Some questions about Synthesis of Ferrate(VI) Ions?
926027
The strong base NaOH will be fully dissociated in water, and you're adding quite a bit (looks like 5 grams/10 mLs).  The water actually consists mostly of hydroxide anions at this point, and hypochlorite ions (your oxidiser) from the bleach.  The pH is 14 or so.  These are the only conditions you'll find Fe(OH)3 at.\n\nIt Ferric hydoxide is  instable due to its loose bonding (easy dissociation) with the hydroxide ions.
752683
a) sulfur tetrafluoride\nb) molybdenum tetrachloride\nc) cobaltous sulfide or cobalt sulfide\nd) SnO\ne) ferric chloride\nf) CrO3\n\nComment: the above names are the ones you're most likely to encounter in industry or in a real-world laboratory.  Few chemists out here would call FeCl3 "iron three chloride"!\n\nOh, I know, your teacher probably expects "sulfur (IV) fluoride, molybdenum (IV) chloride, cobalt (II) sulfide, and iron (III) chloride..."
869137
2H3PO4 + 3Ba(OH)2 ---> 6H2O + Ba (PO4)2 +  2Ba
975769
Al(s) + MnO4- (aq) -----> MnO2 (s) + Al (OH)4- (aq)\n\nSplit the above into two half equations:\n\n1) Al(s)  --->  Al (OH)4- (aq)  and\n2) MnO4- (aq) -----> MnO2 (s)\n\nBalance 1) as follows:\n\n1) 4OH^-  +  Al(s)  --->  Al (OH)4- (aq) now balance for charge:\n1) 4OH^-  +  Al(s)  --->  Al (OH)4- (aq)  +  3e^-\n\nBalance 2) as follows:\n2) MnO4- (aq) ---> MnO2 (s)  +  2H2O (l) (turn xs O into H2O)\nNow balance with H^+:\n2) 4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)\nnow balance charge:\n2) 3e^-  +  4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)\n\nNow add 1) and 2) (electrons happen to be equal):\n4OH^-  +  Al(s)  +  4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)  Al (OH)4- (aq)\nnow OH^-  +  H^+  ---> H2O:\n\n4H2O  +  Al(s)    +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)  Al (OH)4- (aq)\n\nSubtract 2H2O from both sides:\n2H2O (l)  +  Al(s)    +  MnO4- (aq) ---> MnO2 (s)  +    Al (OH)4- (aq)\n\nI hope that helps. If you need more help, please email me or re post.
777814
Q1)\nCa3 (PO4)2  (MW)molecular weight = 3*40 + 2 * (95) = 310 the oxygen weight is 16*4 = 64 .so % of O is (64/310) *100 = 20.6%\n\nQ2)\nSO4 (NH4)2 ---> MW = 32+4*16+ 2*18 =132 Mass of sulfur 32 \nhence  % S = 100*32/132 =24.2\n\nQ3) \n\nFe2 (SO4)3 ----> MW = 56*2+96*3=400 \n\n% of iron 100*112/400 = 28%
607593
K2PtCl6 only reacts with Ferric ions of Iron i.e. Fe3+ this is rate determining reaction
825401
First of all I assume  Na2Co3 should be Na2CO3.\n\nFe(CO)5 + 4NaOH = Na2Fe(CO)4 + Na2CO3+ 2H2O\n\nFirst, balance out the metals (Na and Fe) then C, which was already balanced. Then I chose to balance out H and which automatically balanced out O.
1120784
Well to start off any reaction that produces oxygen is potentially combustible, especially one that requires energy (endothermic) so be careful.\n\nTwo suggestions: \nFirst, electrolysis, this is where you break the covalent bond between oxygen and hydrogen with electricity (energy).  The by products are oxygen and hydrogen gas.\n\nMy second idea is simple rust.  Put iron (Fe) into water (H2O) and you will produce Iron-oxide or rust.  The Oxygen combines with the iron.\n\nGood luck.
799504
FeS + 2HBr --> FeBr2 + H2S\nFe2+ + S2- + 2H+ + 2Br- --> Fe2+ + 2Br- + H2S\n2H+ + S2- --> H2S\n\nK2CO3 + CuCl2 --> 2KCl + CuCO3\n2K+ + CO32- + Cu2+ + 2Cl- --> 2K+ + 2Cl- + CuCO3\nCu2+ + CO32- --> CuCO3\n\npeace\nvixklen
138630
It is a tetrahydrate of Iron III salicylate.  The structure can be found in the paper listed below:
861343
Ferrous means fe (II)\n1) 3Fe2+ + MnO4- + 4H+ to give  3Fe3+ + Mn2+ + 2H2O \n\nThere are two half equations\nthe first is pretty simple: 3 Fe2+ to give 3 Fe3+. That is the oxidation reaction\n\nthe other is the reduction. \nMn +7 is reduced to Mn+2\nthe secret to balancing this is to add water to the side with no oxygen, and H+ to the other side to balance it out. Check that the charges are balanced\n\n2) Fe3+ + Sn2+ to give Fe2+ + Sn3+\n \n3) cant help u here u need a textbook!
582466
1 goes in front of CO2.\n\nFe2O3 + CO ---> 2FeO + CO2\n\nAluminium chloride, AlCl3 because\nAl3+ --- group 3\nCl- --- group 7
844446
AsF3=Arsenic tri fluoride\nAg2CO3=silver carbonate\nFeSO4=Iron[II] Sulfate\nP5O10=Phosphorus Di oxide or penta phosphorus deca oxide\nPb(OH)4=Lead Hydroxide
881429
Fe2O3(s) + 3 CO(g) ----> 2 Fe(l) + 3 CO2(g)\n\nCount up the oxygens; the left side and right side oxygens should match.
946990
FeCl3 + NaCl -> FeCl2 + NaCl3\n\n\nFeCl3 + NaCl -> FeCl2 + NaCl3\n\nthese reactions are not possible \nas this reaction can take place if iron(fe) was more reactive than sodium or fe was less electropositive than na\nhence \nthese reactions are not possible
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 .
832814
Well this is concept of a  'in situ' reaction. Putting it simply, we use a freshly prepared solution so that the solution does not go any external effects(like reduction/oxidation with air) or internal effects(settling down). The reaction of Fehling solution is likely to continue anyway but the effectivity would be reduced over time as for the above mentioned reasons....I could get more 'chemical' but for a basic understanding this should suffice!