Answer:
1383.34 kJ/mol is the energy released on combustion of the organic compound.
Explanation:
Mass of an organic compound = 0.6654 g
Molar mass of organic compound = 46.07 g/mol
Moles of an organic compound = ![\frac{0.6654 g}{46.07 g/mol}=0.01444 mol](https://tex.z-dn.net/?f=%5Cfrac%7B0.6654%20g%7D%7B46.07%20g%2Fmol%7D%3D0.01444%20mol)
Let heat evolved during burning of 0.6654 grams of an organic compound be -Q.
Heat absorbed by calorimeter = Q' = -Q
The total heat capacity of the calorimeter all its contents = C
C = 3576 J/°C
Change in temperature of the calorimeter =
ΔT = 30.589°C - 25.000°C = 5.589°C
![Q'=C\times \Delta T](https://tex.z-dn.net/?f=Q%27%3DC%5Ctimes%20%5CDelta%20T)
![Q'=3576 J/^oC\times 5.589^oC=19,975.536 J=19.975 kJ](https://tex.z-dn.net/?f=Q%27%3D3576%20J%2F%5EoC%5Ctimes%205.589%5EoC%3D19%2C975.536%20J%3D19.975%20kJ)
Q' = 19.975 kJ
Q = -19.975 kJ (negative sign; energy released)
0.01444 moles of an organic compound gives 19.975 kilo Joule.
The 1 mole of an organic compound will give : ![\Delta H_{comb}](https://tex.z-dn.net/?f=%5CDelta%20H_%7Bcomb%7D)
![\Delta H_{comb}=\frac{-19.975 kilo Joule}{0.01444 mol}](https://tex.z-dn.net/?f=%5CDelta%20H_%7Bcomb%7D%3D%5Cfrac%7B-19.975%20kilo%20Joule%7D%7B0.01444%20mol%7D)
![=-1383.34 kJ/mol](https://tex.z-dn.net/?f=%3D-1383.34%20kJ%2Fmol)
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Answer:
Kindly check the attached picture for the diagram showing the titration curve.
Explanation:
The term "titration" is an important part of chemistry because it is being used in the Determination of the concentration is the unknown.
In this question, the titration is done between a polyprotic weak acid, that is Phosphoric acid (H3PO4) and a strong base, NaOH.
H3PO4 + OH^- -----------------> H2O + H2PO4^-.
H2PO4^-2 + OH^- ---------------> H2O + HPO4^2-
HPO4^- + OH^- ------------------> H2O + PO4^3-.
Answer:
He
Explanation:
From Graham's law of diffusion, lighter or less dense gas tends to diffuse faster than denser gas amongst the list, He is the lightest gas and so will diffuse faster