what is the reaction equation to show that cadmiun selenium quantum dots slowly photo oxidize to yeild SO2?
94062
I wanted to help you but I just wonder where the sulfur atom in SO2 comes from? Sulfuric acid, maybe, somewhere in your process? Or is it SeO2? These have to be clarified first to get a reliable answer.
47432
3AgC2H3O2 + Na3PO4 -----> Ag3PO4 + 3NaC2H3O2
394775
2Cu(NO3)2 ----> 2CuO + 4NO2 + O2\nOBSERVATION: brown, choking fumes of NO2 evolved; black solid i.e CuO settles at the bottom\n\n2FeSO4 ----> Fe2O3 + SO2 + SO3 \nOBSERVATION: white fumes of SO3 evolved and reddish brown solid i.e. Fe2O3 formed\n\nEDIT: The above equation for iron(II)sulphate may be a supposed characteristic for its thermal decomposition relating to that of copper(II)nitrate above, but Chaos, O2 will not be evolved here, because SO3 will not get reduced to O2, even on strong heating.
394069
3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n                                        +11H2O\n\nFirst you must calculate the (average)oxidation number of central atom of main compounds in reactants.\nEx;Cr oxidation no=+6 in K2Cr2O7\n   C  oxidation no=-2 in C2H5OH\nThen in the products.\nEx;Cr oxidation no=+3 in Cr2(SO4)3\n   C  oxidation no=0 in CH3COOH\n\nThen you must evalute how many electrons must be transferred.\nEx;Oxidation no of Cr is changing from +6 to +3 \n   So it is reduced.\n   SO each atom of Cr must receive 3 electrons.\n   So totally 6 electrons must be received by K2Cr2O7\n   \n   Oxidation no of C is changing from -2 to 0\n   SO each atom of C must donate 2 electrons.\n   So totally 4 electrons must be donated by C2H5OH\n\nThen you must write now a explantion reaction.\nEx;you must write oppsite (simple)propotion of electrons in front of each compound\n  3C2H5OH + 2K2Cr2O7\n \nThen you can calculate how many Cr,C(,K) atoms should be there\n So, 3C2H5OH + 2K2Cr2O7=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n\nThen you can calculate the medium(SO42-) should be there\n i think 8 SO4 should be there\n\nSo,3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n\nFinally you must equilize H,O atoms(balanced by water)\nSo,3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n                                        +11H2O\n\nSecond one ................\nC2H5OH+4I2+6NaOH=CHI3+HCOONa\n                  +5NaI+5H2O
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
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.
412530
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)
760435
You are looking at a displacement reaction, in which the more reactive metal (Cu) is displacing the less reactive metal (Ag) from its salt solution (AgNO3).\nCu (s) + 2AgNO3 (aq) --> 2Ag (s) + Cu(NO3)2 (aq)\n\nHope this helps. :)
890894
C12H22O11+12O2-------12CO2+11H2O+E
116069
The balance is :\n\n     H2O2(aq) -> O2(g) + 2H+(ag) + 2e-\n MnO2(s) + 4H+(aq)  +  2e-  ->  Mn2+(aq)  + 2H2O\n \n Combine the two half-reactions and cancel on both sides yield the     complete balance equation.\n\n MnO2(s) + H2O2(aq) + 2H+(aq) -> Mn2+(aq) + O2(g) + 2H2O\n\n this is an oxidation-reduction reaction\ntype  decomposition reaction.
571388
Cu + 2H2S04 --> CuSO4 + SO2 + 2H20\n\nCu (copper) is oxidized, because it loses 2 electrons to enter into the +2 state.\n\nThe sulfuric acid is the oxidizing agent, because it oxidizes the copper and is itself reduced to form sulfur dioxide. (Compare the oxidation states. Half of it remains as SO4- because it takes 2 equivalents of H2SO4 to react with Cu, due to the mole to mole ratio.)\n\nI'd do this reaction in the fume hood, though. SO2 is an irritant to your lungs.
702491
Mn2+ +2NH3 +2H2O  <-->  Mn(OH)2 + 2NH4+ (+:sup, 4:sub)\n4Mn(OH)2 + O2 + 2H2O  <-->  4Mn(OH)3\n2Mn(OH)2 + O2  <-->  2(MnO2 . H2O)\n\nMn(OH)2 white sediment\nMn(OH)3 brown sediment\nThe oxidization is faster if we add H2O2
328575
reduction:  2e- + Cu2+ --> Cu      Eo = +0.34 V  cathode (+)\n\noxidation:   Zn   --> Zn2+ + 2e-   Eo = +0.76 V    anode (-)\n________________________________________\n\nZn  + Cu2+ --> Cu  +  Zn2+         Eo = +1.10V therefore cell spontaneous as written.
949985
Reactant, oxidation of molybdenum:\n\n2Mo(s) + 5Cl2(g) → 2MoCl5(s)\n\nProduct\n\nMoO3(s) + 3H2(g) ----> Mo(s) + 3H2O(g)
810244
If the reaction is correct it should be :\n  \n              2ZnS + 3O2  -------->  2ZnO + 2SO2
1185900
H2SO4 + 2KOH --> K2SO4 + 2H2O
958410
1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen