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.
Answer:
V = 43.95 L
Explanation:
Given data:
Mass of CH₄ decomposed = 15.63 g
Volume of H₂O produced at STP = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → 2H₂O + CO₂
Number of moles of CH₄:
Number of moles = mass/molar mass
Number of moles = 15.63 g/ 16 g/mol
Number of moles = 0.98 mol
Now we will compare the moles of H₂O with CH₄.
CH₄ : H₂O
1 : 2
0.98 : 2×0.98 = 1.96 mol
Volume of hydrogen:
PV = nRT
1 atm × V = 1.96 mol × 0.0821 atm.L/mol.K × 273.15 K
V = 43.95atm.L / 1atm
V = 43.95 L
Answer:
semiconducting and tellurium
Explanation:
u did the test hope this helps babes
Answer:
FALSE
Since 0.385 < 0.526, the value for week 3 is accepted.
Explanation:
Qexp = (|Xq - Xₙ₋₁|)/w
where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data
First, the data are arranged in decreasing order, from highest to lowest:
3. 5.6
2. 5.1
8. 5.1
1. 4.9
6. 4.9
5. 4.7
7. 4.5
4. 4.3
Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3
Qexp = (|5.6 - 5.1|)/1.3 = 0.385
From tables, at 95% confidence level, for n = 8, Qcrit = 0.526
Since 0.385 < 0.526, the value for week 3 is accepted.
Answer is (3) both mass number and atomic number.
The notation is ₅₅¹³⁷Cs. The Cs represents the chemical symbol of Caesium element. The subscript number at the left hand side of the symbol indicates the atomic number. Hence, atomic number of Cs is 55. The superscript number at the left hand side of the symbol shows the mass number. Hence, the mass number of the Cs is 137.
