the number of neutrons in an atom can be found by subtracting the atomic number from the what?

The equilibrium vapor pressure with respect to water (eow) is 259.9 Pa. assume that saturation vapor pressure is same as equilibrium vapor pressure.

Therefore, the RH at the frost point is  

RH = (eow / saturation vapor pressure) × 100

= (259.9 Pa / 259.9 Pa) × 100

= 100%

b) At T = -11 °C, we need to compare the equilibrium vapor pressure with respect to water (eow) and the equilibrium vapor pressure with respect to ice (coi) to determine if ice particles will form. From the given table, at T = -11 °C, the equilibrium vapor pressure with respect to water (eow) is 237.7 Pa, and the equilibrium vapor pressure with respect to ice (coi) is 165.3 Pa.

The air is supersaturated with respect to ice, and the presence of Kaolinite particles can provide surfaces for water droplets to condense onto, leading to the formation of ice particles.

c) At T = -12 °C, we compare the equilibrium vapor pressure with respect to water (eow) and the equilibrium vapor pressure with respect to ice (coi). From the given table, at T = -12 °C, the equilibrium vapor pressure with respect to water (eow) is 217.3 Pa, and the equilibrium vapor pressure with respect to ice (coi) is 181.2 Pa.

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Explanation:

If the solution has a pH of 2, then the pOH is 12; the concentration of hydrogen ions would be 10-2 M and the concentration of hydroxide ions would be 10-12 M.

The number of moles of water will combine with 1 mole of CoCl₂.2H₂O is 4 moles.

The balanced chemical equation for the reaction, which is reversible, is CoCl₂.2H₂O + 4H₂O ⇌ CoCl₂.6H₂O.

Pink cobalt(II) chloride crystals can be changed back to blue cobalt(II) chloride crystals by heating or keeping them in a dry environment.

The mass of water lost when 0.50 moles of pink cobalt(II) chloride is turned completely into blue cobalt(II) chloride is 36.03 grams.

When blue cobalt(II) chloride crystals (CoCl₂.2H₂O) become damp, they turn pink and form CoCl₂.6H₂O. This means that 1 mole of CoCl₂.2H₂O combines with 4 moles of water (H₂O) to form 1 mole of CoCl₂.6H₂O. The balanced chemical equation for this reversible reaction is:

CoCl₂.2H₂O + 4H₂O ⇌ CoCl₂.6H₂O

To change pink cobalt(II) chloride crystals (CoCl₂.6H₂O) back to blue cobalt(II) chloride crystals (CoCl₂.2H₂O), you need to remove water from the crystals, which can be done by heating or keeping them in a dry environment.

When 0.50 moles of pink cobalt(II) chloride is turned completely into blue cobalt(II) chloride, 4 moles of water are lost for every mole of CoCl₂.6H₂O converted. Therefore, the mass of water lost can be calculated as follows:

0.50 moles of CoCl₂.6H₂O × 4 moles of H₂O / 1 mole of CoCl₂.6H₂O = 2 moles of H₂O

The molar mass of water is 18.015 g/mol, so the mass of water lost is:

2 moles of H₂O × 18.015 g/mol = 36.03 g

So, 36.03 grams of water is lost when 0.50 moles of pink cobalt(II) chloride is turned completely into blue cobalt(II) chloride.

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Two possible sources of systematic (determinate) error in the experiment are; Incorrect calibration of volumetric glasswareIncorrect mass of NaCl

To calculate the concentration of NaCl in the resulting solution in mg/L NaCl, we can use the formula; Concentration (mg/L) = (Mass of solute ÷ Volume of solution in L) × 1000 g / 1 mg NaCl is present in the stock solution of 25 mL. So, the mass of NaCl in the solution would be;0.0842 g ÷ 25 mL = 0.00337 g/mL. Now, in the resulting solution, a 1.00 mL aliquot of the stock solution is transferred to a 50.00 mL volumetric flask and diluted to volume. Therefore, the volume of the resulting solution is 50.00 mL. We will substitute these values in the formula, Concentration (mg/L) = (0.00337 g/mL ÷ 50 mL) × 1000 g / 1 mg concentration (mg/L) = 67.4 mg/L. Therefore, the concentration of NaCl in the resulting solution in mg/L NaCl is 67.4 mg/L.7.  Concentration = 67.4 mg/LTolerance = 4.28 mg/LTotal concentration = 67.4 + 4.28 mg/L = 71.68 mg/LWe round off this value to one decimal place; Total concentration = 71.7 mg/LTherefore, the concentration of NaCl in the resulting solution using propagation of error through the calculation is 67.4+0.4 mg/L.8. Two possible sources of random (indeterminate) error in the experiment are; Errors in temperature measurement. Errors in measurement of water volume. Two possible sources of systematic (determinate) error in the experiment are; Incorrect calibration of volumetric glasswareIncorrect mass of NaCl.

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Answer:

Yes

Beryllium is a chemical element with the symbol Be and atomic number 4. It is a steel-gray, strong, lightweight and brittle alkaline earth metal. ... Notable gemstones high in beryllium include beryl (aquamarine, emerald) and chrysoberyl.

Calcium is a mineral that is necessary for life. In addition to building bones and keeping them healthy, calcium enables our blood to clot, our muscles to contract, and our heart to beat. About 99% of the calcium in our bodies is in our bones and teeth.

In Ezra 10, the article likely refers to a specific passage from the biblical book of Ezra, chapter 10. However, providing with a general understanding of how repeated keywords or themes can be important in any given text.

Repeated keywords or themes in a text can serve as indicators of the central ideas or concepts discussed. They help reinforce key messages, emphasize important points, and create a sense of cohesion within the text. By identifying and analyzing these repetitions, readers can gain insights into the author's intentions and the overall focus of the article. In the case of Ezra 10, if there were any repeated keywords or themes, they would likely point to significant elements in the narrative or themes within the chapter. By identifying and examining these repetitions, readers can uncover recurring motifs, emphasize character development or conflicts, and gain a deeper understanding of the overall message or lessons conveyed by the author.Please note that without the specific text or access to the article you mentioned.

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magnesium hydroxide is an osmotic laxative, that is that it works by pumping water into the intestines, which helps in producing bowel movement.

This medicine is used to relieve sporadic constipation. It is an antacid that reduces the stomach's acid production. Magnesium hydroxide lowers stomach acid and raises gut water levels, which may encourage bowel motions. Magnesium hydroxide is a laxative used to treat sporadic constipation. Magnesium hydroxide is also employed as an antacid to treat heartburn, indigestion, and sour stomach.

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Explanation: Hey Sonia, I see I'm not the only one at school struggling in Chem. My best guess is to minimize the likelyhood and size-amount of a chemical reaction.

Answer:

The hydronium ion concentration increases to 100 times the original concentration

Explanation:

The pH of a solution is defined as the negative logarithm of the hydrogen or hydronium ion concentration of that solution. It is given by the expression below:

pH = -log[H₃O⁺] = log[H₃O⁺]⁻¹

Assuming the solution was at neutral with original pH = 7;

The new pH of the solution will be = 7 - 2 = 5

At pH = 7;

log[H₃O⁺]⁻¹ = 7

[H₃O⁺]⁻¹ = 10⁷

[H₃O⁺] = 10⁻⁷

At pH = 5

log[H₃O⁺]⁻¹ = 5

[H₃O⁺]⁻¹ = 10⁵

[H₃O⁺] = 10⁻⁵

10⁻⁵ = 10⁻⁷ * 10²

But 10² = 100

Therefore, the hydronium ion concentration increases to 100 times the original concentration

Answer:

B

Explanation:

On Edge

The process that correctly describes the growth in the number of ionized nitrogen molecules over time is D. The number of ionized molecules increases exponentially with each ionized molecule ionizing 3 other molecules every nanosecond.

Let's analyze the equation tN2 = 3⋅N, where N is the number of ionized molecules and t is the time in nanoseconds.

The equation states that the number of ionized molecules at time t is equal to 3 times the number of ionized molecules at the previous time step.

This implies that each ionized molecule ionizes 3 other molecules every nanosecond.

A physicist is studying the nature of static discharge by applying a voltage across a microscopic tube filled with nitrogen.

Static discharge is an event that takes place when there is a sudden flow of electric charge between two objects with different electrical potentials. This electric charge transfer results in a brief burst of electromagnetic energy (EMI or EMF), which can sometimes be observed as a visible spark in low-light conditions or heard as a crackling sound in high-quality audio systems.

The nature of static discharge is dependent on a variety of factors, including the composition of the material, the temperature and humidity, the electrical potential between the two objects, and the distance between the objects. Typically, static discharge occurs more frequently in environments with low humidity, such as during the winter months or in arid regions, as moisture acts as an insulator and prevents the buildup of electrical charge. In addition, materials that are good electrical conductors, such as metals, are more likely to experience static discharge than materials that are poor conductors, such as plastics or rubber.

Therefore, the correct description of the growth in the number of ionized nitrogen molecules over time is:

D. The number of ionized molecules increases exponentially with each ionized molecule ionizing 3 other molecules every nanosecond.

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Complete question is:

" physicist is studying the nature of static discharge by applying a voltage across amicroscopic tube filled with nitrogen molecules.Every nanosecond from then on, any ionized molecules willionize an additional number of molecules not already ionized, and ionization does not get lost.The equationtN2 3shows N, the number of ionized molecules, t nanoseconds after initiating a voltage.Which of thefollowing correctly describes the growth in the number of ionized nitrogen molecules over time?A.The number of ionized molecules increases linearly with each ionized molecule ionizing 2 other moleculesevery nanosecond.B.The number of ionized molecules increases linearly with each ionized molecule ionizing 3 other moleculesevery nanosecond.C.The number of ionized molecules increases exponentially with each ionized molecule ionizing 2 othermolecules every nanosecond.D.The number of ionized molecules increases exponentially with each ionized molecule ionizing 3 othermolecules every nanosecond."

For a general reaction aA + bB → cC, let's sketch the three graphs based on the integrated rate laws. we can say that the slope of the linear graph gives the second-order rate constant, k.

We need to draw pictures for the graphs that are second order in A, and to label each axis, and to show what information the slope of the linear graph gives.

Graph 1: [A] versus t

For a second-order reaction, 1/[A] versus time is a linear plot with a slope of k and an intercept of 1/[A]0.

The equation for the second-order reaction rate law is:

1/[A] = kt + 1/[A]0, where k is the second-order rate constant, t is time, and 1/[A]0 is the reciprocal of the initial concentration of A. \(\frac{1}{[A]} = kt+\frac{1}{[A]_{0}}\)Graph 2: t versus [A]This is a plot of the concentration of A versus time, which is linear for a second-order reaction. The slope of the linear plot is -k and the y-intercept is [A]0. The equation for the second-order reaction rate law is:

[A] = [A]0 / (1 + kt [A]0),

where k is the second-order rate constant, t is time, and [A]0 is the initial concentration of A.

\([A] = \frac{[A]_{0}}{1+kt[A]_{0}}\)

Graph 3: 1/[A] versus 1/tThis is a linear plot of 1/[A] versus 1/t.

The slope of the linear plot is k and the y-intercept is 1/[A]0. The equation for the second-order reaction rate law is:

1/[A] = kt + 1/[A]0, where k is the second-order rate constant, t is time, and 1/[A]0 is the reciprocal of the initial concentration of A.

\(\frac{1}{[A]} = kt+\frac{1}{[A]_{0}}\)

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The standard enthalpy change for reaction (3) is 117.1 kJ.

The standard enthalpy change for reaction (3) can be calculated by using the enthalpy changes of reactions (1) and (2) and applying Hess's Law.

To do this, we need to manipulate the given equations so that the desired reaction (3) can be obtained.

First, we reverse reaction (1) to get the formation of C2H4(g) from C2H6(g):

C2H4(g)C2H6(g) ΔH° = -52.3 kJ

Next, we multiply reaction (2) by 2 and reverse it to obtain 2 moles of C2H6(g) reacting to form 3 moles of H2(g):

2C2H6(g)2C(s) + 3H2(g) ΔH° = 169.4 kJ

Now, we add the two modified equations together:

C2H4(g)C2H6(g) ΔH° = -52.3 kJ
2C2H6(g)2C(s) + 3H2(g) ΔH° = 169.4 kJ

When adding these equations, the C2H6(g) on the left side cancels out with the C2H6(g) on the right side, leaving us with the desired reaction (3):

C2H4(g) + H2(g)C2H6(g) ΔH° = -52.3 kJ + 169.4 kJ = 117.1 kJ

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Answer:

an organism consisting of a cell or cells in which the genetic material is DNA in the form of chromosomes contained within a distinct nucleus. Eukaryotes include all living organisms other than eubacteria and archaebacteria.

Explanation:

you could have just looked it up yourself on go.ogle but thx for the points!

Ionic Equilibrium is the type of equilibrium that occurs between the ions of the solution of any weak electrolyte.

Equilibrium is referred to as the state when in a reaction the amount of reactants that are consumed becomes equal to the amount of the products that are formed or in other words we can say that the rate of change of reactants equals the rate of formation of products within a reaction. Ionic equilibrium exists in case of solutions of weak electrolytes where ions are involved.

For example, If we mix salt in water, so there exits a state where the rate at which salt is mixing in the water becomes equal to the formation of salt solution. This stage is called Ionic Equilibrium. If we write the equation of this reaction, we get

NaCl(solid) + H₂O(liquid) → Na⁺(aqueous) + Cl⁻(aqueous)

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After being mixed with 50.0 L of NaCl solution, the KOH solution's final molarity is 0.45 M.

We can apply the principle of dilution to ascertain the ultimate molarity of the KOH solution. The dilution equation is:

M₁V₁ = M₂V₂

Where:

M1 is the solution's initial molarity (KOH).

V1 is the solution's starting volume (in KOH).

M2 is the solution's final molarity (KOH).

V2 is the solution's total volume (in KOH).

Given:

Initial KOH solution volume (V1) is 10.0 L.

KOH solution's initial molarity (M1) is 2.7 M.

After incorporating NaCl solution, the KOH solution's final volume (V2) equals 10.0 L plus 50.0 L, or 60.0 L.

These values are substituted in the dilution equation:

(2.7 M)(10.0 L) = (M₂)(60.0 L)

27.0 = 60M₂

Calculating M2:

M₂ = 27.0 / 60 = 0.45 M

Therefore, after adding 50.0 L of NaCl solution, the KOH solution's final molarity or molar concentration is 0.45 M.

It's vital to note that this estimate is based on the supposition that the quantities are additive and that NaCl and KOH do not react. Furthermore, no departures from ideal behaviour are taken into consideration in this calculation, which is predicated on perfect behaviour.

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The correct answer is d. 1,3-cyclohexadiene undergoes dehydrogenation readily because it would gain considerable stability by becoming benzene. Benzene is a highly stable and aromatic compound that possesses resonance energy due to its delocalized pi-electrons.

Dehydrogenation is a chemical reaction that involves the removal of hydrogen from a molecule. In the case of 1,3-cyclohexadiene, the removal of two hydrogen atoms would result in the formation of benzene. This transformation would result in the formation of a highly stable compound, which has much lower energy than its precursor.
Moreover, 1,3-cyclohexadiene is an unsaturated compound that possesses a double bond between two carbon atoms. This double bond makes the molecule reactive towards dehydrogenation. During dehydrogenation, the double bond is broken, and the two hydrogen atoms that were attached to the carbon atoms are removed. As a result, the molecule undergoes a structural change, and a highly stable compound, benzene, is formed.
In conclusion, 1,3-cyclohexadiene undergoes dehydrogenation readily because it would gain considerable stability by becoming benzene. This transformation is a result of the removal of two hydrogen atoms from the molecule, and it occurs due to the reactivity of the double bond that the molecule possesses.

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A chemical known as an acid-base indicator alters color when the pH of a solution changes. this statement is true.

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We can change a molecule into a series of one or more elements by breaking down the chemical bonds that connect atoms in the molecule.

The chemical bonds of a molecule can be defined as the different types of attractive forces that link atoms in a molecule, which include, for example, covalent bonds, ionic bonds, hydrogen bonding, metallic bonding, etc.

Therefore, with this data, we can see that we have to break the chemical bonds of a molecule in order to separate molecules into their simplest constituents, i.e., atoms.

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Answer:

B.Magnesium(Mg)

Explanation:

maybe

the Celsius temperature of 1.50 moles of ammonia contained in a 10.0 L vessel under a pressure of 2.0 atm is approximately -56.15 C. The closest answer choice to this value is -50 C.

To determine the Celsius temperature of 1.50 moles of ammonia contained in a 10.0 L vessel under a pressure of 2.0 atm, we can use the Ideal Gas Law equation:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

T = PV/nR

T = (2.0 atm) x (10.0 L) / (1.50 moles x 0.08206 L atm/K mol)

T = 217 K

To convert this temperature to Celsius, we can simply subtract 273.15 K:

T(Celsius) = 217 K - 273.15

T(Celsius) = -56.15 C

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Answer:

I can't understand your language? can you write in English

I'm sorry

Answer:

     3S₈  +  28Br₂ => 8S₃Br₇

Explanation:

Start with either sulfur (S) or bromine (Br) and balance ...

    3S₈  +  Br₂ => 8S₃Br₇    or      S₈  +  7/2Br₂ => S₃Br₇

Balance the remaining reactant ...

    3S₈  +  56/2Br₂ => 8S₃Br₇    

Remove fractions by multiplying by the fraction's denominator

    2(3S₈  +  56/2Br₂ => 8S₃Br₇)     =>     6S₈  +  56Br₂ => 16S₃Br₇      

Reduce to smallest whole number ratio => standard equation at STP ...

        3S₈  +  28Br₂ => 8S₃Br₇

By drawing the resonance structures of nitromethane and methyl nitrite, you can determine that the N-O bond length in methyl nitrite is shorter than in nitromethane due to the partial double bond character in methyl nitrite.

Nitromethane (CH3NO2) and methyl nitrite (CH3ONO) share the same empirical formula. By drawing the resonance structures of these two molecules, you can obtain information regarding the N-O bond length.
For nitromethane (CH3NO2), there is only one resonance structure, with a single bond between the nitrogen and oxygen atom. This single bond results in a longer N-O bond length.

For methyl nitrite (CH3ONO), there are two resonance structures. In one structure, there is a double bond between the nitrogen and oxygen atom, and in the other structure, there is a single bond between the nitrogen and oxygen atom. The true structure of methyl nitrite is a combination of these two resonance structures, resulting in a partial double bond character and a shorter N-O bond length compared to nitromethane.

In conclusion, by drawing the resonance structures of nitromethane and methyl nitrite, you can determine that the N-O bond length in methyl nitrite is shorter than in nitromethane due to the partial double bond character in methyl nitrite.

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Answer:

As water is boiled, kinetic energy causes the hydrogen bonds to break completely and allows water molecules to escape into the air as gas (steam or water vapor). When water freezes, water molecules form a crystalline structure maintained by hydrogen bonding.

Explanation:

In summary:

(a) No internal irreversibilities: Specific entropy at the exit is equal to the specific entropy at the inlet.

(b) No internal irreversibilities: Specific entropy at the exit is equal to the specific entropy at the inlet.

(c) No internal irreversibilities: Specific entropy at the exit is equal to the specific entropy at the inlet.

(d) Internal irreversibilities: Specific entropy at the exit is greater than the specific entropy at the inlet.

(e) Internal irreversibilities: Specific entropy at the exit is greater than the specific entropy at the inlet.

Let's analyze each case separately:

(a) No internal irreversibilities:

If there are no internal irreversibilities, it means that the process is reversible or internally reversible. In such cases, the specific entropy at the exit will be equal to the specific entropy at the inlet. This is because reversible processes conserve entropy within a control volume.

(b) No internal irreversibilities:

Similarly to case (a), if there are no internal irreversibilities, the specific entropy at the exit will be equal to the specific entropy at the inlet. This holds true regardless of the heat transfer rate or location.

(c) No internal irreversibilities:

Once again, if there are no internal irreversibilities, the specific entropy at the exit will be equal to the specific entropy at the inlet.

(d) Internal irreversibilities:

When internal irreversibilities exist, the specific entropy will generally increase within the control volume. Internal irreversibilities could result from factors such as friction, turbulence, or non-equilibrium processes. In this case, the specific entropy at the exit is expected to be greater than the specific entropy at the inlet.

(e) Internal irreversibilities:

Similar to case (d), the presence of internal irreversibilities indicates that the specific entropy at the exit will likely be greater than the specific entropy at the inlet.

In summary:

(a) No internal irreversibilities: Specific entropy at the exit is equal to the specific entropy at the inlet.

(b) No internal irreversibilities: Specific entropy at the exit is equal to the specific entropy at the inlet.

(c) No internal irreversibilities: Specific entropy at the exit is equal to the specific entropy at the inlet.

(d) Internal irreversibilities: Specific entropy at the exit is greater than the specific entropy at the inlet.

(e) Internal irreversibilities: Specific entropy at the exit is greater than the specific entropy at the inlet.

These conclusions are based on general principles, and the specific behavior of the gas and control volume would depend on the details of the process and the specific properties involved.

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The mass of 0.5 mole of the element is approximately 6.025 grams.

To calculate the mass of 0.5 mole of the element, we need to know the molar mass of the element.

Given that the mass of 2 x 10^21 atoms of the element is 0.4 grams, we can use this information to find the molar mass.

The number of atoms in 1 mole of any substance is given by Avogadro's number, which is approximately 6.022 x 10^23 atoms/mol.

First, we calculate the molar mass of the element using the given information:

Molar mass = Mass of 2 x 10^21 atoms / Number of moles of 2 x 10^21 atoms

Molar mass = 0.4 g / (2 x 10^21 atoms / (6.022 x 10^23 atoms/mol))

Molar mass ≈ 0.4 g / (3.32 x 10^-2 mol)

Molar mass ≈ 12.05 g/mol

Now that we know the molar mass of the element is approximately 12.05 g/mol, we can calculate the mass of 0.5 mole of the element:

Mass = Molar mass x Number of moles

Mass = 12.05 g/mol x 0.5 mol

Mass = 6.025 grams

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When the carbonic acid sodium bicarbonate buffer pair buffers lactic acid, the following reaction occurs:$$\text{HCO}_3^- + \text{H}^+ \leftrightarrow \text{H}_2\text{CO}_3$$The carbonic acid/bicarbonate buffer system is one of the most important in human blood. When the pH of the blood decreases (becomes more acidic), the amount of bicarbonate ions in the blood decreases, and the concentration of hydrogen ions increases.

A buffer is a solution of a weak acid and its conjugate base that prevents changes in pH when small amounts of strong acid or base are added. Buffer systems protect organisms from pH changes by regulating and neutralizing acids and bases that enter or are produced by cells.

When the carbonic acid sodium bicarbonate buffer pair buffers lactic acid, the following reaction occurs:$$\text{HCO}_3^- + \text{H}^+ \leftrightarrow \text{H}_2\text{CO}_3$$The carbonic acid/bicarbonate buffer system is one of the most important in human blood. When the pH of the blood decreases (becomes more acidic), the amount of bicarbonate ions in the blood decreases, and the concentration of hydrogen ions increases. To balance the excess hydrogen ions, carbonic acid (H2CO3) is formed from carbon dioxide (CO2) and water (H2O). Carbonic acid then decomposes to form bicarbonate ions and hydrogen ions, and the pH of the blood is returned to normal. The bicarbonate ions act as a base, neutralizing the excess hydrogen ions that cause the blood to become more acidic. This is called the bicarbonate buffer system. Lactic acid is produced during intense exercise when the body doesn't get enough oxygen to meet its energy needs. The buildup of lactic acid in muscles can cause fatigue and muscle soreness. The carbonic acid/bicarbonate buffer system can also help to buffer the excess lactic acid produced during exercise, preventing the blood from becoming too acidic.

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