Heat water; mechanical (the movement of a turbine is based off of mechanical energy, not chemical or potential).
Answer:
<h3>The answer is 0.91 %</h3>
Explanation:
The percentage error of a certain measurement can be found by using the formula
![P(\%) = \frac{error}{actual \: \: number} \times 100\% \\](https://tex.z-dn.net/?f=P%28%5C%25%29%20%3D%20%20%5Cfrac%7Berror%7D%7Bactual%20%5C%3A%20%20%5C%3A%20number%7D%20%20%5Ctimes%20100%5C%25%20%5C%5C%20)
From the question
actual volume = 198.2 mL
error = 200 - 198.2 = 1.8
So we have
![P(\%) = \frac{1.8}{198.2} \times 100 \\ = 0.90817356205...](https://tex.z-dn.net/?f=P%28%5C%25%29%20%3D%20%20%5Cfrac%7B1.8%7D%7B198.2%7D%20%20%5Ctimes%20100%20%5C%5C%20%20%3D%200.90817356205...)
We have the final answer as
<h3>0.91 %</h3>
Hope this helps you
Answer : The pH will be, 3.2
Explanation :
As we known that the value of solubility constant of ferric hydroxide at
is, ![2.79\times 10^{-39}](https://tex.z-dn.net/?f=2.79%5Ctimes%2010%5E%7B-39%7D)
Amount or solubility of iron consumed = (1.800 - 0.3) mg/L = 1.5 mg/L
The given solubility of iron convert from mg/L to mol/L.
![1.5mg/L=\frac{1.5\times 10^{-3}g/L}{56g/mol}=2.7\times 10^{-7}mol/L](https://tex.z-dn.net/?f=1.5mg%2FL%3D%5Cfrac%7B1.5%5Ctimes%2010%5E%7B-3%7Dg%2FL%7D%7B56g%2Fmol%7D%3D2.7%5Ctimes%2010%5E%7B-7%7Dmol%2FL)
The chemical reaction will be:
![Fe(OH)_3\rightarrow Fe^{3+}+3OH^-](https://tex.z-dn.net/?f=Fe%28OH%29_3%5Crightarrow%20Fe%5E%7B3%2B%7D%2B3OH%5E-)
The expression of solubility constant will be:
![K_{sp}=[Fe^{3+}]\times [3OH^-]^3](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BFe%5E%7B3%2B%7D%5D%5Ctimes%20%5B3OH%5E-%5D%5E3)
Now put all the given values in this expression, we get the concentration of hydroxide ion.
![2.79\times 10^{-39}=(2.7\times 10^{-7})\times [3OH^-]^3](https://tex.z-dn.net/?f=2.79%5Ctimes%2010%5E%7B-39%7D%3D%282.7%5Ctimes%2010%5E%7B-7%7D%29%5Ctimes%20%5B3OH%5E-%5D%5E3)
![[OH^-]=1.5\times 10^{-11}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.5%5Ctimes%2010%5E%7B-11%7DM)
Now we have to calculate the pOH.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
![pOH=-\log (1.5\times 10^{-11})](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%281.5%5Ctimes%2010%5E%7B-11%7D%29)
![pOH=10.8](https://tex.z-dn.net/?f=pOH%3D10.8)
Now we have to calculate the pH.
![pH+pOH=14\\\\pH=14-pOH\\\\pH=14-10.8\\\\pH=3.2](https://tex.z-dn.net/?f=pH%2BpOH%3D14%5C%5C%5C%5CpH%3D14-pOH%5C%5C%5C%5CpH%3D14-10.8%5C%5C%5C%5CpH%3D3.2)
Therefore, the pH will be, 3.2