The given question is incomplete, the complete question is:
Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ΔG° = 349 kJ/mol
A) –309 kJ/mol
B) –329 kJ/mol
C) None of the above
D) –349 kJ/mol
E) –369 kJ/mol
Answer:
The correct answer is option D, that is, -349 kJ/mol.
Explanation:
Based on the given information, the reaction is:
NaBr (s) ⇔ Na (s) + 1/2 Br₂ (l), the ΔG° of the reaction given is 349 kJ per mole. In the given question, it is clearly mentioned that there is a need to determine the free energy of the formation of NaBr. Thus, there is a need to keep Na (s) and Br₂ (l) at the reactant side and NaBr (s) at the product side.
Therefore, there is a need to reverse the reaction and change the sign on ΔG.
Now the reaction will become,
Na (s) + 1/2 Br₂ (l) ⇔ NaBr (s), and the ΔG° will now become -349 kJ per mole. Hence, -349 kJ per mole is the free energy of the formation of NaBr (s).
Answer:
Yes atoms can accept the electrons from another atom like
Chlorine accept electron from hydrogen.
Explanation:
Answer:trifluoromethanesulfonic acid (CF3SO3H).
Explanation:
The trifluoromethanesulfonic acid (CF3SO3H) has a halogen atom which stabilizes the leaving group by withdrawal of charge from the SO3- moiety. The methanesulfonic acid (CH3SO3H) contains an electron pushing group which tends to destabilize the charge centre. The better leaving group will be the stabilized anion which in this case is trifluoromethanesulfonic acid (CF3SO3H). This typifies the role of stabilizing factors in formation of chemical species.
Answer:
No, you can not calculate the solubility of X in water at 17 0C.
Explanation:
Solubility refers to the amount of a substance that dissolves in 1000 L of water.
To calculate the solubility of a solute in water, all the water is evaporated and the solid is carefully collected, washed, dried and weighed. The mass of solid obtained can now be used to calculate the solubility of the solute in water as long as there was no loss in mass of solid during the experiment.
In this case, the student threw away part of the solid that precipitated. As a result of this, the mass of solid obtained at the end of the experiment is not exactly the total mass of solute that dissolved in the solvent. Hence, the solubility of X in water at 17 0C can not be accurately calculated.