write the net ionic equation for the reaction between solutions of potassium hydroxide and nitric acid?
822075
For any reaction between a strong base (potassium hydroxide) and a strong acid (nitric acid) the end result is the neutralization of the acid and base to produce water.  The net ionic reaction will always be the same and show the reaction of the acid (H+) with the base (OH-) to form water.  Therefore,\n\nH+  +  OH-  --->  H2O
998581
The answer is A\n\nCoBr2 + Pb(NO3)2 ==> PbBr2  + Co(NO3)2\nPb(C2H3O2)2 + 2KBr ===> PbBr2 + 2KC2H3O2\n\nIn both cases, the net ionic equation is:\n\nPb+2  + 2 Br-  ====> PbBr2(s)
506202
Write out the balanced reaction which occurs.\n\nSince both NaOH and H2SO4 are strong electrolytes (strong base, strong acid), we can assume both dissassociate completely and the reaction produced Sodium Sulfate and water,\n\n2NaOH + H2SO4 --> 2H2O + Na2SO4\n\nFind limiting reactant in order to find which  will be left over (either NaOH or H2SO4).\nIn order to do this, you need to know both the concentrations of the NaOH and the H2SO4 in order to find the initial number of moles of each.\n\nOnce you know how many moles of either OH- or H+ are left over, you can then find the pH, or pOH like this,\n\npH = -log ([H+])\npOH = -log ([OH-])\n\npH + pOH = 14\n\n*Remember to take into account that the final volume of solution is the sum of the intial volumes of each othe reactant solutions when calculating pH.\n\nEDIT:\nYou know, it just occured to me, I dont know why I even brought up pH, that was not what the question was asking.  but everything I said still stands in order to calculate the molarity...you just dont have to go the one extra step and calculate the pH by taking the negative log of the molarity concentration of H+ ions.\n\nAlso, for all of you out there who are just given answers with not other explanation, it would be nice if you stated what the concentration was of....salt? NaOH?  H2SO4?\nBecause of this ommision, I would highly inclined to disreguard their answers as numerically incorrect as well.
906651
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)
318573
The reaction between Hydrochloric acid and Sodium Hydroxide is as follows,\n\nHCl + NaOH --> NaCl + H2O\n\nYou can see that there is a 1:1 ration between the number of moles of HCl needed to completely reacted with NaOH.\n\nSince you are given both a volume and concentration of the NaOH solution, you should be easily able to calculate the number of moles of NaOH availible.\n# mole = volume * concentration\n# moles = .1 Liters * .25 moles/L = .025 moles of NaOH\n\nIf you know you have .025 moles of NaOH, you know you need .025 moles of HCl to react it completely.\n\nIf you know that you have a .15 mole/Liter solution of HCl, you can calculate the volume required to get .025 moles lie this,\nLiters = moles / oncentration\nLiters = .025 moles / .15 moles/Liter = .01667 Liters or about 167 mL of solution.\n\n\nNext time do your own homework.
59751
Write the balanced reaction.  From that write the equilibrium constant equation.  Plug in the starting numbers plus/minus the change and solve for x.\n\nClue - if the loss of N2 and O2 are x, the formation of NO will be 2x.
742926
Cu(s) + 2HNO3(aq) = Cu(NO3)2(aq) + H2(g)\n\nCu + 2H+ + 2NO3- = Cu+2 + 2NO3- + H2 (leaving off the s,aq,g)\n\nCu + 2H+ = Cu+2 + H2 (net ionic equation, the NO3-'s just cancel out, they don't change in the reaction).
145111
The reaction formula is\n\nCH2COOH                           \nCH(OH)COOH (aq) + NaHCO3 (aq)----> \nCH2COOH                            \n\nCitric Acid\n\n       CH2COONa\n CH(OH)COONa    (aq)+ H20 (l) + CO2(g)\n       CH2COONa
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
803450
MgCl2(aq) + 2NaOH(aq)  --> Mg(OH)2(s) + 2NaCl(aq)\n\nNet ionic equation:\n\nMg2+(aq) + 2OH-(aq) --> Mg(OH)2(s)
714201
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
225436
Here is one :\n\nAn aqueous solution of hydrazine can be prepared by reacting ammonia with a solution of sodium hypochorite:\n\nthe net ionic equation:\n\n2NH3(aq) + OCl-(aq) ---> N2H4(aq) +  H2O(l) + Cl-(aq)\n\n""" CAUTION""""\n\nMixing these chemicals together is very dangerous. Please on paper  analysis only .
725081
The answer is Cu + Cu(NO3)2 -> Cu + Cu(NO3)2\n\nThere is no reaction.
1049588
I think that you are trying to find the net ionic equation. \nWhich in this case would be \nAg+ (aq) + Br- arrow AgBr (solid)\nif you are trying to find a single replacement reaction?\nthen you need to put Br- instead of NO3- and cancel out the NO3 - on the product side? \nTried my Best.
727458
2. 2 Na + 2 H2O --> H2 + 2 NaOH\n3. 2 HNO3 + Ca(OH)2 --> Ca(NO3)2 + H2O\n4. C5H12 +  8 O2 --> 6 H2O + 5 CO2
866210
2NO+O2--->2NO2\n\nCaO+H2O--->Ca(OH)2\n\nCaCO3--->CaO+CO2
734957
Na+ Cl-\nC +4 O-2 each\nK+ Br-\nN-3 H+1 each\nMg+2 O-2