1.1 Moles / 0.5 Liters = 0.22 Molarity
I have attached the answer here. hope this helps!
let me know if you have any question
<span>The concentration of pb2+ = 1.00mg/ml
Diluted Solution is 6.0 x 102 ml = 612 ml
Volume of the concentration of pb2+ is 0.054 mg/l is v
(vL)(1.00mg/ml) = (.612L)(0.054mg/l)
Volume = 0.033048L
Volume of the concentration of pb2+ is 0.054 mg/l = 33.048 ml.</span>
H2SO4 + Na2CO3 → Na2SO4 + CO2 + H2O
The molarity of sulfuric acid if 1.78 L were used in the above reaction is
0.453 M (answer 2)
Calculation
find the moles of water produced = mass/molar mass
= 14.5 g /18 g/mol = 0.806 moles
by use of of mole ratio between H2So4 to H2O which is 1:1 the moles of H2SO4 is also = 0.806 moles
Molarity of H2SO4 is therefore = moles/volume in liters
= 0.806 mol/ 1.78 L = 0.453 M (answer 2)
