For the reactions shown, express Kc in terms of Kp, R and T.\n(a) CO(g) + ½O2(g) CO2(g)\n(b) 2Cl2(g) + 2H2O(g) 4HCl(g) + O2(g)\n(c) C(s) + CO2(g) 2CO(g)\n(d) NH4HS(s) NH3(g) + H2S(g)
199487
Write the Kc and the Kp\n\nChange PV=nRT to P/RT=(n/V) to P/RT=(concentration) note n/V is concentration.\n\n\nSubstitute P/RT for each concentration in the Kc.  You will see the Kp in the equation with extra R's and T's.  Finaih solving.
108468
4 OH-(aq) --->  O2(g) + 2H2O(l) + 4e-
481578
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.
397846
CH4 + 2 O2 ---> CO2 + 2 H2O
78615
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
244640
a) 4 C3H5N3O9 --> 6 N2 + 12 CO2 + 10 H2O(g) + O2(g)\n\nb)start off by using the fraction of 50 grams of C3H5N3O9 over the total weight of C3H5N3O9 which is 227 g C3H5N3O9(I got this from adding 3(12) for the 3 atoms of C, 5(1) for the atom of H, 3(14) for the 3 atoms of N, and 9(16) for the 9 atoms of O)\n\nso for the nitrogen:\n50 g C3H5N3O9/ 227 g C3H5N3O9 x the ratio in the reaction of 6 moles N2 / 4 moles C3H5N3O9 x 28 g N2/ 1 mole N2\nIn grams, you get 9.2511 of N2\nwithout that last fraction of the equation, you can solve for the # of moles, which is .3304 moles of N2\n\nFor the CO2\n50 g C3H5N3O9/ 227 g C3H5N3O9 x the ratio in the reaction of 12 moles CO2 / 4 moles C3H5N3O9 x 44 g CO2/ 1 mole CO2\nIn grams, you get 29.0749.0748 grams of CO2\nwithout that last fraction of the equation, you can solve for the # of moles, which is .6608 moles of CO2\n\nFor the H2O\n50 g C3H5N3O9/ 227 g C3H5N3O9 x the ratio in the reaction of 10 moles H2O / 4 moles C3H5N3O9 x 18 g H2O/ 1 mole H2O\nIn grams, you get 9.9119 grams of H2O\nwithout that last fraction of the equation, you can solve for the # of moles, which is .5506 moles of H2O\n\nFor the O2\n50 g C3H5N3O9/ 227 g C3H5N3O9 x the ratio in the reaction of 1 moles O2 / 4 moles C3H5N3O9 x 32 g O2/ 1 mole O2\nIn grams, you get 1.7621 grams of O2\nwithout that last fraction of the equation, you can solve for the # of moles, which is .0551 moles of O2\n\n(whew!)\n\nC)use the ideal gas law, PV= nRT\nFirst, lets find "n": add the number of moles of N2, CO2, H2O, and O2. this is (.3504 moles of N2 + .6608 moles of CO2 + .5506 moles of H2O + .0551 moles of O2 = 1.6191 moles of gas produced) so n= 1.6169 moles\nR= 0.08206 (L*atm)/(mol*K)\nT= temperature in Kelvin, so add 273 to your temperature in C. (125 + 273 = 398 K), thus T= 398 K\nV= Volume, which you are trying to find\n\nSo, by plugging in these numbers, your equation looks like:\n(1.50 atm)V = 1.6169 moles [0.08206 (L*atm)/(mol*K)] 398 K\n1.5 V = 52.8077\nfinally: V= 35.2051 Liters!\n\nSorry about the editing, my addition in my original answer was off, making the rest wrong as well! Know that the mass of C3H5N3O9 was 227 g not 299 g.
288263
NaHCO3(s) + CH3COOH(aq) → CH3COONa(aq) + CO2(g) + H2O(l) \nThis is the real chemical reaction.
439030
H2CO3 + 2KOH ===> K2CO3 + H2O\n2HNO3 + Ca(OH)2 ===> Ca(NO3)2 + 2H2O
375486
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
172052
2Ca(s) + O2(g) -> 2CaO(s)
353554
NaOH + NaH2PO4 ===> Na2HPO4 + H2O\n\nH2S + 2KOH ===> K2S + 2H2O\n\n(NH4)2SO4 + 2KOH ===> 2NH3 + K2SO4 + 2H2O\n\nSO2 + Ca(OH)2 ===> CaSO3 + H2O\n\nIn the above equation, further oxidation in air could occur by\n\n2CaSO3 + O2 ===> 2CaSO4
394818
CoCl2(aq) + FeSO4(aq) = CoSO4 (aq) + FeCl2(aq)
193857
This is simple-\n3Mg(S) + N2(G) (2 is subscript) Yields (arrow) Mg3N2(S)
494033
(NH4)2CO3+MgSO4->(NH4)2SO4+MgCO3
371445
CH4 + 2O2-> CO2 + 2H2O
466328
6HCl +3 I2 + KClO3 --->6 ICl+ KCl + 3H2O\n\n\nMnO4 - +2 NO2- +4 H+ ---> Mn2+2 NO3- +2 H2O
531671
C3H8 + 5O2 --> 3CO2 + 4H2O