The correct answer is option 3. The IUPAC name is Iron(II) sulfide. It is the less stable amorphous form. When this is powdered, it is pyrophoric or it ignites spontaneously in air. It readily reacts with hydrochloric acid producing hydrogen sulfide.
Answer:
the highest is "A" and the lower is "C"
Answer:
7.5 moles of CaBr2 are produced
Explanation:
Based on the equation:
2AlBr3 + 3CaO → Al2O3 + 3CaBr2
<em>2 moles of AlBr3 produce 3 moles of CaBr2 if CaO is in excess.</em>
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Using this ratio: 2 moles AlBr3 / 3 moles CaBr2. 5 moles of AlBr3 produce:
5 moles AlBr3 * (3 moles CaBr2 / 2 moles AlBr3) =
<h3>7.5 moles of CaBr2 are produced</h3>
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Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?
Solving:
Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-402.7 = 67.99*ΔT
If: ΔT (T final - T initial) = ?
Answer:
The water ligands surrounding the cobalt metal center are being replaced by ethylenediamine and chloride ligands which results in a different crystal field splitting. Thus, the energy associated with electron transitions between the do-orbitals will differ for the two compounds showing a color change.
Explanation:
The five d-orbitals are usually degenerate. Upon approach of a ligand, the d-orbitals split into two sets of orbitals depending in the nature of the crystal field.
The magnitude of crystal field splitting is affected by the nature of the ligand. Ligands having filled p-π orbitals such as ethylenediamine lead to greater crystal field splitting.
The change in the colour that takes place when ethylenediamine is added to the solution of cobalt(II) chloride occurs due to a different crystal field splitting pattern. Thus, the energy associated with electron transitions between the d-orbitals now differ for the two compounds showing a color change.
