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3 years ago

An atom has the following electron configuration.

Chemistry

2 answers:

Drupady [299]3 years ago

8 0

Answer: A. 2

Explanation: The outermost energy level shell is 4s2.

mihalych1998 [28]3 years ago

8 0

Answer:

This atom have 2 valence electrons.

Explanation:

The Electronic Configuration of the elements is the arrangement of all electrons of an element in energy levels and sub-levels (orbitals).

There are 7 energy levels, numbered from 1 to 7, and in which electrons are distributed, logically in order according to their energy level. Electrons with less energy will be spinning at level 1. Each level is divided into sub-levels. These sub-levels into which each level is divided can be up to 4. These 4 sub-levels are called: s, p, d, f. In the sub-level s there can only be a maximum of 2 electrons, in p there can be a maximum of 6 electrons, in the sub-level d 10 electrons and finally in the sub-level f there can be a maximum of 14 electrons.

In level 1 there is only one sub-level, which will be the s. In level 2 there are 2 sub-levels, the s and the p. At level 3 there are 3 sub-levels s, p and d. And at level 4 there are 4 sub-levels, the s, the p, the d and the f.

Then, you have the following electronic configuration:

$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{3}$

Each number in front of each letter indicates the level to which the electrons belong, each letter (s, p, d) indicates the sub-level and each super-index indicates the number of electrons at that level and that sub-level.

Valencia electrons are the electrons found in the last electronic layer (called valence orbitals). These electrons are what determine the ability of the atom to form bonds. When an element joins another it does so through its valence electrons. Remember that the last electronic layer belongs to the furthest level.

Looking at the previous electronic configuration, it can be seen that the last layer belongs to the last level, this being level 4. In this case this level has only one sub-level (sub-level s), which has 2 electrons.

So, it is possible to say that <u><em>this atom have 2 valence electrons</em></u>

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In a percentage composition investigation a compound was decomposed into its elements: 20.0 g of calcium, 6.0 g of carbon, and 2

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The percentage composition of this compound : 40%Ca, 12%C and 48%O

<h3>Further explanation</h3>

Given

20.0 g of calcium,

6.0 g of carbon,

and 24.0 g of oxygen.

Required

The percentage composition

Solution

Total mass of compound :

=mass calcium + mass carbon + mass oxygen

=20 g + 6 g + 24 g

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Percentage composition :

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3 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

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