mM to mg/L of glucose solutions?
969572
M (molarity) is defined as "mole per L (mole/L)" of solution\n\nglucose = C6H12O6, its molecular weight is 180.16g/mole (say about 180g/mole)\n\n1000mM = 1000mmole / L\n--> (1000mmole / L ) * (180g / mole)\n--> cancel out the "mole"\n--> 180000 mg/L\n\nsame logic for 970mM = 970mmole / L\n--> (970 mmole / L) * (180g / mole) = 174600 mg / L\n\nnote: you can leave the "m (milli)" there in your calculations, because, remember, "m" simply means 10^(-3), which is 0.001
810837
Just remember glucose (C6H12O6). Then using the basics of balancing any chemical equation you can figure out the rest of it.
753121
Ok, now this one I can do:\n\nMg3N2  ---> 3Mg (s) + N2 (g)
864884
1: You wouldn't measure a dry powder in milliliters.  You can't really know exactly how much water to add unless the know the mls the dissolved drug take up.  You could add 10mls then QS up to 15 mls once it dissolved in the orginal 10mls.\n\n2 If the total volume of 2 grams is 15mls.  Then 7.5 mls added to 50 mls of NS or D5W would be the answer.\n\n3.  If the total volume is 100 mls then 100ml/60 minutes= 1.66ml/min. Of course, you would set up the pump in mls/hr which in this case is 100ml/hr...duh.  If the total volume is 107.5 mls then 1.8mls/min.
583606
maltose it is glucose+glucose
776842
Time takes to dissovle same amount of sugar(sugar cubes and powdered) in water.  \nKeep the mass and temperature of the water constant and do not stir the solution.  Since powdered sugar has more surface area, it will dissolve faster.
205076
Since the density is 1.04 grams/mL and there are 410 mL...there is 426.4 grams of solution (Multiply 410mL by 1.04 grams/mL to give you the mass).  Since it contains 9.50% sugar BY WEIGHT, you can simply multiply 426.4 by 9.50%.  This gives you 40.508 grams.\n\nCALCULATIONS:\n1.04 g/mL * 410 mL = 426.4 grams\n426.4 grams of solution * 9.50% sugar = 40.508 grams of sugar
481495
1. Perform vacuum distillation.  Place the solution in a vacuum flask; apply vacuum (removing air from vessel); apply heat and "boil" the water off.  Collect the condensate and measure its volume.  Weigh the remaining sugar that is left behind in the vacuum flask.\n  -or -\n2. There may be a way to measure the refractive index of the sugar solution.\n-or-\n3. Heat a sample of the solution to determine the boiling point elevation; if you have a table that compares molarity vs. boiling point elevation, you may be able to determine the molarity of the solution by knowing at what temp. the solution boils.\n\n4. You may be able to extrapolate by weighing exactly one cubic centimeter (cc) (volume) of the solution at 25 degrees C.  There may be a table comparing weight of a 1 cc solution at that temp (as the density of 1 cc of water is different than a 1cc sugar solution.
397236
1mg /L is called  "ppm" ---Parts per million\n\nie.,  1ppm solution having 1mg of solute in 1L solvent
808549
M=molesof solute/Liter of solution\n\nM= molarity
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)
850118
Here is a tip.  Whenever you have a question like this just go to Google's search engine.  Your question can then be typed as "1 mcg in mg" in the search box.\n\nInstead of searching for websites, google does the conversion for you.  This is one of many secret tools you can use on Google.\n\nThe answer Google gives is this: 1 mcg = 0.001 milligrams
691897
Have teaspoon of sugar.
644701
Summary: convert your concentration to a mole, mass, or volume fraction and multiply by 1 million.\n\nDiscussion:\nParts per million must be measured in consistent units:\n\nmass per 1 million mass units\nvolume per 1 million volume units\nmoles per 1 million mole units\n\nFor example, if you add 1 milligram of substance to 1 kilogram of solution, the concentration is:\n\n1 mg/1 kg = 0.001 g / 1000g = 1e-6\nMultiply the resulting mass ratio by 1 million and you convert this to ppm\n1e-6*1million = 1 ppm\n\nI don't think it is correct to directly use molarity or molality times 1 million, since both of these are ammount/volume: the units are not consistent.  You have to convert the "volume" in the concentration to a "mass" or "molar" amount consistent with the concentration measurement.\n\nIf you have a concentration in terms of a mole fraction, it is just the mole fraction times 1 million.  A mole fraction of unity (one) is the same as 1 part per part (pure!).\n\nIf you have a concentration in terms of a volume fraction or mass fraction, you just multiply by a million.  It is important to note that whether you do parts by mole, volume, or mass you will get different numbers for a given system.  It is important to be clear which one you are talking about.  You can read more on wikipedia, look up ppm.
807684
Yes, you are correct.\nEveryone else who has posted so far is mistaken.\n\nGiven:\nM = mol of solute / liters of solution.\n\nFirst Step:\n\nSolving for x (liters of solution), you get:\n.40 / .096 = x\nx = 4.167 L\n\nSecond step.\nL => mL\n4.167 / 1000 = 4,167 ml or app. 4000 mL\n\nYour answer is approx. correct because you were asked for mL of water, which you found, not L of solution, which everyone else seemed to think.  You may lose points for not solving to the correct amount of significant figures though.
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.
105947
0.5%solution means 500mg.( 1/2gms )of solute in one liter of solvent.