Answer:
Kb = 0.428 m/°C
Explanation:
To solve this problem we need to use the <em>boiling-point elevation formula</em>:
- <em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m
Where <em>Tsolution</em> and <em>Tpure solvent</em> are the boiling point of the CS₂ solution (47.52 °C) and of pure CS₂ (46.3 °C), respectively. Kb is the constant asked by the problem, and m is the molality of the solution.
So in order to use that equation and solve for Kb, first we <em>calculate the molality of the solution</em>.
molality = mol solute / kg solvent
- Density of CS₂ = 1.26 g/cm³
- Mass of 410.0 mL of CS₂ ⇒ 410 cm³ * 1.26 g/cm³ = 516.6 g = 0.5166 kg
molality = 0.270 mol / 0.5166 kg = 0.5226 m
Now we <u>solve for Kb</u>:
<em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m
- 47.52 °C - 46.3 °C = Kb * 0.5226 m
Answer:
If the pressure of the system increases then the boiling point will increase.
If the pressure of the system decreases then the boiling point will decrease. If there is no change in pressure then the boiling point will remain constant.
Explanation:
Answer: All of the statements are true.
Explanation:
(a) Considering the system mentioned in the equation:-
The sum of total moles in the flask will always be equal to 1 which leads to confirmation of this statement as for 60 secs= 0.16 mol A and 0.84 mol B
(b) 0<t< 20s, mole A got reduced from 1 mole to 0.54 moles while at 40s to 60s A got decreased from 0.30 moles to 0.16 moles.
0 to 20s is 0.46 (1 - 0.54 = 0.46)mol whereas,
40 to 60s is 0.14 (0.30-.16 = 0.14) mol
(0.46 > 0.14) mol leading this statement to be true as well.
(c) Average rate from t1 = 40 to t2 = 60 s is given by:
which is true as well
Hi, you have not provided structure of the aldehyde and alkoxide ion.
Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.
Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.
The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.
After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.
All the structures are shown below.
Answer:
C2H5Cl + H20 ⇆ C2OOH4
Explanation:
The left over acetyl chloride which reacted with water to produce acetic acid