the correct answer is, it will rise, i just answered that question on studyisland.
<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
8 companies i looked it up on google and thats what it said hope its helpful
I believe you can simply add more cells in series to get a higher voltage. Not 100% sure on that.