how to convert chlorides to sodium hypochlorites?
613040
Treat the chloride solution with ozone.
755177
1. Make sure that HCl is standardized.\n2. Write the balance equation:\n\nCa(OH)2 + 2HCl ---> CaCl2 + 2H2O\n\n- this means that 1 mole of Ca(OH)2 reacts for every 2 moles of HCl\n\n3. Get the number of moles of HCl reacted by multiplying the volume used to react Ca(OH)2 by the actual concentration of HCl (value from standardization). Divide the result by 2 - and this is now your moles Ca(OH)2.\n\n4. Get the molar solubility of Ca(OH)2 by dividing its number of moles by its volume. Unit is in gram per liter.
1123465
Na2S + 2 HCl > H2S + 2 NaCl  \n\nH2SO4 + BaOH > H20 + BaSO4  \n\nNH4Cl + NaOH > NaCl + NH4OH\n\nAgNO3 + NaCl > AgCl + NaNO3\n\nnot sure about complete or reversible..:)
489210
Potassium Hydroxide and Hydrochloric acid.\n\nKOH + HCl = KCl + H2O
709021
All chlorides are water soluble\n\nEXCEPTIONS: AgCl, Hg2Cl2
318573
The reaction between Hydrochloric acid and Sodium Hydroxide is as follows,\n\nHCl + NaOH --> NaCl + H2O\n\nYou can see that there is a 1:1 ration between the number of moles of HCl needed to completely reacted with NaOH.\n\nSince you are given both a volume and concentration of the NaOH solution, you should be easily able to calculate the number of moles of NaOH availible.\n# mole = volume * concentration\n# moles = .1 Liters * .25 moles/L = .025 moles of NaOH\n\nIf you know you have .025 moles of NaOH, you know you need .025 moles of HCl to react it completely.\n\nIf you know that you have a .15 mole/Liter solution of HCl, you can calculate the volume required to get .025 moles lie this,\nLiters = moles / oncentration\nLiters = .025 moles / .15 moles/Liter = .01667 Liters or about 167 mL of solution.\n\n\nNext time do your own homework.
170895
Sodion Hypochlorite and Sodium Thiosulfate?  Ive never heard of Na2(s2o3)2?!  Anyways heres the balanced equation:\n\n2NaOCl + Na2S2O3 ---> Na2S2O3 + 2NaOCl\n\nPretty straight forward I guess.  Double Replacement reaction.
476397
bubbling the O2/Cl2 mixture through aqueous NaOH will remove the Cl2.  it reacts to give bleach ... NaOCl ... (and salt and water).  assuming you didn't want the Cl2 ;-)
593991
Another name for the electrolytic dissociation of sodium chloride solution is electrolysis of brine.\n\nThe electrolysis would generate hydrogen gas, chlorine gas, and a solution of sodium hydroxide.
801118
CuCl2 + 2NH4OH --> Cu(OH)2 + 2NH4Cl\n\nCu(OH)2 (s) + 4NH3 (aq)     ----->     Cu(NH3)4^2+
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
958951
Sodium Chloride = Table Salt\nPotasium Chloride = buffering salt used by all animals and plants\nSodium hypochlorate = Bleach\nSodium BiCarbonate =  Baking soda\nSodium Hydroxide = lye   (strongly caustic .. basic solution used in manufacturing and synthesis of many compounds\non, and on, and on
484197
Salt - sodium chloride.\n\nIf you need something a bit more unusual, I believe there is imitation salt which is potassium chloride.
1136961
Are you preparing a buffer soln with NaHCO3 and NaCl ? Then take equimolar amounts of both and dissolve them in 100 ml of water and then dilute them to the desired strength.
250762
2 HCl + CaCO3 ----> CaCl2 + H2O + CO2
502725
That second reaction should be Pb(s) + Cl2(g) ---> PbCl2(s).  Reverse the second reaction and then add the two together and you will get the resultant reaction.\n\nPbCl2(s) ---> Pb(s) + Cl2(g)    H = +359.4 kJ\nPb(s) + 2Cl2(g) ---> PbCl4(s)   H = -329.2 kJ\n\nPbCl2(s) + Cl2(g) ---> PbCl4(s)  H = +30.2 kJ
223710
Start by balancing Al, then Cl. Al is already  balanced, so when you balance Cl everything is easy from their. \nYou final amswer will be:\n3HCl + Al(OH)3 ----> 3H2O + AlCl3