[Ham-Computers] Need information.
Rolly NF7B c
Rolly.NF7B.club at comcast.net
Tue Dec 27 02:36:47 EST 2011
Ken
Here is the math;
Starting out with 1 part of solution at 35% concentration
1 part of Solution contains
0.35 parts of h2o2
0.65 Parts of water
3.00% Desired new % solution
A) Divide the # parts oh h2o2 by the desired new % of 3% to determine the
the # parts total of solution at the desired New %
= 1 / 0.03 = 11.66667 = parts of solution
B) Subtract the # parts of h2o2 from the # parts total of solution to
determine
the total # parts of water in the desired percentage solution
= 11.66667 - 0.35 = 11.31667 = the total # parts of water in the desired
percentage solution
C) Subtract Starting # parts or water from the total # parts of water to
determine
the total # of parts of water to be added.
= 11.31667 -0.65 = 10.66667 or 10 2/3=the total # of parts of water to be
added for each part of the original 35 % solution.
Might want to use 11 to be safe.
Repeating the procedure at desired new % at 8 %
A) = 1 / .08 = 4.375 = the # parts total of solution at the desired New %
B) = 4.375 - 0.35 = 4.025 = the total # parts of water in the desired
percentage solution
C) = 4.025 -0.65 = 3.375 or 3 3/8 = the total # of parts of water to be
added.
Might want to use 3.5 to be safe.
I hope this is helpful.
Rolly NF7B
-----Original Message-----
From: Ken
Sent: Monday, December 26, 2011 5:42 AM
To: boatanchors at mailman.qth.net ; boatanchors at theporch.com ;
ham-computers at mailman.qth.net
Subject: [Ham-Computers] Need information.
Hi Folks,
Please excuse but I desperately need information. My math ain't what it used
to be!
Can someone tell me how to mix 35% solution with clean water to get a 3%
solution.
I bought a quantity of "food grade 35% solution h2o2 and need the 3%
solution.
While at it, how about a 35% solution down to an 8% solution for agricultral
purposes?
My heartfelt thanks. My math never was anything to brag about and far worse
now tho I am only 94 years of age!
Ken N5CM
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