A lab technician uses a 0.25-mol/L sodium carbonate stock solution, Na2CO3(aq), to prepare 250 mL of 0.010-mol/L solution. How much of the stock solution should be used?

pH = 14 - pOH = 14 + lg(0.004) = 11.6

с (Na₂CO₃) = n(Na₂CO₃)/V(Na₂CO₃ solution);n (Na₂CO₃) = c (Na₂CO₃) * V (Na₂CO₃ solution);V = 400 cm3 = 0.4 dm3;n =  0.25 mol/dm3 * 0.4 dm3 = 0.1 mole;answer: 0.1 mole.

1 mol =1000cm30.25=?0.25×1000V=250.0cm3

n(Na2CO3) = C(Na2CO3) × V(Na2CO3) = 0.035 mol L-1 × 0.25 L = 0.00875 mol.m(Na2CO3) = n(Na2CO3) × M(Na2CO3) = 0.00875 mol × (2×23 + 12 + 3×16) = 0.9275...

can potassium carbonate remove calcium hardness from water instead of Sodium Carbonate,I have a KCl solution 20g/l with Calcium 500mg/l, I would normally use Na2CO3 to remove the calciumSolution:yes you...

n=m/Mr n=1.06/(23*2+12+16*3)=1.06/106=0,01 mol

Na2CO3 + 2AgNO3 = Ag2CO3 + 2NaNO3

Na2CO3 + Ca(NO3)2  CaCO3 + 2NaNO3 1. Find chemical amount of sodium carbonate (Na2CO3): m= Mn, then n= m/M; M(Na2CO3) = 22.992 + 12.01+16.003 = 105.99 (g/mol); n(Na2CO3) =...

Concentration = 0.75 mol / 0.5 L = 1.5 mol/L = 1.5 M

As both sodium carbonate and lead carbonate both are solid but their solubility are different due to presence of cations. As in sodium carbonate sodium is the cation which is...