Answer:
74.0 g/mol
Explanation:
Step 1: Write the generic neutralization reaction
HA + NaOH ⇒ NaA + H₂O
Step 2: Calculate the reacting moles of NaOH
At the equivalence point, 33.83 mL of 0.115 M NaOH react.
0.03383 L × 0.115 mol/L = 3.89 × 10⁻³ mol
Step 3: Calculate the moles of HA that completely react with 3.89 × 10⁻³ moles of NaOH
The molar ratio of HA to NaOH is 1:1. The reacting moles of HA is 1/1 × 3.89 × 10⁻³ mol = 3.89 × 10⁻³ mol.
Step 4: Calculate the molar mass of the acid
3.89 × 10⁻³ moles of HA have a mass of 0.288 g.
M = 0.288 g / 3.89 × 10⁻³ mol = 74.0 g/mol
Answer: The standard enthalpy of formation of 
is -252.1 kJ/mol.
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactantss}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f_%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f_%7Breactantss%7D%5D)
Putting the values we get :
![\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B2%5Ctimes%20H_f%7BFe%7D%2B3%5Ctimes%20H_f%7BH_2O%7D%5D-%5B1%5Ctimes%20H_f%7BFe_2O_3%7D%2B3%5Ctimes%20H_f%7BH_2%7D%5D)
![67.9kJ=[(2\times 0)+(3\times H_f{H_2O})]-[(1\times -824.2kJ/mol)+3\times 0kJ/mol)]](https://tex.z-dn.net/?f=67.9kJ%3D%5B%282%5Ctimes%200%29%2B%283%5Ctimes%20H_f%7BH_2O%7D%29%5D-%5B%281%5Ctimes%20-824.2kJ%2Fmol%29%2B3%5Ctimes%200kJ%2Fmol%29%5D)
Thus standard enthalpy of formation of is -252.1 kJ/mol.
This combination in non polar.
POH = - log [ OH⁻ ]
pOH = - log [ 1 x 10⁻⁹ ]
pOH = 9
Answer C
hope this helps!
