PH of acidic buffer = pKa + log [CH₃COONa - HCl] / [CH₃COOH + HCl]
pKa of CH₃COOH = 4.74
Concentration of acetic acid in buffer = 2.0 M
Concentration of sodium acetate = 1.0 M
Concentration of HCl must add = x
pH = 4.74 + log (1-x) / (2+x) = 4.11
x = concentration of HCl must be added = 0.43 M
number of moles of HCl = M * V = 0.43 * 1 = 0.43 mol
mass of HCl must be added = 0.43 * 36.5 = 15.7 g
to neutralize 1 mole of H2 S o4 we need one mole of any if we are having 50 grams of H2 S o4 it means the mole of H2 S o4 in 50 gram will be 50×40)/98 hence
utilising 50 grams of H2 S o4 we need approximately 20. 5 gm of Naoh
A. Radiation
Air in a car is heated up with the radiator and then blown out with an air pump.
Answer:
1.915e23 Atoms
Explanation:
There are 6.02214076e23 is the mass of a mol in amu. Since the amu of lead is 207.2, and there is 65.89 grams of it, you can find that it creates approximately 1.914 atoms.
first we find the mass of the final solution
we know that
density=mass/volume
we know the density and volume
1=m/100
m=100g
now we calculate the mass ofNa2CO3
2% of 100 is equal with 2g of Na2CO3
