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.
I believe so! If not tell me So I can delete this
Answer:
Explanation:
GIven that:
The activation energy = 250 kJ
k₁ = 0.380 /M
k₂ = ???
Initial temperature
1001 K
Final temperature
298 K
Applying the equation of Arrhenius theory.

where ;
R gas constant = 8.314 J/K/mol





/M .sec
Half life:
At 1001 K.


1.82368 secc
At 298 K:


Answer:
Observe: To model how hydrogen and oxygen react to make water, type “H2+O2” into the Reactants box and “H2O” into the Products box.
As the equation is written, which element is not in balanced?
oxygen isn't balanced...
Explanation:
in reactants there is 2hydrogens and 2 oxygen
but in product there is balanced no if hydrogen bt there is only one no of oxygen
Answer:
Moles of BCl₃ needed = 0.089 mol
Explanation:
Given data:
Moles of BCl₃ needed = ?
Mass of HCl produced = 10.0 g
Solution:
Chemical equation:
BCl₃ + 3H₂O → 3HCl + B(OH)₃
Number of moles of HCl:
Number of moles = mass/molar mass
Number of moles = 10.0 g/ 36.46 g/mol
Number of moles = 0.27 mol
Now we will compare the moles of HCl with BCl₃.
HCl : BCl₃
3 : 1
0.27 : 1/3×0.27 = 0.089 mol