Tampilkan postingan dengan label chemistry class 11. Tampilkan semua postingan
Tampilkan postingan dengan label chemistry class 11. Tampilkan semua postingan

7. Consider the reaction: Cr2O72 + 14H+ + 6e- —> 2Cr3+ + 7H2O what is the quantity of electricity in coulombs needed to reduced 1 mol of Cr2O72 ?

7. Consider the reaction: Cr2O72 + 14H+ + 6e- —> 2Cr3+ + 7H2O what is the quantity of electricity in coulombs needed to reduced 1 mol of Cr2O72 ?SOLUTIONasked: what is the quantity of electricity in coulombs needed to reduced 1 mol of Cr2O72 ?answer:Cr2O72 + 14H+ + 6e- —> 2Cr3+ + 7H2OQ/F = mol . Vf . FQ/F = mol. Vf. FQ = 1 . 6 . 96500Q =...

Calculate the pH of a buffer prepared by mixing 300 cc of 0,3 M NH3 and 500 cc of 0,5 M NH4CI. Kb for NH3 = 1,8 × 10^-5

Calculate the pH of a buffer prepared by mixing 300 cc of 0,3 M NH3 and 500 cc of 0,5 M NH4CI. Kb for NH3 = 1,8 × 10^-5SOLUTIONKnown:NH3 = 0,3 MV NH3 = 300 ccNH4Cl = 0,5 MV NH4Cl = 500 ccAsked: Calculate the pH of a buffer ?Answer:Given, kb NH3 = 1.8x10^-5Number of moles of NH3 =...

Ionisation constant of CH3COOH is 1,7×10⁻⁵and concentration of H+ ions is 3,4×10-4. Then find out initial concentration of CH3COOH molecule ?

Ionisation constant of CH3COOH is 1,7×10⁻⁵and concentration of H+  ions is 3,4×10-4. Then find out initial concentration of CH3COOH molecule ?SOLUTIONKnown:CH3COOH = 1,7x10-5H+ = 3,4×10-4Asked: concentration of CH3COOH molecule ?CH3COOH —> CH3COO⁻ + H+x                   3,4x10-4         3x10-4K = [CH3COO-] [H+]/[CH3COOH]1,7x10-5 = [3,4x10-4] [3,4x10-4]/xx = 6,8x10-3so, the concentration of CH3COOH molecule...

A solution consists of 0.2M NH4OH and 0.2M NH4Cl. If Kb of NH4OH is 1.8 × 10^5, the [OH-] of the resulting solution is ?

A solution consists of 0.2M NH4OH and 0.2M NH4Cl. If Kb of NH4OH is 1,8x10-5, the [OH-] of the resulting solution is ?A. 3,6x10-5B. 3,2x10-5D. 1,8x10-5C. 0,9x10-5SOLUTIONNH4OH and NH4Cl is a buffer solution:POH = PKb + log [salt]/[base]Concentration of NH4OH = 0,2 MConcentration of NH4Cl = 0,2 MPKb of NH4OH = 1,8x10-5      POH = - log 1,8x10-5 +...

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