What effect will adding HNO3 to the solution in part c have?  Write the chem equation to explain answer also?
921558
It will be clearer if you write the reaction as the sum of two reactions.\n\nWhen HN3 is dissolved in water, it reacts with the water to form ammonium ion (NH4+) and hydroxide ion (OH-).  Omitting the reaction NH3(g) = NH3(aq), we have:\n\nNH3 (aq) + H2O (L) = NH4+(aq) + OH-(aq)\n\nThe equilibrium constant for this reaction is:\n\nKb = [NH4+]*[OH-]/[NH3]\n\nThe hydroxide produced in this reaction reacts with the manganese ion to form insoluble manganese hydroxide (Ksp ~ 2*10^-13)\n\nMn++(aq) + 2OH-(aq) = Mn(OH)2(s)\n\nThe equilibrium constant for this reaction is:\n\nK = 1/Ksp = 1/([Mn++]*[OH-]^2) \n\nThe reaction you wrote is the sum of 2 times the first reaction plus the second reacton.  Note that as written, the first reaction produces OH- ions while the second one consumes them.  \n\nIf we add an acid like HNO3 to the solution, we are adding H+ ions.  Adding H+ ions is really like subtracting OH- ions because the concentrations of H+ and OH- are inversely proportional to one another through the water dissociation reaction (H2O = H+ + OH-, [H+]*[OH-] = Kw = 10^-14).\n\nWe can therefore write the two equilbrium constants above in terms of the concentration of H+ instead of OH-:\n\nKb = [NH4+]*[OH-]/[NH3] = Kw*[NH4+]/([NH3]*[H+])\n\nK = 1/([Mn++]*[OH-]^2) = ([H+]^2)/(Kw^2 * [Mn++])\n\nThis means that if we add acid (i.e., HNO3) to the solution, we will drive the first reaction in the forward direction; the concentration NH4+ will increase and the concentration of NH3(aq) will decrease.  Adding H+ will drive the second reaction in the reverse direction; Mn(OH)2(s) will dissolve, and the concentration of Mn++ will increase.
215729
K2IC4H6 + HNO3 ---  K2HI + NO3C4H6
892172
44 A) 2AgO --> 4Ag + O2\n     B)NiCO3 --> NiO + CO2\n     C)NH4NO3 --> N2O + 2H2O\n48)A)2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n     B)3Ag2SO4 + 2AlCl3--> 6AgCl + Al2(SO4)3\n     C)&D) I've no idea, sry\n50)A)C4H8 + 6O2 --> 4CO2 + 4H2O\n     B)2C8H18 + 25O2 --> 16CO2 + 18H2O\n     C)2C3H8O3 + 7O2 --> 6CO2 + 8H2O\n     D)2C3H6O + 5O2 --> 6CO2 + 6H2O\n52)A)2Al + 3H2SO4 --> Al2(SO4)3 + 3H2\n     B)2HCl + Ba(OH)2--> BaCl2 + 2H2O\n     C)Au + HCl--> No Reaction(gold is below hydrogen in the metal reactivity series)\n54)A)2Cl2 + 2NaI --> 2NaCl2 + I2\n     B)2NH3--> N2 + 3H2\n     C)4Na + O2--> 2Na2O\n     D)Mg + H2SO4--> MgSO4 + H2\n     E)MgCl2 + Ca(OH)2--> Mg(OH)2 + CaCl2\n     F)H2 + Cl2--> 2HCl
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
744031
Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2\n\n0.1L * 0.2 mol/L = 0.02 mol HNO3\n\n0.02 mole HNO3 * 1 mol CO2/2 mol HNO3 = 0.01 mol CO2\n\nNow plug 0.01 mol CO2 into PV=nRT, using STP conditions.\n\nPV=nRT\n\nV = nRT/P\n\nV = (0.01)(.0821)(298)/(1)\n\nV = 0.24 L\n\n\n*** The person above me did it WRONG!  He confused sodium carbonate with sodium bicarbonate.  Also, he completely forgot to do the stoichiometric conversions.  He solved for moles of HNO3, and then plugged it into PV=nRT.  You must first convert moles of HNO3 to moles of CO2 with stoichiometric conversions, using the BALANCED equation.
866145
HCl + AgNO3-----> AgCl +HNO3\nPb(OOC-CH3)2  + 2LiCl -------> PbCl2  + 2Li(OOC-CH3)\nNa3PO4 + CrCl3 -------> Cr(PO4) + 3NaCl\nH2SO4 + BaCl2 ----------> BaSO4 + 2HCl\nAl2(SO4)3  + 6NH4OH ------> 2Al(OH)3  + 3(NH4)2SO4\n2AgNO3 + H2S -------> Ag2S  + 2HNO3\nCaCl2   + Pb(NO3)2 -------> Ca(NO3)2  + PbCl2\nCa(NO3)2 + Na2CO3 ------>CaCO3  + 2 NaNO3
940335
In your crucible, you have a mixture of mostly MgO and some Mg3N2.  That combined mass is 0.470 g.\n\nWhen you add water, all you are really adding is oxygen, since the hydrogen leaves as part of the ammonia gas.\n\nBasically, you are adding oxygen (atomic weight 16) and removing nitrogen (atomic weight 14).\n\nBecause of the formula Mg3N2(s) + H2O(l) -> MgO(s) + NH3(g),\nyou are 'taking out' 2 nitrogen atoms for every 3 oxygens you add (because there are 2 nitrogens to every 3 magnesium atoms, and each magnesium atom uses 1 oxygen to make MgO)\n\nSo you are gaining (0.486 - 0.470) = 0.016 g from the reaction of Mg3N2 to MgO\n\nThe mass from adding the oxygen is 16 g/mol * (3) * X\nThe mass from taking out the nitrogen is -14 g/mol * (2) * X\n\nWhere X is the number of moles of Mg3N2\n\nSo, 16 g * 3 * X - 14 g * 2 * X = 0.016 g\n\nSolving for X, X = 0.016 / (48 - 28) = 0.0008 mol\n\nThe molecular weight of the Mg3N2 is 100.9 g/mol, so the mass of Mg3N2 is 100.9 g/mol * 0.0008 mol = 0.0807 g\n\nSo, the mass percent is .0807 / 0.470 = 17.2 % Mg3N2\n\n.............\n\nAn another note, one reason you might be confused is your second reaction:\n\nThere are actually 2 reactions going on:\nMg + 1/2 O2 -> MgO\n3Mg + N2 ->Mg3N2\n\nThe first reaction reacts much faster than the second, meaning that more MgO will be produced than Mg3N2.\n\nHope this helps.
621890
CaCl2+ 2NaHCO3 + 2H2O------> 2NaCl + 2H2CO3 + Ca(OH)2 \n\n\nyou have to add water molecules on left hand side         \n                                                                                  \nwill this help you?
787375
1. AgCl(s) + NaNO3\n\n2. AgOH(s) + NaNO3\n\n3. They would not react as both are nitrates already
720117
5.  K = [NH3][HCl] = (0.066 M)^2 = 4.36 x 10^-3\n\n9.  I'm assuming that equation should be H2(g) + CO2(g) <---> H2O(g) + CO (g)\na.  Add H2O will drive equlibrium to the left\nb.  Removing CO will drive equilibrium to the right\nc.  Adding H2 will drive equilibrium to the right\nd.  Adding something to absorb CO2 will drive equilibrium to the left\n\n6. \na.  Increasing the volume of the reaction vessel will reduce the pressure, favoring the production of more gaseous products, so the equilibrium will go to the right\nb.  Equal number of moles of gas on both sides, so increasing the volume will have no effect.\n\n1.  \na.  Decreasing the volume increases the pressure, favoring the side with fewer moles of gas.  Equilibrium would go to the left.\nb.  Equilibrium would go to the left, as there are fewer moles of gas on that side.
294143
add:\n\nH2S2O8(ag)+ NH3(aq) + ? -->(NH4)2S208(aq)  + ?\n\nAmmonia can be prepared in the laboratory by treating ammonium chloride with sodium hydroxide:\n\nthe net ionic equation to this reaction is:\n\n NH4+(aq) +  OH-(aq) ---> NH3(g) + H2O(l) in system 1\n \nand\n\n NH3(g)   = NH3(aq) in solution in system 2\n\nyou finish it ,if you need more help just e-mail me
550972
CO2 + H2O --> CH2O + O2\n read the attached webpage
954955
HCl_aq + NaOH_aq -> NaCl_aq + H2O_l\n(H+)_aq + (Cl-)_aq + (Na+)_aq + (OH-)_aq -> (Na+)_aq + (Cl-)_aq + H2O_l\n(H+)_aq + (OH-)_aq -> H2O_l\n\nFor all the other equations, the (Cl-), (K+), (NO3-), (Na+) are spectator ions, and the n.i.e is the same:\n(H+)_aq + (OH-)_aq -> H2O_l
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
543314
1.) Q = (nh3)^2/(n2*h2^3) = 0.16\n\nAs Q>Kp, reaction will move backward.\n\n2.) Kc = Kp*(RT)^(1+3-2) = 3.1*10^-4*(8.314*700)^2 = 3.55*10^11\n\n3.)\n\nGiven reaction is 2*(ii)+(i)\n\nSo, Kp = (8.3*10^-3)^2*3.1*10^-4=2.13*10^-8\n\nLet original P of HCN = 1\n\nso, partial pressure of C2N2 and H2 = p (assuming 2p of HCN decomposes at equilibrium)\n\nEqn is\n\np*p/(1-2p)^2 = 4*10^-4\n\nSo, p/(1-2p) = 2*10^-2\n\nSo, p = 0.019 atm.\n\nEnjoy.............
942319
This is what you should be looking at:\n\nHCl(gas) + NH3(gas)  -->  NH4Cl (solid)
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)