Answer:
Hybridization: sp
Electron geometry: linear
Molecular geometry: linear
Explanation:
H₃CCCH can also be written as its Lewis structure which is shown in the figure attached. The figure shows that the central carbon atom makes a single bond with CH₃ and a triple bond with CH. This means that the hybridization of the carbon is sp and both the electron and molecular geometry are linear with an 180° bond angle.
Answer: one simple distillation column is required to separate the stream into five pure products. With four different flat bottom flask, for collection of the distilled products
Explanation: simple distillation works with the difference in boiling points of the liquid to be separated. For the separation of five different constituent to be possible, we have to know the boiling points of the constituents.
For your understanding, let's define constituents in the liquid to be A, B, C, D, E. And the boiling points increases respectively. Start by heating the liquid to the boiling point of A to extract A. After a while check if the constituents A is still dropping in the flat bottom flask, if it has stopped dropping, it simply means that we have extracted all A constituents in the liquid, label the Flask A. Get another flask to extract constituent B.
Heat the mixture to the boiling point of B, after a while check if constituent B is still dropping in the flat bottom flask, if it has stopped dropping,it means that we have extracted all B constituent in the liquid, label the Flask B. Get another flask for C.
Repeat the same process for C and D.
After Extracting D we don't need to distillate E because we already have a pure form of E inside to the conical flask.
SEE PICTURE TO UNDERSTAND WHAT A SIMPLE DISTILLATION LOOKS LIKE
A curve of temperature vs. time for the entire heating process.
The sample is heated up to 100.°C, therefore, the heat and time required to heat the sample to its boiling point, the heat and time required to boil the sample, and the heat and time required to heat the sample from its boiling point to 100.°C are needs to be calculated.
i ) Calculating the heat and time required to heat the sample to its boiling point:
Boiling point = 85°C
C(liquid) = 2.5 J/g °C
The heat required up to melting the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from -20°C to 85°C can be calculated as,
Therefore, T f = 85°C and T i = - 20°C
Plug in the values in the specific heat formula to calculate the heat energy required to heat the sample to its melting point,
q3 = 25 g × 2.5 J/g °C × [85 - (-20)]°C
= 25 J/°C ×[85+20]°C
= 6562.5 J
The total heat energy required for heating the sample from initial temperature to boiling point is:-
q1 + q2 + q3 = 500 J + 4500 J + 6562.5 J
= 11562.5 J
The Rate of heating = 450 J/min
450. J = 1 min
11562.5 J = ? min
11562.5 J × 1min/450 J = 25.69 min
ii) Calculating the heat and time required to boil the sample:
∆H Vap = 500 J/g
The boiling is the phase change from liquid to gas at 85°C, therefore, the heat required to boil the sample can be determined
q4= m × ∆Hvap
= 25 g × 500 J/g
= 12500 J
Thus, total heat required to this phase change is q1 + q2 + q3 + q4 = 500 J + 4500 J +6562.5 J + 12500 J = 24062.5 J
The Rate of heating = 450 J / min
450 J = 1 min
24062.5 J = ? min
24062.5J × 1min / 450 J = 53.47 min
iii) Calculating the heat and time required to heat the sample from its boiling point to 100°C
C gas = 0.5 J / g °C
The heat required to boil the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from 85°C to 100°C can be calculated as,
Therefore, T f = 100.°C and T i = 85°C
q5 = 25 g × 0.5 J / g °C × [100 - 85] °C
= 25 J / °C ×15 °C
= 187.5 J
The total heat energy required for heating the sample from initial temperature to 100°C is
q1 + q2 + q3 + q4 + q5 = 500 J + 4500 J + 2625J + 12500 J + 187.5 J
=24250 J
The Rate of heating = 450 J / min
450. J = 1 min
24250 J=? min
Thus, heating the sample to 100.°C takes a total of 53.89 min.
iv) Draw a curve of temperature vs. time for the entire heating process:-
Temperature °C Temperature K Heat energy (J) Time (min)
-40 °C 233 0 0
-20 °C 253 500 1.11
Melting -20 °C 253 5000 11.11
85 °C 358 11562.5 25.69
Boiling 85 °C 358 24062.5 53.475
100 °C 373 24250 53.89
Hence, the graph for the result is in the image.
Learn more about temperature here:-brainly.com/question/24746268
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