How is this terminating the reaction (more info inside)?
931530
NH3 reacts with both HCl and NaOCl. With HCl it forms NH4Cl, with NaOCl it forms cloramine (NH2Cl - gas) and lye (NaOH).
551973
TO CAUSE THE CHANGE CHEMICAL REACTION
418418
Hopefully this diagram can answer your questions:\nhttp://fachschaft.cup.uni-muenchen.de/~schleifi/reaktion/reaction/stobbe.html
934206
2\nthe reaction is\nCa(OH)2+2HNO3=2H2O+Ca(NO3)2
744648
Use Gibbs free energy equation\n\nDeltaG=DeltaH - TDeltaS\n\nYou want Delta G to be negative so if Delta H positive then you need a greater TDeltaS \nDelta S is -25 so in order for the reaction to be spontaneous the T has to be less than zero
744530
It is an electrophilic aromatic substitution reaction.  Because the -COO- acts as an electron withdrawing group to one phenyl group and a moderate electron donating group to the other pheny group the nitration will occur on the pheny group where the COO- is acting as an electron donating group.  The major product of the nitration will have the -NO2 in the para position because the ortho position is partially blocked by the large COPh group.\n\nI hope this helps!
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
469693
Nothing. The two reactants have the same anion, so no reaction will take place. If you create a hypothetical double displacement reaction, you get:\n\nFeS + CS2 --> FeS + CS2\n\nAnd, so, there is no reaction.
130958
The equation is correct, if I perceive it right. You have 3 ethoxy groups on the boron atom, which does correspond to that formula. It looks like what might happen if you reacted borane (BH3), possibly tethered in a THF complex, with ethanol (EtOH) to get the product and H2 gas, but a lot of heat too. Such a substituted Borane would be rather reactive, as a Lewis acid.\n\nB(OC2H5)3 + 6O2 ---> H3B03 + 6CO2 + 6H2\n\nis my best guess
250273
Enough to cancel out the H's. For the reaction with 2H's, add 2OH's to both sides. For the reaction with 3H's, ad 3OH's to both sides.\n\nH + OH --> H2O\n\nEDIT: Oh, well in that case, the H's cancel out, so you have just one H on the side that previously had 3. Then, just add one OH to both sides.
887797
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.
1116694
the reaction leads to production of hydrogen gas ...\n\n\n2Al(s) + 6HCl(aq) → 2Al3+(aq) + 6Cl-(aq) + 3H2(g)
82368
U get precipitate of Lead Iodide. The reaction is already mentioned!!!!\n\nPb(NO3)2 (aq) + 2KI (aq) --> 2KNO3 (aq) + PbI2 (s)
563324
There are a couple of methods but the one I like to use is the method of half reactions.  First we split the reaction into an oxidation half reaction and a reduction half reaction as follows:\n\nCr2O7(-2) -> Cr(+3)   and    CH3OH -> HCO2H\n\nWe know that the left half reaction is the reduction half because the oxidation number of the Cr changes from (+6) to (+3).  Likewise the right reaction is the oxidation because the oxidation number of the carbon changes from (-2) to (+2).\n\nNow that we have our half reactions the next step is to balance everything EXCEPT oxygen and hydrogen.  In this case the only other elements involved besides O and H are Cr and C.  Doing this give us: (the right one is already balanced for carbon; one on each side).\n\nCr2O7(-2) -> 2Cr(+3)   and    CH3OH -> HCO2H\n\nStep three: We balance oxygen be adding water molecules to the side with less oxygen atoms.  Doing this give us:\n\nCr2O7(-2) -> 2Cr(+3) + 7H2O   and   H2O + CH3OH -> HCO2H\n\nStep four: Since we are in acid we add H+ ions to balance the hydrogen atoms on both sides.\n\n14H(+1) + Cr2O7(-2) -> 2Cr(+3) + 7H2O      and         \n\nH2O + CH3OH -> HCO2H + 4H(+1)\n\nStep five: We balance the charges on both sides by adding electrons [e(-1)] to the side with the most positive charge.  Looking at the first half reaction we see that there is a total charge of +14-2 = +12 on the left side, and a total charge of 2*(+3) = +6 on the right side.  Therefore adding 6e(-1) to the left side should balance out the charges giving us: (do the same for the second half reaction)\n\n6e(-1) + 14H(+1) + Cr2O7(-2) -> 2Cr(+3) + 7H2O    and\n\nH2O + CH3OH -> HCO2H + 4H(+1) + 4e(-1)\n\nYou know you have made a mistake if the electrons don't end up on different sides of the two half reactions.\n\nStep six: We must balance the two half reactions to each other i.e. the number of electrons produced in one reaction must equal the number of electrons consumed in the other reaction.  If we multiply everything in the first reaction by 2, and everything in the second reaction by 3 this will give us 12 electrons in both half reactions.  Doing this gives us:\n\n12e(-1) + 28H(+1) + 2Cr2O7(-2) -> 4Cr(+3) + 14H2O    and\n\n3H2O + 3CH3OH -> 3HCO2H + 12H(+1) + 12e(-1)\n\nStep seven: Add the two half reactions together and simplify the result as much as possible by cancelling things that appear on both sides, and by finding the smallest set of whole number coefficients for the reactants and products.  Adding the half reactions together gives us:\n\n12e(-1) + 28H(+1) + 2Cr2O7(-2) + 3H2O + 3CH3OH -> \n4Cr(+3) + 14H2O + 3HCO2H + 12H(+1) + 12e(-1)   \n\nCanceling the electrons on both sides gives:\n\n28H(+1) + 2Cr2O7(-2) + 3H2O + 3CH3OH -> \n4Cr(+3) + 14H2O + 3HCO2H + 12H(+1)\n\nCanceling the extra H+ ions on both sides gives:\n\n16H(+1) + 2Cr2O7(-2) + 3H2O + 3CH3OH -> \n4Cr(+3) + 14H2O + 3HCO2H\n\nCanceling the extra H2O molecules on both sides gives:\n\n16H(+1) + 2Cr2O7(-2) + 3CH3OH -> 4Cr(+3) + 11H2O + 3HCO2H\n\nIt looks like this is the most simplified version.\nLast step: Check to see that the final reaction is balanced for both atoms and charges.  It looks like this is the final answer.
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)
713038
the other product is zinc chloride.\nand the reaction is single replacement.