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.
16.7 g H₂O
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
<u>Chemistry</u>
<u>Stoichiometry</u>
<u>Step 1: Define</u>
[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)
[Given] 1.85 mol NaOH
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol NaOH → 1 mol H₂O
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
16.6685 g H₂O ≈ 16.7 g H₂O
The products will be
Zn is higher than hydrogen in the reactivity series. Thus, it will be able to displace hydrogen from the acid.
The equation of the reaction becomes:
Hydrogen gas is released as a result. In fact, it is one of the ways of preparing hydrogen gas in the laboratory.
More on chemical reactivity can be found here: brainly.com/question/9621716
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