How many moles of H,O must be decomposed to form 200 moles of H,?
24,0 2H +0,

Answer:1 Answer. A chemical equation is balanced when the number of each kind of atom is the same on both sides of the reaction,,

Answer:

If each side of the equation has the same number of atoms of a given element, that element is balanced. If all elements are balanced, the equation is balanced.

Answer:

probably A that's my best guess

Answer:

Hello There!!

Explanation:

The answer is A.H20.This is water which isn't an ionic compound.

hope this helps,have a great day!!

~Pinky~

Given that the x-intercepts of the parabola are -5 and 5 and we need to find out which equation from the given options is that of the parabola.  Option (b) y = (x + 5)(x - 5)

is the equation of a parabola with x-intercepts -5 and +5 in the standard (x, y) coordinate plane.

We know that the x-intercepts of a parabola are the points at which y is zero.

We have two x-intercepts: x = -5 and x = 5. Let's find which equation is correct.

a. y = x² + 25

For x = -5, we have

y = (-5)² + 25 = 50.

It does not satisfy the equation for the x-intercept of -5.

For x = 5, we have

y = (5)² + 25 = 50.

It does not satisfy the equation for the x-intercept of 5.

b. y = (x + 5)(x - 5)

For x = -5, we have

y = (0) (10) = 0.

This satisfies the equation for the x-intercept of -5.

For x = 5, we have

y = (10) (0) = 0.

This satisfies the equation for the x-intercept of 5.

c. y = x(x + 5)(x - 5)

For x = -5, we have

y = (-5) (0) (10) = 0.

This satisfies the equation for the x-intercept of -5.For x = 5, we have

y = (5) (10) (0) = 0.

This satisfies the equation for the x-intercept of 5.

d. y = (x + 5)² - 25

For x = -5, we have

y = (0) - 25 = -25.

It does not satisfy the equation for the x-intercept of -5.

For x = 5, we have

y = (10) - 25 = -15.

It does not satisfy the equation for the x-intercept of 5.

y = (x + 5)(x - 5)

is the equation of a parabola with x-intercepts -5 and +5 in the standard (x, y) coordinate plane.

Hence, the correct answer is option B.

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To calculate the pH at each stage of the titration, we need to consider the reaction between HClO and KOH. The balanced chemical equation for the reaction is:

HClO(aq) + KOH(aq) → KClO(aq) + H2O(l)

Before the addition of any KOH, we have only the HClO solution. HClO is a weak acid, so we can use the ionization constant (Ka) to calculate its initial concentration of H⁺ ions. Since HClO is the only acid present initially, the initial concentration of H⁺ ions is equal to the initial concentration of HClO. Therefore, [H⁺] = 0.220 M.

To calculate the pH after each addition of KOH, we need to determine the amount of HClO that reacts with KOH. From the balanced equation, we can see that the stoichiometric ratio between HClO and KOH is 1:1. This means that for every mole of HClO that reacts, an equal number of moles of H⁺ ions are consumed.

Before addition of any KOH:

[H⁺] = 0.220 M (given)

pH = -log10(0.220) ≈ 0.66

After addition of 25.0 mL KOH:

The amount of HClO reacted can be calculated using the initial concentration and the volume of KOH added. Since the concentration of KOH is the same as HClO, the concentration of HClO remaining is (0.220 M - 0.220 M/4) = 0.165 M. The volume of HClO solution remaining is (50.0 mL - 25.0 mL) = 25.0 mL = 0.025 L. Therefore, [H⁺] = 0.165 M/0.025 L = 6.6 M.

After addition of 35.0 mL KOH:

Following the same calculations as above, [H⁺] = 0.110 M.

After addition of 50.0 mL KOH:

[H⁺] = 0.055 M.

After addition of 60.0 mL KOH:

[H⁺] = 0.022 M.

Keep in mind that pH is a logarithmic scale, so as the concentration of H⁺ ions decreases, the pH value increases.

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Answer: 2 moles

Explanation:

The observed emission spectrum of elemental hydrogen gas is; Series of Lines, colors, Balmer Series, and Ultraviolet and Infrared Lines.

The emission spectrum of elemental hydrogen gas consists of a series of discrete and distinct lines of different colors.

Series of Lines; The emission spectrum of hydrogen gas consists of a series of sharp, discrete lines rather than a continuous spectrum. Each line corresponds to a specific transition between energy levels in the hydrogen atom.

Colors; The lines in the hydrogen emission spectrum are of different colors, representing different wavelengths of light. The colors observed in the Balmer series include red, blue-green, violet, and other shades in between.

Balmer Series; The Balmer series is the most prominent and well-known part of the hydrogen emission spectrum. It corresponds to transitions where the electron in the hydrogen atom jumps from higher energy levels (n ≥ 3) to the second energy level (n = 2). The visible lines in the Balmer series include Hα (red), Hβ (blue-green), Hγ (violet), and so on.

Ultraviolet and Infrared Lines; In addition to the visible lines, the hydrogen emission spectrum also includes ultraviolet and infrared lines. The ultraviolet lines belong to the Lyman series (transitions to the first energy level, n = 1), while the infrared lines belong to the Paschen series (transitions to higher energy levels, n > 2).

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1 grams F2 to mol = 0.02632 mol

10 grams F2 to mol = 0.26318 mol

20 grams F2 to mol = 0.52636 mol

30 grams F2 to mol = 0.78954 mol

40 grams F2 to mol = 1.05272 mol

50 grams F2 to mol = 1.3159 mol

100 grams F2 to mol = 2.6318 mol

200 grams F2 to mol = 5.2636 mol

Answer : The ΔHo for the given reaction is -244.5 kJ/mol.

Explanation:

Given, the following equations and the corresponding ΔHo values:N2(g) + 2 O2(g) → 2 NO2(g) ΔHo = -107.0 kJ/mol2 NO(g) + O2 → 2 NO2(g) ΔHo = -351.5 kJ/mol

The reaction given is ½ N2(g) + ½ O2(g) → NO(g)

To determine the value of ΔHo for the above process, we can use the given thermochemical equations as follows:

What is meant by thermochemical equation?

Thermochemical equations: The chemical equation which includes the term 'Heat' are referred to as thermochemical equations. They include chemical equations for endothermic reactions and exothermic reactions.

ΔHo for the first equation isΔHo = [2ΔHo (NO2(g))] - [ΔHo (N2(g))] - 2[ΔHo(O2(g))]

We haveΔHo (NO2(g)) = - 107.0 kJ/molΔHo (N2(g)) = 0 kJ/molΔHo (O2(g)) = 0 kJ/mol

To find: ΔHo for the process ½ N2(g) + ½ O2(g) → NO(g)Solution:To obtain the required reaction, we need to subtract equation (1) from equation (2).

The obtained equation is:½ N2(g) + ½ O2(g) → NO(g) ΔHo = [2 NO(g) + O2 → 2 NO2(g)] - [N2(g) + 2 O2(g) → 2 NO2(g)] ΔHo = (-351.5 kJ/mol) - (-107.0 kJ/mol) ΔHo = -244.5 kJ/mol.

Therefore, the ΔHo for the given reaction is -244.5 kJ/mol.

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A compound that consists of two non-metal elements in its structure and compound is called a binary molecular compound. They can have two different elements in their structures bonded by various bonds like carbon dioxide, sodium chloride etc.

The naming of [tex]\rm H_{2}O[/tex] is dihydrogen oxygen and of [tex]\rm H_{2}O_{2}[/tex] is dihydrogen dioxygen.

Thus, the naming of water is dihydrogen oxygen and hydrogen peroxide is dihydrogen dioxygen.

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The solubility of silver carbonate in water is approximately 1.26 × 10⁻³ mol/L. The solubility of silver carbonate in 0.22 M AgClO₃ is approximately 2.92 × 10⁻⁵ mol/L. The solubility of silver carbonate in 0.41 M Na₂CO₃ is approximately 1.20 × 10⁻⁶ mol/L.

To calculate the solubility of silver carbonate (Ag₂CO₃) in different solutions, we need to compare the solubility product (Ksp) with the concentrations of relevant ions in the solution. The balanced equation for the dissociation of silver carbonate is:

Ag₂CO₃(s) ⇌ 2Ag⁺(aq) + CO₃²⁻(aq)

(a) Solubility in water:

Since water does not contain any common ions, the concentration of Ag⁺ and CO₃²⁻ ions in water is initially zero. Therefore, we assume x as the solubility of Ag₂CO₃ in mol/L.

The equilibrium concentrations of Ag⁺ and CO₃²⁻ ions are both 2x (as the stoichiometric ratio is 1:1).

The Ksp expression is:

Ksp = [Ag⁺]²[CO₃²⁻] = (2x)²(2x) = 8x⁵

Since the Ksp is given as 8.1 × 10⁻¹², we can set up the equation:

8x⁵ = 8.1 × 10⁻¹²

Solving for x, we find:

x = (8.1 × 10⁻¹² / 8)^(1/5) ≈ 1.26 × 10⁻³ mol/L

Therefore, the solubility of silver carbonate in water is approximately 1.26 × 10⁻³ mol/L.

(b) Solubility in 0.22 M AgClO₃:

In this case, the Ag⁺ ions are already present in the solution due to the presence of AgClO₃. The concentration of Ag⁺ is given as 0.22 M. Since the Ksp expression is [Ag⁺]²[CO₃²⁻], we assume x as the solubility of Ag₂CO₃ in mol/L.

The equilibrium concentration of Ag⁺ ions will be 0.22 + 2x, and the concentration of CO₃²⁻ ions will be 2x.

The Ksp expression is:

Ksp = (0.22 + 2x)²(2x) = 8x³ + 0.88x² + 0.088x

Since the Ksp is still 8.1 × 10⁻¹², we can set up the equation:

8x³ + 0.88x² + 0.088x = 8.1 × 10⁻¹²

Solving for x, we find:

x ≈ 2.92 × 10⁻⁵ mol/L

Therefore, the solubility of silver carbonate in 0.22 M AgClO₃ is approximately 2.92 × 10⁻⁵ mol/L.

(c) Solubility in 0.41 M Na₂CO₃:

In this case, the CO₃²⁻ ions are already present in the solution due to the presence of Na₂CO₃. The concentration of CO₃²⁻ is given as 0.41 M. Since the Ksp expression is [Ag⁺]²[CO₃²⁻], we assume x as the solubility of Ag₂CO₃ in mol/L.

The equilibrium concentration of Ag⁺ ions will be 2x, and the concentration of CO₃²⁻ ions will be 0.41 + 2x. The Ksp expression is then:

Ksp = (2x)²(0.41 + 2x) = 4x³ + 1.64x² + 0.328x²

Again, setting Ksp equal to 8.1 × 10⁻¹², we can solve for x:

4x³ + 1.64x² + 0.328x² = 8.1 × 10⁻¹²

x ≈ 1.20 × 10⁻⁶ mol/L

Therefore, the solubility of silver carbonate in 0.41 M Na₂CO₃ is approximately 1.20 × 10⁻⁶ mol/L.

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the element with highest melting point among Tc, Ag, Rb is Tc that is technetium, while the element which highest melting point among Hg, Os, Ba is Os that is osmium. both of these elements belong to the d block also known as transition elements.

there is no regular trend in the melting point among the d block elements and tungsten is the elements with the highest melting point among the transition elements. the s block elemts on the other hand have relatively low melting points.

Rb and Ba belong to s block thus have lower melting points in comparison to d block elemts like technetium, osmium, silver, and mercury. s block metals are soft and have low melting points due to very week inter-metallic bonding between its atoms.

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Answer:

2.268

Explanation:

5 lb × 0.45359237 = 2.26796185 kg

How to convert Pounds to Kilograms

1 pound (lb) is equal to 0.45359237 kilograms (kg).

1 lb = 0.45359237 kg

The mass m in kilograms (kg) is equal to the mass m in pounds (lb) times 0.45359237:

m(kg) = m(lb) × 0.45359237

Example

Convert 5 lb to kilograms:

m(kg) = 5 lb × 0.45359237 = 2.268 kg

Hope this helped!!!

Answer:

2.26796 kg

Explanation:

In an acid-base reaction involving a neutral base B, the conjugate acid will be H₃O⁺. Option B is correct.

In an acid-base reaction involving a neutral base B, the conjugate acid will be formed by the addition of a proton (H⁺) to the base.

H₂O; Water (H₂O) is not a base but can act as an acid in certain reactions. It can donate a proton to form the hydroxide ion (OH⁻), making it a conjugate base rather than a conjugate acid.

H₃O⁺; The hydronium ion (H₃O⁺) is formed when a proton (H⁺) is added to water (H₂O). It is commonly found in aqueous acidic solutions and can act as an acid by donating a proton. Therefore, H₃O⁺ is the correct answer as it represents the conjugate acid in this acid-base reaction.

OH⁻; The hydroxide ion (OH⁻) is a base, not an acid. It accepts a proton (H⁺) to form water (H₂O) in basic solutions. OH⁻ would be the conjugate base, not the conjugate acid.

Hence, B. is the correct option.

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--The given question is incorrect, the correct question is

"In an acid-base reaction involving neutral base B, what will be the conjugate acid Select the correct answer below , A) H₂O B). H₃0 C). OH⁻."--

Pathogenic bacteria are the most problematic in the food industry. The term "pathogenic" refers to any bacteria that are capable of causing illness when consumed by humans or animals.

Option A is correct

Pathogenic bacteria are often found in foods that are not cooked or handled correctly. This includes raw meat, unpasteurized dairy products, and foods that are stored at incorrect temperatures for an extended period of time. It is crucial for food industry professionals to follow strict safety protocols to minimize the risk of pathogenic bacteria contaminating their products, leading to foodborne illnesses.Foodborne illnesses can be caused by a variety of microorganisms, including bacteria, viruses, and parasites. Pathogenic bacteria are a significant concern in the food industry, as they can cause serious illness and even death in vulnerable populations such as young children, pregnant women, and the elderly. The most problematic bacteria in the food industry are those that are pathogenic and are capable of multiplying quickly in the food environment. Pathogenic bacteria can cause food poisoning, which is a severe illness that can result in dehydration, fever, vomiting, and diarrhea.

The symptoms can be severe and can lead to hospitalization or even death if left untreated. The most common pathogenic bacteria found in food are Salmonella, Listeria, and E.coli. These bacteria are often found in raw or undercooked meat, unpasteurized dairy products, and fruits and vegetables that are not washed correctly. It is essential for food industry professionals to follow strict safety protocols to prevent the growth and spread of these bacteria. Proper handwashing, storage, and cooking techniques can help minimize the risk of foodborne illnesses caused by pathogenic bacteria. In conclusion, pathogenic bacteria are the most problematic in the food industry. Food safety protocols should be followed to minimize the risk of foodborne illnesses caused by these bacteria.

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The number of moles of tin (II) fluoride in 908 grams of tin (II) fluoride is approximately 2.65 moles.

To determine the number of moles of tin (II) fluoride in a given mass, we need to use the concept of molar mass. The molar mass of tin (II) fluoride (SnF₂) is calculated by adding up the atomic masses of its constituent elements: tin (Sn) and fluorine (F).

The atomic mass of tin is 118.71 g/mol, and the atomic mass of fluorine is 18.998 g/mol. By adding these values together, we find that the molar mass of tin (II) fluoride is 156.71 g/mol.

To calculate the number of moles in a given mass, we use the formula:

Number of moles = Mass (in grams) / Molar mass.

In this case, we have 908 grams of tin (II) fluoride. Plugging the values into the formula, we get:

Number of moles = 908 g / 156.71 g/mol = 2.65 moles.

Understanding the relationship between mass, moles, and molar mass is fundamental in chemistry. This concept allows us to convert between different units and make quantitative calculations.

The molar mass plays a crucial role in determining the number of moles in a given mass and vice versa. Exploring further applications of moles and molar mass, such as stoichiometry and chemical reactions, can provide a deeper understanding of chemical processes and their measurements.

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The Right to Know Law is a law that mandates access to information held by the government. It applies to all states and localities in the United States. However, there are exceptions to this rule. In addition to public safety and privacy concerns, there is a category of information that is explicitly excluded from the Right to Know Law. (a) A professional cleaning product is an example of a material that is excluded from the Right to Know Law.

As per the given options, a professional cleaning product is an example of a material that is excluded from the Right to Know Law. In 1984, the federal government amended the Right to Know Law to require businesses to provide information about hazardous chemicals in the workplace to employees. This law, known as the Hazard Communication Standard (HCS), requires employers to make information about hazardous chemicals available to employees in the form of Safety Data Sheets (SDSs) and labels.The HCS applies to all employers with hazardous chemicals in their workplace and requires them to provide their employees with 100-word descriptions of the hazards associated with those chemicals, as well as information on how to protect themselves from exposure. Therefore, a professional cleaning product is an example of a material that is excluded from the Right to Know Law.

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Answer:

8.11

Explanation:

pH = -log[H+]

pH = -log[7.8 x [tex]10^{-9}[/tex]]

pH = 8.107905...

pH = 8.11

Answer:

See explanation

Explanation:

The use of Uranium - 234 to generate electricity depends on a fission reaction. The uranium nuclide is bombarded by fast moving neutrons leading to a chain reaction. Control rods and moderators are used to keep the nuclear reaction under control.

As the nuclear reaction proceeds, heat is generated and steam is consequently produced. This steam is used to turn a turbine and electricity is thereby generated.

The net ionic equation for the mixing of sodium iodide and lead(II) nitrate is:

Pb2+ (aq) + 2I- (aq) → PbI2 (s)

The net ionic equation for the mixing of sodium iodide (NaI) and lead(II) nitrate (Pb(NO3)2) can be determined by examining the dissociation of the compounds and identifying the ions involved in the reaction. Here's the breakdown of the reaction:

Sodium iodide (NaI) dissociates in water to form sodium ions (Na+) and iodide ions (I-):

NaI (aq) → Na+ (aq) + I- (aq)

Lead(II) nitrate (Pb(NO3)2) dissociates in water to form lead(II) ions (Pb2+) and nitrate ions (NO3-):

Pb(NO3)2 (aq) → Pb2+ (aq) + 2NO3- (aq)

When these two solutions are mixed, a double displacement reaction occurs, leading to the formation of a precipitate. The iodide ions (I-) from sodium iodide react with the lead(II) ions (Pb2+) from lead(II) nitrate to form solid lead(II) iodide (PbI2):

2Na+ (aq) + Pb2+ (aq) + 2I- (aq) + 2NO3- (aq) → PbI2 (s) + 2Na+ (aq) + 2NO3- (aq)

The net ionic equation is obtained by removing the spectator ions, which do not participate in the reaction. In this case, the spectator ions are the sodium ions (Na+) and nitrate ions (NO3-):

Pb2+ (aq) + 2I- (aq) → PbI2 (s)

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Answer: I got 15.56

Explanation: I am so sorry if that's wrong but I used the equation Q=m×c×ΔT where

m=mass

ΔT= temp change that occurs

c= specific heat capacity

good luck, again sorry if i totally biffed that

Answer:

B

Explanation:

Pressure is directly proportional to temperature

Answer:

2

Explanation:

The equation for the autoionization of water is H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq), hence option A is correct.

A proton is moved from one water molecule to another during the autoionization of water, resulting in the formation of the hydronium ion ([H₃O⁺) and the hydroxide ion (OH). Kw is the water's autoionization constant, and [H₃O⁺][OH⁻] is the equilibrium expression for this process.

A substance's capacity to interact with itself to produce ions is known as autoionization. [H₃O⁺ and OH⁻ ions are created when a water molecule interacts with another one.

These ions are present in trace concentrations in pure water and have an impact on its chemistry.

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Answer:

You would find starch in a plant leaf.

Explanation:

This is due to the fact that when there is excess energy, it will be stored in plant tissue as starch.

Most inorganic substances do not burn because they do to contain carbon. Organic substances do contain carbon.

Scenario C: Visible light is in the middle of the yellow region of the visible spectrum. Here, we have to estimate its wavelength, frequency, and energy per photon. The wavelength of visible light in the middle of the yellow region of the visible spectrum is approximately 575 nm.

The frequency of the given light can be calculated by using the formula c = νλ where ν is the frequency of light, λ is the wavelength of light, and c is the speed of light. Hence the frequency is given by,ν = c / λν = 3.0 x 10^8 m/s / 575 x 10^-9 mν = 5.22 x 10^14 Hz. To calculate the energy per photon, we use the formula E = hc/λ where h is Planck's constant and c is the speed of light. E = hc/λE = (6.63 x 10^-34 J s) x (3.0 x 10^8 m/s) / (575 x 10^-9 m)E = 3.45 x 10^-19 J per photon.

Scenario D: Visible light has a photon energy of 4.160 x 10^-19 J. Here, we have to determine its wavelength, frequency, and colour. We can use the formula E = hc/λ to find the wavelength of light, where E is the energy of the photon. λ = hc/Let's substitute the given values.λ = (6.63 x 10^-34 J s) (3.0 x 10^8 m/s) / 4.160 x 10^-19 Jλ = 4.8 x 10^-7 m.

The frequency of light can be calculated using the formula c = νλ, where c is the speed of light.ν = c / λν = 3.0 x 10^8 m/s / 4.8 x 10^-7 mν = 6.25 x 10^14 Hz.

To determine the colour of visible light, we can use a chart that maps wavelength to colour. From the chart, it can be seen that the visible light of wavelength 480 nm is blue. Therefore, the visible light in scenario D is blue.

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The bond angle 1 of creatine is approximately 120 degrees while bond angle 2 is approximately 180 degrees. These angles are consistent with the expected angles for a trigonal planar and linear arrangement of atoms, respectively. The shape of creatine, determined by these bond angles, is important for its function in the body.

Creatine is an amino acid derivative that occurs naturally in the body. It is made up of three amino acids: arginine, glycine, and methionine. The structure of creatine, including the bond angles, can be determined using computational tools like molecular modelling software. The bond angles in creatine are determined by the arrangement of the atoms around each bond. Bond angle 1 is the angle between the two nitrogen atoms and the central carbon atom. This angle is approximately 120 degrees, which is consistent with the expected angle for a trigonal planar arrangement of atoms.

Bond angle 2 is the angle between the nitrogen atom and the carbon atom in the guanidine group. This angle is approximately 180 degrees, which is consistent with the expected angle for a linear arrangement of atoms. The bond angles in creatine are important because they determine the shape of the molecule, which in turn affects its properties and functions in the body. For example, the shape of creatine allows it to bind to ATP, which is a molecule that provides energy for cellular processes.

In conclusion, the bond angle 1 of creatine is approximately 120 degrees while bond angle 2 is approximately 180 degrees. These angles are consistent with the expected angles for a trigonal planar and linear arrangement of atoms, respectively. The shape of creatine, determined by these bond angles, is important for its function in the body.

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a)The IUPAC name for the compound (CH3CH2)3CCH2CH2CH(CH3)2 is 2,2,3-trimethylpentane.

b) The IUPAC name for the compound (CH3)2CHC(CH3)2CH2CH2CH3 is 3,3-dimethyl-2-methylbutane.

a) The compound (CH3CH2)3CCH2CH2CH(CH3)2 is named 2,2,3-trimethylpentane. The name is derived by identifying the longest continuous carbon chain, which is five carbons in this case.

The substituents are then listed in alphabetical order, preceded by their respective position on the main chain. The compound contains a tert-butyl group [(CH3)3C-] attached to the second carbon atom, and a methyl group (CH3) attached to the fourth carbon atom.

The presence of the three ethyl groups [CH2CH3] on the third carbon and two methyl groups [CH3] on the fifth carbon is indicated by the prefix "trimethyl" and "pentane" signifies the parent alkane with five carbon atoms.

b) The compound (CH3)2CHC(CH3)2CH2CH2CH3 is named 3,3-dimethyl-2-methylbutane. The name is determined by identifying the longest continuous carbon chain, which is six carbons in this case.

The substituents are then listed in alphabetical order, preceded by their respective position on the main chain. The compound contains two methyl groups (CH3) attached to the third carbon atom, denoted as "3,3-dimethyl."

Additionally, there is a methyl group (CH3) attached to the second carbon atom and an ethyl group (CH2CH3) attached to the fourth carbon atom. The name "2-methylbutane" indicates the parent alkane with four carbon atoms, bearing the methyl group on the second carbon atom.

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Helium, hydrogen, and chlorine are obtained from sources other than the atmosphere. The gases primarily obtained from the atmosphere are nitrogen, oxygen, and argon.

Nitrogen, oxygen, and argon are the main components of Earth's atmosphere and are commonly obtained from the air. They exist in significant quantities in the atmosphere and are often extracted for various industrial and commercial purposes.

On the other hand, helium, hydrogen, and chlorine are not primarily obtained from the atmosphere. Helium is typically extracted from natural gas wells, hydrogen is usually produced from fossil fuels or electrolysis of water, and chlorine is obtained through chemical processes such as electrolysis or from chloride-containing compounds.

The gases primarily obtained from the atmosphere are nitrogen, oxygen, and argon. Helium, hydrogen, and chlorine are obtained from sources other than the atmosphere.

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