Answer:
The correct appropriate will be Option 1 (Acid anhydrides are less stable than esters so the equilibrium favors the ester product.)
Explanation:
- Acid anhydride, instead of just a carboxyl group, is typically favored for esterification. The predominant theory would be that Anhydride acid is somewhat more volatile than acid. This is favored equilibrium changes more toward the right of the whole ester structure.
- Extremely responsive than carboxylic acid become acid anhydride as well as acyl chloride. Thus, for esterification, individuals were most favored.
The other options offered are not relevant to something like the scenario presented. So, the solution here is just the right one.
First calculate number of Nitrogen atoms contained by 1 mole of Zn(NO₃)₂,
As one formula unit of Zn(NO₃)₂ contains 2 Nitrogen atoms, then one mole of Zn(NO₃)₂ will contain,
= 2 × 6.022 × 10²³
= 1.20 × 10²⁴ Atoms of Nitrogen / mole
Now, calculate number of Nitrogen Atoms in 0.150 moles of Zn(NO₃)₂.
As,
1 mole of Zn(NO₃)₂ contained = 1.20 × 10²⁴ atoms of N
Then,
0.15 mole of Zn(NO₃)₂ will contain = X atoms of N
Solving for X,
X = (0.15 mol × 1.20 × 10²⁴ atom) ÷ 1 mol
X = 1.80 × 10²⁴ Atoms of Nitrogen
Result:
So, 0.15 mole of Zn(NO₃)₂ contains 1.80 × 10²⁴ atoms of Nitrogen.
Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. ... When an equal number of atoms of an element is present on both sides of a chemical equation, the equation is balanced.
<h3>
Answer:</h3>
7.4797 g AlF₃
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃
[Given] 15.524 g KF
<u>Step 2: Identify Conversions</u>
[RxN] 6 mol KF = 2 mol AlF₃
Molar Mass of K - 39.10 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of Al - 26.98 g/mol
Molar Mass of KF - 39.10 + 19.00 = 58.1 g/mol
Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 5 sig figs.</em>
7.47966 g AlF₃ ≈ 7.4797 g AlF₃
