A gaseous mixture contains 5.0 moles of nitrogen and 10.0 moles of helium . What is the total pressure if the temperature is 25 degrees C and the volume is 3.0 L? a) 122 atm
B) 56 atm
d) 2.0 atm
C)3.0 atm

The amount of the compound [tex]XY_2[/tex] that can be generated would be 11.28 grams

From the equation of the reaction, the mole ratio of X and Y is 1:1.

Mole of 3.5 grams of X = 3.4/24 = 0.1417 moles

Mole of 4.2 grams of Y = 4.2/35 = 0.12 moles

Thus, Y is the limiting reactant because it is present in a lower amount than needed.

Mole ratio of Y and  [tex]XY_2[/tex] = 1:1

Equivalent mole of  [tex]XY_2[/tex] = 0.12 moles

Molar mass of  [tex]XY_2[/tex] = 24 + (35x2) = 94 g/mol

Mass of 0.12 moles of  [tex]XY_2[/tex] = 0.12 x 94 = 11.28 grams

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The final answer is  1.8 × 102 L of Rn, measured at STP, are produced per day by 1.08 g of Ra.

In order to determine the amount of radon that is produced per day by 1.08 g of Ra, it is essential to calculate the number of Rn atoms that are produced per second.

In the given question, it is known that 1.00 × 1015 atoms of radium produce an average of 1.373 × 104 atoms of Rn per second.

Therefore, we can use this ratio to calculate the number of Rn atoms that are produced per day by 1.08 g of Ra. Once we know the number of Rn atoms produced per day, we can calculate the volume of Rn produced by using the ideal gas law.

The final answer is  1.8 × 102 L of Rn, measured at STP, are produced per day by 1.08 g of Ra.

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the identity of element E in the nuclear reaction (238)U --> (234)E + (4)He is thorium (Th).

What is Nuclear Reaction?

A nuclear reaction is one in which the nucleus of an atom is transformed into the nucleus of one or more daughter nuclei. This differs from a a chemical reaction in which atoms are changed from one form to another.

To identify the element E in the nuclear reaction (238)U --> (234)E + (4)He, we need to consider the conservation of mass number and atomic number.

In the given reaction, the mass number is conserved, as 238 = 234 + 4. This indicates that the total mass on the left side (238) is equal to the total mass on the right side (234 + 4).

Next, we consider the atomic number. The atomic number is the number of protons in the nucleus of an atom. In the reaction, uranium (U) has an atomic number of 92. Since helium (He) has an atomic number of 2, the sum of the atomic numbers on the right side should be equal to the atomic number on the left side.

92 (atomic number of U) = atomic number of E + 2

Solving this equation, we find:

atomic number of E = 92 - 2 = 90

The element with atomic number 90 is thorium (Th).

Therefore, the identity of element E in the nuclear reaction (238)U --> (234)E + (4)He is thorium (Th).

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The number of such relations will be 2n - 2. Hence, the correct option is 2n - 2.

a and b are distinct elements of S. We need to find out how many relations R are there on S such that no ordered pair in R has a as its first element or b as its second element.So, the number of relations R on S such that no ordered pair in R has a as its first element or b as its second element is:2n - 2Note: There are n2 ordered pairs in S × S and out of them, n pairs have the first element a and n pairs have the second element b. These n + n ordered pairs must not be present in R, reducing the number of ordered pairs that may or may not be present in R by 2n. Therefore, the number of such relations will be 2n - 2. Hence, the correct option is 2n - 2.

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Answer:

butter and water don't mix because water is polar and butter is nonpolar

Explanation:

for them to be able to mix they would need to be both polar or both nonpolar.

Answer:

There are 2 moles of oxygen molecules; there are 4 moles of oxygen atoms.

Explanation:

The half-life for this reaction is approximately 2.56 seconds.

Approximately 0.014 M of I₂ will remain after 5.12 s, assuming no recombination of iodine atoms to form I₂.

(a) The half-life of a first-order reaction can be calculated using the equation:

t₁/₂ = (0.693) / k

Given that the rate constant (k) is 0.271 s⁻¹, we can substitute this value into the equation:

t₁/₂ = (0.693) / 0.271 = 2.56 s

(b) To determine the remaining amount of I₂ after 5.12 s, we can use the first-order integrated rate equation:

[tex][A] = [A_{0} ]* e^{(-kt)}[/tex]

Where [A] is the concentration at time t, [A₀] is the initial concentration, k is the rate constant, and e is the base of natural logarithm.

Given [A₀] = 0.050 M, k = 0.271 s⁻¹, and t = 5.12 s, we can calculate:

[A] = 0.050 * e^(-0.271 * 5.12)

[A] = 0.050 * e^(-1.387)

[A] = 0.014 M

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Answers:

Most of the world’s energy comes from what three sources?

Oil, coal, and gas.

These energy sources are called fossil fuels and they are non-renewable resources.

Fuel type: oil

How it's formed: from the remains of ancient marine organisms

Its uses: transportation, industrial power, heating and lighting, lubricants, petrochemical industry, and use of by-products

Fule type: coal

How it's formed: when dead plant matter decays into peat and is converted into coal by the heat and pressure of deep burial

Its uses: electricity generation, metal production, cement production, chemical production, gasification, and other industrial uses

Fuel type: gas

How it's formed: decomposed organic matter mixed with mud, silt, and sand on the seafloor

Its uses: heating & cooling buildings, cooking foods, fueling vehicles, and electricity generation

From 250.0 g of [tex]SO_{2}[/tex], approximately -772.44 kJ of heat is produced, and approximately 24.21 g of [tex]SO_{3}[/tex] are produced when the observed enthalpy change is -75.0 kJ.

To calculate the heat produced from a given amount of [tex]SO_{2}[/tex]and the grams of [tex]SO_{3}[/tex] produced when the observed enthalpy change is given, we need to use the stoichiometry and molar masses of the compounds involved.

First, let's calculate the moles of [tex]SO_{2}[/tex]present in 250.0 g:

Molar mass of [tex]SO_{2}[/tex]= 32.07 g/mol + 32.07 g/mol = 64.14 g/mol

Moles of [tex]SO_{2}[/tex]= Mass of [tex]SO_{2}[/tex]/ Molar mass of [tex]SO_{2}[/tex]

= 250.0 g / 64.14 g/mol

≈ 3.897 mol

Using the stoichiometry of the balanced equation, we can determine the moles of [tex]SO_{3}[/tex]produced:

From the balanced equation: 2 mol [tex]SO_{2}[/tex] → 2 mol [tex]SO_{3}[/tex]

So, 3.897 mol [tex]SO_{2}[/tex]→ (3.897 mol [tex]SO_{3}[/tex]/ 2 mol [tex]SO_{2}[/tex]) = 1.9485 mol [tex]SO_{3}[/tex]

Next, let's calculate the heat produced using the given enthalpy change:

Heat produced = Moles of [tex]SO_{2}[/tex]* ΔH

Heat produced = 3.897 mol [tex]SO_{2}[/tex]* (-198.2 kJ/mol)

≈ -772.44 kJ

Therefore, approximately -772.44 kJ of heat are produced from 250.0 g of [tex]SO_{2}[/tex].

Finally, let's calculate the grams of [tex]SO_{3}[/tex]produced when the observed enthalpy change is -75.0 kJ:

Given ΔH = -75.0 kJ

Moles of [tex]SO_{3}[/tex]= ΔH / ΔH for the balanced equation

= -75.0 kJ / (-198.2 kJ/mol)

≈ 0.3782 mol

Using the molar mass of [tex]SO_{3}[/tex]:

Molar mass of [tex]SO_{3}[/tex] = 32.07 g/mol + 16.00 g/mol + 16.00 g/mol = 64.07 g/mol

Grams of [tex]SO_{3}[/tex]= Moles of [tex]SO_{3}[/tex]* Molar mass of [tex]SO_{3}[/tex]

= 0.3782 mol * 64.07 g/mol

≈ 24.21 g

Therefore, approximately 24.21 g of are produced when the observed enthalpy change is -75.0 kJ.

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The complete question is:

Given a thermochemical equation: 2SO2 (g) + O2 (g) --> 2SO3 (g), ΔH=-198.2 kJ . How many kJ of heat are made from 250.0 g SO2? . How many grams of SO3 (g) are produced when the observed enthalpy change is -75.0 kJ? Report all answers to 3 significant figures and include signs (where appropriate) and units.

Pretty sure the answer is D.

Answer:

oxygen

Explanation:

plants and animals use oxygen

Electrophilic aromatic substitution reactions have various applications in organic chemistry. They are commonly used to introduce functional groups onto aromatic rings, synthesize pharmaceuticals, produce dyes, and create complex organic molecules.

In organic chemistry, electrophilic aromatic substitution reactions are crucial tools for attaching new functional groups to aromatic rings. In these reactions, an electrophile replaces a hydrogen atom on an aromatic molecule. The end output can be used in a variety of different sectors.

Pharmaceutical synthesis is one area in which electrophilic aromatic substitution is used. Chemists can change the solubility, reactivity, and bioavailability of medicinal compounds by selectively adding functional groups to aromatic rings. This enables the creation of fresh medication candidates or the advancement of current ones.

The manufacture of dyes is an additional use. Due to their conjugated systems, aromatic compounds with particular functional groups can display bright hues. Colorful dyes used in textiles, inks, and other industries are produced through the introduction of chromophores onto aromatic rings using electrophilic aromatic substitution processes.

Additionally, it is essential for the synthesis of complex organic compounds to undergo electrophilic aromatic substitution processes. Chemists can create complex chemical structures with particular functions by carefully swapping various locations on an aromatic ring. This makes it possible to synthesize natural substances, sophisticated compounds, and materials with specific qualities.

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Answer:

decreases

Explanation:

Gravirational force is directly proportional to the mass and inversely proportional to the distance.(Newton's law of gravitation)

In the circular flow diagram, firms provide goods and services to product markets.

A circular flow diagram is a simplified economic model that demonstrates how money, goods, and services flow between households and firms in an economy. This diagram is divided into two markets: the product market and the factor market. The factor market deals with inputs such as labor, capital, and raw materials, while the product market deals with outputs such as goods and services.What do firms provide to product markets?Firms are companies or businesses that generate goods or services. In the circular flow diagram, firms produce goods and services that are sold in the product market. Firms sell products to product markets. This process generates revenue for the firms, which they utilize to pay for inputs such as labor and raw materials.

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The angular spread of the light inside the glass is 132.4° for red light and 132.2° for violet light.

Given data: Red light n_red =1.52Violet light n_violet =1.55. Incident angle i = 28°To find: The angular spread of the light inside the glass.

Solution: The angle of incidence i = 28°The angle of refraction r can be calculated using Snell's law:µ1 sin i = µ2 sin where µ1 is the refractive index of the incident medium (air), µ2 is the refractive index of the refracting medium (glass), i is the angle of incidence, and r is the angle of refraction. For red light,µ1 = 1.0003 and µ2 = 1.52For violet light,µ1 = 1.0003 and µ2 = 1.55.

The angle of refraction can be calculated as follows: Red light: sin r = µ1/µ2 sin i= 1.0003/1.52 × sin 28°= 0.3736r = sin–1 (0.3736) = 22.31°Violet light: sin r = µ1/µ2 sin i= 1.0003/1.55 × sin 28°= 0.3863r = sin–1 (0.3863) = 22.93°The angle of deviation δ for a prism is given by:δ = (µ − 1)A. where µ is the refractive index of the prism and A is the angle of the prism. From the geometry of the prism, the angle of deviation can be related to the angle of incidence and the angle of refraction by the formula:δ = I + r – A where A is the apex angle of the prism.

Since the angle of deviation for red and violet light is the same, we can equate the two expressions for δ to obtain: I + r – A = i’ + r’ – A where i’ and r’ are the angles of incidence and refraction for violet light. Subtracting I + r = i’ + r’ from the above equation, we get: A = (i’ – i) + (r’ – r)Let the angular spread of the light inside the glass be θ. Then,θ = A/2= [(i’ – i) + (r’ – r)]/2For red light: i’ = 180° – 22.31° = 157.69°r’ = 180° – sin–1 (sin r/µ1) = 180° – sin–1 (0.2617/1.0003) = 157.45°θ = [(157.69° – 28.00°) + (157.45° – 22.31°)]/2= 132.41°For violet light: i’ = 180° – 22.93° = 157.07°r’ = 180° – sin–1 (sin r/µ1) = 180° – sin–1 (0.2491/1.0003) = 157.24°θ = [(157.07° – 28.00°) + (157.24° – 22.93°)]/2= 132.19°

Therefore, the angular spread of the light inside the glass is 132.4° for red light and 132.2° for violet light.

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Explanation:

List of All Elements That Are Nonmetals

Hydrogen (sometimes)

Carbon.

Nitrogen.

Oxygen.

Phosphorus.

Sulfur.

Metals:-

Gold.

Silver.

Iron.

Copper.

Nickel.

Aluminum.

Mercury( Liquid metal)

Titanium.

Fluorine.

Therefore, after 300.0 seconds, approximately 0.0553 M of N₂O₅ will remain. The correct answer is option b: 0.0553 M.

To determine how much N2O5 will remain after 300.0 seconds, we can use the first-order rate equation:

ln(N₂O₅t/N₂O₅₀) = -kt

Where N₂O₅t is the concentration of N₂O₅ at time t, N₂O₅₀ is the initial concentration of N₂O₅, k is the rate constant, and t is the time.

Substituting the given values:

N₂O₅t = N₂O₅₀ ₓ e^(-kt)

N₂O₅t = 0.150 M * e^(-4.82 x 10⁻³ s⁻¹ ˣ 300.0 s)

N₂O₅t ≈ 0.0553 M

Therefore, after 300.0 seconds, approximately 0.0553 M of N₂O₅ will remain.

The correct answer is option b: 0.0553 M.

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1-hexanol, 2-hexanol, and 3-hexanol are chiral because they possess a chiral center. 2-methylpentanol is achiral because it lacks a chiral center.

To determine which isomeric alcohols with the molecular formula C₆H₁₄O are chiral and which are achiral, we need to examine their structural features, specifically the presence or absence of a chiral center.

A chiral center is a carbon atom bonded to four different substituents, resulting in non-superimposable mirror images. If a molecule has a chiral center, it is chiral; otherwise, it is achiral.

Let's examine the structural isomers of C₆H₁₄O

1-Hexanol (CH₃(CH₂)₄OH)

This molecule contains a chiral center at the carbon atom bonded to the hydroxyl group (OH). Since it has four different substituents (CH₃, CH₂, CH₂, and H), 1-hexanol is chiral.

2-Hexanol (CH₃CH₂CH(OH)CH₂CH₃)

This molecule also contains a chiral center at the carbon atom bonded to the hydroxyl group (OH). It has four different substituents (CH₃, CH₂, CH₂, and CH₃), making 2-hexanol chiral.

3-Hexanol (CH₃CH₂CH₂(OH)CH₂CH₃)

Similarly, this molecule contains a chiral center at the carbon atom bonded to the hydroxyl group (OH). It has four different substituents (CH₃, CH₂, CH₂, and CH₃), making 3-hexanol chiral.

2-Methylpentanol (CH₃CH(CH₃)CH₂CH₂OH)

In this molecule, there is no chiral center present since the carbon atom bonded to the hydroxyl group (OH) is also bonded to two identical methyl groups (CH₃). Therefore, 2-methylpentanol is achiral.

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Answer:

BECAUSE OF DENSITY

Explanation:

Answer:

It's DENSITY

Explanation:

YOURE WELCOME

Answer:

Static Electricity

Explanation:

Answer:

static electricity

Explanation:

Answer:

By Bayer process.

Explanation:

In the Bayer process, bauxite ore is heated in a pressure vessel along with a sodium hydroxide solution (caustic soda) at a temperature of 150 to 200 °C. At these temperatures, the aluminium is dissolved as sodium aluminate (primarily [Al(OH)4]−) in an extraction process.

Answer:

Gaseous nitrogen has unique chemical and physical properties that make it suitable for use in food processing. Nitrogen is inert which means it will not react with prepared food materials, which can alter their aromas or flavors. Also, gaseous nitrogen will effectively displace oxygen minimizing oxidation and the growth of microorganisms that cause foods to lose their freshness and deteriorate faster.

Explanation:

Source: https://www.generon.com/using-nitrogen-gas-in-food-packaging/

"Because it deflects oxygen, nitrogen is just a common gas for food packing." Because oxygen will carry moisture, this is critical. Bacteria use oxygen to grow and survive on organic matter. Bacteria find it much harder to develop while as much oxygen as possible is removed.

Food packaging is the process of packaging food. A package provides security, resistance to manipulation, and particular physical, chemical, and even biological requirements.  Food packaging may include a nutrition label and other information about the product being sold.

Bacteria are commonly found, mostly free-living organisms with  one biological cell. They belong to the prokaryotic organisms category. Bacteria, which have been typically a few micrometers long, were one of the first living species to originate on Earth and may be found in practically every ecosystem.

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We can use the concentration of OH- to calculate the pH of the solution. pOH = -log [OH-]= -log (5.87 × 10^-4)= 3.23pH + pOH = 14pH = 14 - 3.23= 10.77. Therefore, the pH of the solution is 10.77.

To calculate the pH of a solution that is 0.16 M NH3 and 0.22 M NH4Cl, we need to use the Kb expression of NH3.Kb = [NH4+][OH-]/[NH3]. The reaction between NH3 and H2O produces NH4+ and OH-.NH3 + H2O ⇌ NH4+ + OH-In this reaction, one NH3 molecule produces one NH4+ ion and one OH- ion, so the concentration of OH- will be equal to the concentration of NH4+.

Therefore,[NH4+] = 0.22 M[OH-] = 0.22 M. We know that the concentration of NH3 is 0.16 M. To calculate the concentration of OH-, we can use the Kb expression. Kb = [NH4+][OH-]/[NH3]1.79 × 10^-5 = (0.22) (0.22)/0.16[OH-] = 5.87 × 10^-4 M. Now, we can use the concentration of OH- to calculate the pH of the solution. pOH = -log [OH-]= -log (5.87 × 10^-4)= 3.23pH + pOH = 14pH = 14 - 3.23= 10.77. Therefore, the pH of the solution is 10.77.

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Answer: d) The volume occupied by the molecules themselves is no longer negligible. This most be the answer because a, b and c don't make any sense at all.

Explanation:

The volume occupied by the molecules themselves is no longer negligible. Hence, option D is correct.

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

The ideal gas law fails at low temperature and high pressure because the volume occupied by the gas is quite small, so the inter-molecular distance between the molecules decreases.

Hence, the volume occupied by the molecules themselves is no longer negligible.

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Answer:

See Explanation

Explanation:

The experiment of Ernest Rutherford involved the bombardment of a thin gold foil with alpha particles from a radioactive source. A zinc sulphide screen was used to follow the movement of the alpha particles.

It was discovered that most of the alpha particles  followed a  straight path through the gold foil. Some of them were scattered through large angles and few were even scattered in the backward direction.

Since alpha particles were heavier than electrons, they must have been deflected by very strong forces.

These experimental evidences led to the conclusions stated in the question.

After considering the given data we conclude that the concentration of [tex]SO_3^2^-[/tex]must exceed [tex]6.18*10^{-11} M[/tex] to generate a precipitate of [tex]MgSO_3[/tex].

To describe the concentration of [tex]SO_3^{2-}[/tex]required to generate a precipitate of [tex]MgSO_3[/tex], we can apply the following steps:
State the balanced chemical equation for the dissolution of [tex]MgSO_3[/tex] in water.
[tex]MgSO_3(s)--- > Mg_2+(aq) + SO_3^{2-} (aq)[/tex]
Write the equation for the solubility product constant, Ksp, for [tex]MgSO_3[/tex].
[tex]Ksp = [Mg^{2+} ][SO3^{2-} ][/tex]
Stage the given concentration of [tex]Mg^{2+}[/tex] into the expression for Ksp.
[tex]Ksp = (8.9*10^{-11} M)(x)[/tex]
Here,
x = concentration of [tex]SO_3^{2-}[/tex]required to reach the saturation point and generate a precipitate.
Stage the given value of Ksp into the expression for Ksp.
Evaluate for x.
[tex]x = Ksp/[Mg^{2+} ] = (5.5*10^{-21})/(8.9*10^{-11} M) = 6.18*10^{-11 }M[/tex]
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The expected initial pH for Buffer 1 and Buffer 2 is 4.54 and 4.94 respectively.

Buffer 1 is made of Acetate buffer and buffer 2 is made of Acetate buffer as well. We know the volume and concentration of the acid and salt in both buffers. The expected initial pH of Buffer 1 and Buffer 2 can be calculated using the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given as: pH = pKa + log {[A-]/[HA]}where[A-] is the concentration of the salt (NaC2H2O2),[HA] is the concentration of the acid (HC2H2O2),pKa is the acid dissociation constant of the acid (HC2H2O2).

Buffer 1: HC2H2O2 + H2O ⇔ C2H2O2- + H3O+No. of moles of HC2H2O2 = 0.10 M x 0.015 L = 0.0015 mol. No. of moles of NaC2H2O2 = 0.10 M x 0.01 L = 0.001 mol. Total volume = 0.015 L + 0.01 L = 0.025 L[HA] = 0.0015 mol/ 0.025 L = 0.06 M[A-] = 0.001 mol/0.025 L = 0.04 M. The pKa for HC2H2O2 is 4.76pH = 4.76 + log {[0.04]/[0.06]}= 4.76 - 0.22 = 4.54Thus, the expected initial pH for Buffer 1 is 4.54.

Buffer 2: HC2H302 + H2O ⇔ C2H3O2- + H3O+No. of moles of HC2H302 = 0.50 M x 0.01 L = 0.005 molNo. of moles of NaC2H2O2 = 0.50 M x 0.015 L = 0.0075 mol. Total volume = 0.01 L + 0.015 L = 0.025 L[HA] = 0.005 mol/ 0.025 L = 0.20 M[A-] = 0.0075 mol/0.025 L = 0.30 M. The pKa for HC2H302 is 4.76pH = 4.76 + log {[0.30]/[0.20]}= 4.76 + 0.18 = 4.94Thus, the expected initial pH for Buffer 2 is 4.94.

Therefore, the expected initial pH for Buffer 1 and Buffer 2 is 4.54 and 4.94 respectively.

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Answer:

[tex]2.18\%[/tex]

Explanation:

Volume of isopropyl alcohol = 24 mL

Volume of water = 1.1 L

Percentage of isopropyl alcohol is given by

[tex]\dfrac{\text{Volume of isopropyl alcohol}}{\text{Volume of isopropyl alcohol+Volume of water}}\times 100\\= \dfrac{24\times 10^{-3}}{1.1+2.4\times 10^{-3}}\times 100\\ =2.18\%[/tex]

The percent by volume of isopropyl alcohol in the given solution is [tex]2.18\%[/tex].

The given statement "A highly positive charged protein will bind a cation exchanger and elute of with a high salt buffer" is true.

A cation exchanger is an adsorbent that is used to extract charged molecules from a solution and purify them. This adsorbent binds to molecules that are charged positively because it bears a negative charge. The cation exchanger is typically a negatively charged polymer that is insoluble in water and which is a negatively charged polymer. The cation exchanger will bind positively charged protein molecules in a similar manner. Cation exchange chromatography is a type of chromatography that separates molecules based on their charge. The cation exchanger is an example of a stationary phase in cation exchange chromatography. Proteins with positive charges will stick to the stationary phase, and they will elute when a high salt buffer is passed through the column to compete for the binding sites on the stationary phase.

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