<span>It throws it off, you have to re-do the experiment. The NaOH M will be higher because you poured to much inside and made it pink!</span>
<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
Answer:
Q= Magnesium R= Chlorine
Explanation:
The element Q should be magnesium and R is chlorine.
An ionic compound is a compound that is formed by the combination of a metal and non-metal. Such bonds forms when there is a transfer of electrons from the metals to the non-metals. This leaves a net positive charge on the metal and a negative charge on the non-metal.
The electrostatic attraction leads to the formation of the bond.
To solve this problem, the hypothetical compound is QR₂
Mg Cl
2 8 2 2 8 7
So, Mg transfers 2 electrons to two atoms of chlorine.
This leads to the formation of the compound MgCl₂
This problem is providing the initial volume and pressure of a gas in an engine cylinder and asks for the final pressure once the volume of the gas has decreased due to a compression. At the end, the result turns out to be 11.7 atm.
<h3>Boyle's law</h3>
In chemistry, gas laws allow us to calculate pressure, volume, temperature or moles depending on a specified change and based on the concept and equation of the ideal gas, which derives the well-known gas laws; Boyle's, Charles', Gay-Lussac's and Avogadro's.
Thus, since this problem provides initial and final volume and initial pressure for us to calculate the final pressure, we understand we need to apply the Boyle's law as a directly proportional relationship between these two:
![P_1V_1=P_2V_2](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2)
Thus, we solve for the final pressure, P2, to get:
![P_2=\frac{P_1V_1}{V_2}=\frac{1.0atm*175mL}{15mL}\\ \\ P_2=11.7atm](https://tex.z-dn.net/?f=P_2%3D%5Cfrac%7BP_1V_1%7D%7BV_2%7D%3D%5Cfrac%7B1.0atm%2A175mL%7D%7B15mL%7D%5C%5C%20%5C%5C%20P_2%3D11.7atm)
Learn more about ideal gases: brainly.com/question/8711877