a gas with a volume of 525ml at a temp. of -25C is heated to 175C. what is the new volume of the gas?
927436
P1T1/V1 = P2T2/V2\nIn your case there is (apparently) no change in pressure, so the above equation will simplify to:\n\nT1/V1 = T2/V2\n\nV2 = T2V1/T1.\nPlug the numbers you have and find V2.  Remember to convert temperatures to Kelvin.
874229
Heat released = 480*2.34 J\n\nSo 0.02 mol releases this much \n\n1 mol will release 480*2.34/0.02 = 56.16 kJ
754271
2H2 + O2 --> 2H2O\nPV = nRT\nP = 2 + 3 = 5\nT = 25 + 273 = 298\n5.00 atm x 55.0 L = n x .08206 x 298\nn = 11.25 mol H2 and O2 mixture\n11.25 mol x (2 mol H2O / 3 mol) = 7.50 mol H2O\n\nP x 55.0 = 7.50 x .08206 x 298\nP = 3.33 atm
963278
1) Determine how many moles of CH4 we have:\n175 g/ (12.01+4*1.008) =10.9 mol\n\n2) Write up to reaction to determine the amount of O2 in moles:\nCH4 + 2 O2 ->  CO2 + 2 H2O\nWe need 2*10.9 = 21.8 mol of O2\n\n3) The volume of 1 mol gas is 22.4 liter at STP. At the given conditions:\nV(1mol) = 22.4 * (290/273) = 23.79 liter\n\nTherefore 21.8 mol = 519 liter = 0.519 m^3
739100
I can't answer the question, but you need Boyle's Law, the volume of a gas is inversely proportional to it's pressure.  Also, are you assuming there is no temperature change?
136369
reverse reaction one(and change the sign of delta H), add the two reactions together, cancel out on both sides. then add the delta H for both reactions, should get equals 54.08kJ
635000
The easiest way to work this out is from the Ideal Gas Law, PV=nRT. Assuming that they start at the same temperature, this is even easier because both R and T will be constants.\n\nLets make the first gas 1, the second gas 2 and the final mix f so that :\n\nPressure Gas 1 =P1, Volume Gas 1 = V1, Temperature Gas 1 = T1\nPressure Gas 2 =P2, Volume Gas 2 = V2, Temperature Gas 2 = T2\nPressure final mix =Pf, Volume final mix = Vf, Temperature final mix = Tf\n\nthen :  P1V1/RT1+P2V2/RT2=PfVf/RTf\n\ntake out the constant R\n\nP1V1/T1+P2V2/T2=PfVf/Tf\n\nIf T is different for each mix\n\nPf=(Tf*(P1V1/T1+P2V2/T2))/Vf\n\nif T is constant (which I assume from your question)\n\nP1V1+P2V2=PfVf\n\nPf=(P1V1+P2V2)/Vf\n\nOkay?
856026
Use the gas law\nPV=nRT
595268
Use the formula: \n\nPV(1)=PV(2) where P is pressure and v is volume.\n\n400(1)=(2000)x(P)\n\nSolving, P =.2atm\n\nPS: If you ever forget the formula, use common sense. When you place any gas under pressure, what happens to its volume? It goes down, because the pressure is forcing the gas into a smaller volume. So when you increase your gas volume by 5 (400x5=2000) you must lower that pressure by the same amount. So the pressure goes down by 5 (1/5=.2).
869561
C57H110O6(l)+109O2(g)--heat>\n57CO2(g)+55H2O(l)\n\n\nabout  15.05kg of  CO2
1064976
Q (heat energy) = m(mass in gram)*c(specific heat in J/gm/K)*t (change in temp)\nQ = 25 * 1 * (36-24)\nQ = 300 J
195778
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
676662
What did you use to measure the activation energy?\nYou probably used a thermometer which was probably marked to 0.1 degrees. That means that your measurements are probably accurate to 0.05 degrees. Reactant temp may be measured as 25.1 (+/- 0.05) so this means that the temp could be 25.05 or 25.15.\nDo the same for the transition state.\nDraw small vertical bars through each point, e.g for reactants the bar would go from 25.05 to 25.15, with the cross in the middle.\n\nI can't remember the formula, but errors are a lot more pronounced on small measurements. It may be something like the difference between the highest and lowest possible values divided by the recorded value, all multiplied by 100.
1160713
Convert temperatures to absolute; add 273 and then use Charles'law.0.55atm /173 = x /473 so x = 0.55atm times 473/173 = 1.5 atm.   That's the new pressure.
129942
the ideal gas law is:\n \n PV = nRT \n\nSO,\n\n P= .95 atm\n n=  .87\n T=  280K\n R=  .0821\n V=  ?\n\n solve for V\n\n     V = nRT/P =(.87* .0821* 280) /.95 = 21 liters
707517
The 4g of methane is a quarter of a mole. ( 1 mole of methane is 16g). A mole of ANY gas occupies 22.4 litres at normal temperature and pressure. So a quarter of a mole has a volume of 5.6 litres.\nThe equation for the reaction is:\nCH4 +  2O2  =  CO2 +  2H2O\nThis shows that twice as much oxygen is needed as methane.\nSo to burn 5.6 litres of methane you need 11.2 litres of oxygen.\nThis means that the limiting reagent is oxygen as there isn't enough.\nIf 5dm3 of CO2 is made then more oxygen must have been added as you would only expect to get the half the volume of CO2 than the volume of oxygen you started with. In your case you started with 7 dm3 of oxygen and so the maximum amount of CO2 is 3.5 litres.\nAssuming the 4g of methane were burned in EXCESS oxygen you would expect to get 5.6 litres of CO2 and in this case the % yeild is  5 divided by 5.6 multiplied by 100 = 89.3%\n\nAt room temperature the figures would be slightly different. i.e 6 dm3 of CO2 would be expected and the yeild would be 83.33%\n\nSo you can see that it is important to say at what temperatures the gases were measured as they tend to expand when they are warmed.
1167777
D.  Based on the ideal gas law, PV=nRT\n\nSolve for R to get: R = (P*V)/(n*T)\n\nfor two different conditions, then, you would have:\n\n(P1*V1)/(n1*T1) = R = (P2*V2)/(n2*T2)  or,\n\n(P1*V1)/(n1*T1) = (P2*V2)/(n2*T2)\n\nor P1*V1*n2*T2 = P2*V2*n1*T1\n\nNote how V1 and T1 can never be in the numerator together?  \n\nIn order for the ratio to work, one must be the numerator, the other the denominator, because volume is directly proportional to temperature -- the higher the temperature (keeping number of moles and pressure equal), the larger the volume.