[Ham-Computers] Ham-Computers Digest, Vol 95, Issue 6
djones at rgv.rr.com
djones at rgv.rr.com
Sun Jan 15 14:14:55 EST 2012
Mixture problems as simple, if you work based on the pure stuff you have
Make youself a chart that looks like this # (the pound sign)
one columm is the amount you start with, second columm is per cent, and third columm is the pure amount you have or want
If you are adding water to it, then an X for the amount to add and zero per cent for the second columm, the pure will equal
zero.
The last row will list the amount you want to end up with in the form of the amount minus X, second columm the per cent you
what to end up with. and third columm is that per cent times (the amount-X)
The first row, third columm is equal to the third row, third columm. Solve to X.
--
David Jones
djones at rgv.rr.com
N5VCF
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> Today's Topics:
>
> 1. Re: Need information. (Loren Moline WA7SKT)
> 2. Re: Need information. (Gary Pewitt)
> 3. Re: Need information. (Rolly NF7B c)
> 4. Apology. (Ken)
>
>
> ----------------------------------------------------------------------
>
> Message: 1
> Date: Mon, 26 Dec 2011 16:50:53 -0800
> From: Loren Moline WA7SKT <lmoline at hotmail.com>
> Subject: Re: [Ham-Computers] Need information.
> To: <ham-computers at mailman.qth.net>
> Message-ID: <COL102-W40D57CAE1DDDC232B9FEABAFAF0 at phx.gbl>
> Content-Type: text/plain; charset="iso-8859-1"
>
>
> http://www.using-hydrogen-peroxide.com/peroxide-dilution-chart.html
>
>
> Loren WA7SKT
>
> Member: ARRL and Pacific Northwest VHF Society
> Member: Hearsat Satellite Monitoring Group ( www.hearsat.org )
> Location: CN86bx
>
>
>
>
>
>
>
>
>
> > From: lmoline at hotmail.com
> > To: ham-computers at mailman.qth.net
> > Date: Mon, 26 Dec 2011 16:46:25 -0800
> > Subject: Re: [Ham-Computers] Need information.
> >
> >
> > The proper mix is 1 35% solution to 11 parts water..I don't know if the link I sent came through.
> >
> >
> > Loren WA7SKT
> >
> > Member: ARRL and Pacific Northwest VHF Society
> > Member: Hearsat Satellite Monitoring Group ( www.hearsat.org )
> > Location: CN86bx
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > > From: danki6x at earthlink.net
> > > To: ham-computers at mailman.qth.net
> > > Date: Mon, 26 Dec 2011 16:40:58 -0800
> > > Subject: Re: [Ham-Computers] Need information.
> > >
> > > BUT the original solution is already 35% (can't get 100% peroxide, and only
> > > special approval purchase will get you 70%). So you need about 1/3 of the
> > > water listed by Loren (31/3 = 10 1/3 water). Which is why I said 1 Peroxide
> > > to 10 Water. Gets more complicated to try to explain in writing. That's
> > > why I could never write a textbook. Dan KI6X
> > >
> > > -----Original Message-----
> > > From: ham-computers-bounces at mailman.qth.net
> > > [mailto:ham-computers-bounces at mailman.qth.net] On Behalf Of Loren Moline
> > > WA7SKT
> > > Sent: Monday, December 26, 2011 2:14 PM
> > > To: ham-computers at mailman.qth.net
> > > Subject: Re: [Ham-Computers] Need information.
> > >
> > >
> > > Ken,
> > >
> > > I think you need more water.
> > > Take this example.: if you had 3 cups of 33% solution that would be as 1 cup
> > > of 100% and 2 cups of water.
> > >
> > > so you retain the original figure of 1 cup of 100%
> > >
> > > 1 cup of 100% to 30 cups of water would be total of 31 cups.
> > >
> > > 1/31=3.2%
> > >
> > > Since you already have 2 cups of water mixed with your 1 cup of concentrate
> > > you need 30 cups of water - 2 cups or 28 cups of water.
> > >
> > > You now have your original 1 cup of 100% plus 30 cups of water...
> > >
> > > Am I making sense?
> > >
> > >
> > > Loren WA7SKT
> > >
> > > Member: ARRL and Pacific Northwest VHF Society
> > > Member: Hearsat Satellite Monitoring Group ( www.hearsat.org )
> > > Location: CN86bx
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > > From: n5cm at rtconline.com
> > > > To: ham-computers at mailman.qth.net
> > > > Date: Mon, 26 Dec 2011 15:18:31 -0600
> > > > Subject: Re: [Ham-Computers] Need information.
> > > >
> > > > Hi Dan,
> > > >
> > > > Many thanks for the info!
> > > >
> > > > Now, if I mix 1 part 35% with 10 parts h2o (clean water), I should get
> > > > close to 3.5% solution so if I add a bit more of h2o, I should come
> > > > close to 3%. I'll try that!
> > > > Thanks!
> > > >
> > > > It is for internal use therefore the caution. Too much h2o2 (hydrogon
> > > > peroxide) can kill!
> > > >
> > > > Ken N5CM
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > =======
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> ------------------------------
>
> Message: 2
> Date: Mon, 26 Dec 2011 20:28:39 -0600
> From: Gary Pewitt <garypewitt at centurytel.net>
> Subject: Re: [Ham-Computers] Need information.
> To: "Computers \(or other\) used for amateur radio, communications, or
> experimenting" <ham-computers at mailman.qth.net>
> Cc: Dan Violette <danki6x at earthlink.net>
> Message-ID: <4EF92D57.3030703 at centurytel.net>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
> A quart of 3% in the drug or grocery store is about $1.95.
>
>
> On 12/26/2011 6:32 PM, Dan Violette wrote:
> > A little more WATER to get it down a little from the 3.2% for a 1:10 part.
> > Peroxide breaks down into water fairly quickly (reason for the dark bottle)
> > so it is probably already less than 35% to start. Really close enough. Dan
> > KI6X
> >
> > -----Original Message-----
> > From: ham-computers-bounces at mailman.qth.net
> > [mailto:ham-computers-bounces at mailman.qth.net] On Behalf Of Ken
> > Sent: Monday, December 26, 2011 1:19 PM
> > To: Computers (or other) used for amateur radio, communications, or
> > experimenting
> > Subject: Re: [Ham-Computers] Need information.
> >
> > Hi Dan,
> >
> > Many thanks for the info!
> >
> > Now, if I mix 1 part 35% with 10 parts h2o (clean water), I should get close
> > to 3.5% solution so if I add a bit more of h2o, I should come close to 3%.
> > I'll try that!
> > Thanks!
> >
> > It is for internal use therefore the caution. Too much h2o2 (hydrogon
> > peroxide) can kill!
> >
> > Ken N5CM
> >
> >
> >
> >
> >
> >
> >
> > =======
> > Email scanned by PC Tools - No viruses or spyware found.
> > (Email Guard: 7.0.0.18, Virus/Spyware Database: 6.18940)
> > http://www.pctools.com/ =======
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> >
> > -----
> > No virus found in this message.
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> > Version: 10.0.1416 / Virus Database: 2109/4105 - Release Date: 12/26/11
> >
> >
> >
>
>
> --
> Gary Pewitt N9ZSV
> 1500 French Prairie Rd
> Booneville, AR 72927
>
>
>
> ------------------------------
>
> Message: 3
> Date: Mon, 26 Dec 2011 23:36:47 -0800
> From: "Rolly NF7B c" <Rolly.NF7B.club at comcast.net>
> Subject: Re: [Ham-Computers] Need information.
> To: "Computers \(or other\) used for amateur radio, communications, or
> experimenting" <ham-computers at mailman.qth.net>
> Message-ID: <1BC43D1FDF01460B9F75F75B0E859C97 at RollyName>
> Content-Type: text/plain; format=flowed; charset="iso-8859-1";
> reply-type=original
>
> 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
>
>
>
>
> =======
> Email scanned by PC Tools - No viruses or spyware found.
> (Email Guard: 7.0.0.18, Virus/Spyware Database: 6.18940)
> http://www.pctools.com/
> =======
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> This list hosted by: http://www.qsl.net
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>
> ------------------------------
>
> Message: 4
> Date: Tue, 27 Dec 2011 04:51:54 -0600
> From: "Ken" <n5cm at rtconline.com>
> Subject: [Ham-Computers] Apology.
> To: <boatanchors at mailman.qth.net>, <boatanchors at theporch.com>,
> "Computers \(or other\) used for amateur radio, communications, or
> experimenting" <ham-computers at mailman.qth.net>
> Message-ID: <005901ccc485$8c602f30$6401a8c0 at yourb27fb1c401>
> Content-Type: text/plain; charset="iso-8859-1"
>
> Folks,
>
> I certainly did not intend to stir up such a controversy! WOW! Please excuse.
>
> Ken N5CM
>
> P.S. My sincere THANKS to all concerned! I have a supply now of a very close approximation of 3% solution! THANKS
> to all who participated.
>
>
>
>
> =======
> Email scanned by PC Tools - No viruses or spyware found.
> (Email Guard: 7.0.0.18, Virus/Spyware Database: 6.18940)
> http://www.pctools.com/
> =======
>
>
> ------------------------------
>
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> End of Ham-Computers Digest, Vol 95, Issue 6
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