it is measured at 23.2 degrees cent. and 741mmHg.what is the percent KCLO3 by mass in the mixture?(KCLO3 IN THE MIXTURE IS UNCHANGED.)
200588
Hmmm, a multi-step problem!  Love those!\n\nFirst, you will need to know how many moles of oxygen are generated.  For that, you will need to use the Ideal Gas Law:\n\nPV = nRT\n\nYou need to solve for n = PV/RT, and since you know the volume, pressure, and temperature, and the value of R is a constant, this is an easy calculation.\n\nThen, from the equation:  \n\nKClO3 ===> KCl + 1.5 O2\n\nyou can work backwards to determine the number of moles of KClO3 you started with (one mole of KClO3 generates one and a half moles of oxygen, thus the number of moles of KClO3 you started with will be 2/3 of the number of moles of oxygen generated).  Knowing how many moles of KClO3 you had, you can then determine the starting mass of the KClO3, then find out what percentage that was of your original starting mixture.\n\nGood Luck!
241242
This sounds like homework, so I'll give you the ideas but not the data.\n\nThe enthalpy of the mixture at standard conditions is the sum of standard heats of formation of the components of the mix. It will be given in kJ/mol (Ar is 0 kJ/mol being an element, the other two you can look up in tables or on internet). \n\nThen you need to adjust for the sensible energy difference compared to standard conditions : you will also need to look up the specific heat of each component (usually expressed in kJ / mol ºK) and multiply the specific heat of each component by the amount and by the difference in temperature. Make sure you have everything in the same units.\n\nRemember if you are working in moles its easier because to a first approximation, one mole of any gas occupies the same volume (if you undrestand how to allow for deviations from the ideal gas laws, you wouldn't be asking this question) and I assume that the percentages are volume percent, not by weight\n\nEnjoy
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
173450
Write the equation for K.  You know K, [NO2}, so solve for the only variable you don't know [N2O4].  By the way, these are gasses so it is not a solubility product.  That involves a solid.
534622
The KC value can be answered by knowing the equilibrium concentrations of NH3,N2 and H2.\nSince 2 mole of NH3 was taken initially and 1 mole of NH3 remains unreacted so moles of NH3 reacted :1.\nSince 1 mole of NH3 has reacted it means that 0.5 mole of N2 and 1.5 moles of Hydrogen has been formed in the reaction at equilibrium.\n\n              we have \n                            kc=[N2][H2]^3/[NH3]^2;\n                            [N2]=0.5;\n                            [H2]=1.5;\n                            [NH3]=1;\n                therefore\n                            kc=(0.5)*(1.5)^3/(1)^2;\n                 which gives Kc=1.6875\n                clarification to your doubt:\n                since the reaction taking place is:\n                    2NH3 <----->N2 + 3H2. \nthis means    NH3 <----->0.5N2 + 1.5H2. (divide the equation by 2 \n to get moles of N2& H2 produced per mole of NH3)\n   As 1 mole of NH3 reacted which means that 0.5 mole of N2 and   1.5 mole of NH3 is formed at equilibrium.Just substitute in Kc and find the  value as 1.6875.I think that answers your doubt.
439030
H2CO3 + 2KOH ===> K2CO3 + H2O\n2HNO3 + Ca(OH)2 ===> Ca(NO3)2 + 2H2O
577275
mass of KOH:\n(.0313L)(0.118M)(56.11g/mol) = 0.21g\n\nYour question probably asked for the concentration of KOH:\n(.0313L)(.118M)/(.0287L) = 0.13M
99282
22.49 L = 2 moles of HNO3 will react with one  mole of K2HPO4\n\nyou only have .0014 mol  of K2HPO4 available\n\nSO, YOU WILL NEED 31.45 mls of HNO3 to react with .0014mol of K2HPO4
426597
Yup, you got it.  1/2 the temperature in K will give an average molecular speed of 1/2 the speed
439314
The MW of KClO3 is 122.55 (ref.), so 0.274 g KClO3 is .0022358 M. This will evolve 3/2 * .0022358 = .0033537 M O2 * 22.4 L/M = .075124 L O2 at STP. Correcting for temperature, pressure and partial pressure H2O, multiply by 295/273 * 760 / (726.2 -19.8) and volume at given temp. and press. = 0.087337 L.
247376
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
534677
From the reaction NH4Cl(s) ===> NH3(g) + HCl(g)\n\nKp = pNH3 pHCl Can ignore the solid as it is at unit activity\n\nIf the total pressure is 4.4 atm then by Daltons Law of partial pressure the NH3 and HCl must contrinute 2.2 atm each.\n\nTherfore\n\nKp = 2.2 x 2.2 = 4.84\n\nThus for your reaction Kp = 4.84
387866
1) The first one is wrong on the last person who answered, O2 weighs 32g/mol because there are 2 O's in it, 16 each.  So in 1.33x10^4 (13,300mol) you have 4.256x10^5g of O2 (425.6kg).\n\n2) Recognize that hydrogen, oxygen, and chlorine are diatomic gasses, so they always occur in pairs (H2, O2, Cl2).  Others like this are nitrogen (N2), fluorine (F2), bromine (liquid @ room temp. Br2) and Iodine (solid at room temp, I2).  So your equations would be:\n\n2H2+O2-->2H2O\nH2+Cl2-->2HCl\n\n3) PV=nRT and solve for n, number of moles.  Put everything in correct units, liters, atmospheres, moles, and kelvins, and R is a constant R=0.0821 L*atm/mol K.  Kelvin temp=Celcius temp +273\n\nSolving for n:\nn=PV/RT=(1.09atm)(0.942L)/(0.0821L*atm/mol*K)(287K)\nn=0.0436 mol\n\nSo, when we usually want to know a number of moles given a mass and having a molar mass we can look up, we say,\nn=mass/molar mass\n\nin this case, we need to find molar mass, so:\nmass/n=molar mass\n3.17g/0.0436mol=72.71g/mol
498915
Firstly, please forgive me for not having a periodic table at hand. :)\n\nSecondly, please forgive yourself as even the most advanced scientists in the world sometimes forget how to do this calculation.\n\nThirdly, it is such a simple calculation, that it is no wonder that it is so easily forgotten.\n\n1.5% solution (mass/volume) means 1.5 g of KBr in 100 ml of solution\n\nIt's that simple. :)\n\nEnjoy your science. :)\n\nSincerely,\n\nCurtis
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.
165397
At equilibrium, we have this below relationship between K and the concentration of all species.\n     \n　　　　　　　　　　　　　[H2]^2 [O2]\n　　　　　　　　　　　K= ----------------\n　　　　　　　　　　　　　　[H2O]^2\n\nSo we could get the answer by solving this equation.
386427
V1/T1 = V2/T2\n\nFirst convert temperature to Kelvin. 198+ 273 = 471\n\nNext plug into equation\n\n4.81 L / 471 K = 2.49 L / T2\nT2 = 243.8\nthen put it back into C 243.8 - 273 = -29.2 C