Explanation:
The volume of given lead nitrate solution is:
52.5 mL.
The amount of lead iodide formed is ---0.248 g.
To get the molarity of lead (II) ion follow the below-shown procedure:
The number of moles of lead iodide formed is:

0.000537 mole of lead iodide contains --- 0.000537 moles of lead (II) ion.
Thus, the number of moles are there, volume is there, and to get the molarity of lead (II) ion use the formula:

Molarity of lead iodide is --- 0.0102 M.
This is what is normally termed a single replacement reaction, although don't hold me to that. It could have changed to something more modern.
Answer:
True
Explanation:
Ethers react with HI to form the corresponding alcohols and alkyl iodides.
Similarly, ethyl ether react with excess of HI to form ethanol and ethyl iodide. But in the excess of HI as mentioned in the question, ethanol too undergoes
reaction with HI to form ethyl iodide.
<u>Hence, ethyl iodide is the only product when ethyl ether reacts with excess of HI for several hours.</u>
Answer:
I believe it is Potassium (K)
Explanation:
I did the math on a calculator and it was the closest atomic mass to potassium.