If we have 2 bottles of Whiskies each one contains 40% alcohol and 0.5 letters of Whiskey, while we mix both of them to each other then percentage of alcohol is how much?
124853
You'll have 1 liter of whiskey but still 40% alcohol.
14831
The ratio is 1 part rubbing alcohol to 2 parts water.I usually use 1 cup and 2 cups. Put the mixer in a gallon size zip lock then, place the zip lock into another zip lock to prevent any leakage. You can reuse this ice pack over and over. OR - you can make several smaller ice packs if you don't want a gallon size one.
12312
are you talking about smirnoff vodka or the smirnoff malted drinks? \n\nsmirnoff vodka is 40% by volume and the malt drinks are 5 or 6 percent by volume.
8940
Provided you mix the same amount of each liquid.
390481
If I understand correctly what you need then this is what you do:\n(by the way we will take He as 83.7 because it is over 100% other wise.)\nmultiply the molecular or atomic weight of each species by the fraction (percent divided by 100).\n\nMW(CO2) = 44\nMW(N2) = 28\nAW(He) = 4\n\n44*0.11 + 28*0.053 + 4*0.837 = 9.67 g/mol\n\nThis is the equivalent "molar mass" of the mixture.
651367
Answer to question A :\n420 in 1001\nin percent = 41.958%\n\nThe way I worked this out :\n(8x7)/(2x1) multiplied by (6x5)/(2x1) to get the 420\nand (14x13x12x11)/(4x3x2x1) to get the 1001\n\nI assume the second part (B) you are asking how many you would have to take out to guarantee that you get at least one guppy. The answer would be 9\n\nYou have not made the second part of the question very clear, so I don`t really know what you are after?
518259
You expect to end up with 1L of 120 proof hootch.  Therefor you have an ending percentage of 60% alcohol and 40% water.\nSo the volume of alcohol is .6 L and water is .4 L.\nLet X be the quantity of Bacardi 151 and Y be the quantity of cheap rum.\nYou are then working to figure out X liters * (.755 Alcohol ) and Y liters * (.4 alcohol ) = .6 L alcohol \nIn algebra-speak, that is X (.755) + Y(.4) = .6\nYou also know that X + Y = 1 liter.  Taking this equation and solving for Y we have Y = 1 - X.\nWe can now substitute all occurances of Y in the equation with (1-X)  Therefor\nX (.755) + (1-X)(.4) = .6 and we have a single variable to solve for.\n.755X + .4  - .4X = .6 liters\n.355X = .2 liters\nX = .2/.355 \nX = .56338 liters of Bacardi 151\nThere for Y = .43662 liters of cheap rum.\n--------------------\nAdditional info.\n--------------------\nOK, Shirt   Sorry you got lost.  But I skipped a step.\nIn my solution, this is what happened between\n.755X + .4  - .4X = .6 liters and .355X = .2 liters\n\n.755X + .4  - .4X = .6  \nSubtract .4 from both sides\n.755X  - .4X = .2\nSubtract .4X from .755X\n.355X = .2\n\nI like steph's solution too.\nHe doesnt show all the steps either when solving for A.\n(0.755A + 0.4(1-A))/1 = 0.6.\nGet rid of the parenthesis by multiplying .4 \n(0.755A + 0.4 - 0.4A)/1 = 0.6.\nDivision by 1 doesn't change anything, So you can ditch the parenthesis.\n0.755A + 0.4 - 0.4A = 0.6.\nSubtract 0.4A from 0.755A\n.355A + 0.4 = 0.6\n.355A = 0.2\n\nI assumed you were using rum because Bacardi is the only one that make a 151 proof liquor.   Everclear is 190 or 95%.  That will change your quantities alot.  Use the same equation, but replace the 0.755 with 0.90.\nI made my own version of Goldshlagers which is more than 100 proof.  As you are doing, I too used Everclear to raise the proof.  It ended up tasting better than the goldshlagers.  Disolve a dozen Atomic Fireballs into a liter of vodka and add everclear.
474258
From one Big Daddy to another:\n\n.75(x) + .9(40 - x) = .78(40)\n.75x + 36 -.9x  = 31.2\n-.15x = -4.8\nx = 32\n\nThus, they should use 32L of the 75% mixture, and 8L of the 90%.
175571
C1 will be the original concentration of 10mg/cm3, V1 will be the 5 different volumes you are adding.  C2 will be unknown and V2 will be the 10cm3.  For example:\ntube 2...(10mg/cm3)(1cm3)=C2(10cm3)   C2=1mg/cm3
393378
compute the volume of cork approx and then use it to divide the volume of barrel. It can give some close figure. Other option is to create a barrel and corks and see yourself how many they fill in. But make sure constructing the barrel and cork dont cost you 1000$, after all you need to make a profit also
42703
I think it goes like this:\n20% = .2, 25% = .25\n\nlet x = final volume\n.25(12) = .2(x)\n1.25(12) = x\n15 = x\nif x = final volume, volume added = x - 12\nor 15\n15 - 12 = 3\nadd 3 L
147578
mix 50 % alcohol and 50 % distilled water. Apply with a soft cloth.
115691
Let w be the fraction of licquid in the wine glass consisting of water; initially w=0. The fraction of water in the water glass is 1-w (because the total amount of water in both glasses never changes from 100). \n\nLet s be the volume of a spoonful. As the result of a transfer from the water glass to the wine glass:\n\nw --> w' = (100w + s(1-w))/(100+s)\n\nDefining x = w - 1/2, this equation becomes:\n\nw' = x'+1/2 = (100(x+1/2) + s(1/2-x))/(100+s)\n==>\nx' = [(100-s)/(100+s)]x = ax\n==>\nx' = ax where a = (100-s)/(100+s)\n\nThe initial value of x (i.e. w-1/2) is -1/2, so after r operations x = -(1/2)a^r and so:\n\nw = 1/2(1-a^r)\n\n********\n\nFor example if s = 8cc (about 1 tablespoon), a = .85 and after three operations:\n\nw = (1/2)(1-(.85)^3) = .19  (19% or 19 cc water)\n\n********\n\nThe answer is not exactly proportional to the size of a spoonful, but the rationale for the previous answers is probably that if s << 100, then a ~ (1 - 2s/100)\n==>\na^3 ~ (1-s/50)^3 ~ 1-3s/50\n==>\nw ~ 3s/100 (about three spoonfuls).
461121
measure the cold water temp and take a large amout of cold water than hot water and mix them .\nuse the realation mhot (Thot - T2) = mcold(T2 - Tcold)\nwhere:\nmhot = amount of hot water , Thot = hot water temp\nmcold = amount of cold water , Tcold = cold water temp.\nT2 = the temp. of water after mix.\n(considering the specific heat of water constant at this range of temp.)
156041
Not trying to get some help with a chemistry homework problem, are you?\n\n\nOkay, my guess is that the molecules of one of the liquids fit into the "gaps" between the molecules of the other.  Kinda like mixing a quart of pebbles with a quart of bigger rocks.  Since some of the pebbles with fit between the bigger rocks, the combined volume will be noticeably less than two quarts.
474182
OK, I think I have this one pegged:\nThe original mixture is 10% acid and 90% water and the final mixture is 60% acid and 40% water after pouring some amount of the original mixture into 5 liters of pure 100% acid. Set up two equations:\nfor the acid: x (10) + 5 (100) = y (60) where x = original mixture volume and y = volume of resulting mixture.\nfor the water: x (90) + 0 (100) = y (40)\nthen these can be reduced to:\n10x + 500 = 60 y     and  90x = 40y\nsolve the simultaneous equations to find x and y\nThe volume of water in the final mix is equal to y times .40\nGood luck
432415
54mls\n\n18mls per 100ml of liquid x 3