Please help me with this question in chemical equilibrium?
958813
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.
933519
Part A:  equilibruim to the right because more co3 added\n\nPart B:  Equilibrum to left becausemore co2 added\n\nPart C:  Equilibrium to right to replace removed h20\n\nEquilibruim dynamic because although appears stable ions constantly moving back and forth just at the same time
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
857176
S(s) + O2(g)------>SO2\n\n2Mg(s) + O2(g) -------> 2MgO(s)\n\nH2(g) + Br2 (g) -------> 2HBr(g)\n\n4Fe(s) + 3O2 -------> 2Fe2O3 (s)\n\nThen there are single replacement \n\nA + BC-------> AC + B\nZn(s) + 2HCl(aq)------->ZnCl2 (aq) + H2 (g)\nwhere one element replaces another\n\nthen there is double replacement \n\nAB+CD -------> AD + BC\n\nIn this one the positive ions in the reacting compounds switch places\n\nBaCl2 (aq) + Na2So4 (aq) --------.BaSo4 (s) + 2NaCl (aq)
808352
The equilibrium equation is K = (pNO)^2/(pN2*pO2) = 5.0 e-5\nThe conditions for the reaction are:\ninitial: pN2 = 0.79 atm, pO2 = 0.24 atm, pNO2 = 0 atm\nchange: pN2 = -x, pO2 = -x, pNO2 = +2x\nequilibrium: pN2 = 0.79 - x, pO2 = 0.24 -x, pNO2 = 2x\n\nSo, using the equilibrium conditions in the equilibrium equation, you get:\n\nK = (2x)^2 / ((.79-x)*(.24-x))\nK = 4*x^2 / (0.1892 - 1.03x + x^2) = 5.0 E-5\nsolving for x (I'll admit, I used the solver on my calculator)\nx = 0.0015314 or -.0015443\n\nSince the pressure of NO can't be negative, x = 0.0015314 atm\n\nThe final pressures are:\nN2: 0.788 atm\nO2: 0.239 atm\nNO: 0.003 atm
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
832296
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
677826
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
288770
The Equilibrium Law\nIf the rates of the forward and backward reactions are equal at equilibrium, then it means that the concentrations of the reactants and products does not change.  The equilibrium constant Kc, is given by the product of the concentrations of the products (each raised to the power equal to that as the number of moles in the equation), divided by the product of the concentrations of the reactants (each raised to the power equal to that as the number of moles in the equation).  \n\nEg N2 (g)    +    3H2 (g)    2NH3 (g) \n\nKc =          The square brackets mean concentration in moldm-3\n\nor Kp =  The p is the partial pressure in atmospheres-1\n\nNB For equations involving a solids, the solid component is not included in the Kc expression.\n\nThis reaction is an example of a homogeneous equilibrium - all the reactants and products are the same state.\n\nKc is a constant at a given temperature and pressure.  If either of these factors are changed then the Kc value will also change.\n\n[click here for some worked examples]\ni am not in chemistry yet
136146
The 1st answer is wrong. The reaction is\n\nNa2CO3(aq) + SrCl2 (aq)--> 2NaCl (aq) +SrCO3 (s)
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
728397
You need to first balance the equation.\n\n2CO + O2 --> 2CO2\n\nUsing this you find the answer to be D.
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
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.