analytical estimation of cis and trans 2-methyl cyclohexyl acetate?
599905
i think you are correct on that.the only problem is that the substances of this byproduct could effect the particle generator with the obvious results on the galactic balance.
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
1154935
methyl3-benzoyloxy-8-methyl-8-azabicyclo [3.2.1]octane-4-carboxylate  (thats the systematic name)\n\nFormula: C17H21NO4
414562
2-methylheptane\n3-methylheptane \n4-methylheptane \n\n2,2-dimethylhexane \n2,3-dimethylhexane \n2,4-dimethylhexane \n2,5-dimethylhexane \n3,3-dimethylhexane \n3,4-dimethylhexane \n3-ethylhexane \n\n2,2,3-trimethylpentane \n2,2,4-trimethylpentane \n2,3,3-trimethylpentane \n2,3,4-trimethylpentane \n3-ethyl-2-methylpentane \n3-ehtyl-3-methylpentane\n\n2,2,3,3-tetramethylbutan\noops n octane
434220
tert-Butyl isobutyrate ~ 143.7g/mol\nethyl isocrotonate = 114g/mole isocroronate acid = 86g/mole, ethyl = 29, -1 for the H on the acid.\n\nPlease check my math
904652
CH3CH2C(CH3)3.\nI guess you forgot to type the C before the parenthesis. If you didn't forget, then none of the answers is correct (you would be missing a carbon atom).\n\nChem-man I would have contacted you in private about this but you do not give us the option to e-mail you. I have seen several of your answers prompting askers to verify your answer through echemistryhelp. I checked the website and it is not a free one; you post a question and pay for the answer. I'd say that this is a form of indirect advertisement  which is against the community guidelines. Please refrain from doing that or I will have to report you.
125676
a fairly lengthy description can be found at\n\nhttp://apps3.fao.org/jecfa/additive_specs/docs/0/additive-0465.htm
498348
LIQUID-VAPOR EQUILIBRIUM IN BINARY SYSTEMS\nApparatus\nBoiling apparatus with heating mantle and variac (S), 1/100o thermometer (S),\nrefractometer (L), glass beads or boiling chips (S), weighing bottles (D), small\nstoppered vials (D), 250 ml n-hexane (S) and 250 ml absolute ethanol (S).\nOther binary systems which may be investigated are (1) cyclohexane (S)-\nethanol (S); (2) ethyl acetate (S)-ethanol (S); (3) Methanol (S)-ethyl acetate (S);\n(4) isopropanol (S)-n-hexane (S); (5) isopropanol (S)-cyclohexane (S); (6)\nisopropanol (S)-ethylacetate (S); (7) Acetone (S)-cyclohexane (S); (8) Acetone\n(S)- hexane (S); and (9) cyclohexane (S)-2-propanol (S), (10) 1 Propanol-water.\nCONSULT YOUR LAB INSTRUCTOR TO FIND OUT WHICH SYSTEM YOU\nARE TO INVESTIGATE.\nCAUTION: The ORGANIC LIQUIDS are TOXIC and FLAMMABLE! Use all\nnecessary precautions to avoid accidents!\nCAUTION: DISPOSE OF THESE ORGANIC LIQUIDS by placing in the\nWaste Containers provided. DO NOT POUR DOWN THE DRAIN!\n\nAssemble the apparatus with care; consult instructor if you are at\nall doubtful how to put the pieces together. A 3-way stopcock (shown in Figure\n2) is provided as a means of withdrawing samples of distillate and residue at the\nappropriate time. Best procedure is to withdraw distillate first, then the residue.\n\nProcedure\nClean the apparatus by rinsing with a few milliliters of pure n-hexane. Make sure\nvials are clean and DRY and have labels for marking. Add glass beads to flask\n(carefully).\nThere are three particularly important requirements in this experiment.\n• Make sure that the condensed distillate and the liquid residue samples are\ncollected under equilibrium conditions\n• Read the equilibrium temperature accurately\n• Analyze the compositions accurately\nPlace the thermometer in a position near the top and center of the three legs of\nthe first insert as seen in Figure 3. The hole in the second insert should be\nplaced pointed towards the condenser\nPlace 50 ml n-hexane in a flask. Adjust the variac (not more than 100 -110 volts)\nso that liquid boils "reversibly" (very moderate boiling with only very tiny bubbles\nformed) and so that distillate comes over at a rate which will fill the sample tube\nonce every two or three minutes.\nDraw off the first fraction (containing impurities) which comes over and discard\n(use waste container). When a constant boiling mixture is obtained, record the\nboiling point and withdraw samples from distillate and residue into marked vials.\nAdd 1 ml ethanol to the distillate in the flask, obtain equilibrium-boiling, record the\nboiling point, and again withdraw samples of distillate and residue for analysis.\nBoiling points and refractive indices of the distillate and residue are then\ndetermined after successive additions of 0.2, 0.5, 1, 2, 5 and 5 ml of ethanol. The\nrefractive indices are used to obtain the mole fractions of ethanol in these\nsolutions.\nClean the apparatus by rinsing with a few milliliters of ethanol, and repeat the\nprocess starting with 50 ml of ethanol in the flask. Now, however, boiling points\nare to be determined after successive additions of 0.5, 1, 2, 5, 15 and 20 ml of nhexane\nto the flask.\nIn order to construct a plot of refractive index versus mole percent ethanol, the\nrefractive indices are determined for the pure hexane and ethanol and for a\nseries of solutions containing accurately known weights of hexane and ethanol.\nMixtures about 5 ml in volume containing approximately 1 volume of ethanol to 1,\n3, and 6 volumes of hexane are convenient. To obtain a decent standardization\ncurve of refractive index versus mole fraction, a series of 6 to 8 solutions\ncontaining accurately known weights of n-hexane and ethanol must be prepared\nand their refractive indices measured. Only a few milliliters of each of these\nsolutions are required, so do not waste chemicals when preparing them. The\nrequired data for the vapor pressures of the organic liquids are availab
807508
Most of the info provided is superfluous.\nSimply put, you burned something and got 4.161g H2O (MW=18.016), all H coming from the sample.\nSo you formed 4.161/18.016 = 0.23096 mols H2O, and since each mol H2O contains 2 mols H, your sample contained 0.4619 mols of H.\n\nBTW, it is easy to deduce the empirical formula of the compound:\nNo hydrocarbon can have a MW of 60.095, and since only CO2 and H2O were formed, the compound  must contain O...this works out to be C3H8O, which means it is probably propanol (n or iso) or methyl ethyl ether.
747676
2C12H22O11 +24O2 ---> 24CO2 +22H2O\na)F\nb)F\nc)T\nd)F\ne)T\nf)T
125296
use spectrophotometric method(uv-vis spectrometry).Few things u shud remember.....\n1) Use Labert-Beer's law\n2) u shud know the molar extingtion coefficient value of that particular species\n3) concentration of the species\n4) Calculate lambdamax. using LB law\nIt is an indirect method.\n\nUV-VIS. Spectroscopy will give u the lambdamax value directly...alongwith absorbance
139601
This is an esterification, \nButanoic Acid (CH3CH2CH2CO2H)+ ethanol (HO-CH2CH3), in basic media, yields ethyl butyrate. \n\nCH3CH2CH2CO2H + HO-CH2CH3 --> CH3CH2CH2COO-CH2CH3
776359
Methylamine\n\nNext time, just type the chemical formula into http://en.wikipedia.org/ and see if they have your compound documented.
618652
.........CH3\n..........|\nCH3 - C  - CH2 - CH2 - CH2 - CH3\n.........      |\n........      CH3\n\nPlease ignore the full stops in the above image (which i had to insert to get the correct spacing).\n\nI hope that this helps you.
372026
CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g)
802465
carboxylic acid, as represented by -(C=O)-OH. For more info, check the source.
839940
1. 2C6H6 + 15O2 --> 12CO2 + 6HOH\n\n2. C12H22O11 + 12O2 --> 12CO2 + 11HOH\n\n3. CH3COCH3 + 4O2 --> 3CO2 + 3HOH\n\n4. C2H5OC2H5 + 6O2 --> 4CO2 + 5HOH\n\n5. C4H9OH + 6O2 --> 4CO2 + 5HOH