the answer to your question is
volume
In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
<span>2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
</span><span>E = +1.47
</span>
<span>Br(l) + 2e- = 2Br-
</span><span>E = +1.065
</span>
We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V
When the balanced equation for this reaction is:
2Fe + 3H2O → Fe2O3 + 3H2
and according to the vapour pressure formula:
PV= nRT
when we have P is the vapor pressure of H2O= 0.121 atm
and V is the volume of H2O = 4.5 L
and T in Kelvin = 52.5 +273 = 325.5 K
R= 0.08205 atm-L/g mol-K
So we can get n H2O
So, by substitution:
n H2O = PV/RT
= (0.121*4.5)/(0.08205 * 325.5) = 0.02038 gmol
n Fe2O3 = 0.02038 * (1Fe2O3/ 3H2O) = 0.00679 gmol
Note: we get (1FeO3/3H2O) ratio from the balanced equation.
we can get the Mass of Fe2O3 from this formula:
Mass = number of moles * molecular weight
when we have a molecular weight of Fe2O3 = 159.7
= 0.00679 * 159.7 = 1.084 g
∴ 1.084 gm of Fe2O3 will produced
B) ionic bond
(although in reality, every bond is fundamentally the sharing of a pair of electron. but due to unmutal electonegativity, the molecule becomes polar)
Answer:
Amount left after 25 days = 12.5 g
Explanation:
Given data:
Mass of sample = 400 g
Half life of sample = 5 days
Mass left after 25 days = ?
Solution:
First of all we will calculate the number of half lives passes in given time period.
Number of half lives = Time elapsed / Half life
Number of half lives = 25 days/ 5 days
Number of half lives = 5
At time zero = 400 g
At 1st half life = 400 g/2 = 200 g
At 2nd half life = 200 g/2 = 100 g
At 3rd half life = 100 g/2 = 50 g
At 4th half life = 50 g/2 = 25 g
At 5th half life = 25 g/2 = 12.5 g
