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Explanation:
The species or elements which gain electrons and reduces itself are known as oxidizing agent or oxidant.
Ability of an element to act as an oxidizing agent depends on its electrode potential.
The electrode potential of 
is 0.52 V.
The electrode potential of 
is -0.41 V.
The electrode potential of 
is -2.38 V.
Greater is the value of electrode potential, stronger will be the oxidizing agent.
Therefore, rank of these species by their ability to act as an oxidizing agent are as follows.
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Answer:
There are 120 possible ways.To calculate the number of permutations here, where order is important and repetition is not allowed, we use the following formula:Number of permutations = n! / (n - r)! = 5!/0! = 12345 / 1 (note: "!" means "factorial" and 0! equals 1) = 120/1 = 120.For a complete list see below. Let a,b,c,d & e represent the 5 players and their order determine their position:{a,b,c,d,e}
Explanation:
Given the temperature 746 K and activity of Pb equal to 0.055. The mole fraction of Pb is 0.1. So, the mole fraction of Sn = 0.9.Activity coefficient, γ = 0.055 / 0.1 = 0.55.The expression for w=ln〖γ_Pb x RT〗/(X_Sn^2 )=(-0.5978 x 8.314 J/(mol K ) x 746 K)/(0.9 x 0.9)= -4577.7 J= -4578 J
Now we use the computed value above and new temperature 773 K. The mole fraction of Sn and Pb are 0.5 and 0.5 respectively. Calculate the activity coefficient in the following manner.lnγ_Sn=w/RT X_Pb^2=(-4578 J)/(8.314 J/mol x 773 K) x 0.5 x 0.5= -0.718lnγ_Sn=exp(-0.178)=0.386The activity of Sn= γ_Sn x X_Sn=0.386 x 0.5=0.418
w of the system is -4578 J and the activity of Sn in the liquid solution of xsn at 500 degree Celsius is 0.418
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