Answer:
Angle ABE measures 27°.
Explanation:
Refer to the attached diagram related to this question.
The given values are ∠ABE=2n+7 and ∠EBF=4n-13.
Clearly seen in the diagram, ∠ABE and ∠EBF are equal in measure.
Move variable components to one side of the equation.
Split both sides by 2.
The solution for n arrives at 10.
The next step is to calculate ∠ABE.
Consequently, the measurement of angle ABE is 27°.
Answer: Rearrange the lone pairs of electrons from the outer atom(s) to create double or triple bonds with the central atom.
Explanation:
Response:
The pKa value is 13.0.
Clarification:
pKa + pKb = 14
For trimethylamine, Kb = 6.3 × 
Calculating pKb: pKb = - log (6.3 × )
= 1.0
Thus, pKa = 14 - pKb = 14 - 1.0
pKa = 13.0
Verification: The typical range for pKa in weak acids is from 2 to 13.
Answer:
The enthalpy of the second intermediate equation is altered by halving its value and changing the sign.
Explanation:
Let's examine both the first and second intermediate reactions alongside the overall equation concerning the examined process;
First reaction;
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
Second reaction;
2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ
Thus, the overall reaction becomes;
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH =?
According to Hess's law, which states that the total heat change in a reaction is equal to the sum of the heat changes for each step, we cannot simply sum the enthalpies for this overall reaction. Instead, we obtain the overall enthalpy by halving the second intermediate reaction's enthalpy and changing its sign before adding, as illustrated below;
Enthalpy of Intermediate reaction 1 + ½(-Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction
