Answer:
See the explanation
Explanation:
In this case, in order to get an <u>elimination reaction</u> we need to have a <u>strong base</u>. In this case, the base is the phenoxide ion produced the phenol (see figure 1).
Due to the resonance, we will have a more stable anion therefore we will have a less strong base because the negative charge is moving around the molecule (see figure 2).
Finally, the phenoxide will attack the <u>primary carbon</u> attached to the Cl. The C-Cl bond would be broken and the C-O would be produced <u>at the same time</u> to get a substitution (see figure 1).
A catalyst will speed up the activation energy and therefore speed up the reaction. The products will form fast because of this.
Answer:
The solutions are ordered by this way (from lowest to highest freezing point): K₃PO₄ < CaCl₂ < NaI < glucose
Option d, b, a and c
Explanation:
Colligative property: Freezing point depression
The formula is: ΔT = Kf . m . i
ΔT = Freezing T° of pure solvent - Freezing T° of solution
We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.
Option d, which is glucose is non electrolyte so the i = 1
a. NaI → Na⁺ + I⁻ i =2
b. CaCl₂ → Ca²⁺ + 2Cl⁻ i =3
c. K₃PO₄ → 3K⁺ + PO₄⁻³ i=4
Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.
Answer:
Solid metal
Explanation:
The reduced form of metal ions is the metal in elemental state (simple substance). So, if you have a solution with metal ions and they are reduced, you probably will see the deposition of the metal. For example: if you have a solution with sodium ions (Na⁺), and the ions are then reduced, you will see the aparition of a solid phase of metallic sodium (Na(s)), according to the following half-reaction:
Na⁺ + e- → Na(s)
<span>In each case, the same bond gets broken - the bond between the hydrogen and oxygen in an -OH group. Writing the rest of the molecule as "X"
</span>
The factors to consider
Two of the factors which influence the ionisation of an acid are:
<span>the strength of the bond being broken,the stability of the ions being formed.</span>
In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to be similar.
