Hey there!
Na + H₂O → NaOH + H₂
First, balance O.
One on the left, one on the right. Already balanced.
Next, balance H.
Two on the left, three on the right. Let's add a coefficient of 2 in front of NaOH and a coefficient of 2 in front of H₂O, so we have 4 on each side.
Na + 2H₂O → 2NaOH + H₂
Lastly, balance Na.
One on the left, two on the right. Add a coefficient of 2 in front of Na.
2Na + 2H₂O → 2NaOH + H₂
This is our final balanced equation.
Hope this helps!
Answer:
the correct answer to your question is b
Carboxylic acids and alcohols have higher boiling point
than other hydrocarbons due to their polarity and from the fact that they form
very strong intermolecular hydrogen bonding. This is due to the large
difference in their electronegativity that forms between the oxygen and the
hydrogen atom.
Answer: 72.4 kJ/mol
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%28n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%2B%28n_%7BC_3H_8%7D%5Ctimes%20%5CDelta%20H_%7BC_3H_8%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]](https://tex.z-dn.net/?f=-2220.1%3D%5B%283%5Ctimes%20-393.5%29%2B%284%5Ctimes%20-241.8%29%5D-%5B%285%5Ctimes%200%29%2B%281%5Ctimes%20%5CDelta%20H_%7BC_3H_8%7D%29%5D)
Therefore, the heat of formation of propane is 72.4 kJ/mol
