Technically this is a VERY tough probem and can't be answered with the information given. However, I assume that the person making up the problem meant that 0.1 mol HCl was added to enough water to make 1L of solution. (H^+) = (HCl) since HCl is 100% ionized. (H^+) = 0.1 m...
In what? Reaction? procedure?
% by mass = (mass solute/g soln)*100 let x = mass solute [(x)/(400+x)]*100 = 5 Solve for x.
BaCO3 + 2HNO3 ==> H2O + CO2 + Ba(NO3)2 mols BaCO3 = grams/molar mass Using the coefficients in the balanced equation convert mols BaCO3 to mols CO2. Now convert mols CO2 to grams. g = mols x molar mass. Convert mols CO2 to L. L = mols x 22.4.
Science (chemistry) help please
I don't see any data. Hwo many mL EDTA did it take? Where you divide by 0.001--that is where L EDTA goes.
PV = nRT and solve for n = number of mols. Then n*molar msss = grams.
Write and balance the equation. Convert grams given to mols P and mols Cl2. Convert mols P to mols PCl5. Convert mols Cl2 to mols PCl5. The SMALLER number of mols is the correct value for mols PCl5 formed and the reagent providing that number is the limiting reagent.
mols HCl = M x L = 0.01 x 1 = 0.01 mols NaOH = 0.12 mols NaOH after neutralization = 0.11. M NaOH = M OH^- = mols/L. Then pOH = -log(OH^-) and pH + pOH = pKw = 14. Given pOH and pKw, solve for pH.
2H2 + O2 ==> 2H2O 6.82E24 H2 molecules x (1 mol/6.02E23) = ? mols H2. 1 mol H2 makes 1 mol H2O which weighs 18 grams.
This is the corrected copy. An equal sign I typed as a - sign. millimoles HCl = 6M x 51.2 = 3072. mmoles NaOH solution (75 mL) = 3072. M NaOH soln = mmoles/mL = 3072/75 = 4.096M That's the molarity of the diluted solution. The diluted solution started out as 6M and it was ...
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