1.What mode of decay is probable for 41/18 Ar ? What mode of decay is probable f
ID: 990767 • Letter: 1
Question
1.What mode of decay is probable for 41/18 Ar ? What mode of decay is probable for ? a.electron capture b. beta emission C.positron emission D.alpha emission
2.
The standard cell potential, E°cell, for a reaction in which two electrons are transferred between the reactants is +1.33 V. Calculate the standard free energy change, G°, in kJ for this reaction and determine if it is spontaneous or nonspontaneous at 25°C.
A.2.57 × 102 kJ, nonspontaneous B. 2.57 × 102 kJ, spontaneous C. 1.28 × 102 kJ, spontaneous D. +2.57 × 102 kJ, nonspontaneous
3. List the given metal ions in order of increasing strength as oxidizing agents, based on the following standard reduction potentials, E°red .
A. Al3+ < Fe2+ < Zn2+ < Ag+ B. Ag+ < Fe2+ < Zn2+ < Al3+ C. Zn2+ < Fe2+ < Ag+ < Al3+ D. Al3+ < Zn2+ < Fe2+ < Ag+Explanation / Answer
Answer 1)
41/18Ar23 disintegrates by beta minus decay to excited levels and ground state levels of K-41, 41/19K22.
Correct optionis B: beta emission
Answer 2)
Given that, E°cell = +1.33 V and n = 2, T = 25oC = 298.15K,
F = Faraday's constant = 96,485 C/mol
DeltaGo = -nFE°cell
DeltaGo = -2 × 96,485 C/mol × (+1.33 V)
DeltaGo = -256650.1 J = -256.650 KJ
DeltaGo = -256.650 KJ = -2.56 × 102 KJ
Since delta G is negative reaction is spontaneous. Option B is correct.
Answer 3)
Correct answer for this question is option D
Al3+ < Zn2+ < Fe2+ < Ag+
Ag+ + e- à Ag(s) -------------Eo = 0.80V
Fe2+ + 2e- à Fe(s)------------- Eo = -0.45V
Zn2+ + 2e- à Zn(s)------------- Eo = -0.76V
Al3+ + 3e- à Al(s)------------- Eo = -1.66V
A.
Al3+ < Fe2+ < Zn2+ < Ag+
B.
Ag+ < Fe2+ < Zn2+ < Al3+
C.
Zn2+ < Fe2+ < Ag+ < Al3+
D.
Al3+ < Zn2+ < Fe2+ < Ag+
A.
Al3+ < Fe2+ < Zn2+ < Ag+
B.
Ag+ < Fe2+ < Zn2+ < Al3+
C.
Zn2+ < Fe2+ < Ag+ < Al3+
D.
Al3+ < Zn2+ < Fe2+ < Ag+
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