Answer:
Volatile
Explanation:
Nonmetal are usually volatile, which means it evaporate easily.
Answer:
To determine if the forces acting upon an object are balanced or unbalanced, an analysis must first be conducted to determine what forces are acting upon the object and in what direction. If two individual forces are of equal magnitude and opposite direction, then the forces are said to be balanced.
Answer:
D.) Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.
Explanation:
For the reaction:
3H₂(g) + N₂(g) → 2NH₃(g)
The enthalpy change is ΔH = -92kJ
This enthalpy change is defined as the enthalpy of products - the enthalpy of reactants. As the enthalpy is <0, The enthalpy of products is <em>lower </em>than the enthalpy of reactants.
Also, it is possible to obtain the enthalpy change from the bond energies of products - bond energies of reactants, thus, The total bond energies of products are <em>lower</em> than the total bond energies of reactants.
The rate of the reaction couldn't be determined using ΔH.
As the bond energy of ammonia is lower than bonds of nitrogen and hydrogen, <em>D. Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.</em>
I hope it helps!
Answer:
ΔG° = 41.248 KJ/mol (298 K); the correct answer is a) 41 KJ
Explanation:
Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+(aq)
⇒ Kf = 1.7 E7; T =298K
⇒ ΔG° = - RT Ln Kf.....for aqueous solutions
∴ R = 8.314 J/mol.K
⇒ ΔG° = - ( 8.314 J/mol.K ) * ( 278 K ) ln ( 1.7 E7 )
⇒ ΔG° = 41248.41 J/mol * ( KJ / 1000J )
⇒ ΔG° = 41.248 KJ/mol
Answer:
77.98 g/mol ≅ 78.0 g/mol.
Explanation:
- The molar mass of any compound can be calculated by the sum of the atomic masses of different atoms in the compound multiplied by its no. in the compound.
Molar mass of Al(OH)₃ = (Atomic mass of Al) + 3(Atomic mass of O) + 3(Atomic mass of H)
Atomic mass of Al = 26.98 g/mol. & Atomic mass of O = 16.0 g/mol & Atomic mass of H = 1.0 g/mol.
<em>∴ Molar mass of Al(OH)₃</em> = (26.98 g/mol) + 3(16.0 g/mol) + 3(1.0 g/mol) = <em>77.98 g/mol ≅ 78.0 g/mol.</em>
