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2 years ago

How many total valence electrons are available for the Lewis Dot structure of H2S?

Chemistry

1 answer:

Anettt [7]2 years ago

7 0

Answer:

8 Valence Electrons

Explanation:

There are 2 atoms of Hydrogen and 1 atom of Sulfur in the compound.
Each Hydrogen atom has 1 electron (which is the valence electron).
Sulfur has six valence electrons.
Therefore, there is a total of eight valence electrons in H2S.

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Group the following electron configurations in pairs that would represent similar chemical properties at their atoms;

marishachu [46]

Answer: Pairs are: (a) and (d), (b) and (f), (c) and (e)

Explanation:

In a periodic table, elements are arranged in 18 vertical columns known as groups and 7 horizontal rows known as periods.

Elements arranged in a group show similar chemical properties because of the presence of same number of valence electrons.

Valence electrons are defined as the electrons which are present in the outermost shell of an atom. Outermost shell has the highest value of 'n' that is principal quantum number.

For the given options:

For a:

The given electronic configuration is: $1s^22s^22p^63s^2$

The number of valence electrons in the given configuration are 2

For b:

The given electronic configuration is: $1s^22s^22p^63s^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

For c:

The given electronic configuration is: $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For d:

The given electronic configuration is: $1s^22s^2$

The number of valence electrons in the given configuration are 2

For e:

The given electronic configuration is: $1s^22s^22p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For f:

The given electronic configuration is: $1s^22s^22p^63s^23p^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

Electronic configuration of (a) and (d) will form a pair, (b) and (f) will form a pair, (c) and (e) will form a pair and will have similar chemical properties.

Hence, the pairs are: (a) and (d), (b) and (f), (c) and (e)

4 0

2 years ago

Details about outer ear

Verdich [7]

Outer ear is the is external part of the ear, consists of the auricle ( also pinna), and the ear canal. It gathers sound energy and focuses it on the ear drum.

7 0

3 years ago

Write the condensed electron configurations for the Ca atom. Express your answer in condensed form as a series of orbitals. For

umka21 [38]

Answer:

[Ar] 4s²

Explanation:

Ca is the symbol for Calcium. It is the 20th element and it has 20 electrons.

The full electronic configuration for calcium is given as;

1s²2s²2p⁶3s²3p⁶4s²

The condensed electronic configuration is given as;

[Ar] 4s²

6 0

2 years ago

What is the absorbance of the permanganate in the diluted waste solution?

Snowcat [4.5K]

Not to be confused with tetration. This article is about volumetric titration. For other uses, see Titration (disambiguation). Acidâ€“base titration is a quantitative analysis of concentration of an unknown acid or base solution. Titration, also known as titrimetry,[1] is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of an identified analyte. Since volume measurements play a key role in titration, it is also known as volumetric analysis. A reagent, called the titrant or titrator[2] is prepared as a standard solution. A known concentration and volume of titrant reacts with a solution of analyte or titrand[3] to determine concentration. The volume of titrant reacted is called titration volume

7 0

2 years ago

The CRC Handbook, a large reference book of chemical and physical data, lists two isotopes of indium (). The atomic mass of 4.28

ivanzaharov [21]

Answer: The mass of second isotope of indium is 114.904 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

.....(1)

Let the mass of isotope 2 of indium be 'x'

For isotope 1:

Mass of isotope 1 = 112.904 amu

Percentage abundance of isotope 1 = 4.28 %

Fractional abundance of isotope 1 = 0.0428

For isotope 2:

Mass of isotope 2 = x amu

Percentage abundance of isotope 2 = [100 - 4.28] = 95.72 %

Fractional abundance of isotope 2 = 0.9572

Average atomic mass of indium = 114.818 amu

Putting values in equation 1, we get:

$114.818=[(112.904\times 0.0428)+(x\times 0.9572)]\\\\x=114.904amu$

Hence, the mass of second isotope of indium is 114.904 amu

4 0

2 years ago