1) UO2+4HF -> UF4+2H2O. How may liters of 3.55M HF will react with 3.47 kg of UO

ID: 789600 • Letter: 1

Question

1) UO2+4HF -> UF4+2H2O. How may liters of 3.55M HF will react with 3.47 kg of UO2 ??

2) If 0.4893 gms of a mix of Al(OH)3 and Mg(OH)2 is neutralised by 17.28 ml of 1M HNO3, what is the mass % of Al(OH)3 in the mixture?

3)Find the H3O+ and D3O+ concentrations of water and heavy water in the followig cases:

a] Kw = 1.139 x 10^-15 at 0 degree celcius, and 5.474+10^-14 at 50 degree celcius

Find the concentration and pH of H3O+ at 0 degree celcius and 50 degree celcius

b] autoionimzation constant for Deuterium oxide is 3.64 x 10^-16 at 0 degree celcius and 7.89 x 10^-15 at 50 degree celius

Find tge concentrations and pH of D3O+ at 0 degree celcius and 50 degree celius

1)3.47kg of UO2= 3.47*1000/270= 12.5926 moles

moles of HF required= 4* moles of UO2= 50.37 moles

volume of HF= 50.37/3.55 = 14.188 litres

2) moles of HNO3 used= 17.28*1 millimoles= 17.28 millimoles

let weight of Al(OH)3 be x gms

weight of Mg(OH)2= (0.4893- x)gms

one mole of Al(OH)3 requires 3 moles of HNO3 and one mole of Mg(OH)2 requires 2 moles of HNO3

3*(x/78) + 2*(0.4893- x)/58 = 17.28 millimoles

solving we get x= 0.10244

mass % of Al(OH)3= 20.936%

3) Kw= [H3O+][OH-] , [H3O+]=[OH-] for ionization

[H3O+]= sqrt(Kw)

at 0 degrees [H3O+] = 3.3749* 10^-8 M

pH= -log[H3O+] = 7.4717

at 50 degrees [H3O+] = 2.339*10^-7 M

pH= 6.6308

for D3O+ Kw= [D3O+][OD-] , [D3O+]=[OD-] for ionization

[D3O+]= sqrt(Kw)

at 0 degrees [D3O+] = 1.9078* 10^-8 M

pH= -log[H3O+] = 7.719

at 50 degrees [D3O+] = 8.882* 10^-8 M

pH= -log[D3O+] = 7.051

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