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Zarrin [17]
3 years ago
6

Choose the correct mechanistic pathway for each of the following questions:

Chemistry
1 answer:
Evgesh-ka [11]3 years ago
8 0

Answer:

Chlorination of a 1° alcohol using Thionyl Chloride - an SN2 process

Alkene formation via POCl3 reaction with a 2º alcohol - an El process

Explanation:

For a primary alcohol, the chlorination occurs by SN2 mechanism. Remember that the order of SN2 mechanism is methyl > primary > secondary > tertiary. This means that a primary alkyl halide will undergo nucleophillic  substitution by SN2 mechanism.

For a secondary alkyl halide, we normally expect that the mechanism will be E2. When we use POCl3 and pyridine, the alkyl halide passes through a carbocation intermediate which is characteristic of an E1 mechanism.

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What is the entropy of this collection of training examples with respect to the positive class B. What are the information gains
Alenkinab [10]

The data set is missing in the question. The data set is given in the attachment.

Solution :

a). In the table, there are four positive examples and give number of negative examples.

Therefore,

$P(+) = \frac{4}{9}$   and

$P(-) = \frac{5}{9}$

The entropy of the training examples is given by :

$ -\frac{4}{9}\log_2\left(\frac{4}{9}\right)-\frac{5}{9}\log_2\left(\frac{5}{9}\right)$

= 0.9911

b). For the attribute all the associating increments and the probability are :

  $a_1$   +   -

  T   3    1

  F    1    4

Th entropy for   $a_1$  is given by :

$\frac{4}{9}[ -\frac{3}{4}\log\left(\frac{3}{4}\right)-\frac{1}{4}\log\left(\frac{1}{4}\right)]+\frac{5}{9}[ -\frac{1}{5}\log\left(\frac{1}{5}\right)-\frac{4}{5}\log\left(\frac{4}{5}\right)]$

= 0.7616

Therefore, the information gain for $a_1$  is

  0.9911 - 0.7616 = 0.2294

Similarly for the attribute $a_2$  the associating counts and the probabilities are :

  $a_2$  +   -

  T   2    3

  F   2    2

Th entropy for   $a_2$ is given by :

$\frac{5}{9}[ -\frac{2}{5}\log\left(\frac{2}{5}\right)-\frac{3}{5}\log\left(\frac{3}{5}\right)]+\frac{4}{9}[ -\frac{2}{4}\log\left(\frac{2}{4}\right)-\frac{2}{4}\log\left(\frac{2}{4}\right)]$

= 0.9839

Therefore, the information gain for $a_2$ is

  0.9911 - 0.9839 = 0.0072

$a_3$     Class label      split point       entropy        Info gain

1.0         +                        2.0            0.8484        0.1427

3.0        -                         3.5            0.9885        0.0026

4.0        +                        4.5            0.9183         0.0728

5.0        -

5.0        -                        5.5            0.9839        0.0072

6.0        +                       6.5             0.9728       0.0183

7.0        +

7.0        -                        7.5             0.8889       0.1022

The best split for $a_3$  observed at split point which is equal to 2.

c). From the table mention in part (b) of the information gain, we can say that $a_1$ produces the best split.

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Would you weigh more or less on the moon than you do on Earth, and why?
VladimirAG [237]

Answer:You would weigh less on the moon because there is less gravity on the moon.

Explanation:

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3 years ago
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An ionic compound has a solubility of 30 g per 100 mL of water at room temperature. A solution containing 70 g of the compound i
Llana [10]

Answer: A. unsaturated.

Explanation:

Unsaturated solution is defined as the solution in which more solute particles can be dissolved in the solvent.

Saturated solution is defined as the solution in which no more solute particles can be dissolved in the solvent.

Supersaturated solution is defined as the solution in which more amount of solute particles is present than the solvent particles.

Given:  Solubility = 30g/100ml

If 100 ml can dissolve ionic compound = 30 g

300 ml can dissolve ionic compound  =\frac{30}{100}\times 300=90g

Thus solubility is 90g/300 ml and dissolved salt is only 70 g , the solution is said to be unsaturated.

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Room temperature is around 25°C. What is the<br> state of matter of bromine?
goldfiish [28.3K]

Explanation:

I THINK LIQUID

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