I need some help with a few reactions if anyone knows their stuff?
356281
MgCO3 + 2HCl ---> MgCl2 + CO2 + H2O\n\n2HCl + MgO ---> H2O + MgCl2\n\nBarium hydroxide is a base so can be neutralised with any acid.  An example being HCl.\n\nBa(OH)2 + 2HCl ---> BaCl2 + 2H2O\n\nor H2SO4\n\nBa(OH)2 + H2SO4 ---> BaSO4 + 2H2O
988889
understanding the reaction meachanism \neffect of catalysts etc....\ndetermining the amounts needed
867786
Let's take potassium representing group 1.  All the others will do the same thing.  Here is the reaction:\n\n2K(s) + 2HCl(aq) --> 2KCl(aq) + H2(g)\n\nThe reaction for group 2 will be similar.  Let's take calcium  representing group 2:\n\nCa(s) + 2HCl(aq) → CaCl2(aq) + H2(g)\n\nHope this helps.
835717
Hmmm.. I'm not trying to be stingy, but I still prefer using Yahoo! Answer since that there is a bunch of good chemist in this community, to asking for a help from there.\n\n\npeace\nvixklen
63455
What result? Need more detail!!
82368
U get precipitate of Lead Iodide. The reaction is already mentioned!!!!\n\nPb(NO3)2 (aq) + 2KI (aq) --> 2KNO3 (aq) + PbI2 (s)
660043
1.) do your own homework\n2.) this is the computer forum, not high-school chemistry.  Next time post to the right furom.
899611
Ba(NO3)2 and sodium phosphate\n\n1.  Ba(NO3)2 + Na3PO4 --> NaNO3 + Ba3(PO4)2\n\n2.  Ba(+2) + NO3(-1) + Na(+1) + PO4(-3) --> Na(+1) + NO3(-1) + Ba(+2) + PO4(-3)\n\n3.  No reaction (All spectator ions cancel out)\n\nBa(NO3)2 and sodium carbonate\n\n1.  Ba(NO3)2 + Na2CO3 --> BaCO3 + NaNO3\n\n2.  Ba(+2) + NO3(-1) + Na(+1) + CO3(-2) --> BaCO3 + Na(+1) + NO3(-1)\n\n3.  Forms BaCO3 (Barium Carbonate)\n\n\nBa(NO3)2 and pottasium iodide\n\n1.  Ba(NO3)2 + KI --> BaI2 + KNO3 \n\n2.  Ba(+2) + NO3(-1) + K(+1) + I(-1) --> Ba(+2) + I(-1) + K(+1) + NO3(-1)\n\n3.  No reaction (All spectator ions cancel out)\n\nBa(NO3)2 and sodium sulfate\n\n1.  Ba(NO3)2 + Na2SO4 --> NaNO3 + BaSO4\n\n2.  Ba(+2) + NO3(-1) + Na(+1) + SO4(-2) --> Na(+1) + NO3(-1) + BaSO4\n\n3.  Forms BaSO4 (Barium Sulfate)\n\nHopes this helps you!
747348
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?
555396
try this\n\nhttp://antoine.frostburg.edu/chem/senese/101/inorganic/faq/iron-reactions.shtml
881204
Reaction, success, failure, happiness, disapointment.\n\nCome on man, have a guess eh? When you take action, what are the likely results?\ni don't even know if this is what your looking for but  I hope it helps.
656755
great stuff, stevia! I use it alot. I haven't had any odd reactions myself, but everyone reacts differently to different things. I guess if you don't use too much it shouldn't hurt even if it does. I use the liquid, and only a few drops at a time.
811688
I would have to guess that it involves the neutralization reaction of more than 2 substances.\n\nJust a guess
800085
I went through the same thing myself...I would suggest getting a tutor. I think some schools offer tutoring services for free...you might want to check with your schools learning center or the chem/physics department. My college allowed people needing help to sign up for tutors and the school paid the tutors to help students out for an hour a week. If this kind of service isn't available, you could try getting a tutor on your own. I think in the case of orgo, its really best to have someone there to talk through concepts and reactions to make sure you're really understanding it. Books that try to dumb things down just dont help because they can't tell you if youre actually understanding things correctly. Good luck!
799202
for those questions, all you have to do is translate them into the equation\ncheck chemtutor.com for hellp, it explains it much clearer than i can
1109007
sounds like there are a few answers... maybe your text book will help?
209054
Effect of enzyme on reaction