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.
Answer:
The answer to your question is 8.74 g of He
Explanation:
Data
V = 2.4 x 10² L
P = 99 kPa
T = 0°C
mass = ?
Process
1.- Convert kPa to atm
P = 99 kPa = 99000 Pa
1 atm --------------- 101325 Pa
x --------------- 99000 Pa
x = (99000 x 1) / 101325
x = 0.977 atm
2.- Convert temperature to °K
°K = 273 + 0
°K = 273
3.- Substitution
PV = nRT
- Solve for n
n = PV / RT
n = (0.977)(2.4 x 10²) / (0.082)(273)
n = 24.48 / 22.386
n = 1.093 moles
4.- Calculate the grams of He
8 g -------------------- 1 mol
x -------------------- 1.093 moles
x = (1.093 x 8) / 1
x = 8.74 g
Answer:
the answer is k? lol thanks
The best answer choices can be A.because it reduction he eletrolytic from positive