A soft drink is made by dissolving CO2 at 3.00 atm in a flavoured sugar solution and sealed in\nan aluminium can at 20oC. What volume of CO2 is released when a 355 mL can is opened to\n1.00 atm at 20oC and all of the CO2 is allowed to escape. (KH= 2.3 x 10-2)
202051
P1 x V1 / T1 = P2 x V2 / T2\n\nYou have all you need.  Calculate it.
336832
1) By definition, one mole of any gas at STP occupies 22.4 liters.  So, if you have 100 liters of CO2 at STP, you have 100/22.4 = 4.464 moles of CO2\n\n2) 28 grams of N2 = one mole of N2\nuse the ideal gas law: PV = nRT, so V = nRT/P\nn = 1 mole, \nT = 180C + 273.16 (to convert to Kelvins) = 453.16\nP = 620 kPa = 620 kPa * (1 atm/101.325 kPa) = 6.11 atm\n\n3) P1*V1/T1 = P2*V2/T2\nor V2 = P1*V1*T2/(P2*T1)\nP1 = 1 atm, P2 = 3.8 atm; T1 = 293K, T2 = 346K, V1 = 80L\nsubstitute those values into the equation to get V2\n\n4) P1*V1 = P2*V2, or (P1*V1)/P2 = V2; T1 = T2\nP1 = 1 atm, V1 = 66L, P2 = 6atm, and substitute into the equation\n\n5) 19 moles of any gas at STP occupies 19moles*22.4L/mole = 425.6 liters
341792
Na2CO3.xH2O + 2HCl --> 2NaCl + CO2 + (x+1)H2O ---(1)\nNa2CO3 + 2HCl --> 2NaCl + CO2 + H2O         --- (2)\n\nMass of Na2CO3.xH2O is 2.835g\nMean titration result (HCl) was 22.03cm3\n\n25.0cm3 of Na2CO3 solution was neutralised by 22.03cm3 of HCl\n\ncalculating no of moles of HCl,\nno of moles\n= (concentration) x (volume)\n= (concentration) x 0.2203 dm3\n= a [this is a variable]\n\nno of moles of Na2CO3.xH2O = half no of moles of HCl (according to the second chemical equation)\nso, no of moles of Na2CO3 = 0.5a\n\nI assume that the 25.0cm3 of Na2CO3 solution that you used for titration was measured from 250cm3 of solution. The latter had been made by dissolving solid Na2CO3.xH2O in water, then adding water until the total volume was 250cm3.\n\nSo since in 25cm3 there are 0.5a moles, in 250cm3 there'd be 10 times the no of moles, which is 5a.\n\nThus, the no of moles of Na2CO3 is 5a.\n\nYou can calculate the mass of Na2CO3:\nmass of Na2CO3\n= (no of moles) x (molar mass)\n= 5a x (2x23 + 12 + 3x16)\n= 5a x 106\n= b grams\n\nYou had 2.835g of hydrated Na2CO3 in the first place, so if you subtract the mass of Na2CO3 from the total mass, the remaining mass is the mass of water.\nMass of water = (2.835 - b) g\n\nSo then, you can find the no of moles of water:\nno of moles of water\n= (mass of water) / (molar mass)\n= (2.835 - b) / (2x1 + 16)\n= (2.835 - b) / 18\n= x\n\nAnd there you have it! :D You just have to substitute the concentration of the acid and everything will fall into place.
21949
Dalton's law states that the total pressure is equal to the sum of the individual partial pressures of the gases in the mixture: therefore if the vapor pressure of water @ 21.5 degrees C is 2.6kPa and the total pressure in the flask is 99.4kPa then your answer should show 99.4-2.6=96.8kPa\n\nHope that helps...
444098
If you are asking what your words exactly are, then speaking in terms of an analytical chemist, you would be able to take only 3 aliquotes of exact 25 mL. After that, there is no doubt that 25 mL liquid is still in flask - there is no way by which you can squeez out the remaining 25 mL COMPLETELY from the flask. This is due to that a small amount , may be 1/2 an mL, would still be sticking to the walls of the flask. \nDont believe in me? Try it yourself !!!
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.
266815
Well, let's see here.  You have a "2-liter" bottle that is really 2.10 liters, with 250 ml of water in it.  You want to add enough Alka Seltzer tablets to generate 4.50 atm at 28°C.  I presume you're thinking of making some kind of soda bottle rocket . . .\n\nThis involves using the Ideal Gas Law:  PV = nRT, and what you need to solve for first is the number of moles (n) of CO2 that you need to generate to make the pressure in the bottle 4.5 atm.\n\nWe'll assume that the bottle is rigid -- the volume won't change -- and that the temperature is constant at 28°C.  So there's three of the four variables in the equation:\n\nn = PV/RT\n\nP = 4.50 atm\nV = 2.1 liters - 0.25 liters (the volume of the water), so \nV = 1.85 liters\nT = 28°C = 291K\nR = 8.20568x10^-2 liter-atm/K-mole\n\nso n = (4.50 atm)*(1.85 liter)/(291K)*(8.20568x10^-2)\nn = 0.35 moles of CO2 are required\n\nNow, you have Alka Seltzer tablets that are 35% sodium bicarbonate -- that is where the CO2 will come from.  Assuming all of your sodium bicarbonate will react (100% return), according to the reaction:\n\nHCO3- + H+  =====> H2O + CO2\n\nyou would need 0.35 moles of HCO3- to generate 0.35 moles of CO2.  All of that would come from NaHCO3:\n\n0.35 moles of NaHCO3 = 0.35 * 84.007 grams/mole = 29.4 grams of sodium bicarbonate\n\nIf one Alka Seltzer tablet has a mass of 3.065 grams, and 35% of it is bicarbonate, then you have 1.07 grams of bicarbonate per tablet.  \n\nSo you need *at least*\n\n29.4 grams/(1.07 grams per tablet) = 27.48 tablets\n\nto generate enough CO2 to get a pressure of 4.5 atm.  Make it 27.5 just to be sure.\n\nThat's a lot of Alka Seltzer!  Stand back, 'thar she blows!"
436322
Let's assume that the flask is upside down in a water bath.\nAt 100 C, it contains some water (volume = x), and a volume of hot gas (VH).  When cooled to 0 C, the volume of the gas decreases, and some water (214 mL) is pulled into the flask. The volume of the cool gas is now VC. \n\nSome relationships to get you started:\n\nVH - x = 300 mL\nVH - VC = 214 mL\n\nCharles' Law\n\nVC / TC  =   VH / TH\n\nor\n\nTH /TC    =    VH / VC
227604
Lol, trying to get people to do your chemistry homework?\nUse PV=NRT \nP = Pressure in ATM (1 ATM in STP)\nV = Volume in liters (convert 65cc to liters)\nN = number of moles (will be constant)\nR = gas constant (0.0821 L.ATM/ Mol.K)\nT = temp in Kelvin (273 + degrees C) (STP = 273 K or 0 C)\n\nwith moles being held constant, and the gas constant being constant, your equation should look like;\n\nPV/T (initial) = PV/T (final). solve for P (final)\n\nIf you can't solve this in an hour, i'll post the answer.\n\n....... i was hoping you'd try to solve this and post your answer.....\n\ni got .9383 ATM as the new pressure. mind, that answer is not in sig fig.
576840
ok.. first off the equilibrium expression is the product of concentration of products over the product  concentration of reactants, with any coefficients taken to the power of the of the coefficient For\n\nA+2B <=> C \n\nthe expression would be [C] /(over) [A] x [B](to the second)\n\n(that's C divided by A times  B-squared)\n\n____________________________________________________\n\nNow, in this particular problems the solutions are given to you. (Btw, in this solution i'm assuming your carbon concentration is carbon dioxide because there is no carbon the equation).\n\nYou should have [H2O] X [CO] /  [H2] X [CO2]\n\nSub in your contrations and you should get: .7864\n\nWARNING: I HAVEN'T DONE THIS IN A LITTLE WHILE... I'M A CHEM MAJOR, BUT I'M NOT GOING TO 100% GUARANTEE THAT
485995
(P1V1)/T1 = (P2V2)/T2\n\nP1 = 1 atm\nV1 = 1 m3\nT1 = 0°C = 273 K\n\nP2 = 1 atm\nV2 = x m3\nT2 = 105°C = 378 K\n\nV2 = T2(P1V1)/(P2T1)\n     = 378[(1)(1)]/[(1)(273)]\n     = 1.38 m3\n\nOriginal density is 0.179 kg/m3\nnew desity (0.179kg/m3)/1.38\n\n=1.30 kg/m3
107966
The aqueous solubility of AgNO3 at 20C is 67.8 (solubility is 100*m2/(m1+m2) where m2 = mass of solute and m1 = mass of solvent).\n\nRearraging the equation gives us m2 = (m1*67.8/100)/(1-67.8/100).  or\nm2 = (m1*.678)/(1-.678) = 210.56g of AgNO3 @ 20C\n\nAt 50C the aqueous solubility of AgNO3 is 79.2 so\nm2 = (m1*.792)/(1-.792) = 380.77g of AgNO3 @ 50C\n\nTo find the amount needed to make the second solution saturated we must subtract answer 1 from answer 2.\n\nWe find that 170.21g of AgNO3 must be added to make the solution saturated at 50C.
225868
Actually, the CO2 is not in liquid form.  The CO2 molecules are trapped in the liquid.  That is why you can actually see the CO2 rise to the top of the soda.\n\nWhen you shake the bottle, you are releasing a large volume of the CO2 molecules.  However, you might wonder why the pressure just stays the same since nothing has been added to the bottle.  What's the difference between the CO2 molecules being in the liquid or in the air space above it?  Quite a bit of difference.\n\nWhen the CO2 is in the liquid, it is not able to expand like  a gas.  It does take up space in the liquid, but the liquid keeps it in a more confined space.  When the CO2 escapes the liquid, it now fills the air space as a gas pushing in all directions.  It is now able to expand like a normal gas which is why you get more pressure.
533710
When you do dilutions you need to use M1V1=M2V2 to figure out the concentration of each solution.  You know how much of the stock solution was put into each, and the volume of stock, and the total volume so use algebra and figure out the concentration of the diluted solution.\n\nYou haven't gotten to the phase where you use Beer's law.  that comes after you test your dilutions and test an unknown.
524469
This is a "system of equations" problem, where you have to solve for two different equations.  Denoting A as the number of liters of antifreeze A and B for antifreeze B, the first equation is...\n\n--- A+B=20\n\nwhich should be obvious.  The second equation (concerning only the alcohol) would then be...\n\n--- 0.18A+0.1B=0.15(A+B)\n\nbecause for every liter of A, there's 0.18 liters of alcohol and for every liter of B, there's 0.1 liters of alcohol and you're looking to set it up such that 0.15 of the total liters is alcohol.  Note that since we already know A+B to be 20 we can simplify the second equation to...\n\n--- 0.18A+0.1B=0.15(20)\n--- 0.18A+0.1B=3\n\nSo all you need to do is to find A and B so that they fit that equation and that they add up to 20.  If A+B=20, then B=20-A, which we can substitute in the latest equation to get...\n\n--- 0.18A + 0.1(20-A)=3    (then just do some algebra)\n--- 0.18A + 2 - 0.1A=3\n--- 2 + 0.08A=3\n--- 0.08A=1\n--- A= 1/0.08 = 12.5\n\nThe mixture should have 12.5 liters of A and 20-12.5, or 7.5 liters of B.  You can then check the answer if you wish...\n\n--- 0.18(12.5) + 0.1(7.5) = 2.25 + 0.75 = 3
66488
At STP, 1 mole of any ideal gas will have a volume of 22.4 Liters.\nUsing this logic, we can deduce that at STP, and equal volumes of (what we will assume are ideal gasses) gasses will each contain the same number of moles.\nIf each flask contains the same number of moles, then they much each contain the same number of molecules.\n\nBy saying that the flasks are at STP, we are given all the information we need to solve this.  STP (Standard temperature and pressure) implies a temperature of 273 K and a pressure of 1 atm.
465989
To solve this question we need to know the specific heat of the resulting KI solution.  To make things easier, we can probably assume that the specific heat of the solution is the same as the specific heat of water (c = 4.184 J/g degree C).\n\nThe energy (Q) released when a given mass, m, cools a given number of degrees C, (delta T), with a known specific heat, c, is:\nQ = mc(delta T)\n\nWe know the mass of the water, we know its specific heat, we know how many degrees its temperature changed, now just plug in and get an answer,\nQ = (60 g)(4.184 J/g degree C)(14 degree C)\nQ = 3514.56 Joules\nQ = 3.51 kJ\n\nBut this amount of heat energy was due to 28.7 grams of KI dissolving, and we want to know the heat given off by 1 mole, so we need to convert 28.7 grams of KI into moles of KI using the molar mass (166 g/mol).\n28.7 g KI / 166 g/mol KI = .173 moles KI\n\nSo if 3.51 kJ of energy was absorbed when .173 moles of KI dissolved, if one mole of KI dissolves, 20.3 kJ would be absorbed....the answer which you were looking for.\n\nBy the way....\nthe "(triangle) H" you are referring to is the change in the enthalpy of the reaction.  The "triangle" is the Greek letter "Delta".  It is more common to see it written out as "delta" or "change in" if one cannot use the actual Greek alphabet in typing, as opposed to "triangle".  Δ