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spin [16.1K]
3 years ago
13

Why does a hydrogen atom look like it does

Chemistry
2 answers:
MatroZZZ [7]3 years ago
7 0
A hydrogen atom is number 1 on the periodic table so therefore a neutral hydrogen atom has 1 electron. since it’s mass is 1.00784 amu there is no neutron due to its size.
Ksju [112]3 years ago
5 0

Answer:

Lily me and logan caught u cheating

Explanation:

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What did Aristotle believe?
34kurt

Answer:

He believed in biology

Explanation:

he believed the world was made up of individuals.

8 0
4 years ago
Phosphorus-32 has a half-life of 14.0 days. Starting with 6.00 g of 32P, how many grams will remain after 42.0 days ?
aleksandrvk [35]
6g...........3g..........1,5g........0,75g
0d..........14d.........28d.........42d

After 42 days 0,75g of Phoshorus-32 will remain.
4 0
3 years ago
Read 2 more answers
Which of these equations is balanced?
aev [14]
B is the answer. For an equation to be balanced there must be the same amount of each element on the two sides. The only one where this is the case is B.
6 0
3 years ago
Read 2 more answers
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
Oxycodone (C18H21NO4), a narcotic analgesic, is a weak base with pKb = 5.47. Calculate the pH of a .00500 M oxycodone solution.
Ganezh [65]

Answer:

pH = 10.11

Explanation:

Hello there!

In this case, since it is possible to realize that this base is able to acquire one hydrogen atom from the water:

C_{18}H_{21}NO_4+H_2O\rightleftharpoons C_{18}H_{21}NO_4H^+OH^-

We can therefore set up the corresponding equilibrium expression:

Kb=\frac{[C_{18}H_{21}NO_4H^+][OH^-]}{[C_{18}H_{21}NO_4]}

Which can be written in terms of the reaction extent, x:

Kb=\frac{x^2}{0.00500M-x}=3.39x10^{-6}

Thus, by solving for x we obtain:

x_1=-0.000132M\\\\x_2=0.0001285M

However, since negative solutions are now allowed, we infer the correct x is 0.0001285 M; thus, the pOH can be computed:

pOH=-log(x)=-log(0.0001285)=3.89

And finally the pH:

pH=14-pOH=14-3.89\\\\pH=10.11

Best regards!

5 0
3 years ago
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