You should take note that the question is about stability. A compound is stable if it does not easily react with other elements. Hence, its reactivity must be low. As you move down the group, reactivity decreases. So, the halide at the very bottom is the least reactive. It would then be logical that the most stable conjugate base is I⁻ and the least stable conjugate base is the most reactive which is F⁻.
Iron (iii) chloride is obtained by vapor condensation from the reaction between chlorine gas and iron fillings.
<h3>How can iron (iii) chloride be formed from iron fillings?</h3>
Iron (ii) chloride can be formed from iron fillings in the laboratory as follows:
- Iron fillings + Cl₂ → FeCl₃
Chlorine gas is introduced into a reaction vessel containing iron fillings and the iron (iii) chloride vapor formed is obtained by condensation.
In conclusion, iron (iii) chloride is formed by the the direct combination of iron fillings and chlorine gas.
Learn more about iron (iii) chloride at: brainly.com/question/14653649
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A feature that an iron metal has is a sea of electrons
Answer:
Mass = 55.52 g
Explanation:
Given data:
Number of atoms of Li = 4.81×10²⁴ atom
Number of grams = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
For Li:
4.81×10²⁴ atom × 1 mol / 6.022 × 10²³ atom
8 moles
Mass in gram:
Mass = number of moles × molar mass
Mass = 8 mol × 6.94 g/mol
Mass = 55.52 g
Answer: Theoretical Yield = 0.2952 g
Percentage Yield = 75.3%
Explanation:
Calculation of limiting reactant:
n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol
pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol
- n-trans-cinnamic acid is the limiting reactant
The molar ratio according to the equation mentioned is equals to 1:1
The brominated product moles is also = 9.584*10⁻⁴ mol
Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)
= (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g
Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952
= 75.3%
