The rate of this reaction is markedly increased if a small amount of sodium iodide is added to the reaction mixture. The sodium
1 answer:
Answer:
See explanation
Explanation:
SN2 reaction is a synchronous reaction in which the leaving group departs as the nucleophile is being attached to the substrate in a single step. This makes it imperative to use a good leaving group when carrying out an SN2 reaction such as the one shown in the question.
The addition of a small amount of NaI acts as a sort of external nucleophile which assists the departure of the leaving group thereby enabling the reaction to proceed faster. This occurs because the iodide ion is very good leaving group.
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Taking into account the definition of calorimetry, the specific heat of metal is 0.165 .
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
So, the equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where:
- Q is the heat exchanged by a body of mass m.
- C is the specific heat substance.
- ΔT is the temperature variation.
In this case, you know:
For metal:
- Mass of metal = 50 g
- Initial temperature of metal= 45 °C
- Final temperature of metal= 11.08 ºC
- Specific heat of metal= ?
For water:
- Mass of water = 250 g
- Initial temperature of water= 10 ºC
- Final temperature of water= 11.08 ºC
- Specific heat of water = 1.035
Replacing in the expression to calculate heat exchanges:
For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)
For water: Qwater= 1.035 × 250 g× (11.08 C - 10 C)
If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.
Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:
- Qmetal = + Qwater
- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 × 250 g× (11.08 C - 10 C)
Solving:
- Specific heat of metal× 50 g× (-33.92 C)= 1.035 × 250 g× 1.08 C
Specific heat of metal× 1696 g×C= 279.45 cal
Specific heat of metal=
<u><em>Specific heat of metal= 0.165 </em></u>
Finally, the specific heat of metal is 0.165 .
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