Air is made up of different gases, such as oxygen, nitrogen, and carbon dioxide. Which statement best describes these three components of air? A)They are all classified as pure substances.

B)They cannot react with another substance.

C)They are chemically bonded to one another.

D)They can be classified as elements.

Answers

Answer 1

Answer: I think the correct answer from the choices listed above is option A. The three components of air are all classified as pure substances since they are not chemically bonded so they can be separated by certain processes and be present as a pure substance. Hope this answers the question.

Answer 2

Answer:

The statement that best describes the three component of air is they are all classified as pure substances.

What are pure substances?

Pure substances are those substances which are made up of only one material and has specific and fixed structure. They are also called as elements.

Thus, the correct option is A) They are all classified as pure substances.

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Which materials typically make up the a horizon in soil?

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The A horizon is a surface horizon that largely consists of minerals (sand, silt, and clay) and with appreciable amounts of organic matter. This horizon is predominantly the surface layer of many soils in grasslands and agricultural lands.These materials typically accumulate through a process termed illuviation, wherein the materials gradually wash in from the overlying.

In which of these substances are the molecules probably moving most quickly?solid iron at −45°C
solid iron at 300°C
liquid iron at 1600°C
liquid iron at 1800°C

Answers

The substance with molecules that are probably moving most quickly is liquid iron at 1800°C. This is because the particles of the liquid iron can move or slide past each other due to the small space available for movement. It cannot be compressed further due to the repulsive forces that acts between them. And also, the temperature is increases which causes for more thermal vibration of atoms.

Answer:

The answer is D

Explanation:

Which barium salt is insoluble in water?A) BaCO3. B) BaCl2
C) Ba(ClO4)2. D) Ba(NO3)2

Answers

$\boxed{{\text{A) BaC}}{{\text{O}}_{\text{3}}}}$ is insoluble in water.

Further Explanation:

Solubility rules

These help in predicting whether the given compound is soluble or insoluble in nature. Some of the solubility rules are as follows:

1. Group 1A compounds are soluble in nature.

2. All the common compounds of ammonium ion and all acetates, chlorides, nitrates, bromides, iodides, and perchlorates are soluble in nature. But the ions whose chlorides, bromides, and iodides are not soluble are ${\text{A}}{{\text{g}}^ + }$ , ${\text{P}}{{\text{b}}^(2 + )}$ , ${\text{C}}{{\text{u}}^ + }$ and ${\text{Hg}}_2^(2 + )$ .

3. All common fluorides, except for ${\text{Pb}}{{\text{F}}_{\text{2}}}$ and group 2A fluorides are soluble. Also, all sulfates except ${\text{CaS}}{{\text{O}}_{\text{4}}}$ , ${\text{SrS}}{{\text{O}}_{\text{4}}}$ , ${\text{BaS}}{{\text{O}}_{\text{4}}}$ , ${\text{A}}{{\text{g}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ and ${\text{PbS}}{{\text{O}}_{\text{4}}}$ are soluble.

4. All common metal hydroxides except ${\text{Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ , ${\text{Sr}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ , ${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ and hydroxides of group 1A and that of transition metals are insoluble in nature.

5. All carbonates are insoluble in nature, except for sodium, potassium and ammonium.

6. Salts having ions like $\text{Cl}^-$ , $\text{Br}^-$ or $\text{I}^-$ are generally soluble except those of $\text{Ag}^+$ , $\text{Pb}^(2+)$ and ${\left( {{\text{H}}{{\text{g}}_2}} \right)^{{\text{2 + }}}}$ .

7. Group 1A and group 2A perchlorates are soluble in nature.

8. All sulfates of metals are soluble, except for lead, mercury (I), barium, and calcium sulfates.

9. The salts having nitrate ions are soluble in nature.

According to the solubility rules, only carbonates of sodium, potassium and ammonium are soluble in nature. So barium carbonate $\left( {{\text{BaC}}{{\text{O}}_{\text{3}}}} \right)$ is insoluble in nature. But chlorides, nitrates and perchlorates of barium are soluble in water. Therefore option A is the correct answer.

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Answer details:

Grade: Senior School

Chapter: Chemical reaction and equation

Subject: Chemistry

Keywords: barium, water, insoluble, solubility rules, soluble, insoluble, water, BaCO3.

BaCO₃ is generally considered insoluble in water. Carbonates are often insoluble, except for some alkali metals and ammonium carbonates. Therefore, option A is correct.

The solubility of salt depends on its nature and the interactions between its ions and water molecules. In this case, determine the solubility of each salt by considering the common solubility rules.

BaCl₂ is soluble in water. Chlorides (Cl-) are mostly soluble except for a few exceptions, such as silver chloride, lead chloride, and mercury(I) chloride.

Ba(ClO₄)₂ is soluble in water. Perchlorates (ClO4-) are typically soluble.

Ba(NO₃)₂ is soluble in water. Nitrates (NO3-) are mostly soluble.

Based on the solubility rules, the salt that is insoluble in water is BaCO₃ (barium carbonate).

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A polar commercial antacid contains both Al(OH)3 and Mg(OH)2. Are these compounds Arrhenius bases, Brønsted-Lowry based, or both? Explain your answer.

Answers

Answer:

Magnesium hydroxide, Mg(OH)2, is an Arrhenius base and therefore also a Brønsted-Lowry base.

However, water can also act as a base by accepting a proton from an acid to become its conjugate base, H3O+. where Al(OH)3 is acting as a Lewis Base.

Explanation:

Arrhenius acids and bases

An Arrhenius acid is any species that increases the concentration of H+start text, H, end text, start superscript, plus, end superscript in aqueous solution.

An Arrhenius base is any species that increases the concentration of OH−start text, O, H, end text, start superscript, minus, end superscript in aqueous solution.

How do you find the name of the chemical formula Pb3N4 (the 3 and 4 are subscripts)

Answers

This is a salt because there is a cation and an anion. Because lead can have multiple charges, its charge must correspond to the charge of the anion. The charge on the N is -3, so the 4 N's equal a total charge of -12. Therefore, the total charge of the 3 Pb's must be 12, so each lead should have a charge of +4. So the name of the compound is Lead (IV) Nitride.

Final answer:

The name of the chemical compound Pb3N4 is Lead (IV) Nitride. The number of atoms of each element in the compound is represented by the subscript numbers, and the '-ide' suffix is used for the non-metal (Nitrogen). The Roman numeral indicates the oxidation state of Lead.

Explanation:

The chemical formula Pb3N4 corresponds to a compound made up of Lead (Pb) and Nitrogen (N). The subscript numbers (3 and 4) denote the number of atoms from each element present in the compound. Considering that Lead is a metal and Nitrogen is a non-metal, we use the '-ide' suffix for the non-metal in the name of the compound. Therefore, the name of Pb3N4 is Lead (IV) Nitride. The Roman numeral (IV) indicates the oxidation state of Lead in the compound.

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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.

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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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