chem.103 if you have a 3gals. and a 5gal. jug thats full, how can you get 4gals out of them?
149069
since water seeks its own level, just connect a tube at the bottom of two 5 empty gal jugs so that water can flow in one another then plug the tube opening with cotton . fill one with 5gal of water and the other with 3gal of water.Now remove the plugs GRAVITY will do the rest dividing the two volume of water into   two eqaul 4gal jug of water . No water is wasted.
354060
0.2 gal. or 25.6 fluid oz.
270274
ur equation is wrong....\nu will get H2O + C...\ncheck again\nif it's correct, u have to find out which one is the limiting reagent by calculating the number of moles of each reactant used...thus figuring out the number of moles of product and multiplying with the molecular mass of methanol...
778647
By fractional distillation.\nDEG distills at 93 'C, you can collect first and then collect water at 100 'C.
342141
ok well, the equation is balanced so you need equal moles of Ag and Cl to make the products. They give you the concentration and volume of Ag, so you can find the amount of moles by multiplying 0.104mol/L * 0.0272L = 0.0028288 mols of Ag. so 0.0028288 mols of Cl were there originally. and you have the volume of water, so you divide moles by volume and you get the concentration. 0.0028288 mols/0.025L = 0.113152 M. \n\nanother way to do it is the easy way, use c1v1 = c2v2. you have c1 = 0.104M, v1 = 27.2mL, v2 = 25mL, so you can find c2. you dont have to worry about units in this case so its easier. \n\nhope this helped. let me know if u need more help. laterz
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps
798989
1) Calculate the mass of acid in 1 liter:\nIn one liter we have 0.01 mol = 1.321 gram\n\n2) Calculate the dilution factor:\nWe have to dilute by a factor 5/1.321 = 3.785..  \n\n3) Calculate the volume of the 5g/litre solution we have to dilute to obtain 100 ml:\nAs the dilution factor was 3.785, we need 100/3.785 = 26.42 ml of the original solution.
623690
u need about 8.264 gal of jam
109042
13 and a third hours is the number i came up with...\n\nJust plug some fake numbers in, say, for example, the tank that these two pipes are filling is 500 gallans.  then you can figure out how much each pipe put into it, take that five hundred, divide by 8, then multiply by 3, that's how many gallens the smaller pipe put into the tank in 5 hours...   now, take that number and divide by the 5 hours, that's how much that smaller pipe can pour out into the tank in one hour...  to take that 500 gallens and divide by the amount that can be poured in one hour and it'll give you how many hours it'd take to fill at that slower rate...\n\nThis is not a standard way of figuring out this problem, but it seemed like it worked to me, it should have taken more then double the 5 hours, but less then 3 times...   does this help at all??
761139
Fill the 3 and dump into the 5\nFill the 3 again and then fill the 5 with 2 litres\nEmpty 5 and put the last litre in the 3 bucket into the 5\nFill the 3 again and dump into 5 for a total of 4 litres.
872004
First, you need to know 3 grams of WHAT.  If the ? doesn't tell you, then assume it is water and state this in your answer.  Since 1g of H2O=1ml of H2O, 3g=3ml.  Next, convert the ml to English units (ounces or gal), and then convert to teaspoon and tablespoon.
898328
First look at the reaction.  You need 4 moles of BF3 and 3 moles of 3 H2O. or 4/3 = 1.333\n\n5.63 grams of water are 0.313 moles of water.  In order for H2O not to be the limiting reagent you will need at least 0.313 * 1.333 moles = of BF3 or 0.418.  You have more than that (0.496 moles).  So if all your water reacts you will still be left with some BF3.  So water is the limiting factor.\n\nFor the second part, you remember that all water is used (because it is the limiting reagent).  Note from the reaction that for each 3 moles of water, 3 moles of HBF4 are produced (or 1 to 1).\n\nUsing all your water (0.313 moles) will then give you the same moles of HBF4, or 0.313 moles of HBF4.\n\nThat's it.
1047482
First find out the SAFE amount for one gallon of water. \nThen, multiply by 500.
245982
Sorry, can't help. You use different units than I do.\n but concentration(mol per dm3) = no. of moles(mol) divided by volume(dm3)
602334
PV = nRT\n\nV = 3 L\nn = g CO2/MW CO2 + g O2/MW O2\nT = 273 K\nR = 0.082057 L-atm / mol-K\n\nsolve for P
145334
Multiply cubic feet by 7.481 to get gallons.
1003334
Assuming D is diameter, calculate the volume of the tank, subtract the volume of the cylinder and convert the remaing volume to gallons.  There are 128 cubic inches (0.0741 cubic feet) per gallon.\n\nyou do the math, that is obviously the intent of your assignment