Infinite because Sodium Chloride is neutral and will not neutralize sulfuric acid.
The answer is 1,2 mol.
100ml / 1000 = 0.1mol *1M = divide by 1000 moles = mass / molecular mass mass = mole x molecular mass of nacl mass = 0.1 x 58.5 mass = 5.85 answer is 5.85 grams
0.2% of anthrone was dissolved in 100ml of conc H2SO4.
h
3100-3200 kg
NaCl doesn't neutralize sulfuric acid.
The answer is 1,2 mol.
100ml / 1000 = 0.1mol *1M = divide by 1000 moles = mass / molecular mass mass = mole x molecular mass of nacl mass = 0.1 x 58.5 mass = 5.85 answer is 5.85 grams
0.2% of anthrone was dissolved in 100ml of conc H2SO4.
h
14.1 mL is required to titrate 10.00 ml of 0.526 M H2SO4.
3100-3200 kg
40g urea+ 80ml of 40%H2SO4! add 2g stannous chloride, heat untill solution is clear and make the volume upto 100ml. it is called foulger's reagent!
Grams NaOH?? Balanced equation. 2NaOH + H2SO4 --> Na2SO4 + 2H2O 4.9 grams H2SO4 (1 mole H2SO4/98.086 grams)(2 mole NaOH/1 mole H2SO4)(39.998 grams/1 mole NaOH) = 4.0 grams NaOH needed =================
Balanced equation first. 2Al + 3H2SO4 >> Al2(SO4)3 + 3H2 ( only onfo on H2SO4 and it does say; ' completely reacted. ' So, H2SO4 limits and drives the reaction )( if it did not say this you would have to find mol Al and see what limits ) 250 grams H2SO4 (1mol H2SO4/98.086g )(1mol Al2(SO4)3/3mol H2SO4 )(357.17g/1mol Al2(SO4)3 ) = 303 grams of Al2(SO4)3
ACTUALLY The answer is: 3H2SO4 + 2K3PO4 = 3K2SO4 + 2H3PO4 Because the last answer was not balanced correctly.
Strong base is added to neutralize the strong acid (H2SO4).