Answer:
<u><em>Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.</em></u>
A chemistry student weighs out 0.0941 g of hypochlorous acid (HClo) into a 250. ml. volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits mL.
Explanation:
1 mole HClO = 74.44g
0.0941g =
= 0.00126 moles
Concentration = no. of moles/volume in L
Hence, Concentration of HClO = 0.00126/ 0.250L
= 0.005M.
C1V1 =C2V2
0.005 × 250 mL = 0.2 × V2
<u><em>Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.</em></u>
Answer : The internal energy change is -2805.8 kJ/mol
Explanation :
First we have to calculate the heat gained by the calorimeter.
![q=c\times (T_{final}-T_{initial})](https://tex.z-dn.net/?f=q%3Dc%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29)
where,
q = heat gained = ?
c = specific heat = ![5.20kJ/^oC](https://tex.z-dn.net/?f=5.20kJ%2F%5EoC)
= final temperature = ![27.43^oC](https://tex.z-dn.net/?f=27.43%5EoC)
= initial temperature = ![22.93^oC](https://tex.z-dn.net/?f=22.93%5EoC)
Now put all the given values in the above formula, we get:
![q=5.20kJ/^oC\times (27.43-22.93)^oC](https://tex.z-dn.net/?f=q%3D5.20kJ%2F%5EoC%5Ctimes%20%2827.43-22.93%29%5EoC)
![q=23.4kJ](https://tex.z-dn.net/?f=q%3D23.4kJ)
Now we have to calculate the enthalpy change during the reaction.
![\Delta H=-\frac{q}{n}](https://tex.z-dn.net/?f=%5CDelta%20H%3D-%5Cfrac%7Bq%7D%7Bn%7D)
where,
= enthalpy change = ?
q = heat gained = 23.4 kJ
n = number of moles fructose = ![\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BMass%20of%20fructose%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20fructose%7D%7D%3D%5Cfrac%7B1.501g%7D%7B180g%2Fmol%7D%3D0.00834mole)
![\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole](https://tex.z-dn.net/?f=%5CDelta%20H%3D-%5Cfrac%7B23.4kJ%7D%7B0.00834mole%7D%3D-2805.8kJ%2Fmole)
Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole
Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.
Formula used :
![\Delta H=\Delta U+\Delta n_gRT](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5CDelta%20U%2B%5CDelta%20n_gRT)
or,
![\Delta U=\Delta H-\Delta n_gRT](https://tex.z-dn.net/?f=%5CDelta%20U%3D%5CDelta%20H-%5CDelta%20n_gRT)
where,
= change in enthalpy = ![-2805.8kJ/mol](https://tex.z-dn.net/?f=-2805.8kJ%2Fmol)
= change in internal energy = ?
= change in moles = 0 (from the reaction)
R = gas constant = 8.314 J/mol.K
T = temperature = ![27.43^oC=273+27.43=300.43K](https://tex.z-dn.net/?f=27.43%5EoC%3D273%2B27.43%3D300.43K)
Now put all the given values in the above formula, we get:
![\Delta U=\Delta H-\Delta n_gRT](https://tex.z-dn.net/?f=%5CDelta%20U%3D%5CDelta%20H-%5CDelta%20n_gRT)
![\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K](https://tex.z-dn.net/?f=%5CDelta%20U%3D%28-2805.8kJ%2Fmol%29-%5B0mol%5Ctimes%208.314J%2Fmol.K%5Ctimes%20300.43K)
![\Delta U=-2805.8kJ/mol-0](https://tex.z-dn.net/?f=%5CDelta%20U%3D-2805.8kJ%2Fmol-0)
![\Delta U=-2805.8kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20U%3D-2805.8kJ%2Fmol)
Therefore, the internal energy change is -2805.8 kJ/mol
360 seconds?
i’m guessing that is the answer as the question is unreasonable
Answer: 30.06% is your answer.
I hope this helps.
Stay safe and have a good day :D