i have a problem on my hw and am not sure how to do it.\n\nit says to calculate the unknown quantity of\n\nP = 782.4 mm Hg; V = ?; N = 0.1021 mol; T = 26.2 C
135483
PV=nRT\nP=pressure (1 atm = 760 mmHg)\nV=volume\nn=moles\nR=constant that is determined by units of pressure [0.082057 (L atm/mol K)]\nT=Temp (must be in Kelvin)
374975
Well, mv = h/λ\nm = h/(λv)\n= 6.626E-34 / (3.31E-15 x 0.01 x 3E8)\n= 6.673E-26\nConvert this to Da\n= 6.673E-26 / 1.66E-27\n= 40.2\nYou know, this is Ca. (look for it in the periodic table)
56977
Pay more attention in class! And assuming it is an ideal gas, use PV=nRT. Since there is not change in volume you can basically drop it from the equation and make P=nRT. If you like algebra you can leave it in but it will cancel out eventually. You could solve for 'n' first using initial P and T, and then plug and chug.
450619
This may not be the way your teacher taught it, but trust me it works.\n\nYou can use the ideal gas law for each bulb:\nPV = nRT\nWith a constant T, you can cancel out RT and just solve PV = k, where k is a function of n, the number of moles.\n\nP1 * V1 = k1 and P2 * V2 = k2\nk1 = 672 mm Hg * 1.00 L = 672. mm Hg * L\nk2 = 442 mm Hg * 0.500 L = 221 mm Hg * L\n\nYou can add these together: k1 + k2 = k3 = 893 mm Hg * L\n\nThe final volume is 1.5 L\n\nP3 = k3/V3 = 893 mm Hg * L / 1.5 L = 595.3 mm Hg\n\nconvert that to atmospheres, multiply by 0.001316 atm/mm Hg = 0.783 atm
308586
I think this is how you would solve this problem. Use the equation PV=nRT, re-arrange to solve for V. \n\n(STP = standard temperature and pressure = 1 atm of pressure and 273 Kelvins)\n\nV = nRT/P = (1 mol)(0.0821 L*atm/k*mol)(273)/1 atm = 22.4 L
396592
w = -(P x deltaV). Delta V is final V minus inital V (which you get by using the PV=nRT equation at each temperature). so -(3 x -11.8) = +35.39.\n\ndelta E = q + w. So add the q you got and the w you get.\n\ndelta H = delta E + (P delta V). \n\nThere ya go. hope this helps
520031
Given:\nPV = nRT, 862 torr x (1 bar/750torr) = 1.15 bar, 117 C +273 = 390 K, CH4 = 16 g/mol\n\nn/V = P/(RT)\n\nn/V = (1.15bar)/[(0.08315 L bar/K mol)x(390K)\nn/V = 0.035 mol/L x (16 g/mol) x (1L/1000mL)x(1mL/1cm^3)\ng/cm^3 = 5.67 x 10^-4 g/cm^3
8480
There's a formula for volumes of gasses:\n\nPV = nRT.\n\nIf I remember right, P = pressure, V = volume, n = number of moles, T = temperature, and R = molar density, or something molar that is specific to the gas.
474092
These are GAS LAW problems, so you need to use PV=nRT.  However, for these problems, nRT will be constant so you only need to use P(pressure) and V(volume).  Therefore, for both of these questions, use the equation:\nP1V1 + P2V2 = PtVt   where P1 is the pressure in the first vessel, P2 is the pressure in the second vessel, and Pt is the total pressure of the connected system.\n\nProblem #1\nP1V1 + P2V2 = PtVt  \n(7.0atm)(15.0L) + (P2)(8.4L) = (10.0atm)(23.3L)  \nP2 = 15.4 atm\n\nProblem#2\n(7.2atm)(15.0L) + (4.1atm)(8.3L) = (Pt)(23.3L)  \nPt = 6.1 atm
506992
Just let the units tell you what to do.  H=mass* T*sp heat.  with T in K
468279
I dont understand what r u doing with "p*v/t"; but if assume ideal gas behavior, PV = nRT, so P = nRT/V [n is number of moles, R is ideal gas constant, T is temperature, V is volume of the flask, P is the pressure]
382600
V= d/t\n18=5/t\nt=5/18\nt=0.27777 -----> 0.3 h
332890
M1V1=M2V2\nie... 0.24*40.11=M2*0.55\nM2=17.50\nMOLARITY=                                                (MOLECULAR WT/GIVEN WT)*(1000/VOLUMWEOF THE SOLUTION IN MILILITERS)\nIF U CAN SUSBSTITUTE THE VOLUME OF THE SOLUTION THEN U CAN FIND OUT THE MOLECULAR WT\nhope u know how to find out the molar mass fr molecular wt.
504910
See if this helps:\nPV = nRT = NkT, where P is pressure, V is volume, n is the number of moles, N is the total number of molecules, T is temperature, k = 1:38 £ 10¡23 J/K is the Boltzmann constant and R is a universal constant, R = 8:314 J/mol.K.
533582
Use PV = n RT, RT = constant. Therefore n = PV * constant in the three cases.\nWe have \nH2 :  P = 372 torr V = 2 liter  n_H2=  744 * constant\nN2:  P = p torr V = 1 liter  n_N2 = p * constant\nCombined: P = 303 torr V=3 liter  n_combined = 909 * constant\n\nTherefore  744 + p = 909 => p = 165 torr
510144
use the formula\nA=2pir(h+r)\n=2pi*7*21\n=294pi
480181
we have PV = m/M * R * T\nT = t(celsius) + 273\nR = 0.082 = const\nM is the molecular mass\nP is the pressure (1atm = 760 mmHg)\n\nso M = (m*R*T)/(P*V) = 3.6 * 0.082 * (23.4+273)/(4042.4/760 * 0.183 )= 90 g/mol\n\nwith the percentages we can find the compound \n\nthe formula is CxHyOz (x,y,z is the number of atoms)\n\nx : y : z = %C/M(C) : %H/M(H) : %O/M(O) = 40/12 : 6.71/1 : 53.28/16 = 3.33 : 6.71 : 3.33 = x : 2x : x\n\nwe can use this table to find the answer\nx=1 ; M = 30 : CH2O\nx=2 ; M = 60 : C2H4O2\nx=3 ; M = 90 : C3H6O3\nso the formula is C3H6O3