conversion of absorbance to nmol p-nitrophenol/ml?
920352
Did you measure the absorbance of a standard in addition to the unknown sample?\n\nIf so, you could plug the values into (As / Au) = (Cs / Cu)\nWhere\nAs = Absorbance of the Standard\nAu = Absorbance of the Unknown\nCs = Concentration of the Standard\nCu = Concentration of the Unknown?\n\nAlternatively, did you get a value for the molar absorptivity of PNP at the wavelength you used? \nIf so, you could plug it into\n\nA = (epsilon)(b)(c)\n\nWhere\nepsilon is the molar absorptivity,\nb is the length of the path through which the light passed,\nc is the concentration of the chromogenic substance.\n\nAnd you could rearrange that to solve for c.\n\nHope it helps
741946
Very easy:  \nPPM is equivalent to μg/mL (That is micrograms per milliliter)\n1 PPM = 1 μg/mL\n(the "μ" in micrograms is the greek letter mu)\n\nSo, a 10 PPM solution is 10 micrograms dissolved in 1 mL of water.  (or 10 milligrams dissolved in one liter (which is the same as10,000 micrograms in 1,000mL))\nSo, a 50 PPM solution is 50 micrograms dissolved in 1 mL of water.  (or 50 milligrams dissolved in one liter)\nEtc.\n\nPPM concentrations are just simple ratios, μg to mL.\n1 microgram = 0.000001grams\n\nThe volume (mL) always refers to WATER.  \nYou'll dissolve X amount of each of you reagents in X amount of water.
1013801
Molar mass of NH4NO3 = 14+4+14+48 = 80\n\nSo, 1.4 g is equal to 1.4/80 = 0.0175 mol\n\nNow, mass of soln = 74+1.4 g = 75.4 g\n\nSo, heat absorbed = 75.4x4.18x(25-23.71) = 404.415 J\n\nSo, 0.0175 mol absorbs 404.415 J\n\nSo, 1 mol absorbs 23109.429 J\n\nSo, heat of dissolution of NH4NO3 is 23.109 kJ/mol (+ve as heat is absorbed)\n\nb) Lattice energy = 630+23.109 = 653.109 kJ/mol
385735
The molecular mass of p-nitrophenylphosphatase is 33 kDa.
23140
http://www.esmalloffice.com/SBR_template.cfm?docNumber=PL12_0700.htm
245982
Sorry, can't help. You use different units than I do.\n but concentration(mol per dm3) = no. of moles(mol) divided by volume(dm3)
878297
Assuming 1 mole (28.01348 g) N2 = 22.4 liters at STP, I get (28.01348*.00220102/22.4) =  0.0027526 g if collected at STP. Correcting by 273/293 * (1.5*760-17.5)/760 (for temp., press. and water) = 0.003788 g. \nSo I basically agree with your results. What's the ans. supposed to be?
780066
V = nRT/P\nV = (25g/44g/mol)(0.08206Latm/molK)(298K)/1atm\nV = 138.9 L
1115546
You can experimentally calculate the molar absorption coefficient for methylvilolet. Prepare one molar (1M) solution of methylvilolet (c). Measure the maximal absorbance (A) for methylvilolet in a specific wavelenght on a spectrophotometer.Use the equation were given in below.\n\n                   A=e. l . c\n A=absorbance\n e=molar absorption coefficient (L/mol . cm)\n l= 1 cm (path of light)\nc=concentration (1M)
842140
Since phenol is a (very) weak acid, it might be possible to make a quantitative measurement by titration of phenol solution with a standard solution of alkali using phenolphthalein indicator.\nAnother possibility is to make use of the fact that phenol reacts readily with bromine to form tribromophenol. This would involve a rather complicated process:\n1. Add the phenol to a measured excess of bromine solution;\n2. Then add the solution with the excess bromine to potassium iodide solution (making sure that there is more than enough KI); this produces an equal amount of iodine to the excess bromine.\n3. Titrate the iodine with standard sodium thiosulphate solution (starch indicator).
852232
Pv = nRT\nn = m/M (M Molecular weight)\nPV = (m/M)RT\nP = (m/V)RT/M\nPM/RT = m/v = Density.\nPlug the numbers, watch the units and get the density.\n\nOh a numerical answer, yes, ...42.
139895
2 ways come to mind.\n\nYou make a series of solutions of increasing concentration as well as a blank (deionized water) and measure their absorbance in a spectrometer. You find a regression of the data and figure out the methylene blue based on its absorbance.\n\nStandard addition. You take unknown absorbance and add increasing concentrations of methylene blue to it. Graph the data and figure out the concetration corresponding to the y intercetp.
958410
1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen
278097
Pressure Swing Absorption: a special method for purifying gases (there is a very good presentation on UOP home page)
889419
1) Decide what solvent you want to use: water or organic?\n2) Decide how much solution you want: 1L?\n3) Check the solubility of phenol in the solvent you choose. And calculate how much phenol you'll need by weight to make a saturated solution\n4) Find out the relationship between solubility and temperature.\n5) I assume you want to make a saturated phenol solution at room temperature. If the solubility increases as temperature goes higher, add a bit more phenol than needed (from the calculation above) to the solvent. \n6) Heat the flask or beaker that contains the solvent and phenol until all phenol dissolves.\n7) Let the container cool down to room temperature.\n8) If no phenol crystalizes the solution is over-saturated. You may rub the side of the flask or beaker to help phenol crystalize.\n9) You have your saturated phenol solution!!
471742
Ummm...You will have to figure out which is faster... but I can give an idea as to proceedure...First, try to answer this question: What are all the colored species in grape juice? You need to know that first. Hopefully there is only one. That would simplfy things. Then you need an extinction coefficient for that compound, if you can find one. Using that, you can take several absorbance measures and use Beer's law to establish the concentration of that species in your stock solution. Beer's law states that Absorbance = extinction coefficient x concentration x path length. Your path length is the width of your cuvette, which is usually 1 cm. Beer's Law will not be valid if you use a solution that is so concentrated that it gives you an absorbance value greater than 2. At Absorbance = 2, well over 99% of light is being absorbed. A reading over 2 is thus considered analytically unusable. Your juice can be diluted, if necessary, with DEIONIZED (aka distilled) water. Once you do that, I would roughly time each reaction by just looking at it with my eye, adding just a drop or so of bleach/peroxide initially, and increasing the amount only if necessary. Make sure to add exactly the same amount of bleach/peroxide to exactly the same volume of grape juice. Once you have the reaction basically set up to your satisfaction, then place your juice sample in the colorimeter and get ready to add your reagent. Remember to take a measurement before you add any bleach/peroxide for EACH of the two tests. And take your measurements on the same time frame (ex: every 30 sec.)Good Luck!!! :-)
851336
Since 1g=1mL, 1ppm=1mg/L.\n\n1. Convert ppm to mg/L\n\n2. Convert mg to g (divide by 1000)\n\n3. Convert g/L to moles/L (use the molecular weight)