What is the perfect chemical anlysis of the sulfated water?
452959
please be careful with what you understand to be "natural" in thinking such things have a therapeutic value.\n\nturds are a natural product, you wish to bathe in those?
751720
chromium (II) sulfate fo sho
777672
S2O3(2-)-->SO4(2-)          balance S first\nS2O3(2-)-->2SO4(2-)         balance O with water\n5H2O+S2O3(2-)-->2SO4(2-)        balance H with H+\n5H2O+S2O3(2-)-->10H++2SO4(2-)       balance charge with e-\n\n5H2O+S2O3(2-)-->10H++2SO4(2-)\n                                        +8e-\nthe 8e- is with the product side
708361
Arsenic (III) oxide and water are the correct reactants.  They will form Arsenious acid (H3AsO3):\n\nAs2O3 + 3 H2O --> 2 H3AsO3\n\nThe proof of acidity is:\n\nH3AsO3  -->  H+  +  (H2AsO3)-\n\nThis shows the possibility of donating a proton.
983710
H2SO4 will dehydrate and remove water.\n\n\n(CH3)2 CH - CH2-CH=CH-(CH3)2
1103908
you mean into a chloride?\neasy just add concentrated hydrochloric acid.the chloride will precipitate. and the sulfate will come out as hydrogen sulfide gas and water. maybe you need hydrogen to be passed in the mixture.\nna 2(so4)   + hcl  =      nacl   + h2s  + h2o
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
529624
its dilute h2so4 (38%) sulphuric acid.
1029176
There is a school of thought that there is no such thing as sulphurous acid, that what is referred to as sulphurous acid is merely a solution of sulphur dioxide in water,\n\nHaving said that, "sulphurous acid" is a mild reductant and might reduce aqua regia, which is a strong oxidant due to the nitric acid content.  In practice, dilution would be more important than than the reduction using sulphurous acid but you could get the effect by bubbling gaseous sulphur dioxide through the aqua regia.\n\nIn point of fact, I have my doubts whether sulphur dioxide would act as a reductant under these conditions; I suspect the kinetics would be very slow.  If it did, you might see reduction of gold to metal or possibly precipitation of gold chloride.
1129845
You are right. The rest of the options are ionic where they will ionize into the cation and the SO42- ion but in the case of NaHSO4, the solution will first ionize into Na+ and HSO4- and then into H+ and SO42- i.e. there will still be some ions trhat remain HSO4- ions so it is the right answer
241693
chemical process for the manufacture of paper pulp that employs an acid bisulfite solution to soften the wood material by removing the lignin from the cellulose. Sulfite cooking liquor used in the process consists of free sulfur dioxide obtained by the burning of sulfur or by the roasting of iron pyrites, dissolved in water at a concentration of four to eight percent
788847
The column elutes sulfuric acid, H2SO4\n\nMolarity is mol/L (mol/dm^3) and is also mmole/mL\n\nYou used 25 mL of a KOH solution containing 10^-2 mmoles/mL, or a total of 0.25 mmoles\n\nThe neutralization reaction is   H2SO4  +  2 KOH  ---->  K2SO4  +  2 H2O\n\nThe column reaction is  CaSO4 + 2 i-EH+  ---->  i-E2Ca+2  +  H2SO4   (i-E is ion-Exchange)\n\nBottom line is that there is a sulfate for every TWO KOH's, and there is a Calcium for each ONE sulfate.\n\nHope this helps.
1029892
Na2S (aq) + 2H2O (l) = 2NaOH + H2S (g)
814262
H2SO4 + H20 -> H2SO4:H2O (water decreased)\nHNO2 + H2O-->NH3 + O2   (you balance it)\nNaHSO3 + H2O-->H2SO3 + NaOH(you balance it)
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
1016946
Pb(SO4)2\nPbSO4  is lead (II) sulfate!
860022
Here are the formulas:\n\nAluminum Hydroxide = Al(OH)_3\nSulfuric Acid = H_2SO_4\nWater = H_2O\nAluminum Sulfate = Al_2(SO_4)_3\n\nThe correct balanced equation is:\n\n2[Al(OH)_3] + 3[H_2SO_4] + → Al2(SO_4)_3 + 6[H_2O]