Answer: d) -705.55 kJ
Explanation:
Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.
![\Delta H=+1411.1kJ](https://tex.z-dn.net/?f=%5CDelta%20H%3D%2B1411.1kJ)
Reversing the reaction, changes the sign of ![\Delta H](https://tex.z-dn.net/?f=%5CDelta%20H)
![C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)](https://tex.z-dn.net/?f=C_2H_4%28g%29%2B3O_2%28g%29%5Crightarrow%202CO_2%28g%29%2B2H_2O%28l%29)
![\Delta H=-1411.1kJ](https://tex.z-dn.net/?f=%5CDelta%20H%3D-1411.1kJ)
On multiplying the reaction by
, enthalpy gets half:
![0.5C_2H_4(g)+1.5O_2(g)\rightarrow CO_2(g)+H_2O(l)](https://tex.z-dn.net/?f=0.5C_2H_4%28g%29%2B1.5O_2%28g%29%5Crightarrow%20CO_2%28g%29%2BH_2O%28l%29)
![\Delta H=\frac{1}{2}\times -1411.1kJ=-705.55kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20-1411.1kJ%3D-705.55kJ%2Fmol)
Thus the enthalpy change for the given reaction is -705.55kJ
Answer:
pH = 1.30
Explanation:
The reaction that would take place between HCl and NaOH is:
In this problem, the 0.1 moles of NaOH would react with 0.1 moles of HCl, <u>leaving 0.1 moles of HCl unreacted</u>.
These 0.1 HCl moles are dissolved in 2 liters, thus the concentration of H⁺ species is:
- [H⁺] = 0.1 mol / 2 L =0.05 M
Finally we <u>calculate the pH</u>:
Answer:
A) It tends to increase from top to bottom of a group