For the reactions shown, express Kc in terms of Kp, R and T.?
444216
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.
948592
Kc=2 = conc products / conc reactants\n\nKc = 2 = [CO][H2O]/[CO2][H2]\n\nStarting concs\n[CO] and [H2O] = 0\n\n[CO2] and [H2] = 0.7 moles/10L = 0.07M\n\nEquilibrium concs\n\nFrom the balanced equation, if x CO2 reacts, it reacts with x H2 to form x H2O and x CO\n\n[CO] and [H2O] = x\n\n[CO2] and [H2] = 0.7 moles/10L = 0.07M-x\n\n\nKc=2=(x)(x) / (0.07-x)(0.07-x)\n\nsolve for x and plug back into the final concentration equations.
827688
a) 2C2H6 + 7O2  --> 4CO2 + 6H2O\n\nb) C6H12 + 9O2  --> 6CO2 + 6H2O\n\nc) C5H12 + 8O2  --> 5CO2 + 6H2O\n\nYou must apply Markovnikov rule for the next addition reactions:\n\na) CH2=CHCH2CH3 + HCl  --> CH3CH(Cl)CH2CH3\n\nb) CH2=C(CH3)CH3 + HBr  --> CH3-CBr(CH3)CH3
731401
Correct.\nD. when delta G is positive the reaction is non-spontaneous
919460
its actually pretty simple. U multiply by -1 when u flip over an equation...so the reactants become the products and the products the reactants. that way instead of having a negative number u will have a positive number for the kJ. as in this problem. U have to flip #1 so u can cancel a few things with #2. so it becomes..\n1)3 H2O(l) + 2 CO2(g) AH---->CH3CH2OH(l) + 3 O2(g) (multiply by -1 gives u a postive +1367kJ. part two stays the same.\n2) C2H4(g) + 3 O2 ---> 2 H2O(l) + 2 CO2 (g) AH= -1411 kJ\n\nNow the 3 O2 from #1 and #2 cancel since they are on opposite sides. Now the 2 H20 cancel from #2 and you are left with only one in #1 since they are also on opposite sides...which after canceling..leaves u with your original equation.\n\nC2H4(g) + H2O(l) ------> CH3CH2OH(l)\n\nnow since u flipped over #1.....and u have a positive +1367kJ and\n#2 is left the same...its still at negative -1411kJ. u add them together and u get an enthalpy change of -44kJ. sorry its pretty confusing to explain..if anything message me and i willl try to help.
919726
Sodium and Potassium are correct.  They are extremely reactive with water, forming Hydroxides and releasing Hydrogen gas, which is usually ignited from the exothermic heat of the reaction.\n\n2 Na + 2 H2O -> 2 NaOH + H2\n\n2 K + 2H2O -> 2 KOH + H2
979402
reactant(s)\nSynthetic
706351
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
842680
2 KBrO3 + 3 N2H4 -> 2 KBr + 3 N2 + 6 H2O\n\nYou can check by finding out the total number of atoms of both sides:\n 2 Potassium\n 2 Bromine\n 6 Oxygen\n 6 Nitrogen\n12 Hydrogen
945429
Equation becomes CuCl2 + 2KI --> CuI2 + 2KCI\n\nProducts are Cuprous Iodide and Potassium Chloride
748741
2K + I2 -> 2KI\n\nBa + S -> BaS\n\nMgBr2 + MnI6 -> No Reaction\n\nC7H16 + H2O -> No Reaction\n\n2KI + Br2 -> 2KBr + I2
694411
IAnd for your other request, I would request you to send a seperate mail to me through yahoo Answer so that I am able to send it part by part.These are long subjects and would take some time,The balanced equation would be                                           H3PO4+3KOH=K3PO4+3H2O
968320
here, you just use the ideal gas law, PV=nRT.  since molarity is equal to moles of solute per liters of solution, M=n/V, so P=MRT.\ntherefore,\nKp=Pco*Ph2o/(Ph2*Pco2)\n  =[CO]RT*[H2O]RT/ ([H2]RT*[CO2]RT)\nall the RT coefficients end up canceling out, so:\nKp=[CO][H2O]/[H2][CO2]\n  =Kc\nsince there were equal numbers of gas molecules on each side of the equation, Kp was equal to Kc.  as a general rule, Kp=Kc*(RT)^N, where N is the change in the number of gas molecules.
192443
Write down the equation of reaction: K + Cl2 -> KCl (not balanced)\n\nBalance the equation of reaction: 2K + Cl2 -> 2KCl (balanced)\n\nFind the moles of Potassium: 5 g K / 39.10 g/mol K = 0.128 mol\n\nFind the moles of Chlorine gas: 7 g Cl2 / (2 x 35.45 g/mol) = 0.098 mol\n\nDetermine the stoichiometric ratios: \nCl2 -> 2 KCl\nK ->   KCl   \n\nFind the moles of product: \n\nFor potassium: 0.128 mol K x (1 mol KCl / 1 mol K) = 0.128 mol KCl\n\nFor chlorine:  0.098 mol Cl2 x (2 mol KCl / 1 mol Cl2) = 0.196 mol KCl\n\nSo potassium is the limiting reactant. \n\nMass of KCl that can be formed: 0.128 mol K x (1 mol KCl / 1 mol K) x (39.10 g/mol K +  35.45 g/mol Cl) =  9.54 g KCl
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
288242
The reaction should be written as follows,\n2 KClO3(s) --> 2 KCl(s) + 3 O2(g) H = -89.4 kJ\n\nYou either did not include the arrow or it is not displaying properly.\n\nPotassium Chlorate decomposes with heat to form Potassium Chloride and Oxygen gas, no where in this reaction is elemental Potassium involved.\n\nThe standard enthalpy of formation of any pure element in its natural state is zero. Oxygen is a diatomic molecule and thus, Oxygen gas (O2) has a enthalpy of formation of 0 kJ in this reaction, no matter how much of it you have, it is still zero.\n\nSince the Potassium Chloride is the only product left that has any non-zero enthalpy of formation, we know that the over all change in enthalpy is equal to the enthalpy of formation of KCl * the number of moles produced.\nIf the over all change in enthalpy is -89.4 kJ and 2 moles of KCl are produced, then the enthapy of formation of the KCl must be 44.7 kJ/mole less than the enthalpy of formation of the Potassium Chlorate.
482555
AgBr-------->Ag+ + Br-\n(a) Ksp=[Ag][Br] solubility=s=[Ag]=[Br]=Ksp^0.5\ns=(6*10^-7)^0.5\n(b) [Br]=0.01 (because all of kBr was dissuciated)\ns=Ksp/[Br]=(6*10^-7)/0.01 mol/lit