 1.025 g/mL
V3 = V1 + V2

C1V1 + C2V2 = C3V3

Since we're actually talking about density, not concentration, the dilution water has a non-zero mass that must be taken into account.

The chemist in me agrees with all of this. The units of mL and gallons are bothering me even though there wouldn't be a difference in the calculation

ETA: Where's the update??
This post was edited on 12/7/15 at 11:35 pm
Posted by Pectus
Internet
Member since Apr 2010
67302 posts
Posted on 12/11/15 at 8:12 pm to
Well...

What was the result?!
Posted by Hopeful Doc
Member since Sep 2010
15003 posts
Posted on 12/11/15 at 8:52 pm to
Why don't they just buy a house that's already painted?
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