How do I make a 125nmolar solution of Glucose. Help !!!!!?
117205
The molecular weight of glucose is 120.\n120 grams of glucose in 1 liter of water is 1 molar.\ndivide 120 grams by 10^-9, mix that in 1 Liter if water and the result is 1nM. Multiply 120*10^-9 by 125 and dilute to 1L, thats 125nM. About 15mG glucose in 1ml H20.
115621
since glucose is a monosaccharide, it is a simple sugar with a formula of C6H12O6. Now, to make glucose very easily, here's a hint. Since glucose is a product of photosynthesis (along with oxygen) just get a plant, put it under a UV light, and voila, glucose. The chemical formula for photosynthesis is (6)H2O + (6) CO2 + Light = C6H12O6 + (6)O2
241388
This is a simple conversion problem.\n\nWe need one fact, the number of milliliters in a tablespoon. The key here is that milliliters and tablespoons are both units of volume, so you can easily convert between the two. Google answers that question for me (see link). Your textbook probably also has a conversion chart:\n1 US tablespoon = 14.7867648 milliliters\n\nSince you have 125 mg per 5 ml or 125mg/5ml, you divide that to determine that you have 25 mg per 1 ml or 25mg/1ml. This step isn't always necessary, but it can help to make the final equation easier to solve.\n\nSimply take that number and multiply by the conversion factor (rounded to 14.8 ml/tbsp). \n\n25 mg/1 ml * 14.8 ml/1 tbsp = 370 (mg * ml)/(ml * tbsp) = 370 mg/tbsp\n\nThe trick to these problems is to keep track of and look at the units (always write them out). See how the ml units cancel out in the equation above? Hope that helps explain how to do the problem!
583606
maltose it is glucose+glucose
109521
I remember learning about this very complex process, its called Gluconeogenesis and if you do a simple search you can find the step by step processes!
552494
Hey that's worth a shot although i think ill need a couple kilo's of sugar, couple of slices of bread, some vitamin B to get the process going. but then again the carbon dioxide produced by the ferment will kill me. oh well back to the drawing board.
914456
okay...100 g*(mol/12.01g)=8.33 mol C\n\ntake the aforementioned 1mol *(-113kj/2mol)=-56.5 kJ released for 1 mol\n\nthen multiply that -56.5 by 8.33= -470.65 kJ released\nnote it is exothermic and thus negative
967067
that's so weird, we're doing something similar in my bio class...\n\nI would agree with you....glucose should be broken down faster...perhaps theres some error in ur results.\n\nI did the same experiment...and I had glucose produce a little more CO2 than sucrose.
788768
Just the usual stuff.  Use an electrolyte (H2SO4/HNO3), a couple of electrodes, a reduction table of half cell potentials, and some electricity.
217873
Measure concentration. Use a dummy (with water only) measuring that with a LabPro probe. Then compare that to a certain volume of a variable soft drink. I did this in high school, hope it works!
582636
Try dissolving sugar in water then freezing it this is a solid solution. The solute is the sugar and the water is the solvent
851560
solution...Bicarbonate and sodium. :)
102422
Hopefully by "Stoichiometry Concept Map" you want to know the route you take from a given point A to point B.\nBasic "MAP" :\n\nGrams of Substance A divided by molar mass of A ===> Moles of A times the coefficients of A and B from balanced equation to get moles of substance B===> Moles of substance B times molar mass of substance B ===> Grams of substance B\n\nExample:  How many grams of water produced in the oxidation of 1.00 g of glucose (C6H12O6)\nThe balanced equation is:\n\nC6H12O6(s) + 6 O2(g) ----> 6 CO2(g) + 6 H2O(l)\n\nSubstance A = Glucose\nSubstance B = Water\n\nOkay to solve lets convert 1 g of glucose into moles by dividing 1.00 g by C6H12O6 mol weight of 180.0 g/mol\n\n1.00g/180.0g/mol = 5.56 * 10^-3 mol\n\nNow multiply by 6/1 because the coefficient in front of glucose is 1, so put one in the denominator, the coefficient in front of the water is 6, so put six in the numerator.\n\n(5.56 * 10^-3 mol * (6/1)) = 3.33 * 10^-2 mol of H2O \n\nNow take that result and multiply it by the molar mass of Substance B\n\n(3.33 * 10^-2 mol *(18.0g/mol)) = 0.600g of H2O\n\nSo your final answer is 0.600 grams of H20 is produced when 1.00 gram of glucose is oxidized.
663838
I agree with Thinkstoomuch's Answers in most part.  \n\nSolid Cu, C will not be soluble in water.  However glucose and  CuBr2 will be soluble in water.  At STP solubility of glucose (C6H12O6) is about 90g/ml.  That eliminates choice 1 and 3.\n\nWhen a compound ionizes in soln it makes electrolyte.  Glucose will not ionize but CuBr2 will ionize to Cu++ and 2Br-, choice 4 can be eliminated.\n\nThe third clue is just there to confirm your answer and CuBr2 being an ionic compound will indeed have a relatively high melting point.  498°C to be exact.  \n\nClearly, Cupric Bromide should be your Answer.\n\nI hope this helps.
327703
The amount is diluted by a factor of 25.  So the resulting solution would have a concentration of 11.6M divided by 25 or \nor 0.464M.
106576
You'd need 4 L of dehydrated alcohol and 1 L of water in order to get a 80% solution. Getting a dry solution of ethanol would be a challenge, though, so I would expect it to be closer to 75-77% in actuality (but theoretically 80)
25334
Apart from the reactions you will get when you dissolve glucose in water, which are mostly reversible when you crystallize the glucose, and the reactions when you dissolve salt in water (formation of a water molecule "coat" aka complex around the sodium ion), there will be no reaction between glucose and salt. \n\nSolubility of either compound will be reduced somewhat by the presence of the other compound. The addition of salt to an over-saturated glucose solution might start a crystallization process, but any over-saturated sugar solutions are hard to crystallize. Look at breakfast sirup or honey for an everyday occurrance.