Empirical formula of ionic compound is FeO. In which the composition of atoms is 1 : 1.
Empirical formula of an ionic compound is defined as the formula which gives whole number ratio of atoms of various elements present in molecule of compund.
mass of iron in compound = 34.95 g
molar mass of iron = 55.8 g
mass of oxygen in compound = 15.05 g
molar mass of oxygen = 32 g
number of moles of iron present in the compound are ratio of mass of iron in compound/ molar mass of iron
number of moles of iron in compound= 34.95 / 55.8 = 0.6263 ~ 1
number of moles oxygen in compound= 15.05/ 32 = 0.473 ~ 0.5
the ratio of the number of oxygen atoms to number of iron atoms present in one formula unit of iron compund is 2×0.5 / 1 = 1 : 1
Hence , the required empirical formula of iron compound is FeO.
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Physical change- evaporation , condensation
Chemical change- combustion , neutralization
Answer:
(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔2Na[Al(OH)4](aq) + 3H2(g)
∴ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11
(2) H2O(l) + SO3(g) ↔ H2SO4(aq)
∴ Kc = [ H2SO4 ] / PSO3 = 0.0123
(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)
∴ Kc = Kc = 1 / PO2∧6
Explanation:
(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔ 2Na[Al(OH)4](aq) + 3H2(g)
∴ O / Al: 0 → +2 ≡ 2e-
Na: +1 → +2
∴ R / H: +1 → 0
2 - Al - 2
2 - Na - 1
8 - O - 8
14 - H - 14
⇒ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11
(2) H2O(l) + SO3(g) ↔ H2SO4(aq)
1 - S - 1
4 - O - 4
2 - H - 2
⇒ Kc = [ H2SO4 ] / PSO3 = 0.0123
(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)
8 - P - 8
12 - O - 12
⇒ Kc = 1 / PO2∧6
Answer:
In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.
Explanation:
The equation used to calculate the constant for first order kinetics:
.....(1)
Rate law expression for first order kinetics is given by the equation:
![t=\frac{2.303}{k}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2.303%7D%7Bk%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
......(2)
where,
k = rate constant
=Half life of the reaction = 
t = time taken for decay process = ?
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the reactant = 0.163 M
[A] = amount left after time t = 66.8% of
[A]=
t = 1,409.19 s
1 minute = 60 sec
In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.
