If you were helping a friend do stoichiometry problems, what would you tell them about how they might use subscripts and coeffic
2 answers:
<u>Answer:</u> By using law of conservation of mass.
<u>Explanation:</u>
A balanced chemical equation always follow law of conservation of mass.
This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This also means that total number of individual atoms on reactant side must be equal to the total number of individual atoms on the product side.
The chemical reaction for the formation of water follows the equation:
Total mass on reactant side:
Total mass on product side:
Hence, by using law of conservation of mass.
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This question does not contain the structures of the molecules. The structures in Daylight SMILES format are:
I. C1=CC=CC=C1C(=O)C
II. C1=CC=CC=C1CC=O
III. C1=CC(C)=CC=C1C=O
IV. C1=CC=CC=C1CCC
V. C1=CC=CC=C1C(C)C
The structures are also attached
Answer:
The structure of compound IV is consistent with the information obtained analysis
Proposed structures for the ions with m/z values of 120, 105,77 and 43 are (also attached):
C1=CC=CC=C1C(=[OH0+])C |^1:7|
C1C([CH0+]=O)=CC=CC=1
C1[CH0+]=CC=CC=1
C(#[OH0+])C
respectively
Explanation:
The IR peak at 1687 cm⁻¹ is indicative of an α unsaturated carbonyl carbon. While the 1H NMR singlet is of the methyl group next to carbonyl and the multiplet near 7.1 ppm is a characteristic peak of benzene. This data shows points towards structure I.
Mass spectrum peak at 120 m/z is of molecular ion peak. In the case of carbonyl-containing molecule, this peak is observable. The signal at 105 shows the loss of a methyl group next to the carbonyl. m/z value of 77 is the characteristic cationic peak of benzene, while the peak at 43 infers the formation of acylium ion (RCO+) due to α-cleavage. All this data agrees with the structure of acetophenone (Structure 1)
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,
and
The entropy of the training examples is given by :
= 0.9911
b). For the attribute all the associating increments and the probability are :
+ -
T 3 1
F 1 4
Th entropy for is given by :
= 0.7616
Therefore, the information gain for is
0.9911 - 0.7616 = 0.2294
Similarly for the attribute the associating counts and the probabilities are :
+ -
T 2 3
F 2 2
Th entropy for is given by :
= 0.9839
Therefore, the information gain for is
0.9911 - 0.9839 = 0.0072
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 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 produces the best split.
