a) 400.0 mL of 0.10 M NaCl\nb) 200.0 mL of 0.10 M FeCl_3\nc) 300.0 mL of 0.10 M CaCl_2\nd) 200.0 mL of 0.10 M kBr\nHello...I am studying for my chem test tomorrow. I am really not sure on what to do with this problem, help + source + step by step would really be appreaciated. Thanks a bunch!
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Well, look at each species. NaCl dissociates into two ions, giving a strength of 0.20M. In 400 mL, that'd be 0.08 moles.\n\nFeCl3 dissociates into four ions, giving a strength of 0.40M. In 200 mL, that'd also be 0.08 moles.\n\nCaCl2 dissociates into three ions, giving a strength of 0.30M. In 300 mL, that'd be 0.09 moles.\n\nKBr dissociates into two ions, giving a strength of 0.20M. In 200 mL, that'd be 0.04 moles.\n\nSo, the calcium chloride (CaCl2) solution contains the greatest number of ions.\n\n\nHope this helped.
375486
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
360474
Aluminium is Al3+\n Chlorine is CL1-\n Potassium is K1+ \n\nHahaz, I think should be correct. Hahaz, I did your homework.
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a) 3SiO2 +2 Al --->3 Si + 2Al2O3\n\nb)AgNO3 +HCl---> AgCl +HNO3\n\nc) Mg + CuSO4 ---> MgSO4 + Cu\n\nd) Pb(NO3)4 +4 NaI --->4 NaNO3 + PbI4 (s)
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b,\n\nall the others are ionic compounds.
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Start with 2.948g K2Cr2O7. End with how many mL. K=39, Cr=52, O=16, K2Cr2O7=294. Multiply by 1molK2Cr2O7/294gK2Cr2O7. The gK2Cr2O7 cancel, leaving molK2Cr2O7. Multiply by 1000mLK2Cr2O7/0.5molK2Cr2O7. This comes from 0.5M, which means 0.5 mol in 1000mL. The molK2Cr2O7 cancel, leaving mL solution to be evaporated.\n\nStart with 45mL NaCl, end with gNaCl. Na=23, Cl=35.5, NaCl=58.5. Multiply by 0.125molNaCl/1000mLNaCl. This comes from 0.125M. The mLNaCl cancel, leaving molNaCl. Multiply by 58.5gNaCl/1molNaCl. The molNaCl cancel, leaving the gNaCl that is the answer.\n\nStart with 100g glucose, G. End with L of G. C=12, H=1, O=16, C6H12O6=180. Multiply by 1molG/180gG. The gG cancel, leaving molG. Multiply by 1L G/0.5molG. This comes from 0.5M G. molG cancel, leaving L of G. Which is the answer.
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1. AgCl(s) + NaNO3\n\n2. AgOH(s) + NaNO3\n\n3. They would not react as both are nitrates already
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Hi Kitty!  Dr. J to the rescue with a full answer.\n\n1) 4Li + O2 ---> 2Li2O\n\n2) 4HF + SiO2 ---> SiF4 + 2H2O\n\n3) It is simpler if we eliminate the K and the sodium and write the equation ionic because it is a redox reaction and we need to balance.\n\nThe overall redox equation will be:\n\nCr2O7(=) + SO3(=) + H(+) ---> Cr(+3) + SO4(=).\n\nNow you balance by the half reaction method.  The two reactions are:\n\n6e(-) + Cr2O7(=) + 14H(+) ---> 2Cr(+3) + 7H2O\n3SO3(=)  + 3H2O --->  3SO4(=) + 6H(+) + 6e(-)\n\nEliminating the electrons:\n\nCr2O7(=) + 3SO3(=) + 8H(+) ---> 2Cr(+3) + 3SO4(=) + 4H2O\n\nThis one was tough!!!  Thumbs up!\n\nIf you want you could put the K and the Na back.\n\nBest answer?
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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.
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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)
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(1) NaHCO3    --->  Na(+) + HCO3-\n(2) Ca(NO3)2  ---> Ca2(+) + 2NO3-\n(3) Al2(CO3)3  ---> 2Al(3+) + 3CO3(2-)\n\nThis is all taking place in water (ideally), so you should consider that too.
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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
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H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
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bcd\n\n\n\nSolution:\nCaCN2 + 3H2O ---> 2NH3 + CaCO3
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NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
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a)4 Fe + 3O2--->2 Fe2O3\n\nCH4 + 2O2 ---> CO2 +2 H2O
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[Cr(en)3]Cl3  =  Tris(ethylenediamine)chromium (III)\ntrichloride\n\n[Fe(CO)3]  =  Iron carbonate or Iron (II) Carbonate\n\n[CoCl(NH3)5]Cl2  =  pentaamminechlorocobalt(III) chloride but it is [Co(NH3)5Cl]Cl2\n\nK3[Al(C2O4)3]  =  potassium triethanedioatealuminate(III)\n\nK3[Co(NO2)6]  =  potassium hexanitritocobaltate(III)\n\n[Fe(H2O)6]++  =  Hexaaquairon(II) chloride\n\nK[Ag(CN)2]  =  Potassium dicyanoargentate(I)