Answer:
1. pH = 1.23.
2. ![H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)](https://tex.z-dn.net/?f=H_2C_2O_4%28aq%29%20%2BOH%5E-%28aq%29%5Crightarrow%20HC_2O_4%5E-%28aq%29%2BH_2O%28l%29)
Explanation:
Hello!
1. In this case, for the ionization of H2C2O4, we can write:
![H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+](https://tex.z-dn.net/?f=H_2C_2O_4%5Crightleftharpoons%20HC_2O_4%5E-%2BH%5E%2B)
It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the pKa is:
![pKa=-log(Ka)=-log(5.90x10^{-2})=1.23](https://tex.z-dn.net/?f=pKa%3D-log%28Ka%29%3D-log%285.90x10%5E%7B-2%7D%29%3D1.23)
The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:
![pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23](https://tex.z-dn.net/?f=pH%3D1.23%2Blog%28%5Cfrac%7B0.347M%7D%7B0.347M%7D%20%29%5C%5C%5C%5CpH%3D1.23%2B0%5C%5C%5C%5CpH%3D1.23)
2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:
![n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol](https://tex.z-dn.net/?f=n_%7Bacid%7D%3D0.347mol%2FL%2A1.00L%3D0.347mol%5C%5C%5C%5Cn_%7Bacid%7D%5E%7Bremaining%7D%3D0.347mol-0.070mol%3D0.277mol)
It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:
![H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)](https://tex.z-dn.net/?f=H_2C_2O_4%28aq%29%20%2BOH%5E-%28aq%29%5Crightarrow%20HC_2O_4%5E-%28aq%29%2BH_2O%28l%29)
Which is also shown in net ionic notation.
Best regards!
They should identify the confounding variable.
Some condition that is not examined by the scientist might alter the experiment result. That condition is called confounding variable. If the method of the experiment same but result is very different, there should be unidentified confounding variable. It could be air humidity, temperature, ventilation, light, time of the year or anything that might not be seen by naked eye.
Try to redo the experiment with controlling variable as much as possible.
Answer: 0.100 m ![K_2SO_4](https://tex.z-dn.net/?f=K_2SO_4)
Explanation:
Elevation in boiling point is given by:
![\Delta T_b=i\times K_b\times m](https://tex.z-dn.net/?f=%5CDelta%20T_b%3Di%5Ctimes%20K_b%5Ctimes%20m)
= Elevation in boiling point
i= vant hoff factor
= boiling point constant
m= molality
1. For 0.100 m ![K_2SO_{4}](https://tex.z-dn.net/?f=K_2SO_%7B4%7D)
, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be ![3\times 0.100=0.300](https://tex.z-dn.net/?f=3%5Ctimes%200.100%3D0.300)
2. For 0.100 m ![LiNO_3](https://tex.z-dn.net/?f=LiNO_3)
, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be ![2\times 0.100=0.200](https://tex.z-dn.net/?f=2%5Ctimes%200.100%3D0.200)
3. For 0.200 m ![C_2H_8O_3](https://tex.z-dn.net/?f=C_2H_8O_3)
, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be ![1\times 0.200=0.200](https://tex.z-dn.net/?f=1%5Ctimes%200.200%3D0.200)
4. For 0.060 m ![Na_3PO_4](https://tex.z-dn.net/?f=Na_3PO_4)
, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be ![4\times 0.060=0.24m](https://tex.z-dn.net/?f=4%5Ctimes%200.060%3D0.24m)
Thus as concentration of solute is highest for
, the elevation in boiling point is highest and thus has the highest boiling point.
calculate moles of both reagents given and the moles of FeS that each of them would form if they were in excess
moles = mass / molar mass
moles Fe = 7.62 g / 55.85 g/mol
= 0.1364 moles
1 mole Fe produces 1 mole FeS
Therefore 7.62 g Fe can form 0.1364 moles FeS
moles S = 8.67 g / 32.07 g/mol
= 0.2703 moles S
1 mole S can from 1 moles FeS
So 8.67 g S can produce 0.2703 moles FeS
The limiting reagent is the one that produces the least product. So Fe is limiting.
The maximum amount of FeS possible is from complete reaction of all the limiting reagent.
We have already determined that the Fe can form up to 0.1364 moles of FeS, so this is max amount of FeS you can get.
Convert to mass
hope this helps :)