Calculate the mass percentage of chloride in Nickel (II) Chloride?
954876
NiCl2. mass percentage would be same as relative molecular mass percentage.\n\nCl2=71g/mol\nNiCl2=130g/mol\n\n71/130 x 100% = ??
66513
http://en.wikipedia.org/wiki/Copper%28I%29_chloride
836848
first you'll start with Q=m*c*delta t to get the total energy of the aluminum. divide the grams of aluminum by its molar weight and then use Q=m*c*delta t where the Q's are the same, c for the water is 4.18 J/gC and then solve for the mass. good luck!
876667
Fe + CuCl2 --> FeCl2 + Cu\n\nmole of CuCl2 = 8.00 / 134.5 = 59.47 mmole\n\nmole of Fe = 5.40 / 56 = 96.42 mmole\nmole of unreacted Fe = 4.08 / 56 = 72.85 mmole\nmole of reacted Fe = 96.42 - 72.85 = 23.56 mmole\n\nmole of unreacted CuCl2 = 59.47 - 23.56 = 35.09 mmole\n\nYep, there will be around 35 mmole CuCl2 left in the beaker.\n\npeace\nvixklen
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
1013656
Cut up some steel into small piecies.\nWeight out a set amoult of the steel, say 1 g.\nAdd a set amoult of the HCl acid and stir untill the steel dissiloves.\nNow do a titiration using a base of set molarity.\n\nNow the maths part\nZn + 2HCl -> ZnCl + 2HX\nX is a countion\nSo for one mole of Zn in the solution will react with 2 moles of HCl\nSo using the amoult of HCl and the molarity you can work out the number of moles of HCl you have in your solution.\nLets call this A\n\nHow for the tiratation\n\nHCl + NaOH -. NaCl + H2O\n\nSo one mole of NaOH reacts with one mole of NaOH\nWork out the number of moles of HCl left in the zinc solution\nBy using the volume of NaOH and the morality of it.\nCall his B\n\nNow the number of Moles of HCl the Zn has reacted with is\nA - B\n\nCall this C\n\nAs Zn reacts with 2 moles of HCl there are\nC/2 moles of Zn in the solution.\nCall this D\n\nFrom where you can work out the number of atoms in the zinc By using D x arggavios contant (sp) to give you the number of\natoms in your sample.\n\nNow share the number of atoms in your sample by the weight you using in forming your sample will give you the atoms per gram.\nNow times this final number by the weight of the starting piece of steel and you have your number of atoms.
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
803450
MgCl2(aq) + 2NaOH(aq)  --> Mg(OH)2(s) + 2NaCl(aq)\n\nNet ionic equation:\n\nMg2+(aq) + 2OH-(aq) --> Mg(OH)2(s)
279553
Find the weight of AgCl per mole, divid it into 71.65 g/l\nwill give u the Molar concentration of Cl, Ag atoms.\nOnce you know the Molar concentration of Cl, it is the same\nas the Molar concentration of HCl, since thats the only source of Cl atom.  \nPH 3 is same as 0.001 moles of H+, from both the HCl and HNO3.\n\nAg = 107.9 grams per mole\nCl = 35.5 grams per mole\nAgCl = 107.9 + 35.5 = 143.4 grams per mole\n\nThere is an error in this question, the acid concentration\nis mostly likely to be 50% of each.\n\nBut a PH 3, or 0.001 mole of acid cant not produce 71.65 g/l of AgCl, assuming you are adding such as AgNO3 to the acids.  Otherwise, you did not give enough information to solve this.\n\nYour might have mean 0.07165 grams/liter of AgCl\nThis will give the answer of 50% of each acid, or 0.0005 mole each.\n\n0.07165 / 143.4 = 0.0005 mole per liter of Cl atoms\nwhich is the same as the mole per liter of HCl
749394
divide the 15.8 g by the gfw ( molecular weight of Lithium chloride )
1132105
% yield = Actual (5.53g) divided by Theoretical (use periodic table)\n                                times 100%.\n\nFind limiting first by finding moles of H2 and Cl2.  \nH2 moles = 5.95\nCl2 moles= 0.17\nsince H2 to Cl2 is 1:1 in the chemical formula(1H2+1Cl2-->2HCL)\nyou can't react all of the H2 present.  THerefore Cl2 is the limiting reactant.  \n\nYou therefore use Cl2 reactant info to find the Total products formed, because when all Cl2 is reacted the reaction stops.  \nSimply convert moles Cl2 all the way to grams HCL.\nYou get::  12.4g HCl\n\nThis is expected but you actually got 5.53, so the percent yield (use first formula) =44.6%\nI'm checking my work to make sure
733447
Br+H20--->HBr+O\n\nNi(II)O+H20--->Ni+H202\n\nKO2+H2O--->KOH + H2\n\nHope that helps.
192443
Write down the equation of reaction: K + Cl2 -> KCl (not balanced)\n\nBalance the equation of reaction: 2K + Cl2 -> 2KCl (balanced)\n\nFind the moles of Potassium: 5 g K / 39.10 g/mol K = 0.128 mol\n\nFind the moles of Chlorine gas: 7 g Cl2 / (2 x 35.45 g/mol) = 0.098 mol\n\nDetermine the stoichiometric ratios: \nCl2 -> 2 KCl\nK ->   KCl   \n\nFind the moles of product: \n\nFor potassium: 0.128 mol K x (1 mol KCl / 1 mol K) = 0.128 mol KCl\n\nFor chlorine:  0.098 mol Cl2 x (2 mol KCl / 1 mol Cl2) = 0.196 mol KCl\n\nSo potassium is the limiting reactant. \n\nMass of KCl that can be formed: 0.128 mol K x (1 mol KCl / 1 mol K) x (39.10 g/mol K +  35.45 g/mol Cl) =  9.54 g KCl
836906
I had to read about this in order to answer your question at this web site:\n\nhttp://home.um.edu.mt/chemistry/CH230/2.Analytical/CP210a.PDF\n\nSo, two moles of DMG combines with each Ni++ ion.\n\n(0.2145 g Ni)* (1 mole Ni/58.71 g Ni) = 3.65*10^-3 mole Ni\n\nIf what you tell me is correct that DMG is 116.1194 then the Ni complex which loses 2 H+ for the Ni++ has a molecular wt of\n172.8134 (please double check)\n\n(3.65*10^-3 mole Ni)* (172.8134 g/mole) = 0.6314 g\n\nplease check for errors
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.
955650
Do you have the percentage abundances of the isotopes?\nE.g. for chlorine, it exists as 2 isotopes, namely Cl-37 and Cl-35.\nCl-37 (with 17 protons and 20 neutrons) has an abundance of 25% with Cl-35 (with 17 protons and 18 neutrons) has an abundance of 75%.\nThus average relative atomic mass of chlorine\n= [(25/100) x 37] + [(75/100) x 35]\n= 35.5\n\nHope this helps :)
906161
Measure the mass of a nickle, then use the atomic weights of each element to determine the number of nickel atoms.  I won't do you're homework for you, but that should get you started.