Answer:
The standard enthalpy of formation of this isomer of 
is -220.1 kJ/mol.
Explanation:
The given chemical reaction is as follows.
The expression for the entropy change for the reaction is as follows.
![\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Brxn%7D%3D%5B8%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28CO_%7B2%7D%29%20%2B9%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28H_%7B2%7DO%29%5D-%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%2B%20%5Cfrac%7B25%7D%7B2%7D%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28O_%7B2%7D%29%5D)
Substitute the all values in the entropy change expression.
![-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]](https://tex.z-dn.net/?f=-5104.1kJ%2Fmol%3D%5B8%28-393.5%29%2B9%28-241.8%29kJ%2Fmol%5D-%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%2B%20%5Cfrac%7B25%7D%7B2%7D%280%29kJ%2Fmol%5D)
Therefore, The standard enthalpy of formation of this isomer of is -220.1 kJ/mol.
Answer:
128.4 m
Explanation:
3.604m + 104.29m + 3.1m + 17.41m
Add all the values
= 128.404 m
The significant figure rule for addition is for the sum to have the same number of decimal places as the value with the least number of decimal places. In the addition sentence 3.604m + 104.29m + 3.1m + 17.41m, the value with the least number of decimal places is 3.1, which has 1 decimal place. Therefore, we round our sum so that it also has 1 decimal place.
128.404 m
= 128.4 m
I hope this helps!
Answer:
Explanation:
125 cm3 -> 0.125 L
M = no. mole / volume (L)
M = 0.25 / 0.125 = 2 M
<span>Enthalpy is regarding the amount of heat that is given off or taken in during the process of a reaction, while entropy is about the disorderliness of a reaction.
Both are related in the equation ∆G=∆H-T∆S, where ∆G is the Gibbs free energy. So we can say that a reaction is both enthalpy and entropy driven. It's like, both of them are interlinked with each other. </span>
Answer:
The products formed are
2-methyl-3-isopropylnonan-3-ol and 2-methyloctan-2-ol
The products should be in equal ratio since an equipmolar amount of acetone and 2,4-dimethylpentan-3-one was initially present as reactants
Explanation:
Ketones reacts with gridnard reagents to form tertiary alcohol, and since acetone and 2,4-dimethylpentan-3-one are both ketones, gridnard reagent reduced both to tertiary alcohols:2-methyl-3-isopropylnonan-3-ol and 2-methyloctan-2-ol
