Close but I think you have a math error. Here is your work. m= 75 mL of water 0.001kg water= o.o75 of warer. 2.37 c = 3*0.512 c/m*m m=1.54297 You are ok to here to get moles 1.59297*0.075/1=0.119473 mol I think you mistyped (transferred from the step above incorrectly) which ...
delta T = 102.37-100 = ? i = 3 Kb = 0.512 m = ?
I assume MCl2 is an ionic chloride so i, the van't Hoff factor, is 3. delta T = i*Kb*m You know delta T, Kb, and i (= 3). Solve for molality, m. Then m = mols/kg solvent. You know kg solvent and m, solve for mols. Then mols = grams/molar mass You know mols and grams, solve...
5 mL? 5*0.1 = 0.1*x solve for x.
I wonder if you meant C2H5OH, ethanol. However, here is the balanced equation for your post. C3H5OH + 4O2 ==> 3CO2 + 3H2O
The idea here is that you have a ppt of BaSO4 + CaCO3. On heating the CaCO3 loses CO2 but CaO remains. How much remains? That will be 2.40 mg (see below). If you have 4.3 mg CaCO3, it will lose CO2 in heating. How much will it lose? CaCO3 ==> CaO + CO2 4.3mg.....2.40..1.89 ...
g Ag2O = g Ag2CO3 x (molar mass Ag2O/molar masss Ag2CO3) = ? at 100% yield. That times 0.50 = g Ag2O actually produced.
Note the correct spelling of celsius. q = mass Fe x specific heat Fe x delta T.
Look up Eo for H2 ==> 2H^+ + 2e Eo = 0 Calculate Eo for the Pb half cell. E = Eo Pb + (0.0592/2)log(1/0.15) Then Ecell = EoH2 + EoPb
chemistry (ionic eq.)
50 mL x 0.2M = 10 millimols Pb^2+ 50 mL x 1.5M = 75 mmols Cl^- ........Pb^2+ + 2Cl^- ==> PbCl2 I.......10......75.........0 C......-10.....-20........10 what's left.0....55.......10 So we have left a PbCl2 solid in a solution with an excess of 55 mmoles Cl^- in a volume...
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