I need help finding the %Composition for %cychohexane and %toluene. The formula (area of peak/total area) * 100% WELSIS(modified after loading)FID1 A, Front Signal (NO SAMPLE NAME (FRONT) 2020-05-16 19-38-24.D)pA800000700000600000-5000004000003000002000001000000.20.40.60.8min=== =Area Percent Renort0.4040.44260S'00.5830.6560.8120.884 Signal 1: FID1 A, Front SignalPeak RetTime TypeWidth[min]-| ----|-------|-----AreaHeight(pA]Area#3[min][pA*s]-|------10.404 BVS0.0353 4.44304e41.65585e42.3562720.442 VV S0.0186 1.90102e41.70425e41.0081630.509 VVS0.0134 7.44826e49.27473e43.9500240.583 VVS0.0322 1.69872e68.78940e590.0879250.656 VBAS0.0611 4.81084e41.31317e42.5513260.812 BVX0.0208591.40356435.478670.0313670.884 VV X0.0159282.09906275.426060.01496Totals :1.88563e61.01913e6

To determine: % composition for Cyclohexane and toluene

Answered: I need help finding the %Composition…

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter15: Electrophoresis And Other Separation Methods

Section: Chapter Questions

Problem 1P

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The concentration of hydronium ions and hydroxide ions can vary widely in a solution, easily from 1 M to 1 x 10-14 M. This variation is too wide to fit on an analog readout. In order to make this into a smaller scale suitable for an analog readout, the log to the base 10 was used so that the read out is focused on powers of 10. Note that the log [1] = 0, and the log of 1 x 10-14 = -14. It was then decided that the numbers should be positive so the sign was changed. The way to measure concentrations of both H+ and OH- was thus defined as -log. This was then referred to as p. Thus, pH = -log[H+], pOH = -log[OH-], and pK = -log[Keq]. You need to figure out how to do this mathematical manipulation on your calculator. The acid-base reaction in pure water is H2O + H2O ⇔ H3O+1 + OH-1 and it was found that in pure water under standard conditions the [H3O+1] = [OH-1] = 1 x 10-7 M Neutral pH is defined as having a [H+] or [H3O+1] of 1 x 10-7 M. The pH of a neutral solution is thus pH = -log[1 x…

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