1,3-pentadiene has two double bonds which are conjugated, which undergo electrophilic addition reaction on reacting with 
.
The structure of 1,3-pentadiene is shown in the image.
When strong acid such as reacts with 1,3-pentadiene, the electrophilic addition reaction can occur either on double bond at 1,2-position or at 3,4-position. The reaction that occurs is shown in the image.
Answer:
NaNO₃
Explanation:
A precipitate is a compound or a salt formed from a precipitation reaction and does not dissolve in water and therefore will exist in solid state.
From the choices given precipitation reaction will occur between;
- Fe(NO₃)₃(aq) + 3NaOH(aq) → Fe(OH)₃(s) + 3NaNO₃(aq)
- Cu(NO₃)₂(aq) + 2NaOH(aq) → Cu(OH)₂(s) + 2NaNO₃(aq)
- FeSO₄(aq) + 2NaOH(aq) → Fe(OH)₂(s) + Na₂SO₄(aq)
Fe(OH)₃, Cu(OH)₂, and Fe(OH)₂ are precipitates.
From the rules of solubility, hydroxides are insoluble except Ca(OH)₂ which is slightly soluble and hydroxides of ammonium and alkali metals.
Answer:
412.1kJ
Explanation:
For the reaction , from the question -
4Fe (s) + 3O₂ (g) → 2Fe₂O₃ (s)
Δ Hrxn = Δ H°f (products) - Δ H°f (reactants)
In case the compound is in its standard state , enthalphy of formation is zero
Hence ,
for the above reaction ,
ΔHrxn =( 2 * Δ H° (Fe₂O₃ )) - [ ( 4 *Δ H° Fe ) + (3 * Δ H° O₂ )]
The value for Δ H°(Fe₂O₃ ) = - 824.2kJ/mol
Δ H° Fe = 0
Δ H° O₂ = 0
Putting in the above equation ,
ΔH rxn = ( 2 * Δ H° (Fe₂O₃ )) - 0
ΔHrxn = 2× - 824.2 kJ / mol = - 1648.4 kJ/mol
- 1648.4 kJ/mol , this much heat is released by the buring of 4 mol of Fe.
Hence ,
for 1 mol of Fe ,
- 1648.4 kJ/mol / 4 = 412.1kJ
A is the reasonable answer
Answer: Special care is required to prepare a solution of sodium hydroxide or NaOH in water because considerable heat is liberated by the exothermic reaction. The solution may splatter or boil. Here is how to make a sodium hydroxide solution safely, along with recipes for several common concentrations of NaOH solution.
Explanation:
Hope this helps!
