the total pressure of gases a, b, and c in a closed container is 4.1 atm . if the mixture is 36% a, 42% b, and 22% c by volume, what is the partial pressure of gas c? view available hint(s)for part a the total pressure of gases a, b, and c in a closed container is 4.1 . if the mixture is 36% a, 42% b, and 22% c by volume, what is the partial pressure of gas c? 1.5 atm 1.7 atm 0.22 atm 0.90 atm

Ideal gas law is valid only for ideal gas not for vanderwaal gas. Therefore, the pressure that will be required for neon at 30°C to have the same density as nitrogen at 20°C and 1.0 atm is 1.44atm.

Ideal gas equation is the mathematical expression that relates pressure volume and temperature. There is no force of attraction between the particles.

Mathematically the relation between Pressure and temperature can be given as

P₁M₁=DRT₁

P₂M₂=DRT₂

Rearranging the two equation we get

P₁M₁÷T₁=P₂M₂÷T₂

where,

P = pressure

M= Molarity

D=density

T =temperature

R = Gas constant = 0.0821 L.atm/K.mol

Substituting all the given values, we get

(P₁×M₁)÷T₁=(P₂×M₂)÷T₂

(P₁×20)÷303=(1.0 ×28)÷293

P₁=1.44atm

Therefore, the pressure that will be required for neon at 30°C to have the same density as nitrogen at 20°C and 1.0 atm is 1.44atm.

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Explanation

Given:

mass of glycol = 0.250g

mass of solvent (water) = 40g = 0.04 Kg

i = 1

Required: The boiling point of the solution

Solution

where Kb is the boiling point elevation constant

m is the molal concentration (molality) of all species

Step 1: Find the moles of glycerol

n = m/M

n = 0.250g/92,09382 g/mol

n = 2.71x10^-3 mol

Step 2: Find the molality

molality (m) = moles of solute/Kg of solvent

molality = 2.71x10^-3/0.04Kg

molality = 0.068 m

Step 3: Find the change in temperature

Kb of water = 0.512 cm^-1

DT = 0.512 x 0.068

DT = 0.035 'C

Step 4: Find the boiling point

Boiling point of water = 100 'C

DT = 0.035 'C

New boiling point =100 + 0.035 = 100.035 'C

Answer

100.035 'C

The reactivity of a substance with oxygen is a chemical, not a physical property. The reason it is called a chemical property is that it relies on its electron configuration to determine how it behaves around other substances.

A chemical property is a property of any material that becomes apparent during or after a chemical reaction; that is, any quality that can only be determined by changing the chemical properties of a substance.

Oxygen is a very reactive element, is highly paramagnetic and readily combines with other elements. One of the most important chemical properties of oxygen is that it promotes combustion. Even at room temperature, oxygen binds to elements and forms e.g. rust.

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The Vapour pressure of the solution is 99.56 torr

The saturated vapour pressure of the solution maybe lower then the pressure of pure solvent in solution. It is equal to the vapour pressure of the pure solvent at the same temperature scaled by mole fraction of solvent stated by Rault's Law.  

P(solution) = mol fraction of solvent × P° ( solvent)

given, pressure of pure toulene = 40.2 torr

pressure of pure cyclohexane = 125 torr

∴ P (solution) = 0.3 × 40.2 + 0.7 × 125

P (solution) = 99.56 torr

hence, vapour pressure is calculated to be 99.56 torr.

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The atomic number of rubidium is 37, it means that its electron configuration is as follows:

Rubidium is an alkali metal. Its ion is formed when it loses an electron. It means that the rubidium ion has 36 electrons instead of 37, and its configuration is as follows:

Cyclobutyne does not exist. A hydrocarbon molecule called cyclobutyne has a triple bond inside of a ring with four carbon atoms.

Due to its high ring strain, this cycloalkyne is extremely unstable and hasn't been isolated in its purest form.

Organic compound butyne has the chemical formula HC2CH2CH3. It is a flammable, colourless gas. 1-Butyne takes part in processes that are common for terminal alkynes, such as hydrogenation, condensation with formaldehyde, and alkyne metathesis.

Due of its extremely low torsional barrier and the difficulty in identifying it using high-resolution infrared spectroscopy, 2-butyne is of interest to physical chemists.

Cycloalkenes include cyclobutene. It offers potential for research but no immediate applications. It is an effortlessly condensible colourless gas. In a contemporary synthesis, cyclobutanol is dehydrated twice. The ammonium salt OH was first thermolyzed to create the chemical.

Cycloalkanes are monocyclic saturated hydrocarbons in organic chemistry. To put it another way, all of the carbon-carbon bonds in a cycloalkane's structure are single and it only contains hydrogen and carbon atoms grouped in a configuration with a single ring.

When a cycloalkane has more than 20 carbon atoms, it is often referred to as a cycloparaffin. The isomers of alkenes make up all cycloalkanes.

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Nothing like cyclobutyne exists. Cyclobutyne, a hydrocarbon molecule, features a triple bond inside of a ring made up of four carbon atoms. This cycloalkyne hasn't been isolated in its purest form because of its high ring strain, which makes it very unstable.

Butyne is an organic molecule with the formula HC2CH2CH3, according to chemistry. It is a flammable, colorless gas. 1-Butyne takes part in processes that are typical for terminal alkynes, such as hydrogenation, condensation with formaldehyde, and alkyne metathesis.

Because of its very low torsional barrier and the difficulty in locating it using high-resolution infrared spectroscopy, 2-butyne is of interest to physical chemists.

One of the cycloalkenes is cyclobutene. Although it might be studied, there are currently no applications in real life. It is an easily condensable colorless gas. In a contemporary synthesis, cyclobutanol passes through two stages of dehydration. The ammonium salt OH was first thermolyzed to create the chemical.

In organic chemistry, cycloalkanes are monocyclic saturated hydrocarbons. To put it another way, a cycloalkane has single carbon-carbon bonds throughout its whole structure.

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Two mole of ideal gas is compressed isothermal from 10 to 10,000 atmospheric pressure at 500 degree kelvin then the entropy change in the process is  -114.86 J/K

Ideal gas is the hypothetical gas composed of molecule which follow a some rules in that ideal molecule do not attract or repel to each other and here given data is

Ideal gas = 2 mole

Pressure = P₁ = 10

Pressure = P₂ = 10,000

Temprature = 500 degree kelvin

We have to calculate entropy change in the process =?

We know the entropy change

ΔS = cv ln(T₂/T₁) +nR ln (P₂/P₁)

For isothermal compression temprature constant,

ΔS = nR ln (P₂/P₁)

ΔS = 2×8.314×ln(10/10,000)

ΔS = -114.86 J K⁻

ΔS = -114.86 J/K

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ANSWER

Option C

EXPLANATION

Given that;

In the reaction above, we have the following data

At the reactants side;

3 atoms of manganese

4 atoms of nitrogen

1 atom of sodium

1 atom of fluorine

At the products side

1 atom of manganese

4 atoms of fluorine

3 atoms of sodium

1 atom of nitrogen

To balance the above equation, apply the law of conservation mass

Law of conservation of mass states that matter can neither be created nor destroyed but can e transformed from one formato another.

To balance the equation, 1 mole of Mn3N4 reacts with 12 moles of Na Tto give 3 moles of MnF4 and 4 moles of Na3N

So, the new equation becomes

The following data can be deduced in the above equation

At the reactants side

3 atoms of Mn

4 atoms of N

12 atoms of Na

12 atoms of F

At the products side

3 atoms of Mn

12 atoms of F

12 atoms of Na

4 atoms of N

Looking atthe vabove data, the number of atoms of each element at the reactants side is equal to the number of atoms of same elements at the products side.

Hence, the correct answer is option Ce

u

Answer:

Tc

Explanation:

You just have to follow the rows with the exponents. Just remember that when we get to d, the number in the front is a period lower. Hope this helps!

A sample of gas occupies a volume of 70.3 ml as it expands, it does 140.6 j of work on its surroundings at a constant pressure of 783 torr and the final volume of the gas is 11.12L

A gas is a sample of matter that conforms to the shape of a container in which it is held and acquires a uniform density inside the container and even in the presence of gravity and regardless of the amount of substance in the container

Here given data is 11.12L

V₁ = Initial volume = 70.3 ml = 0.0703ml =  (1L=1000ml)

w = 140.6 J = 1.387 J = (1Latm=101.3J)

Pressure =  783 torr = 1.03 atm  = (1atm=760 torr)

V₂ = final volume = ?

1.387Latm = 1.03 atm × (V₂ - 0.0703ml)

(V₂ - 0.0703ml) = 11.05

V₂ = 11.12L

The final volume of the gas is 11.12L

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To find the order of a reaction with respect to one reactant, you will monitor the Percentage as the reactant of product is changed.

To determine the reaction order from experimental data, either the differential rate law or the integrated rate law can be used. In the rate law, the exponents are frequently positive integers: 1 and 2 or even 0. As a result, the reactions in each reactant are of zeroth, first, or second order. Changes in the concentration of the reactants or products can be used to calculate the rate. changes in reactant or product mass or volume If the reaction is the result of multiple reaction steps with different orders, the order of the reaction can be changed by adjusting the reagent concentration. Because the order is often ruled by the slowest reaction, you can manipulate the relative reaction speed by varying the concentrations.

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Answer:

Explanation:

You will watch the Percentage when the reactant of the product is changed in order to determine the sequence of a reaction with respect to one reactant.

Either the differential rate law or the integrated rate law can be used to infer the reaction order from experimental data.

The exponents of the rate law are frequently positive numbers, such as 1 and 2 or even 0.

Because of this, each reactant undergoes reactions of the zeroth, first, or second order.

The rate can be calculated using changes in the concentration of the reactants or products.

A change in the reactant, product, or reactant's volume by changing the reagent concentration, the reaction order can be modified if it involves several steps that have been performed in a different sequence.

You can also control the relative reaction speed by changing the concentrations because the order is frequently determined by the slowest reaction.

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Answer:

its Boron i hope this helped.

Explanation:

Answer:

Explanation:

About 1.3 seconds

It takes light about 1.3 seconds to travel from the moon to the Earth at a distance of 384000 km. This is because the speed of light is about 300000 km/s. So, it takes light about 1.3 seconds to travel from the moon to the Earth.

Answer:

1.25 moles(N) of NaOH are present in 1000 gm of water.

and 1000gm = 1L

Answer:

Element XX has a relative atomic mass of 6.86 aum

Explanation:

See attached worksheet.

Answer:

freezing points hope it helps u

Distillation is a process used to separate a mixture of liquids based on different freezing points. Therefore, option C is correct.

Distillation is the process of turning a liquid into a vapor, which is then condensed back into a liquid state. The simplest illustration of it is when steam from a kettle condenses into drops of distilled water that are left on a cold surface.

In general, distillation is the process of vaporizing a liquid to its boiling point, then condensing it back into a liquid to separate it from impurities or other solutes.

Among the many industrial uses of distillation are the production of alcoholic drinks, water purification, and oil refining. Distillation is a physical procedure that removes desired pure compounds from an initial source using heat and other techniques.

Thus, option C is correct.

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When a chemical reaction happens, the atoms in the reactants are rearranged to create compounds with various chemical characteristics. Although identical atoms are present in both the reactants and the products, sodium chloride was produced by rearranging the atoms in the reactants. The reactants lacked sodium chloride.

Sodium is a soft, silver-colored metal that may be sliced with a knife. With water, pure sodium metal reacts violently (and even explosively), creating sodium hydroxide, hydrogen gas, and heat.

2Na(s)  +  2H₂O(l)  ⟶  2NaOH(aq)  +  H₂(g)

Chlorine is a poisonous yellow-green gas with a pungent smell.

But when sodium and chlorine combine, they create sodium chloride, or table salt, which is known to practically everyone in the world.

2Na(s)  +  Cl₂(g)  ⟶  2NaCl(s)

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The pH of buffer solution is 4.72.

1) Molarity of acetic acid = 0.45 M

The volume of acetic acid = 300 ml

Molarity of NaOH = 0.65 M

Volume of NaOH = 100 ml

An acidic buffer solution is calculated by using Henderson's equation:

pH = pKa + ㏒[salt] / [acid]

where pKa = 4.75

pH = 4.75 + ㏒[0.065M] / [0.07M]

pH = 4.75 + ㏒(0.929)

pH = 4.75 - 0.03

pH = 4.72

Therefore, the pH of the buffer solution is 4.72.

2) NaOH is not the conjugate base to acetic acid. As we know NaOH is a strong base and acetic acid is a weak acid. Thus, an acid-base reaction occurs between acetic acid and NaOH, thus forming CH₃COONa, which is the conjugate base of acetic acid.

Acidic buffer solution is formed by mixing of weak acid and its conjugate salt of strong base. Thus, NaOH acts as limiting reactant and only CH₃COOH and CH₃COONa is present in solution. Thus, CH₃COOH is weak acid and CH₃COONa is its conjugate salt of strong base forms buffer solution. In the reaction of NaOH and CH₃COOH, NaOH gets completely reacted and forms CH₃COONa. Thus, NaOH is added to make an effective acetate buffer.

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There is only one sodium ion for every chlorine ion due to its valency.

The number of electrons that an atom needs to lose or gain in order to achieve a stable electron configuration i.e the octet of an element is known as the valency of an element, which serves as a gauge of that element's potential for combining.

The valency of Sodium (Na) is 1 and that of chlorine (Cl) is also one. therefore they combine to form a stable compound. The same rule follows in the compound formed by Potassium (K) and Oxygen(O) also. The valency of oxygen is 2 which means that it should be surrounded by 2 potassium ions to provide electrons to fulfill both the elements' octets.

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Answer:

H202(aq)+2Fe2+(aq)+2H+=>2fe3+(aq)+2H20(l)

Explanation:

H202(aq)+2Fe2+(aq)+2H+=>2fe3+(aq)+2H20(l)

The white precipitate formed is lead iodide (PbI₂) of mass 131.38g

The term "precipitation reaction" may be described as "a chemical response taking place in aqueous solution wherein ionic bonds integrate to shape an insoluble salt". These insoluble salts fashioned in the course of precipitation reactions are referred to as precipitates. Precipitation reactions are normally double displacement reactions with the formation of stable residues referred to as precipitates.

One of the first-class examples of precipitation reactions is the chemical response among potassium chloride and silver nitrate, which precipitates stable silver chloride. This is an insoluble salt fashioned as a made from precipitation reactions. The chemical components for this precipitation response is proven below.

AgNO3(aqueous) + KCl(aqueous) —–AgCl(precipitate) + KNO3(aqueous)

For the given case,

The balanced reaction is as follows:

Pb(NO₃)₂ (aq) + 2KI (aq) → 2KNO₃ (aq) + PbI₂ (s)

This tells us that for every mole of lead iodide (PbI₂) produced, it requires 2 moles of potassium iodide (KI).

If you have 0.57 mole of KI, they will produce: 0.57/2 = 0.285 mol of PbI₂

Number of moles = [tex]\frac{mass}{molar mass}[/tex]

mass of lead iodide (PbI₂)  = Number of moles × molar mass

Mass of lead iodide = 0.285 × 461.01

Mass of lead iodide = 131.38g

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By calculating the response factor for the internal standard first, we can calculate that the concentration of the unknown analyte sample is 2.64 ppm.

When solving problems like this, we need to calculate the appropriate response factor:

A(X) / C(X) = F * A(S) / C(S)

A(X) - response for the analyte (3.47)

C(X) - concentration of the analyte (1 ppm)

F - response factor

A(S) - response for the internal standard (5.23)

C(S) - concentration of the internal standard (1 ppm)

F = A(X) * C(S) / (C(X) * A(S)

F = 3.47 * 1 ppm / (5.23 * 1 ppm)

F = 0.663

Now we can use this in the second calculation to obtain the concentration of the analyte in the sample. One thing to note is that in second measurement 1 mL of a 5.00 ppm internal standard was diluted to 10 mL, so the concentration of the internal standard is actually 5.00 ppm / 10 = 0.500 ppm.

4.26 / C(X) = 0.663 * 2.43 / 0.500 ppm

C(X) = 4.26 * 0.500 ppm / (0.663 * 2.43)

C(X) = 1.32 ppm

Because 5 mL of the analyte solution were diluted to 10 mL, the concentration is actually 2 * 1.32 ppm = 2.64 ppm.

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Bread grows mold more quickly at room temperature than in the refrigerator because more reactants will have the energy necessary to proceed to products at room temperature than at refrigerator temperatures.

Molds are formed on bread due to various reactions that facilitate the living of the molds on the bread and allow them to reproduce on the bread.

However, as molds are living organisms characterized as fungi, they need enzymes for function.

The enzymes and proteins normally function best at room temperature. On the other hand, when the temperatures are low, such as that of a refrigerator the molds will not have enough energy because the proteins and enzymes will be inactivated. Hence, molds grow on bread quickly at room temperature as compared to bread in a refrigerator.

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Answer:

I catalyst promoter is the answer

The work done by the gas that expands at constant external pressure ( isobaric expansion) is 174.8 J

The work done by the gas that expands at constant external pressure is work done under isobaric conditions.

The work done by an isobaric expansion is calculated with the formula below:

Work done = P (Vf − Vi)

where;

P = 1.15 atm

P is converted to N/m² and volume is converted to m³

1.15 atm = 1.15 * 101325 N/m²

P = 116523.75 N/m²

Vf = 1.65 * 10⁻³ m³

Vi = 0.15 * 10⁻³ m³

W = 116523.75 N/m² ( 1.65 * 10⁻³ m³ - 0.15 * 10⁻³ m³)

Work done = 174.8 J

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         Red blood cells contain hemoglobin. When carbonic acid, which is found in tissues, is present, hemoglobin releases the bound oxygen. The oxygen that is attached to hemoglobin is released into the blood's plasma and taken up by the tissues in the capillaries, which are also where carbon dioxide is created.

          If more air is exchanged, the partial pressures of the outside atmosphere will approach but never reach them due to the remaining volume.

      Combustible Oxide : The dissociation curve shifts to the left as a result of one CO molecule binding to hemoglobin increasing the affinity of the other binding locations for oxygen.

       Lower CO2, greater pH, and lower temperature all move oxygen dissociation to the left.

       The pCO2 is significantly influenced by two variables. The person's breathing rate and depth are the first thing to consider. Someone who is hyperventilating will "blow out" more CO2, which will lower pCO2 levels. Holding one's breath causes CO2 to be retained, which raises pCO2 levels.

      Limited pressures : There are two reasons why the alveolar oxygen partial pressure is lower than the ambient O2 partial pressure. First, the upper airway humidifies the air as it reaches the lungs, lowering the partial pressure of oxygen to around 150 mmHg while increasing the partial pressure of water vapour (47 mmHg).

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Answer:

Protects the fish from their environment.

Explanation:

Fish have a mucus on their outer covering that is very slimy and slippery; this serves as a benefit for the fish that will allow it to swim more easily in water.

It also works as a little suit of armor for the fish to protect themselves against predators.

Avogadro's Number:

Avogadro's number, which is equal to 6.02214076 x [tex]10^{23}[/tex], is the quantity of units in one mole of any material. Hence, 1 mole of HBrO3 will contain

6.02214076 x [tex]10^{23}[/tex] atoms.

Then, 0.009567 mole of HBrO3 will contain

= 0.009567 moles x 6.02214076 x [tex]10^{23}[/tex] atoms

=  0.0576138 x    [tex]10^{23}[/tex]  atoms = 5.761 x [tex]10^{21}[/tex] atoms.

Hence, 0.009567 mole of HBrO3 will contain 5.761 x [tex]10^{21}[/tex] atoms.

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