Law of Conservation of Mass:
In a chemical reaction the amount of ___ has to be equal to the amount of ____

At equilibrium, the number of moles of \(Cl_2\) (g) will be 0.2025 mol.

1: Write the balanced chemical equation:

\(C_O\)(g) + \(Cl_2\)(g) ⟶ \(C_OCl_2\)(g)

2: Set up an ICE table to track the changes in moles of the substances involved in the reaction.

Initial:

\(C_O\)(g) = 0.3500 mol

\(Cl_2\)(g) = 0.05500 mol

\(C_OCl_2\)(g) = 0 mol

Change:

\(C_O\)(g) = -x

\(Cl_2\)(g) = -x

\(C_OCl_2\)(g) = +x

Equilibrium:

\(C_O\)(g) = 0.3500 - x mol

\(Cl_2\)(g) = 0.05500 - x mol

\(C_OCl_2\)(g) = x mol

3: Write the expression for the equilibrium constant (Kc) using the concentrations of the species involved:

Kc = [\(C_OCl_2\)(g)] / [\(C_O\)(g)] * [\(Cl_2\)(g)]

4: Substitute the given equilibrium constant (Kc) value into the expression:

1.2 x \(10^3\) = x / (0.3500 - x) * (0.05500 - x)

5: Solve the equation for x. Rearrange the equation to obtain a quadratic equation:

1.2 x \(10^3\) * (0.3500 - x) * (0.05500 - x) = x

6: Simplify and solve the quadratic equation. This can be done by multiplying out the terms, rearranging the equation to standard quadratic form, and then using the quadratic formula.

7: After solving the quadratic equation, you will find two possible values for x. However, since the number of moles cannot be negative, we discard the negative solution.

8: The positive value of x represents the number of moles of \(Cl_2\)(g) at equilibrium. Substitute the value of x into the expression for \(Cl_2\)(g):

\(Cl_2\)(g) = 0.05500 - x

9: Calculate the value of \(Cl_2\)(g) at equilibrium:

\(Cl_2\)(g) = 0.05500 - x

\(Cl_2\)(g) = 0.05500 - (positive value of x)

10: Calculate the final value of \(Cl_2\) (g) at equilibrium to get the answer.

Therefore, at equilibrium, the number of moles of \(Cl_2\) (g) will be 0.2025 mol.

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The enthalpy of combustion for 1kg of urea is -1223525.84 J/mol.

Urea is a compound that is used in fertilizers and in some plastics.The enthalpy of combustion for urea is the amount of energy that is released when urea is burned. In order to calculate the enthalpy of combustion for 1kg of urea, we need to use the information that is provided to us in the question. Let us start by writing down the balanced equation for the combustion of urea: CO(NH2)2 + 3/2 O2 → CO2 + 2H2O + N2
The balanced equation shows that 1 mole of urea reacts with 1.5 moles of oxygen gas to produce 1 mole of carbon dioxide, 2 moles of water, and 1 mole of nitrogen gas. The enthalpy change for this reaction is equal to the amount of energy that is released when 1 mole of urea is burned.
The heat of combustion (ΔHc) of urea is -632.6 kJ/mol. This means that 632.6 kJ of energy is released when 1 mole of urea is burned. We know that 12g of urea raised the temperature of water by 30°C. We can use this information to calculate the amount of energy that was released when 12g of urea was burned.
The specific heat capacity of water is 4.18 J/g°C. This means that it takes 4.18 J of energy to raise the temperature of 1 gram of water by 1°C. Therefore, it takes 4.18 x 1000 = 4180 J of energy to raise the temperature of 1 kg of water by 1°C.
We know that 12g of urea raised the temperature of water by 30°C. Therefore, the amount of energy that was released when 12g of urea was burned is:
Energy = mass x specific heat capacity x temperature change
Energy = 0.012 kg x 4180 J/kg°C x 30°C
Energy = 1497.6 J
We can now use this information to calculate the enthalpy of combustion for 1kg of urea:
Enthalpy of combustion = energy released / moles of urea burned
Enthalpy of combustion = 1497.6 J / (0.012 kg / 60.06 g/mol)
Enthalpy of combustion = - 1223525.84 J/mol
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II. Elements in group 7 show a gradual change in physical properties.

III. The position of an element in period 3 is related to the number of electrons in the highest Occupied energy level.

The periodic table is a table that displays each chemical element in each system. These chemical elements have a very large number, with different properties, from one element to another.

The purpose of making the periodic table of elements is to recognize the element names of chemical particles easily through grouping.

The periodic system is also defined as an arrangement that displays various chemical elements based on their atomic number and similar chemical properties.

In grouping the periodic table, the physical and chemical properties of each of these elements have been outlined by chemists in the form of a table of elements.

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Therefore,  the mass in grams of the dihydrate H2C2O4 • 2 H2O (molar mass = 126.0 g/mol) is needed in order to make 100.0 mL of a 0.356 M solution of H2C2O4 is 4.49g.

To calculate the  mass in grams of the dihydrate H2C2O4 • 2 H2O we first need to find the moles of H2C2O4.

Moles of solute = molarity * litres in solution.

             = 0.356 *0.01L=0.0356moles.

since dihydrate H2C2O4 • 2 H2O  has molar mass of 126.0g/mol it means it has a mass of 126g.

By using mole to mass conversion.

mass of H2C2O4 • 2 H2O= moles of H2C2O4 • 2 H2O * molar mass of H2C2O4 • 2 H2O

mass of H2C2O4 • 2 H2O= 0.0356moles * 126glmol

                                         =4.49g.

Therefore,  the mass in grams of the dihydrate H2C2O4 • 2 H2O (molar mass = 126.0 g/mol) is needed in order to make 100.0 mL of a 0.356 M solution of H2C2O4 is 4.49g.

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You would have 175.32 grams of NaCl after boiling out all the water from 5.0 L of a 3 M NaCl solution.

Salt dissolves in water, forming sodium and chloride ions. If all the water were to be evaporated by boiling, the ions would join once more to produce solid salt. NaCl is safe to boil without harm, nevertheless. 2575 F, or 1413 C, is the boiling point of sodium chloride.

Three moles of NaCl would remain after boiling off all the water from a 5.0 L solution of 3 M NaCl. You can use the molar mass of NaCl, which is 58.44 g/mol, to get the mass of NaCl.

Mass of NaCl = moles of NaCl x molar mass of NaCl

Mass of NaCl = 3 mol x 58.44 g/mol

Mass of NaCl = 175.32 g

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The wavelength οf the spectral line prοduced when an electrοn in a hydrοgen atοm undergοes the transitiοn frοm the energy level n=6 tο n=1 is apprοximately 980 nanοmeters.

Wavelength is a term used tο describe the distance between twο adjacent peaks οr trοughs οf a wave. It is usually denοted by the Greek letter lambda (λ) and is measured in units οf length, such as meters, centimeters, οr nanοmeters.

The wavelength οf the spectral line prοduced when an electrοn in a hydrοgen atοm undergοes the transitiοn frοm the energy level n=6 tο n=1 can be calculated using the Rydberg fοrmula:

1/λ = R (1/n1² - 1/n2²)

where λ is the wavelength οf the spectral line,

R is the Rydberg cοnstant (1.097 × 10⁷ m⁻¹),

n1 is the initial energy level (6 in this case), and

n2 is the final energy level (1 in this case).

1/λ = R (1/n1² - 1/n2²)

= (1.097 × 10⁷ m⁻¹) (1/6² - 1/1²)

= (1.097 × 10⁷  m⁻¹) (1/36 - 1/1)

= (1.097 × 10⁷  m⁻¹) (1/36 - 1)

= (1.097 × 10⁷  m⁻¹) (-35/36)

= -1.02 × 10⁶ m⁻¹

Taking the reciprοcal οf bοth sides οf the equatiοn, we get:

λ = -1/(1.02 × 10⁶ m⁻¹)

= 9.80 × 10^-7 m

Finally, cοnverting this tο nanοmeters, we get:

λ = 9.80 × 10⁻⁷ m × (1 nm / 10⁻⁹ m)

= 980 nm

Therefοre, the wavelength οf the spectral line prοduced when an electrοn in a hydrοgen atοm undergοes the transitiοn frοm the energy level n=6 tο n=1 is apprοximately 980 nanοmeters.

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Answer:

Explanation: Butane can be converted to 2-butano through a process called isomerization. Isomerization is a chemical reaction in which a compound is converted into one of its structural isomers, which are molecules with the same chemical formula but different arrangement of atoms.

To convert butane to 2-butano, one common approach is to use a catalyst such as a metal or metal oxide to facilitate the rearrangement of the atoms in the molecule. This process typically involves heating the butane to a high temperature (around 200-300°C) in the presence of the catalyst, which promotes the isomerization reaction. Other conditions such as pressure and the presence of a solvent may also be used to control the reaction.

It's important to note that this process typically has a low yield, meaning that only a small fraction of the starting butane is converted to 2-butano. As a result, additional purification steps may be needed to separate the 2-butano from the other products of the reaction.

Answer:

A

Explanation:

it has a methyl group, ethyl group and amine group

The correct IUPAC name for the structure shown in the provided image is "ethylamine." The structure consists of a central nitrogen atom bonded to two carbon atoms.

The correct IUPAC name for the structure shown in the provided image is "ethylamine." The structure consists of a central nitrogen atom bonded to two carbon atoms. According to the IUPAC naming rules, the longest carbon chain is selected as the parent chain, which in this case consists of two carbon atoms. The substituent attached to the parent chain is an ethyl group, denoted as "C2H5". The amine functional group, which consists of the nitrogen atom, is named as "amine". Since there is only one amine group attached to the carbon chain, it is referred to as "ethylamine." Therefore, option C) "ethylamine" is the correct IUPAC name for the given structure.
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Answer: false

Explanation:

Answer:

log = ( - 0.51 z 2) / ( 1 + ( / 305)

Explanation:

Answer:

low density

Explanation:

Because hydrogen make a natural choice For one of it The 1st practical use

Hydrogen gas has lower density than air. This is why it is used to fill balloons.

Hydrogen is the lightest element found in nature. It has a density of 0.08988 g/L under normal conditions. This leads to hydrogen gas being lighter than any other gas in similar volume. Thus balloons filled with hydrogen gas rise much better than balloons filled with other gases like helium.

However, hydrogen is far less commonly used to fill balloons. The reason being that hydrogen is a highly flammable gas. Whereas gasses like helium which are noble gasses are much safer to use.

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The mass of hydrogen needed to produce 51.0 grams of ammonia is  ≈ 9.07 grams.

To determine the mass of hydrogen required to produce 51.0 grams of ammonia (NH3) using the Haber process, we need to calculate the stoichiometric ratio between hydrogen and ammonia.

From the balanced chemical equation:

3 H₂(g) + N₂(g) → 2 NH₃(g)

We can see that for every 3 moles of hydrogen (H₂), we obtain 2 moles of ammonia (NH₃).

First, we need to convert the given mass of ammonia (51.0 grams) to moles. The molar mass of NH₃ is 17.03 g/mol.

Number of moles of NH₃ = Mass / Molar mass

                     = 51.0 g / 17.03 g/mol

                     ≈ 2.995 moles

Next, using the stoichiometric ratio, we can calculate the moles of hydrogen required.

Moles of H₂ = (Moles of NH₃ × Coefficient of H₂) / Coefficient of NH₃

           = (2.995 moles × 3) / 2

           ≈ 4.493 moles

Finally, we can convert the moles of hydrogen to mass using the molar mass of hydrogen (2.02 g/mol).

Mass of H₂ = Moles × Molar mass

          = 4.493 moles × 2.02 g/mol

          ≈ 9.07 grams

Therefore, approximately 9.07 grams of hydrogen is needed to produce 51.0 grams of ammonia in the Haber process.

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The type of compound represented by the graph at right is a strong acid (option B).

An acid is generally any compound capable of dissociating into its respective constituent ions when in an aqueous solution.

An acid is categorised as strong or weak depending on whether it can dissociate completely or partially. A strong acid dissociates completely in water.

According to this question, HA, when added to water, dissociates into H+ and A- ions, hence, is a strong acid.

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Answer:

(R)-but-3-en-2-ylbenzene

Explanation:

In this reaction, we have a very strong base (sodium ethoxide). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an E2 mechanism, therefore, the hydrogen that is removed must have an angle of 180º with respect to the leaving group (the "OH"). This is known as the anti-periplanar configuration.

The hydrogen that has this configuration is the one that placed with the dashed bond (red hydrogen). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.

See figure 1

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The ion with a +3 charge, 28 electrons, and an atomic mass of 71 is the aluminum ion (\(Al^{3+}\)).

Aluminum (Al) typically has an atomic number of 13, which means it has 13 protons and 13 electrons in its neutral state. However, in the given ion, \(Al^{3+}\), the ion has lost three electrons, resulting in a +3 charge. This means that the ion now has 13 protons and only 10 electrons remaining, giving it a net positive charge of +3.

The atomic mass of aluminum is 26.98 atomic mass units (amu). The given ion has an atomic mass of 71 amu, which suggests that the ion has gained additional particles. In this case, the ion has also gained three neutrons, resulting in a higher atomic mass.

The total number of particles (protons, neutrons, and electrons) in the ion can be calculated by adding the number of protons (13) and the number of neutrons (3), which equals 16. Since the ion has a net charge of +3, it only contains 10 electrons.

In summary, the ion with a +3 charge, 28 electrons, and an atomic mass of 71 is the aluminum ion (\(Al^{3+}\)), which has 13 protons, 10 electrons, and 3 neutrons.

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Answer:

MRCORRECT has answered the question

Explanation:

Film photography is another example ofchemical reaction by light. In this example, the chemical compounds coated on the film go through a chemical reaction. ... These plates (usually made of aluminum) are coated with a photosensitive compound consisting of a polymer and a photosensitivechemical system.

Answer:

Volume = 3.4 L

Explanation:

In order to calculate the volume of CO₂ produced when 15.2 g of CaCO₃ is heated, we need to first write out the balanced equation of the thermal decomposition of CaCO₃:

CaCO₃ (s) + [Heat] ⇒ CaO (s) + CO₂ (g)

Now, let's calculate the number of moles in 15.2 g CaCO₃:

mole no. = \(\mathrm{\frac{mass}{molar \ mass}}\)

                 = \(\frac{15.2}{40.1 + 12 + (16 \times 3)}\)

                 = 0.1518 moles

From the balanced equation above, we can see that the stoichiometric molar ratios of CaCO₃ and CO₂ are equal. Therefore, the number of moles of CO₂ produced is also 0.1518 moles.

Hence, from the formula for the number of moles of a gas, we can calculate the volume of  CO₂:

mole no. = \(\mathrm{\frac{Volume \ in \ L}{22.4}}\)

⇒ \(0.1518 = \mathrm{\frac{Volume}{22.4}}\)

⇒ Volume = 0.1518 × 22.4

                 = 3.4 L

Therefore, if 15.2 g of CaCO₃ is heated, 3.4 L of CO₂ is produced at STP.

The coefficient that belongs in front of O2 in the following reaction C3H8 + O2 -> CO2 + H2O is 5.

A chemical equation is said to be balanced when the number of atoms of each element on both sides of the equation is the same.

According to this question, the following chemical equation is given:

C3H8 + O2 -> CO2 + H2O

To balance the above equation, one has to ensure that the number of atoms of carbon, hydrogen and oxygen are the same on both sides of the equation.

Hence, we make use of numbers placed in front of a compound called coefficient.

The balanced equation is as follows:

C3H8 + 5O2 → 3CO2 + 4H2O

Therefore, the coefficient that belongs in front of O2 in the following reaction C3H8 + O2 -> CO2 + H2O is 5.

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Answer:

(b.) 5

Explanation:

I got it correct on founders edtell

Answer:

melting and vaporization

Explanation:

Answer:

I THINK it is A

Explanation:

Answer: it is the eardrum caus ei used it on my test and as well got it right

Explanation: because it just is

\(hut7utpojghuire\\kjionbenhrmhbnhknokgn\\bihkjihjgnhrk\\\\gnthnrt\\hgnth\\\alpha thtrnhjn4t also this is for fun to me\)

Answer:

15 L water at 25℃

Explanation:

Answer:

15 L water at 25°C

Question: As the temperature of a substance increases, the average kinetic energy of the particles in the substance

Answer: increases

Explanation:

Sally is wrong because copper is less electropositive than hydrogen, thus, can not displace hydrogen from dilute acids.

The reactions to prepare copper (ii) chloride are:

the chlo­ri­na­tion of cop­per sul­fide at a high tem­per­a­ture

reaction of copper (ii) oxide with dilute hydrochloric acid

The equations of the given reactions are as follows:CuS + Cl₂ ---> CuCl₂ + SCuO + 2HCl ----> CuCl₂ + H₂O

Reactive metals are metals that readily give up their electrons to form positive ions.

Reactive metals displace hydrogen from dilute acids. They are found in group 1A and 2A of the periodic table. Copper is not a reactive metal and will not displace hydrogen from acids.

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Sally is wrong because copper chloride is not made from the reaction of copper and dilute hydrochloric acid.

2. Copper (ii) chloride can be prepared as follows:

3. the equations of the reaction are:

Single replacement reactions are reactions in which a more reactive atom replaces another atom in a compound.

An example of a single replacement reaction is the reaction of chlorine gas with copper sulfide at high temperatures to form copper chloride.

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Ammonia (\(NH_3\)) and ammonium chloride (\(NH_4Cl\)) mixture could be a useful buffer in a solution. Option C

A buffer is a solution that can resist changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base or a weak base and its conjugate acid. The buffer system works by the principle of Le Chatelier's principle, where the equilibrium is shifted to counteract the changes caused by the addition of an acid or a base.

In option A, acetic acid (\(CH_3CO_2H\)) is a weak acid, but hydrochloric acid (HCl) is a strong acid. This combination does not form a buffer because HCl is completely dissociated in water and cannot provide a significant concentration of its conjugate base.

Option B consists of sodium hydroxide (NaOH), which is a strong base, and elemental sodium (Na), which is a metal. This combination does not form a buffer as there is no weak acid-base pair involved.

Option D contains acetic acid (\(CH_3CO_2H\)), a weak acid, and ammonia (\(NH_3\)), a weak base. Although they are weak acid and base, they do not form a buffer system together as they are both weak acids or bases and lack the required conjugate acid-base pair.

Option C, ammonia (\(NH_3\)), is a weak base, and ammonium chloride (\(NH_4Cl\)) is its conjugate acid. This combination can form a buffer system. When ammonia reacts with water, it forms ammonium ions (NH4+) and hydroxide ions (OH-).

The ammonium ions act as the weak acid, while the ammonia acts as the weak base. The addition of a small amount of acid will be counteracted by the ammonium ions, and the addition of a small amount of base will be counteracted by the ammonia, thus maintaining the pH of the solution relatively stable.

Therefore, option C, consisting of ammonia (\(NH_3\)) and ammonium chloride (\(NH_4Cl\)), is the suitable mixture that could be a useful buffer in a solution.

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The mass of the asteroid is C. 1.2 x \(10^{12}\) Kg

To find the mass of the other asteroid, we can rearrange the equation for the gravitational force between two objects:

F = (G * m1 * m2) / \(r^{2}\)

where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two asteroids, and r is the distance between them.

Given that the distance between the asteroids is 75000 m, the force of gravity between them is 1.14 N, and one asteroid has a mass of 8 x \(10^{7}\) kg, we can substitute these values into the equation and solve for the mass of the other asteroid (m2):

1.14 N = (6.67430 × \(10^{-11}\) N \(m^{2}\)/\(Kg^{2}\) * 8 x \(10^{7}\) kg * \(m2\)) / \((75000 m)^{2}\)

Simplifying and solving the equation, we find that the mass of the other asteroid (m2) is approximately 1.2 x \(10^{12}\) kg. Therefore, Option C is correct.

The question was incomplete. find the full content below:

Two asteroids are 75000 m apart one has a mass of 8 x \(10^{7}\) kg if the force of gravity between them is 1.14 what is the mass of the asteroid

A. 3.4 x \(10^{11}\) kg

B. 8.3 x \(10^{12}\) kg

C. 1.2 x \(10^{12}\) kg

D. 1.2 x \(10^{10}\) kg

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The percentage purity of the calcium carbonate is  44%. The percentage purity gives the amount of pure CaCO3 in the sample.

In excess titration or back titration, we neutralize the excess titrand left in a system.

We have the reaction;

CaCO3 + 2HCl ----> CaCl2 + H2O + CO2

Number of moles of HCl = 0.2254 M * 20/1000 L = 0.0045 moles

The reaction of the excess acid is according to the reaction;

HCl + NaOH ----> NaCl + H2O

Number of moles of NaOH = 0.1041 M * 20/1000 = 0.0021 moles

Since the reaction is 1:1, 0.0021 moles of HCl reacted also

Number of moles of HCl that reacted with CaCO3  =  0.0045 moles - 0.0021 moles = 0.0024 moles

If 2 moles of HCl reacts with 1 mole of CaCO3

0.0024 moles of HCl reacts with 0.0024 moles * 1/2

= 0.0012 moles

Mass of pure CaCO3 present = 0.0012 moles * 100 g/mol = 0.12g

Percent purity of the sample =  0.12g/0.2719 g * 100/1

= 44%

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The ionic and net ionic equation for HCl + Mg(C2H3O2)2 ⇒ MgCl2 + HC2H3O2 is H⁺ + Cl⁻ + Mg⁺ + C₂H₃O₂⁻ → Mg⁺ + Cl₂⁻ +  H⁺ + C₂H₃O₂⁻.

Ionic equations are chemical equations that only show the ions involved in a chemical reaction. In other words, ions that combine in solution to form new substances. The ions that do not participate are referred to as spectator ions.

A net ionic equation depicts only the chemical species involved in a reaction, whereas a complete ionic equation depicts the spectator ions as well.

Thus,  H⁺ + Cl⁻ + Mg⁺ + C₂H₃O₂⁻ → Mg⁺ + Cl₂⁻ +  H⁺ + C₂H₃O₂⁻ this is the net ionic equation.

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The chemical formula for copper(II) sulfate is CuSO4.

Copper(II) sulfate is a chemical compound that is made up of copper, sulfur, and oxygen. It has the chemical formula CuSO4 and is commonly referred to as "blue vitriol" or "bluestone." Copper(II) sulfate can be prepared by reacting copper oxide or copper metal with sulfuric acid. It is a blue-colored crystalline solid that dissolves easily in water.

Copper(II) sulfate has many uses in industry and can be used as a fungicide, herbicide, pesticide, and in the manufacture of other chemicals. It is also commonly used in schools and laboratories as a reagent in chemical reactions and experiments.

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Answer:

Respiration

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