calculate the % composition of calcium & chlorine in the anhydrous salt & its 3 hydrates (4 compositions each)
358871
Calcium chloride = CaCl2 = Mwt 110.98\n% Ca = (40.08 / 110.98)*100 = 36.11%\n% Cl = (70.90 / 110.98)*100 = 63.89%\n\nCalcium chloride dihydrate = CaCl2.2H2O = Mwt = 147.01\n% Ca = (40.08 / 147.01)*100 = 27.26%\n% Cl = (70.90 / 147.01)*100 = 72.74%\n\nOther hydrates can be calculated in the same way.\n\nHope this helps.
59648
You'll probably want to know typical values for drinking water and sea water.\n\nNormally, water contains dissolved salts with Na, K, Ca, Mg and depending on quality Fe and Mn as main kations and carbonate (and hydrogen carbonate), chloride, sulfate and to (hopefully) lesser extent nitrate and phosphate ions. There are numerous possibilities to combine these ions into solid salts, making the exercise fairly pointless.\n\nIf your teacher insists, Na and K for all the anions except HCO3(1-) and CO3(2-) which usually are written as Mg and Ca salts.\n\nCommon trace elements/compunds include Al, Si, NO2(1-), Li.\n\nMore or less routinely monitored heavy metal traces include As, Cd, Cr, Co, Cu, Ni, Hg, Pb, Sb, Tl, Zn
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
867182
6HCl +3 I2 + KClO3 --->6 ICl+ KCl + 3H2O\n\n\nMnO4 - +2 NO2- +4 H+ ---> Mn2+2 NO3- +2 H2O
738944
Potassium chloride, Chromium oxide, Barium chlorate, Ammonium chloride, Phosphorus trichloride
1051035
1-. NaCl + H2O Sodium Chloride ( table salt )\n2-. NaSO3 +2 H2O Sodium Sulfate ( or sulfite I always forget which is the 3 and which is the 4 )\n3-. NaC2H3O2 + H2O ( I'm guessing.  You will want to check your periodic table for the charges on this one )\n4-.NH4Cl + H2O.  Ammonium Chloride\n5-.NSO2 + 3H2O\n6-. not sure\n\nI presume you are working salt solutions.  Ammonia was never my strong suit. I'm positive about the first two.
770385
FeCl3+3NaOH>>Fe(OH)3+3NaCl
922294
1) NH(4)Cl - Ammonium chloride\n\n2) HClO(2) - Chlorous acid\n\n3) Ca(BrO(3))2 - Calcium Bromate\n\n4) BeSO(4) - Beryllium sulphate\n\n5) (NH(4)) (3)N - Ammonium nitride\n\n6) RbI - Rubidium iodide
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
920228
CuSO4•5H20\nThe dot notation denotes the hydrate.
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
78988
I2 + 2NaCl ---> 2NaI + Cl2\n\nSodium Iodide and Chlorine
745644
H3PO4 + 3NaOH ----> 3H2O + Na3PO4
866210
2NO+O2--->2NO2\n\nCaO+H2O--->Ca(OH)2\n\nCaCO3--->CaO+CO2
1025655
Think of the net ionic equation which is occuring.\nSince both the HCl and Ca(OH)3 are dissolved in aqueous solution, broken into ions, just focus on the H+ and OH- part of it first.\n\nH+(aq) + OH-(aq) --> H2O(l)\nThe Hydronium ion and the Hydroxide ion react to form water.\nWhat is left over are the Ca+2 and Cl- ion which drop out of the net ionic eqaution since they form the very soluble compound, Calcium Chloride (CaCl2).\n\n2 HCl (aq) + Ca(OH)2 (aq) --> 2 H2O (l) + CaCl2 (aq)
73081
beaver is right. a side point seeing as how it looks like a chemistry class question....\n CARBO HYDRATE contains C H O in a 1:2:1 ratio.
891286
Percentage composition means calculating the percentage (by mass) of an atom in its molecule. For instance, we would say:\n\n"Calcium Hydroxide (Ca(OH)2) is x% Calcium, y% Oxygen, and z% Hydrogen"\n\nTo find the percentage composition, first find the total mass of 1 mole of the molecule:\n\nFor Ca(OH)2, there is 1 Ca atom, 2 O atoms, and 2 H atoms. The atomic masses are as follows:\nCa = 40.08 g/mol. \nO = 16.00 g/mol\nH = 1.01 g/mol\n\nThe atomic mass of 1 mol of Ca(OH)2:\nCa + 2*O + 2*H = 40.08 + 2*16.00 + 2*1.01 = 74.1 g/mol\n\nTo find the percentage of Ca, divide the mass of Ca in the molecule by the total mass of the molecule:\n\n%Ca = 40.08 / 74.1 = 0.54 = 54%\n\nDo the same thing with O and H, but remember that there are TWO molecules of O and H in each molecule of Ca(OH)2:\n\n%O = 32 / 74.1 = 0.43 = 43%\n\n%H = 2.02 / 74.1 = 3%\n\n\nNow, repeat this process with the other four:\n\nNaHCO3 = Na + H + C + 3*O = 22.97 + 1.01 + 12.01 + 3*16 = 83.99 g/mol\n%Na = 22.97 / 83.99 =  0.27 = 27%\n%H = 1.01 / 83.99 = 0.01 = 1%\n%C = 12.01 / 83.99 = 0.14 = 14%\n%O = 48 / 83.99 = 0.57 = 57%\n\nCaO = Ca + O = 40.08 + 16.00 = 56.08 g/mol\n%Ca = 40.08 / 56.08 = 0.71 = 71%\n%O = 16.00 / 56.08 = 0.29 = 29%\n\nCaCO3 = Ca + C + 3*O = 40.08 + 12.01 + 3*16.00 = 100.09 g/mol\n%Ca = 40.08 / 100.09 = 0.40 = 40%\n%C = 12.01 / 100.09 = 0.12 = 12%\n%O = 48.00 / 100.09 = 0.48 = 48%\n\nAl2(SO4)3 = 2*Al + 3*S + 12*O = 2*26.98 + 3*32.07 + 12*16.00 = 342.17 g/mol\n%Al = 53.96 / 342.17 = 0.16 = 16%\n%S = 96.21 / 342.17 = 0.28 = 28%\n%O = 192.00 / 342.17 = 0.56 = 56%\n\nI hope this helps!