How many millilitres of ethyl alcohol are required to prepare 5 litres of 80% alcohol?
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)
946830
Just remember M1*V1=M2*V2 where M is molarity and V is volume. A little Algebraic manipulation gives you:\n\nV1 = (M2*V2)/M1\n     = (0.195*15.5)/18.0\n     = 0.168 L = 168 mL
443106
5-milligrams-per-milliliter, so 15 x 5 = 75. 75 millimeters. Next time pay attention in math.
335996
75 proof*100%ethanol/200 proof=37.2 %ethanol\n\n.798g/ml*1000ml/l=798g ethanol/l\n\n37.2/100 *798 g ethanol/l=299.25g ethanol/l
239849
.7M x 20mL = .25M x (Volume of new IA solution)\nVol = (.7 x 20) / .25\nVol= 56mL
968095
The density of alcohol (ethyl) is 0.789 g/mL, so it would take up more space (less dense).  Water is 1 g/mL
226482
Here goes: your 80/20 gas mixture is flowrate x 1.8\n\nyour 70/30 gas mixture is flowrate x 1.6\n\n***Dana Oakes is a great book to use***
843623
To work out the volume of solvent needed to make a 10mM (0,01M) solution use the formula:\n\nKnown mass of solid\n---------------------------------     = volume of solvent needed (in litres)\n0.01  x  Relative Mass of solid
232453
480,000.0 mg /1000 = 480g, 0.80 L x 1000 = 800 ml, so 480g/800ml.  divide both sides by 800 = 0.6g/ml. m = milli = 1/1000. 1000 mL = 1 L, 1000 mg = 1g, 1000 meter = 1 kilo meter. Kilo = 1000.
752805
The first thing you need to do is figure out which of the three liquids is going to be the one that limits the maximum quantity of the mixture that can be made.  Since the ratios are 5:2:1 you need 5 parts x, 2 parts y, and 1 part z.  You have 25 gallons of x and, if that's 5 'parts' then 1 'part' must be 5 gallons.  Now you need 2 'parts' (or 10 gallons) of y, and you do have that much in stock.  Lastly 1 'part' (5 gallons) of z is required and you have that much in stock as well.  So you'll mix 25 gallons (5 'parts') of x, 10 gallons (2 'parts') of y, and 5 gallons (1 'part') of z for a total of 25+10+5 = 40 gallons of the mixture.\n\nHope that helps.\n\n\nDoug
559540
the formula M=moles/L\n first convert mL into L\n(500)(10^-3) = .5L\nM= .50moles/.5L =  1M
163814
by distilling non-pure alcohol. (you need the proper equipment though - they do this in like high school chemistry class or something like that). But as far as I know, it's impossible to get 100% pure alcohol, because it will absorb some water vapour from the air, so the closest you can get is lika 99,8% or something.
989768
Pure Ethyl alcohol has a hellvua (R+M)/2 octane number at 115!\n\nFirst number is RON, second number is MON\n\nmethanol                    133 : 105         \nethanol                       129 : 102 \niso propyl alcohol       118 :  98 \nmethyl tertiary butyl ether   116 : 103 (this is MTBE the stuff they just took out of gasoline) \nethyl tertiary butyl ether    118 : 102 \ntertiary amyl methyl ether    111 :  98\n\nHope this does the trick for you, and no I am not a chemical engineer, but I do know how to use Google! :^)
840903
First find out how many  moles of NaOH in 5.18 g sample.\nmolar mass for NaOH is 23 + 16 + 1 = 40 g/mol\n\n5.18 g NaOH / 40 g/mol = 0.1295 mols NaOH\n\nNext calculate the molarity of the first solution. Molarity = mol/Volume (in L)   300 mL = 0.3 L\n0.1295/0.3 L = 0.4317 M\n\nFinally that solution is being diluted so find out how many mols you are taking out when you remove 25 mL uisng the same equation Molarity = mol/vol    26 ml = 0.026 L\n\n0.4317M = x mols / 0.026 L  \n  x mols = 0.0112242 mols\n\nNow just one last step. Find the molarity of the new solution using same equation. 100 mL = 0.1 L\n\nM = 0.0112242 mols/0.1 L\nYour final M = 0.112242 M
15951
Well Alcohol is a little lighter than water, and 1 gram of water is 1 mililiter.\n\nThere are 473 mililiters in 1 US pint so there are a little less than 473 grams of alcohol in a pint.\n\nNow keep in mind, your typical vodka is only 40% alcohol, wine is about 12% and most american beer is 3-5%.\n\nSo to get 30 grams of alcohol you would need to drink approximately 2 12 ounce american beers.
105947
0.5%solution means 500mg.( 1/2gms )of solute in one liter of solvent.
1083808
100 mmol/L acetic acid in terms of mass per litre:\n = 100 mmol/L * 60 g/mol = 6000 mg/L or 6g/L which is 0.6% acetic acid.\n\nBut if you start with a 10% solution of acetic acid, it is 10/0.6 or 17 times too concentrated for what you need.  Dilute with 17 volutes of water.\n\nI hope this helps.  There are other ways to show if need be.