Fog is a cloud that hasa. a small vertical extent.
b. no ability to produce rain.
c. particles made of ice crystals.
d. its base at or near the ground

Answers

Answer 1

Answer:

Answer: d. Its base at or near the ground.

Explanation:

Fog is a visible weather condition in which the tiny droplets of water present in the atmosphere come together and which forms a thick cloud which lies close to the surface of the ground (land) or above the sea surface. Fog cuts the visibility down to one kilometers, it makes it difficult to observe the things through it. The fog is influenced by several factors such as wind conditions, nearby water bodies, and topography.

Answer 2

Answer:

Fog is a cloud that has its base at or near the ground. The answer is letter D. Fog results when atmospheric water vapor condenses to the extent that the new forms, water droplets or ice crystals, become visible. Saturation of the air and sufficient condensation nuclei are ordinarily prerequisites for the formation of fog. One of the great hazards associated with fog is a lack of visibility.

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If 0.25 mol of Br2 and 0.55mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of BrCl ?? Br2(g)+Cl2(g) 2BrCl at 400 K, Kc= 7.0

Answers

Answer:

[BrCl] = 0,13M

Explanation:

For the reaction:

Br₂(g) + Cl₂(g) ⇄ 2BrCl

kc = [BrCl]²/[Br₂] [Cl₂] (1)

If initial [Br₂] and [Cl₂] are 0,25mol/3,0L=0,083M and 0,55mol/3,0L=0,18M, the concentrations in equilibrium are:

[Br₂] = 0,083M - x

[Cl₂] = 0,18M - x

[BrCl] = 2x

Replacing in (1)

7 = (2x)²/ (0,083 - x)(0,18 - x)

7 = 4x²/ x²- 0,263x + 0,01494

7x² - 1,841x + 0,10458 = 4x²

3x² - 1,841x + 0,10458 = 0

Solving for x:

x = 0,0633. As [BrCl] = 2x. [BrCl] = 0,13M

I hope it helps!

Elements that are characterized by the filling of p orbitals are classified as _____.a. Groups 3A through 8A
b. transition metals
c. inner transition metals
d. groups 1A and 2A

Answers

Answer : The correct option is, (a) Groups 3A through 8A

Explanation :

The general electronic configurations of :

Group 1A : $ns^1$

Group 2A : $ns^2$

Group 3A : $ns^2np^1$

Group 4A : $ns^2np^2$

Group 5A : $ns^2np^3$

Group 6A : $ns^2np^4$

Group 7A : $ns^2np^5$

Group 8A : $ns^2np^6$

Transition metal : $(n-1)d^((1-10))ns^((0-2))$

Inner transition metal (Lanthanoids) : $4f^((1-14))5d^((0-1))6s^2$

Inner transition metal (Actinoids) : $5f^((0-14))6d^((0-1))7s^2$

From the general electronic configurations, we conclude that the groups 3A through groups 8A elements that are characterized by the filling of p-orbitals.

Elements that are characterized by the filling of p orbitals are classified as $\boxed{{\text{a}}{\text{. Groups 3A through 8A}}}$ .

Further Explanation:

In order to make the study of numerous elements easier, these elements are arranged in a tabular form in increasing order of their atomic numbers. Such a tabular representation of elements is called a periodic table. Horizontal rows are called periods and vertical columns are called groups. A periodic table has 18 groups and 7 periods.

a. Groups 3A through 8A

The elements from group 3A to 8A has the general outermost electronic configuration of $n{s^2}n{p^(1 - 6)}$ . So the added electrons are to be filled in p orbitals.

b. Transition metals

These metals have the general valence configuration of $\left( {n - 1} \right){d^(1 - 10)}n{s^(0 - 2)}$ . This indicates that the added electrons enter either s or d orbitals.

c. Inner transition metals

These are classified as lanthanoids and actinoids. The general outermost configuration of lanthanoids is $4{f^(1 - 14)}5{d^(0 - 1)}6{s^2}$ while that of actinoids is $5{f^(0 - 14)}6{d^(0 - 1)}7{s^2}$ . In both cases, the added electron enters either d or f orbitals.

d. Groups 1A and 2A

The elements of group 1A have the general valence electronic configuration of $n{s^1}$ . It implies the last or valence electron enters in the s orbital. The group 2A elements have a general configuration of $n{s^2}$ . Here also the last electron enters the s orbital.

So elements from groups 3A to 8A are classified by the filling of p orbitals and therefore option a is correct.

Learn more:

Which ion was formed by providing the second ionization energy? brainly.com/question/1398705
Write a chemical equation representing the first ionization energy for lithium: brainly.com/question/5880605

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Periodic classification of elements

Keywords: periodic table, configuration, ns1, ns2, d, p, f, 3A, 8A, transition metals, inner transition metals, lanthanoids, actinoids, orbitals, 1A, 2A.

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nitrous acid
nitric acid
dihydrogen oxide

Answers

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Fog is a cloud that hasa. a small vertical extent. b. no ability to produce rain. c. particles made of ice crystals. d. its base at or near the ground

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Fog is a cloud that hasa. a small vertical extent.
b. no ability to produce rain.
c. particles made of ice crystals.
d. its base at or near the ground