I would like to know if you can accurately measure the pitch of the blub, blub blubs you get when filling a container with liquid!??  If you can measure them, does water create the same pitch as other liquids?? Is the pitch dependent on the size and or shape of the container being filled? So if one bottle was 6cm high and there was another identical bottle but it was 7cm high, would the final blub of the taller bottle be higher?  Help - I need to know!!!!! :-))\n\nI've tried emailing Brainiac but it keeps coming back - any advice on that!??!?
133084
I suspect that blub frequency (I'm referring to a poonk sound) is determined more by air cavity size than characteristics of water, and is affected by the container's maximum dimension. This means the final blub probably wouldn't be consistently different in tall and short (but otherwise identical) bottles since the long dimension of the air cavity has shrunk to a minimum.\nThe reason I believe it's not water vibrations is that water in typical sized containers seems to vibrate in a higher frequency range (example, playing a water-filled stemware glass). This may not be completely relevant, but in some air/liquid situations you perceive more of a boink than a blub or a poonk. That's an upward sweep in frequency which I'd bet indicates a shrinking air cavity.\nAs far as measuring the frequency, I'd suggest first identifying a pitch. You probably can train your ear to find a pitch if you can't already. Then find a musical instrument and try to match the pitch. Assume A natural above middle C is 440 Hz (or you won't be far off if you assume middle C is 256 Hz), and go from there. An octave higher is twice the frequency, and a half-tone is a ratio of 2^(1/12) or about 1.06.
606541
sorry I am with Marianne.\nI could answer if it was in relationship to altering the volume of liquids in gravy. That would be when my spoon can stand up in it.\nnot a science person.
219638
I'm guessing that your question pertains to the series of standard solutions you would prepare using the stock solution and volumetric flasks/pipette.  \n\nYes, you can use the C1V1=C2V2 formula, however there's a more practical and faster method that can be arrived at through common sense.  That is the concept of dilution factor.  If you have a stock solution of .5M iron for instance and you wanted to make a .05M iron solution, just dilute the next solution by a factor of 10%.  One way this can be achieved is by taking 10 mL of this stock solution using a volumetric pipette and adding it to a 100 mL volumetric flask.  To lessen the sampling error, it's better to take a larger sample volume, but usually this becomes tedious since lab equipment can be limited.\n\nSo take the original x molarity stock solution.  And if your final desired concentration is y, then the dilution factor is\n\ny/x \n\nThis percentage, y/x, represents the volume percentage in the final solution, that is multiply y/x by the particular To Deliver volume of the volumetric flask, to determine how much volume of the stock solution you need.......place this solution into the volumetric flask with a volumetric pipette, and then dilute to the mark with the standard solvent.
426855
(Edit: Assume we always pour from the 8 to the 5, from the 5 to the 3, and from the 3 back to the 8. Let k be the number of times 5 quarts is poured from 8 quart jug into the 5 quart jug and k is also the number of times 3 quarts is poured from the 3 quart jug back into the 8 quart jug. Then to end up with 4 quarts in the 8 quart jug we must have:8-k(5-3)=4 so k=2.  In order for there also to be 4 quarts left in the 5 quart jug we must have 5k (i.e. the number of times the 5 quart jug is filled) -3k (the number of times the 3 is emptied into the 8) =4. That is 5k-3k=4. k(5-3)=4. k=2. Since k=2 in both cases, we have a solution.)\n\nIt actually has a fairly simple logical/mathematical solution.\nCall the jugs: J3, J5 and J8\nYou will have to end up with 4 quarts in J8 and 4 quarts in J5\nwith no outside water supply so there can be no spillage.\n\nThere is no sense in filling both J3 and J5 from J8.\nSo fill J5 from J8, and then fill J3 from J5. After this, J5 will have 2 Qs left in it. This could be poured back into J8, but let's see what happens if we complete a J8-J5-J3-J8 cycle (by filling J5 from J8 then J3 from J5 and then pouring J3 back into J8)  In doing so we must end up with 8-5+3 = 6 Qs in J8, and there will be 8-6 = 2 Qs in J5. Now, if you complete another round without first moving the 2 Qs in J5 to J3 then you will get the same result as in the first round(since you'll just be moving 3 Qs around in a circle). However, if the 2 Qs in J5 are poured into J3 before beginning the next round then you must end up with 6-5+3 = 4 Qs in J8 after the 2nd round. And since the round ends by emptying J3, the remaining 4 quarts must be in J5, which is the desired result. Stepwise, it is goes as follows.\n\nFill J5 from J8\nFill J3 from J5\nThere are now 3 Qs in J8, 2Qs in J5 and 3 Qs in J3\nPour J3 into J8 \nPour J5 into J3\nThere are now 6 Qs in J8 and 2Qs in J3\nFill J5 from J8\nFill J3 from J5\nThere is now 1 Q in J8, 4 Qs in J5 and 3Qs in J3\nPour J3 into J8\nThere are now 4 quarts in both J8 and J5\n\n.
160292
The test liquid will sink to the bottom because it is more dense than water. Water has a density of 1.0 g/ml. There fore it can be simply thought of as "lighter."\n\nThere will be two visible separate layers.
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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
493280
I have a serious feeling that you haven't given us all the info.\nIf you know the density of HCl you can get the first one.\nBut we don't have enough info to even begin to tell you the relevance of the NaOH.   More info, or concentrate on your own homework...
274772
Volume of the cylinder= the length of the  cylinder X the cross section area  of the cylinder\n\nTherefore, V of tin = the length of the  tin X π r²\n                    = 8.4 cm X π (7/2) cm²\n                    = 102.9π cm³\n                    = 323.2699 cm³\n\nGiven the volume of soup = 200 litres\n                         = 200 X 1000\n                         = 200000 cm³\n\nThus, the number of tins = 200000 cm³/323.2699 cm³\n                         = 618.678\n\nSince there can't be .678 tin, we round off the number. Therefore the number of tins = 618\n\nU might wonder why 618 and not 619. It is because Bradley surely won't sell a tin that only be filled partially.
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If you are trying to make a 1:80 dilution of some solution, by adding it to 3 L of water,  you need the 3 L of water to be 79 x the solution amount, S.\n\nThis will satisfy the need for S/(S+3) = 1/80, since 3=79S, and S+3=80S.\n\nTherefore, S=3/79, which rounds to 0.037975 L, or 37.975 mL.\n\nIf you are simply trying to combine some amount of solution so that 3 L is 80 times as much as the amount of the solution, then you need 3/80 or 0.0375L or 37.5 mL of the solution and the mixing ratio will be 1/80 or 1:80.  \n\nIt wasn't clear whether you wanted the mixing ratio or the dilution factor (also a ratio), so I've given you both.
676982
I think that's dalton's law.  Whoever came up with it, the pressure is additive.  So convert so everything is in the same units, multiply by the percent, add, and then compare to the answers.  I bet it's more fun than you predict.  Whooo!  Conversion factors!  I don't want to take that kind of enjoyment away, so get cracking!
112078
When in the bottle or can the solution is in equilibrium with the partial pressure of C02 equal to the pressure in the can. When the can is opened the pressure in the can drops (the ffffft or hiss sound you hear) and now the partial pressure of C02 in the soda solution is greater than the atmospheric pressure. This results in C02 coming out of solution, which causes bubbles to form. The bubbles represent a separate phase, which is a heterogeneous mixture rather than the homogeneous mixture prior to opening.\nNote that the CO2 levels does not completely equilibrate upon opening. Once the solution has come nearly to equilibrium (re CO2), the soda is said to have 'gone flat' and is no longer fizzy.\n\nI'm not sure the presence of oxygen has anything to do with this phenomenon.
58722
You all thought way too hard on this one. There is no way to know who has more because the asker did not tell what size the containers were. \nWhoever had the biggest container had more because the numbers he gave are more than 1, you can not fill a container 1 and 2/3 of the way, nor can you fill one 7/4 of the way. Once they get to the top, they are full at 1 each. For example, if you try to pour 1 and 3/4 gallons (which is 7/4) into a one gallon container, you will only have one gallon in the container.\nIf both containers were the same size they both have the same amount.
419467
ooo this will be fun!\n\nok..\nA)\n2.207g of H2O MW. 18.0\nmol = weight of sample (g) / molar weight (g/mol) (n=m/M)\nn = 2.207 / 18\nn = 0.1226 moles of H2O\n\nnow, since there is 2 hydrogens in H2O, then when the molecule is broken down, there is twice as many moles of hydrogen then the overall molecule,\ntherefore,\n\n2x 0.1226 = 0.2452 moles of H.\n\ndetermining number of moles of CO2 is also helpful,\n4.313g, n = 4.313 / (12.01 + 16 +16) = 0.0980moles\napprox, 5(moles H2O) :  4(moles CO2)\n\nB) \nworking the number of moles of this compound will be useful:\nn = 1.816 / 74.122 = 0.0245 moles, which is 1/10 of the moles of H produced. (or 1:5 of the moles produced of water)\n\nif you think about it, the reaction is something like this:\nCH2 + O2 --> CO2 + H2O\nclearly the above doesn't make sense, because 1) it isn't balanced, and 2) the reactant needs a higher molecular wt.. now thinking about it, the following can be deteremined:\nand from the ratio of moles found above:\n       1     : (excess):  4        :    5\nC4H10 + 13/2O2 --> 4CO2 + 5H2O\n\nnow, I amoung most allow fractions in equations, but if you don't, multiply through by 2.\n2C4H10 + 13O2 --> 8CO2 + 10H2O\n\nnow, you want the empirical forumla,\ntherefore,\ndivide by highest common factor,\nC2H5 \n\nthe empircal fomula is just a ratio and doesn't make huge sense, as the formula for an alkane is CnH2n+2\nhope that makes sense :)
334800
The physics is sound, atleast on paper.\n\nbefore you ask the question : how does drink lose its fizz, you should be asking: what is fizz and how is it added to drinks in the first place.\n\nThe process of adding fizz to drinks is called carbonation. Carbon Dioxide (CO2) is bubbled through water/drink under pressure. The gas reacts with water producing a very weak acid called Carbonic acid.\n\nH20 + CO2  --->  H2CO3\n\nCarbonic acid gives the twang or sharp taste to soda pop, beer, sparking mineral water and champagne.\n\nHowever carbonic acid is a very unstable compound and readily decomposes into water and carbon dioxide. The CO2 bubbles out of the liquid, what we observe and call fizz.\n\nWe can prevent the decomposition of carbonic acid, by creating conditions where the compound is stable, hence mantaining the fizz of the drink; viz: \nlow temperature and \nhigh pressure.\n\nRefridgeration helps like some of the other answerers mentioned.\n\nyour device seems to help by re-pressurizing the container, helping the remaining carbonic acid remain stable longer. It should work theortically, but how well it actually does depends on the particular design, on how much pressure it can create and how air-tight it is (it could leak slowly over time and lose the pressure) this would also depend upon how snugly it fits over the neck of bottles of different size. like I said, the design would determine the effectiveness. \n\nyou will only know how good it is from experience :)
660840
It depends on the density of the liquids... think about this... 1 liter of water weights less than 1q liter of quicksilver or one liter of maple syrup... the density changes the weght of the liquid by volume.
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1) Basic\n2) BTB turns yellow in a base\n3) HCI + LiOH -> H2O + LiCI\n\nI think these should be right.
78409
Are you seriously a 4th grader...^\n\nProbably because the liquid may be slightly impure and the only way to get the true boiling point of the liquid is to measure the temperature at the reflux ring.