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re: Chemistry question for the OT
Posted on 12/7/15 at 9:34 pm to FutureMikeVIII
Posted on 12/7/15 at 9:34 pm to FutureMikeVIII
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?
2
Posted on 12/7/15 at 9:39 pm to Bmath
When the water you're adding only has a difference of 0.03 kg/L...yes.
Explain my mistake, I'm all ears.
Explain my mistake, I'm all ears.
This post was edited on 12/7/15 at 9:41 pm
Posted on 12/7/15 at 9:54 pm to Bmath
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
[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
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