Which of these best describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide? Which of these best describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide? F. Nitrogen monoxide has one more atom of nitrogen. G. Nitrogen dioxide has one fewer atom of oxygen H. Nitrogen monoxide has one fewer atom of oxygen J. Nitrogen dioxide has one more atom of nitrogen show more Which of these best describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide? F. Nitrogen monoxide has one more atom of nitrogen. G. Nitrogen dioxide has one fewer atom of oxygen H. Nitrogen monoxide has one fewer atom of oxygen J. Nitrogen dioxide has one more atom of nitrogen

Answers

Answer 1

Answer: Nitrogen monoxide has 1 oxygen atom and
Nitrogen dioxide has 2 oxygen atoms

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Look at the equation below.O2 + CS2 → CO2 + SO2Which of the following shows the equation balanced?A. 3 O2 + CS2 → CO2 + 2 SO2B. 3 O2 + CS2 → 2 CO2 + SO2C. O2 + CS2 → CO2 + SO2D. 4 O2 + 4 CS2 → 4 CO2 + 4 SO2
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Which of the following elements would be classified as a nonmetal? sulfur aluminum silicon calcium

Answers

sulfur is a non metal

Sulfur is considered a non-metal. Thanks for posting your question. Feel free to ask me any question. I might not always know the answer. Although i'll try to answer correctly. Have a good day love.

What is the three-dimensional shape of the molecule with this Lewis structure?

Answers

Trigonal planar is the three-dimensional shape of the molecule with this Lewis structure. Hence, option C is correct.

What is an electron dot structure?

Lewis dot structures also called electron dot structures are diagrams that describe the chemical bonding between atoms in a molecule.

According to the Lewis structure of formic acid, the carbon atom is a central atom and it has three bond pairs without any lone pair of electrons. In VSEPR theory, the double bond is treated as one bond pair for the prediction of the shape of the molecule.

Hence, formic acid will have a trigonal planar geometry around the carbon atom and tetrahedral geometry around oxygen atom as it has two lone pair and two bond pairs

Hence, option B is correct.

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trigonal planar which is answer choice c

Which statement about electric charge is true? A. Electric charge is created by an electric current. B. Electric charge influences whether objects attract or repel one another. C. Electric charge tends to be neutral in the nuclei of atoms. D. Electric charge will increase with an increase in temperature.

Answers

Answer:

Explanation:

Answer:

Electrical forces can act from a distance

Explanation:

A. Because electrical forces act at a distance, charged objects brought near a neutral object will cause electrons on the neutral object to rearrange their position

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.

Cause and effect of local winds?

Answers

Local winds are caused by the different surfaces on each lands which heat up differently based on climate and its temperature. This is a complete difference in temperature, and therefore causes a difference in pressure as well. This is capable of causing the air to move differently based on the actual location.

An element crystallizes in a face centered cubic lattice and has a density of 1.45 g cm-3 . The edge of its unit cell is 4.52 x 10-8cm.a) How many atoms are in each unit cell?
b) What is the volume of a unit cell?
c) What is the mass of a unit cell?
d) Calculate the approximate atomic mass of the element.

Answers

In the given fcc element, a. the number of atoms is 4. b. The volume of a unit cell is $\rm 9.23\;*\;10^-^2^3\;cm^3$ . c. Mass of unit cell is $\rm 1.34\;*\;10^-^2^2\;g$ . d. The approximate atomic mass of the element is 80.7 amu.

The face-centered cubic lattice has 3 atoms from the 6 faces, and 1 atom from the eight corners. Thus, the total atoms in the face-centered lattice are four.

The face-centered lattice has been a cube.

The volume of cube = $\rm (edge)^3$

The volume of unit cell = $\rm (4.52\;*\;10^-^8\;cm)$

The volume of unit cell = $\rm 9.23\;*\;10^-^2^3\;cm^3$

The mass of a unit cell can be calculated from density. Mass can be defined as the ratio of volume to density.

Mass = $\rm (volume)/(density)$

Mass of unit cell = $\rm (9.23\;*\;10^-^2^3\;cm^3)/(1.45\;g\;cm^-^3)$

Mass of unit cell = $\rm 1.34\;*\;10^-^2^2\;g$ .

The approximate atomic mass of the element can be calculated by the mass of the carbon atom.

Mass of 1 carbon atom = $\rm (mass\;of\;1\;mole\;carbon)/(number\;of\;atoms\;in\;1\;mole\;Carbon)$

Mass of 1 carbon atom = $\rm (12)/(6.023\;*\;10^2^3)$

Mass of 1 carbon atom = 1.992 $\rm *\;10^-^2^3$ grams.

atomic mass unit per gram can be given as;

amu/gram = $\rm (12)/(1.992\;*\;10^-^2^3)$

amu/gram = $\rm 6.022\;*\;10^2^3$ amu/gram

1 gram = $\rm 6.022\;*\;10^2^3$ amu

1 amu = 1.661 $\rm *\;10^-^2^4$ gram.

The average atomic mass = mass of unit cell $*$ amu\gram

= $\rm 1.34\;*\;10^-^2^2\;g$ . $*$ 1 amu/ 1.661 $\rm *\;10^-^2^4$ gram.

= 80.7 amu.

In the given fcc element, a. the number of atoms is 4. b. The volume of a unit cell is $\rm 9.23\;*\;10^-^2^3\;cm^3$ . c. Mass of unit cell is $\rm 1.34\;*\;10^-^2^2\;g$ . d. The approximate atomic mass of the element is 80.7 amu.

For more information about the face-centered cubic lattice, refer to the link:

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Final answer:

A face centered cubic lattice consists of 4 atoms. The volume of the unit cell is 9.22 x 10^-23 cm^3 and the mass is 1.34 x 10^-23 g. The approximate atomic mass of the element is 2.02 amu.

Explanation:

The element is said to crystallize in a face centered cubic lattice. This implies that there is one atom at each corner of the cube (8 corners for a total of 1 atom, since each corner atom is shared among 8 adjacent cubes). There is also one atom on each face of the cube (6 faces for a total of 3 atoms, since each face atom is shared among 2 adjacent cubes). Thus, a total of 4 atoms are present in each unit cell. (a)

The volume of a unit cell (edges for a cube) can be calculated by the formula 'volume = side^3', where side in this case is 4.52 x 10-8cm. Hence, the volume equals (4.52 x 10^-8cm)^3 = 9.22 x 10^-23cm^3. (b)

The density of the substance is given as 1.45g/cm^3. The formula for density is 'mass/volume' which implies that mass can be calculated as 'density x volume'. Hence, the mass of the unit cell is (1.45g/cm^3) x (9.22 x 10^-23 cm^3) = 1.34 x 10^-23 g.(c)

The atomic weight of the element can then be calculated by taking this overall mass and dividing by the number of atoms in a unit cell (4). So, the atomic weight is (1.34 x 10^-23 g) / 4 = 3.35 x 10^-24 g. But atomic weights are usually given in atomic mass units (amu), not grams, and 1 amu = 1.66 x 10^-24 g. Therefore, we have an atomic weight of (3.35 x 10^-24 g) / (1.66 x 10^-24 g/amu) = approximately 2.02 amu. (d)

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