We are given the chemical reaction and the amount of reactant used for the process. We use these data together to obtain what is asked. We do as as follows:
0.882 mol H2O2 ( 1 mol O2 / 2 mol H2O2 ) = 0.441 mol O2 produced
Hope this answers the question.
No, they both refer to the ratio of moles in compounds of a chemical reaction. Along with mole ratio and molar ratio, this can also be called the mole-to-mole ratio.
Hope this helped :)
Answer:
![3.28*10^{-7}](https://tex.z-dn.net/?f=3.28%2A10%5E%7B-7%7D)
Explanation:
ΔG° = - 8.03 kJ/mol & ΔG = -45.28 kJ/mol
Thus:
Overall ΔG = (- 8.03 + 45.28) kJ/mol
ΔG = 37.25 kJ/mol
ΔG = 37250 J/mol
Temperature = 27 °C
= (27 + 273)K
= 300 K
Therefore :
ΔG = -RTInK
In K = ![\frac{\delta G}{RT}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cdelta%20G%7D%7BRT%7D)
In K = ![\frac{37250}{300*8.314}](https://tex.z-dn.net/?f=%5Cfrac%7B37250%7D%7B300%2A8.314%7D)
In K = ![\frac{37250}{2494.2}](https://tex.z-dn.net/?f=%5Cfrac%7B37250%7D%7B2494.2%7D)
In K = 14. 93
K = ![e^{-14.93}](https://tex.z-dn.net/?f=e%5E%7B-14.93%7D)
K = ![3.28*10^{-7}](https://tex.z-dn.net/?f=3.28%2A10%5E%7B-7%7D)
the standard free energy of the first reaction at 27 °C = ![3.28*10^{-7}](https://tex.z-dn.net/?f=3.28%2A10%5E%7B-7%7D)
Answer:
(a) 5.04; (b) 5.18; (c) 12.30
Explanation:
(a) pH of buffer
HA + H₂O ⇌ H₃O⁺ + A⁻
![\begin{array}{rcl}\text{pH}& = &\text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\& = &4.74 + \log\dfrac{0.2}{0.1}\\\\& = &4.74 + \log 2\\& = &4.74 + 0.30\\& = & \mathbf{5.04}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Ctext%7BpH%7D%26%20%3D%20%26%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%26%20%3D%20%264.74%20%2B%20%5Clog%5Cdfrac%7B0.2%7D%7B0.1%7D%5C%5C%5C%5C%26%20%3D%20%264.74%20%2B%20%5Clog%202%5C%5C%26%20%3D%20%264.74%20%2B%200.30%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B5.04%7D%5C%5C%5Cend%7Barray%7D)
(b) pH after addition of base
![\text{Moles of NaOH} = \text{0.002 L} \times \dfrac{\text{10 mol}}{\text{1 L }} = \text{0.02 mol}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20NaOH%7D%20%3D%20%5Ctext%7B0.002%20L%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B10%20mol%7D%7D%7B%5Ctext%7B1%20L%20%7D%7D%20%3D%20%5Ctext%7B0.02%20mol%7D)
HA + H₂O ⇌ H₃O⁺ + A⁻
I/mol: 0.1 0 0.2
C/mol: -0.02 +x +0.02
E/mol: 0.08 x 0.22
![\text{pH}& =4.74 + \log\dfrac{0.22}{0.08}\\\\& = & 4.74 + \log 2.75\\& = & 4.74 + 0.44\\& = & \mathbf{5.18}\\\end{array}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%26%20%3D4.74%20%2B%20%5Clog%5Cdfrac%7B0.22%7D%7B0.08%7D%5C%5C%5C%5C%26%20%3D%20%26%204.74%20%2B%20%5Clog%202.75%5C%5C%26%20%3D%20%26%204.74%20%2B%200.44%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B5.18%7D%5C%5C%5Cend%7Barray%7D)
(c) pH of water after addition of base
![\text{[OH$^{-}$]} = \dfrac{\text{0.02 mol}}{\text{1.002 L}} = \text{0.02 mol/L}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%24%5E%7B-%7D%24%5D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.02%20mol%7D%7D%7B%5Ctext%7B1.002%20L%7D%7D%20%3D%20%5Ctext%7B0.02%20mol%2FL%7D)
pOH} = -log0.02 = 1.70
pH = 14.00 - pOH = 14.00 - 1.70 = 12.30