Answer:
- 10.555 kJ/mol.
Explanation:
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).
∆H°rxn is the standard enthalpy change of the reaction (J/mol).
T is the temperature of the reaction (K).
∆S°rxn is the standard entorpy change of the reaction (J/mol.K).
∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants
<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.
∵ ∆S°rxn = ∑∆S°products - ∑∆S°reactants
<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>
<em></em>
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>
The question is incomplete, the complete question is;
AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the formula AlBr3 is —
aluminum bromide
monoaluminum tribromide
aluminide bromine
aluminum tribromide
Answer:
aluminum bromide
Explanation:
Having known that AlBr3 is an ionic compound and aluminium is the central atom here, we now have to ask ourselves if Aluminium exists in other stable oxidation states.
We must take cognizance of the fact that the oxidation number of the central atom in a compound becomes part of the name of that compound when other stable oxidation states for atoms of the same elements exists.
Since the +3 state is the only stable oxidation state for aluminium, the name of the compound is simply aluminium bromide.
Answer:
6.12 L
Explanation:
Given that,
Initial volume, V₁ = 5 L
Initial temperature, T₁ = 7.0°C = 343 K
Final temperature, T₂ = 147°C = 420 K
We need to find its new volume. The relation between volume and temperature is given by :
So, the new volume is 6.12 L.
Kilo
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king Hector died by drinking chocolate milk
25 milligrams equal 0.025 grams
