Answer:
where is the option I can't see plz give the option first the I try to give answer
56Fe3+ is an iron ion with a charge of +3, meaning it has 23 electrons after 3 electrons are lost from the uncharged state. It loses the 4s2 electrons first, then one of the 3d electrons.
The student's question refers to 56Fe3+. 56Fe3+ is an iron ion with a charge of +3, which means it has lost three electrons. Understanding the Electron configuration is vital here. Normally, iron (Fe) has 26 electrons as its atomic number is 26. Each electron has a negative charge. But in 56Fe3+, 3 electrons are lost; hence, it will have 23 electrons.
Let's look further into this. The electron configuration of an uncharged Fe atom is 1s²2s²2p6 3s²3p6 4s²3d6. The iron atom loses the 4s2 electrons first (becoming Fe2+), then one of the 3d electrons (becoming Fe3+ or 56Fe3+ as indicated in the question).
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***a dense substance that is hard and incompressible ***
a high-energy state of matter made up of a swirling, ionized gas
a loose association of particles moving past one another
a dense substance that is hard and incompressible
Answer:
The % yield of the reaction is 73.8 %
Explanation:
To solve this, we list out the given variables thus
Mass of aluminium in the experiment = 2.5 g
mass of oxygen gas in the experiment = 2.5 g
Molar mass of aluminium = 26.98 g/mol
molar mass of oxygen O₂ = 32 g/mol
The reaction between aluminium and gaseous oxygen can be written as follows
4Al + 3O₂ → 2Al₂O₃
Thus four moles of aluminium forms two moles of aluminium oxide
Thus (2.5 g)÷(26.98 g/mol) = 0.093 mole of aluminium and
(2.5 g)÷(32 g/mol) = 0.078125 moles of oxygen
However four moles of aluminium react with three moles of oxygen gas O₂
1 mole of aluminum will react with 3/4 moles of oxygen O₂ and 0.093 mole of aluminium will react with 0.093*3/4 moles of O₂ = 0.0695 moles of Oxygen hence aluminium is the limiting reagent and we have
1 mole of oxygen will react with 4/3 mole of aluminium
∴ 0.078125 mole of oxygen will react with 0.104 moles of aluminium
Therefore 0.093 mole of aluminium will react with O₂ to produce 2/4×0.093 or 0.0465 moles of 2Al₂O₃
The molar mass of 2Al₂O₃ = 101.96 g/mol
Hence the mass of 0.0465 moles = number of moles × (molar mass)
= 0.0465 moles × 101.96 g/mol = 4.74 g
The of aluminium oxide Al₂O₃ is 4.74 g, but the actual yield = 3.5 g
Therefore the Percentage yield = ×100 = × 100 = 73.8 % yield