For the mixing of NH4 NO3 (aq)  and KOH(aq) what is the complete balanced equation?
98695
My answer is,\n \nNH4NO3(ag) + KOH(aq) -> NH4OH(aq) + KNO3(aq)
850392
You have to use the methods for equilibrium and pH of a salt to calculate the concentration of salt that would produce that pH.\nI cannot find my notes from chemistry right now,\nhowever you need to know the definition of Ka(or Kb)\nthen set up the equation as the salt reacts.\n\nnh4+ + h20 <--> nh3 + h30+\n\npH = -log[H30+}, so your [h30+] = 10^-pH\nyou can then calculate the concentration of nh4 using molar ratios and such, and then you will have molarity.
814611
CdSo4+NH4OH ---->Cd(OH)2 + (NH4)2SO4
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
497226
2NH4V03 + H2S04 ---> (NH4)2S04 + 2V02 + H20
959309
#3:   113.6g ammonia is the maxium that can be produced, i think.\n\nBecause:   when you set up the problem you do:   5 moles N2 x 2 moles of NH3 (coming from the balanced equation) x 17.04g NH3 (formula mass) equaling.....170.4g NH3\n\nthen....\n\nyou have to do it again with the other number(10 moles of H2)\n\n10 moles H2 x 2 moles NH3 / 3 moles H2 x 17.04 equaling....113.6.   \n\nyou pick the lowest number becuase that is the maxium you can have.   so your answer is 113.6g NH3.\n\n\n\n#4:   15.33g NO2 is the maxium that can be produced \n\nbecause: 20g of O2 to grams of NO2.  To do that, you do...20g O2 / 32g O2 (Formula mass of O2) x 46.0g NO2 (formula mass of NO2)\nwhich equals......28.76\n\nthen do:\n\n10gNO / 30.0g NO (formula mass of NO) x 46.01(formula mass of NO2) and that equals....15.33\n\nso your answer is the lowest number which is 15.33.\n\n\ni hope that helps and i hope i'm right...
914017
I will suggest that first you write the equation in a compact mode, balance it and then add all the bells and whistles.\n\nCuSO4 +K3PO4 ---> Cu3(PO4)2 + K2SO4\n\n3CuSO4 +2K3PO4 ---> Cu3(PO4)2 + 3K2SO4\n\n3CuSO4(aq) +2K3PO4(s) ---> Cu3(PO4)2(s) + 3K2SO4(aq)\n\nNow that it is balanced it is easier to break down:\n\n3Cu(+2) + 3SO4(-2) + 2K3PO4(s) --->  Cu3(PO4)2(s) + 6K(+) + 3SO4(-2)
979796
2H3PO4+6NaOH-------->6H2O+2Na3PO4\n\nThere has to be two moles of phosphoric acid and six moles of sodium hydroxide for it to balance out
850425
KOH + HNO3 ---> KNO3+ H20\n\nmoles of KOH = 0.025L*0.2mol/L = 0.005 mol\nconcentration of nitric acid solution = 0.005mol/0.030L = 0.167M\nconcentration of KNO3 = 0.005mol/0.055L = 0.091M
1061977
(H+) + NH3 -> NH4+\n It makes an ammonium ion, if you are talking about a hydrogen ion and not H2.
770691
CoCl2(aq) + FeSO4(aq) = CoSO4 (aq) + FeCl2(aq)
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
727004
(NH4)2Cr2O7 -> Cr2O3 + 4*(H2O) + N2\n\n(NH4)2SO4 + 2*(NaOH) -> 2*(NH3) + 2*(H2O) + Na2SO4\n\nFe2(SO4)3 + 6*(NH3) + 6*(H2O) -> 2*(Fe(OH)3) + 3*((NH4)2SO4)
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)
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
708721
Here is the right equation for nitric acid and sodium bicarbonate\n\n\nHNO3 + NaHCO3 ---> NaNO3 + H2O + CO2\n\nIf you do not like the sodium:\n\nHNO3 + HCO3(-) --> NO3(-) + H2O + CO2