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Triss [41]
3 years ago
14

Which one of the following atoms has the largest radius? Sr, Rb, K, Ca, Y

Chemistry
1 answer:
Alex17521 [72]3 years ago
6 0

Answer:   Rb (rubidium) has a radius of 290 pm, making it have the largest of the 5 elements listed.

Explanation:

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Enter your answer in the provided box. how many degrees of unsaturation are present in c7h8o?
Blizzard [7]
(2C+2+N-X-H)/2
(2*7+2+0-0-8)/2
4
4 0
3 years ago
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.

4 0
3 years ago
Which of the following is a molecular formula for a compound with an empirical formula of CH2O and a molar mass of 150. g/mol.
trapecia [35]

Answer:

C₅H₁₀O₅

Explanation:

Let's consider a compound with the empirical formula CH₂O. In order to determine the molecular formula, we have to calculate "n", so that

n = molar mass of the molecular formula / molar mass of the empirical formula

The molar mass of the molecular formula is 150 g/mol.

The molar mass of the empirical formula is 12 + 2 × 1 + 16 = 30 g/mol

n = (150 g/mol) / (30 g/mol) = 5

Then, we multiply the empirical formula by 5.

CH₂O × 5 = C₅H₁₀O₅

7 0
3 years ago
If a bag of cement of mass 50 kg is acted upon by a force which produces acceleration 20m/ s calculate the force if you don't kn
dybincka [34]

Answer:

F=1000kg/m/s²

Explanation:

F=ma

F=50×20

F=1000

8 0
3 years ago
What is the molarity of 1.5 liters of an aqueous solution that contains 52 grams of lithium fluoride, LiF, (gram-formula mass =2
denis23 [38]
The answer is (1) 1.3 M. The first thing you need to do is to convert the unit of gram to mole. The mol number of LiF is 52/26=2 mol. Then using the volume to calculate the molarity: molarity=2/1.5=1.3 M.
5 0
4 years ago
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