A  fuel tank is half filled with petrol A. Then it is filled with petrol B up to half of the remaining volume of the tank. half of the remaining portion is filled with A. half of the rest  is filled with B. the process is continued till the tank is filled. what is the ratio of volumes of A and B,when the tank is  fully filled up ?\n\nhow do i solve this problem ?
136008
Ratio of Volume of A and B when the tank is filled up:\nV(A):V(B)=2:1\nV(A)= 1/2+1/8+1/32 +...+1/2^(2*n-1); n=1-->infinity\nV(B)= 1/4+1/16+1/64 +...+1/2^(2*n);  n=1-->infinity
443802
PV=nRT  n=PV/RT   V=nRT/P\n\nR= 0.08206 L atm mol^-1 k^-1\n\n922 torr/760 torr= 1.213 atm\n\nn= (1.213 atm x 35.3 L)/(0.08206 Latm/molK x 315K)\n\nn=1.6565 mol Propane\n\nSTP = 1 atm, 273K\n\nV = (1.6565 mol x 0.08206 Latm/molK x 273K)/(1 atm)\nV = 37.1095 L\nV = 37.1 L (sig figs)
469906
V = kT/P, where V = volume, T = temp, P = pressure and k is the constant of proportionality.\nYou are given, T = 250, P = 76.2 and V = 750.\nUsing this information, you can solve for k.\na) 750 = k(250)/76.2\nk = 750(76.2)/250\nk = 228.6\nb) Go back to the equation at the top and replace k with 228.6\nV = 228.6T/P\nc) You are given T = 300 and P = 73.06\nPlug these in to find V.\nV = 228.6(300)/73.06\nV = 938.68
106450
The efficiency of an ideal engine is (1- T2/T1) or (T1- T2)/T1 = (600 - 400) / 600 =0.333.\nBut in the problem it is given that it is 0.5 which is not possible.
91451
According to Charle's Law, the volume and temperature of a gas are related as follows,\n\nV1 / T1 = V2 / T2\n\nSo if you half the initial temperature, the volume of the the gas would need to be halfed aswell in order for the above eqaution to stay true.\n\nWhen the temperature of gas is decreased (all other things remaining constant), its volume decreases aswell.  The same for when a gas's temperature increases, the gas expands.
167332
Not the full 100% it will be mixed or another fuel source will be found.
339499
15a) Ahh, lengths, areas, and volumes. Lengths are proportional to each other, areas are proportionatl to the square of the length, and volumes are proportional to the cube of the length.\n\na) The ratio of the area of the windscreen is ⅜^2\nb) The ratio of the capacities of their boots is ⅜^3\nc) The ratio of the widths is ⅜\nd) I suppose the model car has four wheels, just the original. It's a model, after all.\n\n15b) we are looking for a rectangle with area equal to 16. Let the width be w. Then the length is w + 6.\n\n16 = w(w + 6)\nw^2 + 6w -16 = 0\n(w + 8)(w - 2) = 0\n\nw = 2, and the length is 8. The area is 16.
502884
2.17 moles\n\nthe ratio is 3:1, not 2:1 as the above answerer states
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.
133383
height is one dimensional, volume is 3 dimensional.\n\nLet ratio of height of A and B be r:R\n\n2.5 r = R\n\n250 / R^3 = volume of A / r^3\n\n= 250 * r^3/R^3\n\n= 250 * 2.5^3\n\n= 16
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.
547495
I'll assume that Q12 is actually Q13 since you entered 11 twice in Part 1.\n\n13) The density is 3g/2cm^3, so 20 cm^3 has a mass of 30 g.\n15aa) the model is 9/16th the real thing\n15ab) ?\n15ac) the model is 3/8ths the real thing\n15ad) 1:1, since they both will have 4 wheels.\n15b) A=w(w+6)=16 (the area of a square w/ side 4)\nThe dimensions of the rectangle are length 8 and width 2.\n17a) 1, 3, 6, 10, 15, 21\n17b) 1/2n(n-1)\n17c) n\n18a) 5 pipes would fill the tank in 1 hr 18 min\n18b) 60 km/h
388705
The volume does not change because the gas is contained in the tank.  So (P1*V1)/T1=(P2*V2)/T2.  Solving for P2:\nP2=(P1*V1*T2)/(T1*V2)\nV1=V2 so they cancel\nP2=(P1*T2)/T1\nP2=(2*(273+45))/(273+20)=2.17 atm\n\nps: divers use air not pure oxygen and the tank is charged to over 2000 psi (136 atm)
322308
As the volume is constant,\n(P/T)initial = (P/T) final\n1/(25+273.15) = P/(50+273.15)\nP = 1.084 atm\n\n(Taking the pressure of the initial gas was 1 atm.
526522
well, its all in the stoichiometry.\nsince u have the balanced equation, just do it out!\nuse 1 gall C8H12 * the equavalent in liters (convert 1 gallon to liters) and then u can do the stoichiometry going from liters of C8H12 to grams of oxygen using the molar ratio. \ngood luck! hope this helps!
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
501677
There is about 20% oxygen in the air which is more than enough.\n\nAlthough a pound is 454 grams. you asked about gallons. Gasoline is about 6 pounds per gallon.\n\nMost gasoline is burned in internal combustion engines which do not burn it completely.It forms carbon monoxide.\n\nGasoline is not pure octane, C8H18 \n\n 2C8H18 + 25O2 ---> 16CO2 + 18H2O \n\nNow you can work out the rest, 2 moles of octane to 25 moles of O2. Use themolecular weight per mole of each.Figure out how many moles of octane yoy have.