sulfurus acid, carbonate acid, hydrocloric acid, phosphoric acid, or nitric acid?
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Oxidizing agents are reduced during a reaction, meaning they gain electrons.\n\nNitric acid is an extremely powerful oxidizing agent.\n\nOxalic acid, mentioned in a previous answer, is actually a very common reducing agent.
378635
a) Ag2S\nb) SnCl4\nc) Na3N\nd) SrI2\n\nDo us a favor and pay attention during class. That way you'd know how to do these.
547878
a) +1\nb) 0   (in molecule state)\nc) 0\nd) -3\ne) 0\nf) +4 of Carbon\ng) +2 of Mg\nh) +1 of Chlorine (interesting!)\ni) +6 of Sulphur
461016
Sulphuric acid when it ionises in solution, will form 2 H+ ions per molecule of H2SO4 originally present.  Or rather, for each mole of sulphuric acid present, 2 moles of H+ will form.  The strength of an acid obviously depends on the number of H+ present in the solution.\n\nHCl on the other hand will only produce 1 H+ per molecule of H2SO4.\n\nIf you follow the same logic, H3PO4 should have produced 3 H+ per molecule of the acid and hence become the strongest.  Unfortunately it is not as simple.  Although, phosphoric acid can indeed ionise to form 3 H+, it is not energetically profitable.\n\nH3PO4 → H+ + H2PO4–       Ka1= 7.5×10^−3\nH2PO4–  → H+ + HPO42–       Ka2= 6.2×10^−8\nHPO42–  → H+ + PO43–       Ka3= 2.14×10^−13\n\nAs the Ka values decreases, it means that it becomes less easy for the reaction to proceed.  In the case, you can see that it is quite easy for a molecule of phosphoric acid to ionise forming one H+.  But further ionisations are quite difficult due to the smaller values of Ka.
482242
Cu2S + 12HNO3 -> Cu(NO3)2 + CuSO4 + 10NO2 + 6H2O\n\nThere you go.
115880
Well, Sulfuric acid is a diprotic acid. The first proton is a strong acid, and the second proton is a fairly strong weak acid.\n\nBecause it's a strong acid, there will be 18 M of H3O+ ions.\n\n-log(18) = -1.255\n\nTo the correct number of significant figures, the pH is -1.26.\n\nEDIT: dr_futuredentist is very wrong. The second hydrogen\ndoes not dissociate fully at all.
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An acid and its conjugate base differs by a hydrogen cation (or proton) H+, so:\n\na. H2SO4 and HSO4-\nd. HCl and Cl-\ne. HF and F-\n\nare conjugate acid-base pairs
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2BF3 + 3Li2SO3 → B2(SO3)3 + 6LiF\n3CaCO3 + 2H3PO4 → Ca3(PO4)2 + 3H2CO3\n8Ag2S → 16Ag+S8
372647
(a): Si+, P2+\n\n(b) Ar+, S-\n\n(c) Ca+, Cl2-\n\nBut the other answers that you gave are also correct.
375486
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
13644
n-hexanoic acid\n\nThere are a LOT of compounds with that chemical formula.  Check chemfinder.com for more.
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HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
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1. Molarity = concentration and it expressed as mol per dm^3\n    Change the mass to mole and the 851 ml to dm^3. Then find   \n    the ratio of the number of moles to volume in dm^3\n\n2. Soluble ionic compounds breaks up the water molecules in H^+ ions and OH^- ions. Sodium carbonate give Na^+ ions and CO3^2- ions......Other ions that are present are HCO3^- ions called hydrogen carbonate ions. I have found that a solution of sodium carbonate is alkaline: implying that there are more OH^- ions to H^+ ions, due to the formation of HCO^- ions.\n3. OH^- (aq) + H^+(aq) ----> H2O (l)\n4. H^+ ions (ie. hydrogen ions), OH^- ions (ie. hydroxyl ions),\n   Ca^2+ calcium ions and Cl^- chloride ions\n5. Hydrogen ions, hydroxyl ions. ammonium ions and sulphate ions.\n6. Actually the calcium metal reacts with water to give hydrogen gas and calcium hydroxide. At the same time, calcium reacts with nitric acid to give calcium nitrate, water and brown nitrogen dioxide gas.\nHence there is no simple ionic equation for your question\nYou may want to consider this one:\nCa^0  +  2H^+   + 2OH^-  -----> H2(g)  + Ca^2+  + 2OH^-\n7. Hydrogen ions, hydroxyl ions, iron (III) ions Fe^3+  and sulphate ion SO4^2-
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http://en.wikipedia.org/wiki/Cyanuric_acid
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Phosphoric Acid, H3PO4.
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1 - NH4OH\n2 - PbO\n3 - KBr\n4 - PCl5 (five)\n5 - Al(NO3)3\n6 - Ba3(PO4)2  (not sure)\n7 - C4H10\n8 - Sn(OH)4\n9 - NH4NO3\n10 - Li3P\n11 - Ba3ClO2   (not sure)\n12 - MgSO3
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a=oxygen diflouride\nb=dichlorine octoxide\nc= sulfur trioxide\nd=phosphorus pentoxide