Use PV =nRT
so P = nRT/V
= 1 mole(0.08205 L atm/K mol)(1000K) / 2 L
= 41 atm
Oxidation state of I is (-1) and for CO it is zero. Let's assume that the oxidation state of Fe in Fe(CO)₄I₂<span> (s) is x. For whole compound, the charge is zero.
Sum of oxidation numbers in all elements = Charge of the compound.
Here we have 1Fe , 4CO and 2I
hence we can find the oxidation state as;
x + 4*0 + 2*(-1) = 0
x + 0 - 2 = 0
x = +2
Hence the oxidation state of Fe in product </span>Fe(CO)₄I₂ (s) is +2.
Same as we can find the oxidation state (y) of Fe in Fe(CO)₅(s).
y + 5*0 = 0
y = 0
Since oxidation state of Fe increased from 0 to +2, the oxidized element is Fe in the given reaction.
The half life equation is -->P(t) = Pi (0.5) ^ (t/c)
c is equal to the element to reach its half-life (5 seconds)t is equal to the duration of time the element is expose to (20 seconds)Pi is the initial amount (340)0.5 is the base of this exponential function to represent half-life.P(t) is the expression for the function of time
P(20) = 340 (0.5)^20/5P(20) = 340 (0.5) ^4P(20)= 21.25 grams
Fraction = P(t)/Pi = P(20)/Pi =21.25/340 =1/16
Therefore, when given 20 seconds, 340 grams of Fluorine-21 will degrade to 21.25 grams OR 1/16 of its original mass.
Hope this method helps! (This is my answer btw, I think you may have accidentally posted twice?)
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