what is the chemical equation involved in the reaction between concentrated nitric acid (HNO3) and glucose?
116913
I think it is this:\n\n            \nC12H22O11(s) --- HNO3---->  12C(s) = 11H2O(g) + heat
630069
NaOH (Base) + HNO3 (Acid) -> NaNO3 (Salt) + H2O (Water)\n   H  H H\n   |  | |\nH-C-C-C-H\n   | | |\n   H H H\n\nReplace the last H by an -O-H\n\nDraw the carbon skeleton first.
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
957244
2NO + O2 --> 2NO2\nMr of NO = 14 + 16 = 30\nMoles of NO = (29.03/30) = 0.9677 (to 4 s.f.)\nMr of O2 = 32\nMoles of O2 = (19.35/32) = 0.6047 (to 4 s.f.)\n\nComparing the moles of NO and O2 given, you should notice that NO is the limiting reagent.\nHence moles of NO2 produced = moles of NO used = 0.9677\n(since the ratio of NO2 to NO is 2:2)\nMass of NO2 produced = 0.9677 x (Mr of NO2) = 0.9677 x 46 = 44.5 g (to 3 s.f.)\n\nHope this helps :)
550253
This time we use oxygen to burn the sugar:\n\nSugar + Oxygen = Carbon Dioxide + Water + Energy\n\nUsing glucose:\nC_6H_12O_6 + 6 O_2 -> 6 CO_2 + 6H_2O + Energy
888667
Cu2S + 12HNO3 -> Cu(NO3)2 + CuSO4 + 10NO2 + 6H2O\n\nThere you go.
921837
NaHCO3 --> Na+ + HCO3-\nCa(NO3)2 --> Ca(2+) + 2NO3-\nAl2(CO3)3 --> 2Al(+3) + 3CO3 (-2)
326922
Another easy reaction:\n\nMg(s) + HNO3 ===========> Mg(NO3)2 + H2 (g)\n\nMagnesium reacts rapidly with nitric acid to make magnesium nitrate and hydrogen gas
1002183
Concentrated Nitric Acid and copper.  Be careful.  Nasty stuff... really!
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
964051
First you must know the equation of the reaction, so you can establish the mole connections between the involved substances\n\n2HNO3(aq) + Mg(OH)2(aq) → Mg(NO3)2(aq) + 2H2O(l)\n\nNow, you have 60,0 mL of 0,100 M Mg(OH)2(aq)\nTherefore:\n0,100M*(60,0 mL)(1L/1000 mL)= 6,00*10^-3 moles of pure Mg(OH)2\nNow, 2 moles of HNO3(aq) react with 1 mol of Mg(OH)2(aq)\n\n6,00*10^-3 moles Mg(OH)2*(2 moles HNO3/1mol Mg(OH)2 )= 0,0120 moles of HNO3(aq)\nMolarity = moles solute/volume of solution\n0,200M= 0,0120 moles /?\n?= 2,40*10^-3 L of solution= \n2,40 mL of 0.200M HNO3(aq)  are needed to completely neutralize 60.0 ml of 0.100 M Mg(OH)2 solution
954211
Your first conclusion is correct, but your second isn't.  Since you already have the mechanism given to you, you can see that the 2nd step is the slow step.  So, the rate law should be:\n\nrate = k (N2O2) (H2)\n\nSince N2O2 is an intermediate, it can't appear in the overall rate law, so you have to get rid of that.  From the first reaction, you see that N2O2 is formed from the collision of 2 NO.  So, the concentration of N2O2 is proportional to [NO]^2.  That gives you the overall rate law that you had, and the overall order of the reaction is 3rd order.\n\nYour second conclusion is incorrect because the slow step requires the collision of N2O2 and H2.  The slow step is just a second order reaction.  \n\nSo, you've got things pretty well in hand...Hope this helps a little...
610699
Pb(NO3)2(aq) + H2SO4(aq/l)[depending on concentration] --> PbSO4(s) + 2HNO3(aq)\nPb2+(aq) + SO42-(aq) --> PbSO4(s)\n\nI don't see how an redox reaction could occur.
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)
835111
4 NH3  +  5O2  --->  4 NO  +  6H2O\n2 NO  +  O2 ---> 2 NO2  +  N2O4\n3 NO2  +  H2O ---> 2 HNO3  +  NO
1062523
2NaI + Hg(NO3)2 --> HgI2 + NaNO3\nwe have;\n52 ml and a 0.78M Hg(NO3)\ntherefore we can find # of Moles\nn = CV\nn = 0.78 * 0.052                       Concentration should be in Litres\nn = 0.04056 mol\n\naccording to the equation;\n2 moles of NaI react with 1 mole of Hg(NO3)2\ntherefore 2(0.04056) moles of NaI needed in the reaction\ni.e. 0.08112\nwe know V i.e. Volume equals the ratio of #of moles and Concentration \ntherefore ; \nV= n/C\nV= 0.08112 / 1.60\nV= 0.0507 L\ni.e. 50.7 ml of NaI required in the reaction !!
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)