Calculate the period when the frequency is 50Hz.
1 answer:
You might be interested in
Answer:
Explanation:
(a) Part 1: reaction. This is a nucleophilic substitution reaction in which we have two steps. Firstly, chlorine, a good leaving group, leaves the carbon skeleton to form a relatively stable secondary carbocation. This carbocation is then attacked by the hydroxide anion, our nucleophile, to form the final product.
To summarize, this mechanism takes places in two separate steps. The mechanism is attached below.
Part 2: reaction. This is a nucleophilic substitution reaction in which we have one step. Our nucleophile, hydroxide, attacks the carbon and then chlorine leaves simultaneously without an intermediate carbocation being formed.
The mechanism is attached as well.
(b) The rate determining step is the slow step. Formation of the carbocation has the greatest activation energy, so this is our rate determining step for . For
, we only have one step, so the rate determining step is the attack of the nucleophile and the loss of the leaving group.
Answer:
-21 kJ·mol⁻¹
Explanation:
Data:
H₃O⁺ + OH⁻ ⟶ 2H₂O
V/mL: 50 50
c/mol·dm⁻³: 1.0 1.0
ΔT = 4.5 °C
C = 4.184 J·°C⁻¹g⁻¹
C_cal = 50 J·°C⁻¹
Calculations:
(a) Moles of acid
So, we have 0.050 mol of reaction
(b) Volume of solution
V = 50 dm³ + 50 dm³ = 100 dm³
(c) Mass of solution
(d) Calorimetry
There are three energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the water
q₃ = heat to warm the calorimeter
q₁ + q₂ + q₃ = 0
nΔH + mCΔT + C_calΔT = 0
0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0
0.050ΔH + 1883 + 225 = 0
0.050ΔH + 2108 = 0
0.050ΔH = -2108
ΔH = -2108/0.0500
= -42 000 J/mol
= -42 kJ/mol
This is the heat of reaction for the formation of 2 mol of water
The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.