A milkman has two cans of capacities 10L and 20L with milk of concentrations 60% and 70% respectively.  If he takes 3L and 2L out from different cans,not necessarily  in that order of the cans,whats the maximum milk concentration of the mixture taken out ?\n\na)60%\nb)62%\nc)66%\nd)68%\n\nQuestion:\n---------\n\nquestion says ,"whats the maximum milk concentration of the mixture taken out.."\n\nwho made this mixture !!!!!??\n\nin the problem we had  " If he takes 3L and 2L out from different cans...."\n\nso, he only took out . problem did not say that they have been mixed up!\n\nor is it implicit ?
217930
Yes he took 2L from 1 can and 3L from the other and mixed them together to make a 5L mixture. The mixture will have a higher concentration if he takes 3L from the 70% can and 2L from the 60% can. This will give the maximum concentrated mixtue.\nIt will be (2x.6 + 3x.7)/5 x 100% = (1.2 + 2.1)/5 x 100% =\n3.3/5 x100% = 66%
196258
A blonde heard that taking a bath in milk makes you look beautiful. So she left a note to her milkman telling him to leave her 15 gallons of milk. When the milkman received the note, he thought that there must me some mistake, so he knocked on the blondes door to clarify the order.\n\n"Excuse me, miss, but...is this request written correctly?"he asked.\n\n"Yes it is," replied the blonde."I'm going to fill up my bath tub with it.\n\n"Would you like that pasteurized?"inquired the milkman.\n\n"No,"said the blonde. "Just up to my waist."
345637
FW of NaCl is 58.44  then\n\n   58.44 gm in 1000 mL = 1 Molar\n\n   4.78  gm in 200 mL = ?\n\ndo cross multiplication\n\n(4.78 gm x 1000 mL x 1 M) / (58.44 gm x 200 mL) = 0.4089 M\n\n(gm/gm and mL/mL cacels out so you get answer in M)\n\n\nNow you are taking 28 mL of 0.4089 M solution and diluting it to make final volume of 100.  Use M1V1 = M2V2\n\n0.4089 M x 28 mL = ? x 100 mL\n\n\nThen  ? = (0.4089 M x 28 mL) / 100 mL    { mL/mL cancels out and you get answer in Molarity)\n\n\n? = 0.1145 Molar is your answer
402080
you want me to do your homework for you? I think not.\n\nI worked for about 10 years in a clean room and did this very stuff all day long. So perhaps I can provide some hints.\n\nI will tell you how to do the vancomycin question, and the first Pen G question in hopes that it well help you figure out how to do things.\n\nIf the final volume is 20 ml and you are adding 13.8 ml water then the volume of the powder is 20 minus 13.8 ml. The answer is 6.2 ml. I'm not sure where they are getting their vancomycin from, but they got some huge purity issues. Commercially available vancomycin is about 1 gm per ml.\n\n\nOK you have 5 million units diluted to 250,000 units per ml. \n\n5,000,000/250,000 = 20 ml subtract out 18.2 ml equals 1.8 ml powder volume. Oops, that's the answer to all four question\n\nNow let's see if you can answer what the volume for 2 million units would be.
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
459492
Maybe the milk carton was emptied during breakfast.  But how much was drunk, and how much remains?\n\nIf the carton accelerated at 50 m/s/s then (adjusting for gravity), the 20N force caused an acceleration of 59.8 m/s/s\n\n(20N)/x = 59.8m/s/s where x is the total mass of the milk carton, and the remaining milk, after everybody had breakfast.\n\n20/x = 59.8\n20 = 59.8x\nx = 0.334 kg\n\nso the carton now has a mass of 334 grams.  Evidently the family drank 1.66kg of milk for breakfast.
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
99457
If you get the milk for free, why bother with the whole cow?\n\nYou're obviously not too discriminating...what's your number?  I'm looking to finish myself and I won't date you either.
43684
YOur premise for the first set is wrong.\n\nif c + p + s = 100 \nthen 10c + 3p + .5s  cannot = 100, but rather it would be \nXc + Yp + Zs would equal 100 where X is 10, Y is 3 and Z is .5\n\nBUT, according to your work problem, it couldn't be 10c plus anything since cows are $10 and 10 would eat up the whole $100.
632165
The equations are wrong as written.\n12.0Mx V1 = 0.600Mx0.300L   is correct , solve for V1\n\n15%xV1 = 0.10%x0.0200  is correct.,  solve for V1 \n\nNotice that when you multiply concentration times volume the result is an amount.  The units of the amount depend on the units you use for concentration and volume.  In the first case the amount is moles.\n\nSo the equation says  moles1 = moles2  \nThe moles on the left side must equal the moles on the right.\nOtherwise the equation would violate the principle of conservation of matter.\n\nIn the second example, % is strictly speaking not a unit but a label.  The label, in this case, tells you how the number was arrived at.  That is to say, to get % you take a dimensionless fraction (dimensionless because units in the numerator and denominator cancel one another) and multiply it by 100.  Therefore all percents and parts per labels denote dimensionless numbers.  In the second example then, the units on the left and right side of the equation are both Liters.  The amounts on the left and right sides, therefore,  are specified as liters. \n\nYour problem seems to have more to do with algebra than chemistry.  Also, when working numerical problems like these always analyse the units involved.  All the science is in the units.  The numbers are meaningless without units or labels.  Therefore study all equations from this viewpoint and you will soon begin to have a much deeper understanding.
245994
This is an organic chemistry question?  This is general physical science, and easy general science at that!!\n\nYou have 30 drums (I assume they're all full) of identical volume 1.62E6 liters, for a total volume of:\n\n30 * 1.62E6 liters = 4.86E7 liters * 1000mL/L = 4.86E10 mL\n\nbut one mL = one cm3, so you have 4.86E10 cm3\n\n2.25E8 kg = 2.25E11 grams, so\n\n2.25E11 grams/4.86E10 cm3 = 4.63 gram/cm3
407738
For your first, |x - 3| > 10\n\nFor your second, if two cans of soda cost $1.40, how much would 7 cans cost?
529024
I'm not going to solve these for you.  My homework days are far behind me.  I will try to teach you how to solve them, though.\n\n1.  Keep your eye on the moles and don't get confused by the changing volumes.  The final solution is 0.00436 M  and you know the final volume so calculate how much KNO3 is there (=0.00436 M x Vf, in liters).  That will tell you how much was in 25 mL of the second solution.\nSo how much was in the total second solution?  Multiply the moles above by five (25 mL goes into 125 mL five times.)  This is also how many moles was in original 20 mL solution.\nSo, to find the molarity of the original solution, divide the number of moles by the volume, in liters (20 mL = 0.02 L).\n\n2.  Same thing goes for this one.  Multiply 1.66 M by .037 L (remember that M is in terms of liters, not millilieters!) to get the moles in the first solution.  Do the same for the second.  Add the two volumes together to get the resulting volume (in liters!!!)  Divide the sum of the moles in both solutions by the total combined volume and you have your answer.\n\n3.  Same would go for this one if you hadn't left out what volume of solution the Pb++ was in.  Anyway, the way to do this one is to notice that the reaction of Pb+2 ion and SO4-2 ion is a one-to-one proposition.  So, if you figure out how many moles of Na2SO4 you have in 0.009 grams of the stuff, you will also calculate how many moles of Pb(NO3)2 you have.  Divide that by the volume in liters and you have your answer.  [Nice touch by Teach to throw in univalent species to cross you all up.  These are irrelevant to the reaction you're considering.  Still, I like that.  Makes you think.]\n\nGood luck.
247376
I don't know if this is right ... but check it and tell me\n\nV*C KOH = V*C HBr\n\nV KOH = 18.6ml C KOH = 0.752M\nV HBr = 22.1ml C HBr = ?\n\n18.6*0.752 = 22.1*C HBr\n\nC HBr = 0.6329M\nIts seems reasonable cause the difference between concentrations isn't significant
593255
Let x = product 1, y = product 2, and z = product 3\nP = 50x + 20y + 25z\n9x + 3y + 5z <= 500\n5x + 4y + 0z <= 350\n3x + 0y + 2z <= 150\n"SALES POTENTIAL FOR PRODUCT 1,2 EXEEDS THE MAX PRODUCTION AND FOR3. 20PERWEEK"\nI'm not quite understanding this sentence.\nWhat is the max production?\n\n"AND FOR3. 20PERWEEK" I interpret that as z >= 20
656073
enantiomeric excess (%) = \n(specific rotation of the mixture/ specific rotation of the pure enantiomer in excess)*100 \n\nor \n\nenantiomeric excess (%) = \n( | [enantiomer 1] - [enantiomer 1] | ) /([enantiomer 1] + [enantiomer 1] *100 \n\nwhere the square brackets indicate concentrations, and the | -- | in the numerator means "absolute value".
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%.