Products develop much faster than regular reactants
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Answer:
Ge: [Ar] 3d10 4s2 4p2 => 6 electrons in the outer shell
Br: [Ar] 3d10 4s2 4p5 => 7 electrons in the outer shell
Kr: [Ar] 3d10 4s2 4p6 => 8 electrons in the outer shell
Explanation:
The electron affinity or propension to attract electrons is given by the electronic configuration. Remember that the most stable configuration is that were the last shell is full, i.e. it has 8 electrons.
The closer an atom is to reach the 8 electrons in the outer shell the bigger the electron affinity.
Of the three elements, Br needs only 1 electron to have 8 electrons in the outer shell, so it has the biggest electron affinity (the least negative).
Ge: needs 2 electrons to have 8 electrons in the outer shell, so it has a smaller (more negative) electron affinity than Br.
Kr, which is a noble gas, has 8 electrons and is not willing to attract more electrons at all, the it has the lowest (more negative) electron affinity of all three to the extension that really the ion is so unstable that it does not make sense to talk about a number for the electron affinity of this atom.
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<h3>
Answer:</h3>
Limiting reagent: Potassium iodide
Mass of the precipitate (PbI₂) is 4.453 g
<h3>
Explanation:</h3>
We are given;
- 60.0 mL of 0.322 M potassium iodide
- 20.0 mL of 0.530 M lead () nitrate
We are required to identify the limiting reactant and determine the mass of the precipitate formed.
<h3>Step 1: Write the balanced equation for the reaction</h3>
- The balanced equation for the reaction between potassium iodide and lead (II) nitrate is given by;
2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂(s)
<h3>Step 2: Determine the number of moles of the reagents</h3>
Moles of KI
Moles = Molarity × volume
Moles of KI = 0.322 M × 0.060 L
= 0.01932 moles
Moles of KNO₃
Moles = 0.530 M × 0.020 L
= 0.0106 M
From the equation;
- 2 moles of KI reacts with 1 mole of Pb(NO)₂
- Therefore; 0.01932 moles of KI will require 0.00966 moles of Pb(NO₃)₂
- This means, KI is the limiting reagent while Pb(NO₃)₂ is the excess reagent.
<h3>Step 3: Determine the mass of the precipitate PbI₂</h3>
2 moles of KI reacts to produce 1 mole of PbI₂
Therefore;
Moles of PbI₂ = Moles of KI ÷ 2
= 0.01932 moles ÷ 2
= 0.00966 moles
But molar mass of PbI² is 461.01 g/mol
Therefore;
Mass of PbI₂ = 0.00966 moles × 461.01 g/mol
= 4.453 g
Therefore, the mass of the precipitate formed (Pbi₂)is 4.453 g
The formula for density is density =
. Substituting the given, the density is 162.69 g/cm3.