Molarity = moles of solute/volume of solution ( so, not a great molarity expected )
4.60 grams H2SO4 (1mol H2SO4/98.086g) = 0.0469 moles/450ml
= 1.04 X 10^-4 Molarity.
The answer is 0,255 litres.
h2so4 can reduce fehling solution
You need 252 g sulfuric acid.
Initially, the mass of H2SO4 required to prepare 5.8 liters of 1.5 molar solution should be calculated. Number of moles present in 5.8 L of 1.5 molar solution = 1.5 mol L-1 x 5.8 L= 8.7 molMolar mass of H2SO4 = 98 g mol-1Therefore, mass of H2SO4 in the above solution = 8.7 mol x 98 g mol-1= 852.6 gMass of H2SO4 in the original solution per litre = 1.531 g x 32/100= 0.48992 gVolume of sulphuric acid required to prepare 1.5 molar solution = 852.6 g/0.48992 g = 1740.3 L
I think you just use C1V1 = C2V2 , but don't forget the stoichiometric ratio of1 NaOH 0.5 H2SO4NaOH:C1 = .1650 mol/LV1 = 26.48 mLH2SO4:C2 = ? (mol/L of H2SO4)V2 = 28.22 mL26.49 x .1650 (mmoles of OH-) = 2 * C2 x 28.22 (mmoles of H+ from H2SO4) =C2 = 0.5 * 0.1548848335 = 0.07744 (mol/L H2SO4)
The chemical reaction is:CuCl2 + H2SO4 = CuSO4 + 2 HCl(g)The gas is hydrogen chloride.
The answer is 0,526 mL.
You need 49,8 mL H2SO4 6,4M.
Assuming pH on just simple concentration. - log(0.01 M H2SO4) = 2 pH =====
The answer is 7,66 L.
The answer is 293,64 mL.
how will make solution for 0.005 h2so4
100% pure H2SO4 doesn't exist
For a detailed answer visit: algebra.com/algebra/homework/word/mixtures/Mixture_Word_Problems.faq.question.786696.html
The answer is 1,2 mol.
Sulphuric Acid requires, I believe 27.2ml to make a 1N solution.
The first solution is more concentrated because it contains 6 moles of H2SO4 per one liter of solution. The second solution is less concentrated because it contains 0.1 moles of H2SO4 in one liter. In equal amounts of each example, the first would have more H2SO4.
h2so4 can reduce fehling solution