Chemistry NCERT Exemplar Solutions Class 12th Chapter Five
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New answer posted
2 months agoContributor-Level 10
Moles of Fe3O4 =
Moles of CO =
So limiting reagent = Fe3O4
So moles of Fe formed = 60
Weight of Fe = 60 * 56 = 3360
New answer posted
2 months agoContributor-Level 10
Millie q. of H2SO4 used by NH3 = 12.5 * 1 * 2 = 25
So millimoles of N = 25
Moles of N = 25 * 10-3
Wt of N = 14 * 25 * 10-3
% of N =
New answer posted
2 months agoContributor-Level 10
Where, K2 is at 310 K and K1 at 300K
K2 = K1 * 103
K1 = K2 * 10-3
x = 1
New answer posted
2 months agoContributor-Level 10
At anode (oxidation)
2H2O O2 (g) + 4H+ + 4e-
At cathode (Reduction)
2H+ + 2e- H2 (g)
No. of gm- equivalents =
New answer posted
2 months agoContributor-Level 10
Mass = d * v = 1.02 * 1.2 = 1.224gm
Moles of acetic acid = 0.0204 moles in 2L
So molality = 0.0102 mol/kg
i = 1 + a for acetic acid
0.0198 = (1 + a) * 1.85 * 0.0102
α = 0.04928 = 5%
New answer posted
2 months agoContributor-Level 10
85gm NH3 = 5 moles of NH3
Enthalpy change for 1 mol = 23.4 kJ
Then enthalpy change for 5 mol = 23.4 * 5 = 117 kJ
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