Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Explanation:
The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction
.

![\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
We are given:

Putting values in above equation, we get:
![-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]](https://tex.z-dn.net/?f=-1.0940%5Ctimes%2010%5E4%3D%5B%2816%5Ctimes%20-393.5%29%2B%2818%5Ctimes%20-285.8%29%5D-%5B%2825%5Ctimes%200%29%2B%282%5Ctimes%20%5CDelat%20H_f%7BC_8H_%7B18%7D%28l%29%7D%5D)

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Answer:
a. The atom will go from a two-dimensional configuration to a three dimensional configuration.
d. The bond angle will increase.
f. The number of unhybridized p orbitals will decrease.
Explanation:
Sp2 is the atomic bond in which orbitals mixes with only two orbitals. These orbitals form three sp2. When two carbon atoms are overlapped they form sigma bond by overlapping of sp2 bonds. Sp3 bond is created when there is one lone molecule available for combination. When the bonding is updated from sp2 to sp3 then unhybridized orbitals will decrease causing the bond angle to increase.
Answer:1
Explanation:i know cuz I got it right
Answer:
The other signal that would indicate the presence of a C= C bond appears close to 3100
.
Explanation:
Bands that appear above 3000
are often unsaturation diagnoses suggest. The band at 3000-
3100
is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.
Answer:
Here's what I get
Explanation:
1. Balanced equation
HQ⁻ + CH₃-Br ⟶ HQ-CH₃ + Br⁻
(I must use HQ because the Brainly Editor thinks the O makes a forbidden word)
2. Mechanism
HQ⁻ + CH₃-Br ⟶[HQ···CH₃···Br]⁻⟶ HQ-CH₃ + Br⁻
A C B
The hydroxide ion attacks the back side of the carbon atom in the bromomethane (A).
At the same time as the Q-H bond starts to form, the C-Br bond starts to break.
At the half-way point, we have a high-energy intermediate (C) with partially formed C-O and C-Br bonds.
As the reaction proceeds further, the Br atom drops off to form the products — methanol and bromide ion (B).
3. Energy diagram
See the diagram below.