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Molecular weight of vanillin = 152 g/mol
Futher molecular formula of vanillin is C8H8O3
Atomic weight of oxygen = 16 g/mol
Thus, 152 g of vanillin contains 16 g of oxygen
∴ 0.045 g (45 mg) of vanillin contains

= 0.00473 g
Also, number of moles of vanillin in 0.045 g sample =

Now, 1 mole = 6.023 X 10^23 molecules
∴ 2.96 X 10^-4 mole = 1.78 X 10^20 molecules
From molecular formula, it can be seen that 1 molecule of vanallin contain 3 atoms of oxygen
∴1.78 X 10^20 molecules contain 3 X 1.78 X 10^20 = 5.34 X 10^20 oxygen atoms
Answer: The standard enthalpy of formation of this isomer of
is -210.9 kJ
Explanation:
The given balanced chemical reaction is,

First we have to calculate the enthalpy of formation of
.

![\Delta H^o=[n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{O_2}\times \Delta H_f^0_{(O_2)+n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28CO_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%2Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%7D%5D)
where,
We are given:

Putting values in above equation, we get:
![-511.3kJ/mol=[(8\times -393.5)+(9\times -241.8)]-[(\frac{25}{2}\times 0)+(1\times \Delta H_f^0_{(C_8H_{18})}](https://tex.z-dn.net/?f=-511.3kJ%2Fmol%3D%5B%288%5Ctimes%20-393.5%29%2B%289%5Ctimes%20-241.8%29%5D-%5B%28%5Cfrac%7B25%7D%7B2%7D%5Ctimes%200%29%2B%281%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%7D)

Answer:
Explanation:Due to the mental pressure,
Due to peer pressure,
Lack of love and affection from the family members.
Influence from the T.V advertisement.