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Vesna [10]
3 years ago
9

An atom's valence electrons are located in the atom's outermost energy level, true or false?

Chemistry
2 answers:
REY [17]3 years ago
8 0

Answer:

The statement is <u><em>TRUE</em></u>.

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. Each level is divided into sub-levels. These sub-levels in which each level is divided can be up to 4, 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.

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. It is also possible to say that these electrons are the electrons furthest from the nucleus of the atom, which is why they participate in chemical reactions.

And electrons, in addition to determining the reactivity of the elements, allows to determine their location in the Periodic Table. All elements of the same group have the same number of valence electrons (This does not occur in transition metals). The valence electrons increase in number as one advances in a period. Then, at the beginning of the new period, the number decreases to one and begins to increase again.

Finally, <u><em>it is possible to say that the statement is true. </em></u>

Sedaia [141]3 years ago
3 0

Answer: The answer is True

Explanation: I hope my answer helps :)

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The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com
Elanso [62]

Answer:

(a) Ksp=4.50x10^{-7}

(b) Ksp=1.55x10^{-6}

(c) Ksp=2.27x10^{-12}

(d) Ksp=1.05x10^{-22}

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)

Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L}   )^2\\\\Ksp=4.50x10^{-7}

(B) Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)

Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}

(C) NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)

Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}

(D) La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)

Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}

Best regards.

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Explanation:

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