Answer:
Molecular Formula Fe2O12S3·5H2O
IUPAC Name iron(3+);tri sulfate;pentahydrate
Explanation:
Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...
Using an ionic chloride and Phosphoric acid
H3PO4 + NaCl ==> HCl + NaH2PO4
H3PO4 + NaI ==> HI + NaH2PO4
H2SO4 + NaCl ==> HCl + NaHSO4
This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.
The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.
2I-² === I2 + 2e-
Answer:
![\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}](https://tex.z-dn.net/?f=%5CDelta%20S%3D1.8x10%5E%7B-3%7D%5Cfrac%7BkJ%7D%7BK%7D%3D1.8%5Cfrac%7BJ%7D%7BK%7D)
Explanation:
Hello.
In this case, given the heat of fusion of THF to be 8.5 kJ/mol and freezing at -108.5 °C, for the required mass of 5.9 g, we can compute the entropy as:
![\Delta S=\frac{n*\Delta H}{T}](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cfrac%7Bn%2A%5CDelta%20H%7D%7BT%7D)
Whereas n accounts for the moles which are computed below:
![n=5.9g*\frac{1mol}{72g} =0.082mol](https://tex.z-dn.net/?f=n%3D5.9g%2A%5Cfrac%7B1mol%7D%7B72g%7D%20%3D0.082mol)
Thus, the entropy turns out:
![\Delta S=\frac{0.0819mol*8.5 kJ/mol}{(-108.5+273.15)K}\\\\\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cfrac%7B0.0819mol%2A8.5%20kJ%2Fmol%7D%7B%28-108.5%2B273.15%29K%7D%5C%5C%5C%5C%5CDelta%20S%3D1.8x10%5E%7B-3%7D%5Cfrac%7BkJ%7D%7BK%7D%3D1.8%5Cfrac%7BJ%7D%7BK%7D)
Best regards.