Chemistry Help Needed!!!?
964537
Tritium.\n\nThat is the isotope of hydrogen with one proton and two neutron in the nucleus.  It is used as a low grade radio-active activator in phosphorescent paint, replacing radium which is much more dangerous.\n\n(Science Teacher below is wrong: deuterium is a stable nucleon, it is thus not radio active)
866145
HCl + AgNO3-----> AgCl +HNO3\nPb(OOC-CH3)2  + 2LiCl -------> PbCl2  + 2Li(OOC-CH3)\nNa3PO4 + CrCl3 -------> Cr(PO4) + 3NaCl\nH2SO4 + BaCl2 ----------> BaSO4 + 2HCl\nAl2(SO4)3  + 6NH4OH ------> 2Al(OH)3  + 3(NH4)2SO4\n2AgNO3 + H2S -------> Ag2S  + 2HNO3\nCaCl2   + Pb(NO3)2 -------> Ca(NO3)2  + PbCl2\nCa(NO3)2 + Na2CO3 ------>CaCO3  + 2 NaNO3
1100440
This site should help you.\n\nhttp://sparknotes.com/chemistry\n\nGood luck.
72603
Wow, I hope these are right.  I wish i had a solubility chart or something (Which you probably have)  \n\nCuSO4(aq) + Na2S(aq) ---> CuS (s) + NaSO4 (aq)\nNet ionic - Cu2+ (aq) + S2- (aq)  ---> CuS (s)\n\nNiCl2 + 2KNO3 --> Ni(NO3)2 (aq) + 2KCl (aq)\nnet ionic - none (if NiCl2 isn't aq, then this is wrong)\n\nCa(OH)2 + 2HC2H3O2 ---> Ca(C2H3O2)2 (aq) + 2H2O (l) \nnet ionic - H+ (aq) + OH-(aq) ---> H2O (l)\nactually, since the acetic acid is a weak acid, this might be a little different.  If you learned it this way, then your net ionic is:\nHC2H3O2 (aq) + OH- (aq) ---> H2O (l) + C2H3O2- (aq)\n\nHope this helped
1171989
http://en.wikipedia.org/wiki/Chemistry\n \nthis nust be helpful for you
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
429906
Never trust materials science majors ...\n\n1. NaCl (s) --> Na+ (aq) + Cl- (aq)\n2. Ca (s) + 2H2O (l) --> Ca(OH)2 (s) + H2 (g)\n3. Ca(OH)2 (s) --> H2O (g) + CaO (s)
796834
try asking this in the chemistry section next time....
160963
K2SO4 ---> 2K+  +  SO4-\n\nHCl ---> H+  +  Cl-\n\nNaHCO3 ---> Na+  +  HCO3-\n\nNH4OH ---> NH4+  +  OH-\n\nBaCl2 ---> Ba-2  +  2CL-\n\n(NH4)3PO4 ---> 3NH4+  +  PO4-3
725081
The answer is Cu + Cu(NO3)2 -> Cu + Cu(NO3)2\n\nThere is no reaction.
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1) this one is already balanced for you!!\n\n2)Count the number of Mg and O on the right side, and count the number of Mg and O on the right side.  On the left, you have 2 O and 1 Mg, and on the right you have 1 Mg and 1 O.  To make the oxygen on both sides equal, you need to double the MgO.  Now on the left, you have 2 O and 1 Mg, and on the right, you have 2Mg and 2 O.  Now you need to multiply the Mg by two so the Mg on both sides is equal.  and you're done!!!! \n2Mg + 02 --> 2MgO\n\n3) On the left, you have 1 Zn, 1H, and 1 Cl.  on the right you have 1 Zn, 2Cl and 2H.  the Zn are balanced and you just need to multply the HCl on the left to balance the whole equ.\nZn + 2HCl --->ZnCl2 + H2\n\n4) On the left, you have 1 P and 2O, and on the left you have 4 P and 10 O. If you multiply the P on the right by 4, and the O on the right by 10, you have 4P and 10 O on both sides\n4P + 5O2 ---->P4O10\n\n5)  On the left, you have 1 Fe 2 Cl, and pn the left, you have 1 Fe and 3 Cl.  The easiest way to balance this is multiply the Cl on the right by the amount of the left one and vice versa. when you do that, you have 1 Fe and 6Cl on the right, and you have 2Fe and 6Cl on the left. To finish the equation, multiply the Fe by two.\n2Fe + 3Cl2---->2FeCl3\n\nVoila!  You have balanced 5 equations!\n\ni hope this was helpful!
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44 A) 2AgO --> 4Ag + O2\n     B)NiCO3 --> NiO + CO2\n     C)NH4NO3 --> N2O + 2H2O\n48)A)2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n     B)3Ag2SO4 + 2AlCl3--> 6AgCl + Al2(SO4)3\n     C)&D) I've no idea, sry\n50)A)C4H8 + 6O2 --> 4CO2 + 4H2O\n     B)2C8H18 + 25O2 --> 16CO2 + 18H2O\n     C)2C3H8O3 + 7O2 --> 6CO2 + 8H2O\n     D)2C3H6O + 5O2 --> 6CO2 + 6H2O\n52)A)2Al + 3H2SO4 --> Al2(SO4)3 + 3H2\n     B)2HCl + Ba(OH)2--> BaCl2 + 2H2O\n     C)Au + HCl--> No Reaction(gold is below hydrogen in the metal reactivity series)\n54)A)2Cl2 + 2NaI --> 2NaCl2 + I2\n     B)2NH3--> N2 + 3H2\n     C)4Na + O2--> 2Na2O\n     D)Mg + H2SO4--> MgSO4 + H2\n     E)MgCl2 + Ca(OH)2--> Mg(OH)2 + CaCl2\n     F)H2 + Cl2--> 2HCl
47432
3AgC2H3O2 + Na3PO4 -----> Ag3PO4 + 3NaC2H3O2
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This is what you should be looking at:\n\nHCl(gas) + NH3(gas)  -->  NH4Cl (solid)
867182
6HCl +3 I2 + KClO3 --->6 ICl+ KCl + 3H2O\n\n\nMnO4 - +2 NO2- +4 H+ ---> Mn2+2 NO3- +2 H2O
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
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AgBr-------->Ag+ + Br-\n(a) Ksp=[Ag][Br] solubility=s=[Ag]=[Br]=Ksp^0.5\ns=(6*10^-7)^0.5\n(b) [Br]=0.01 (because all of kBr was dissuciated)\ns=Ksp/[Br]=(6*10^-7)/0.01 mol/lit