Calculate the molar solubility of silver(i) bromate with ksp = 5. 5×10-5. also, convert the molar solubility to the solubility. the molar solubility is ___________ and the solubility is ____________

Answer:

4.70 L

Explanation:

Use the basic relationship:

P1V1/T1 = P2V2/T2

Re-arrange to:

P1V1 T2 / (T1P2) = V2   To find the new volume

Sub in the values:

880 * .5 * 327 / ( 300*102) = 4.70 L

Oxygen is the crucial gas required for an organism to carry out the cellular process. The seal can dive for 4 minutes before it runs out of oxygen. Thus, option a is correct.

Oxygen is the most vital element of the organism that is required by almost all cellular and metabolic activities. It is inhaled by the respiratory system and passed to cells and tissues through the circulatory system.

Given,

Mass of seal = 100 kg = 100000 gm

Rate of oxygen consumption (r) = 1 ml/ (gh)

The volume of the stored O₂ = 7000 mL

The time (t) is calculated as,

t = V ÷ mr

Substituting the values above,

t = 7000 ÷ 100000 × 1

= 0.07 hour

In minutes it will be given as,

0.07 × 60 = 4.2 minutes

Therefore, option a. 4 minutes is the time till the seal can remain underwater without oxygen.

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Your question is incomplete, but most probably your full question was, The oxygen consumption rate of a 100 kg seal is 1 ml/(g h). assuming that it has oxygen stores of 7 liters, how many minutes can it dive before running completely out of oxygen?

Answer:

unbalanced

Explanation:

balanced: Al+HCl=AlCl+H

Answer:

Explanation: So, the atomic number is the number of protons present in an atom. For example, Helium's atomic number is 2, meaning it only has 2 protons. In most cases, the number of nuetrons and protons are equal, so Helium (usually) has 2 protons and 2 nuetrons - this is the nucleus. The atomic mass, or mass per nucleon is the proton and nuetron added together, so the atomic mass of Helium is about 4. Therefore, on the periodic table, as the atomic number increases by 1, the atomic mass increases by about 2 (the atomic mass is an average so it's not exact).

Considering the Dalton's partial pressure, the mole fraction of each gas is:

The pressure exerted by a particular gas in a mixture is known as its partial pressure.

So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

[tex]P_{T} =P_{1} +P_{2} +...+P_{n}[/tex]

where n is the amount of gases in the gas mixture.

This relationship is due to the assumption that there are no attractive forces between the gases.

Dalton's partial pressure law can also be expressed in terms of the mole fraction of the gas in the mixture. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of a component to the number of moles of all the components present.

So in a mixture of two or more gases, the partial pressure of gas A can be expressed as:

[tex]P_{A} =x_{A} P_{T}[/tex]

In summary, the total pressure in a mixture of gases is equal to the sum of partial pressures of each gas.

Mole fraction of each gas

In this case, you know that:

Then:

Substituting the corresponding values:

Solving:

In summary, the mole fraction of each gas is:

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Answer:

It's too short. Write at least 20 characters to explain it well.

Explanation:

It's too short. Write at least 20 characters to explain it well.

When formaldehyde and acetone then react with each other( aldol condensation) then it will be formed methyl vinyl ketone.

In organic chemistry, an aldol condensation would be a condensation reaction in which an enol and enolate ion combines with a carbonyl chemical to produce a -hydroxy aldehyde or -hydroxy ketone, that is then dehydrated to produce a conjugated enone.

In aldol condensation, when  formaldehyde and acetone then react with each other then it will be formed methyl vinyl ketone.

It can be written as

[tex]CH_{2} O + CH_{3} COCH_{3}[/tex] → [tex]HOCH_{2} -CH_{2} -CO-CH_{3}[/tex]

When it will be heated then it gives methyl vinyl ketones.

[tex]HOCH_{2} -CH_{2} -CO-CH_{3}[/tex] → [tex]CH_{2} =CH-CO-CH_{3}[/tex]

So, the pair of reactants will be formaldehyde and acetone

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The pH of a  a 0. 10 m solution of NaCN at 25°C is 11.15

Calculation

The reaction in the solution is given below

                   [tex]CN^{-} + H_{2} O[/tex] → HCN + [tex]OH^{-}[/tex]

initial                    0.1

change           ( -x)                (+x)

equilibrium     (  0.1 - x )         x

Kb = [HCN] [[tex]CN^{-}[/tex] ]/[[tex]CN^{-}[/tex] ]

Kb × Ka = Kw = 1.0 × [tex]10^{-14}[/tex]

Kb =  1.0 × [tex]10^{-14}[/tex] / 4.9  × [tex]10^{-10}[/tex] =  [HCN] [[tex]CN^{-}[/tex] ]/[[tex]CN^{-}[/tex] ] = [tex]x^{2}[/tex]/ (  0.1 - x )

Kb =  2.04× [tex]10^{-5}[/tex] = [tex]x^{2}[/tex]/ (  0.1 - x )

Since , [NaCN] /Kb > 100 , we can simplify the above equation to

=  2.04× [tex]10^{-5}[/tex] = [tex]x^{2}[/tex]/ (  0.1 )

x = 1.43 × [tex]10^{-3}[/tex] M = [HCN] = [ [tex]OH^{-}[/tex]]

Then pOH = 2.84

pH + pOH = 14

pH = 14- pOH = 14 - 2.84 = 11.15

Therefore , the pH is 11.15

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When hexan-1-ol is treated with conc. H₂SO₄ at moderate temperatures, Di-n-hexyl ether is formed via SN₂ mechanism.

SN₂ Mechanism is mechanism involving the factors of steric hindrance following the order of 1° > 2° > 3°. It is a Nucleophilic substitution reaction.

Williamson Ether Synthesis is the synthesis of ether through SN₂ mechanism in which an ether is treated with conc. H₂SO₄ at moderate temperature.

When hexan-1-ol is treated with conc. H₂SO₄ at moderate temperatures, Di-n-hexyl ether is formed as below:

CH₃CH₂CH₂CH₂CH₂CH₂OH +Conc. H₂SO₄-->(CH₃CH₂CH₂CH₂CH₂CH₂)₂0

Hexan-1-ol                                                                 Di-n-hexyl ether

Williamson Ether Synthesis is a reaction involving a alcohol that is deprotonated to form an ether.

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The volume occupied by 22. 0 g of helium gas at 26. 0 ° C and 1. 20 atm of pressure is 112.37 L .

Calculation ,

According to ideal gas equation ,

PV = nRT        ....(i)

where P is the pressure = 1. 20 atm

V is the volume of the helium gas = ?

R is the universal gas constant = 0.082 atm L /K mol

T is the temperature of the gas =  26. 0 ° C = 299 K

n is the number of moles

Number of miles (n)  = given mass/ molar mass =22 g/4 = 5.5 moles

By putting the value of pressure , volume , temperature and  universal gas constant  in equation ( i) we get

1. 20 atm ×V = 5.5 moles × 0.082×299

V = 5.5 moles × 0.082×299/1. 20 atm = 112.37 L

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The equilibrium constant of the system is 11.3.

We know that the equilibrium constant shows the extent to which reactants are converted into products.

Now;

2NO2 ⇄ N2O4

[NO2] = 0.500 mol /1 L =  0.500 M

[N2O4] =  0.186 mol  /1 L =  0.186 M

The ICE table is;

          2NO2 ⇄ N2O4

I           0.5           0

C         -2x              +x

E          0.5 - 2x     0.186

The concentration of NO2 at equilibrium = 0.5 - 2(0.186) = 0.128

Keq =  0.186/( 0.128)^2

Keq = 11.3

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In the above question the probability the electron will tunnel through the barrier is 0.011%.

An electron volt is the amount of energy required to move a charge equal to 1e⁻ across a potential difference of 1eV.

To calculate the probability the electron will tunnel through the barrier is calculated as -

Energy of electron

E=80eV

=80×1.6×10⁻¹⁹

=128×10⁻¹⁹

Height of the barrier U=100 eV

=100×1.6×10⁻¹⁹J

Thickness L=0.2×10⁻⁹m

Probability T=e⁻²cl

C=√2m(U-E)/h

=√[2×1.67×10⁻²⁷(160-128)×10⁻¹⁹] / 1.055×10⁻³⁴

=10.34×10⁻²³/1.055×10⁻³⁴

=9.8×10¹¹

2CL=2×9.8×10¹¹×0.2×10⁻⁹

=3.92×10²

T=e⁻²cl

= 1/e⁻³⁹²/¹⁰⁰⁰⁰

=0.0198%

=0.011%

Hence ,the probability the electron will tunnel through the barrier is 0.011%.

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The major product of reaction between 1-butanol and Na2Cr2O7 is butanoic acid.

When a primary alcohol like 1-butanol (OH is bonded to a primary carbon) is begin to oxidize in the presence of strong oxidizing reagent such as sodium dichromate (Na2Cr2O7) and H2SO4, sulfuric acid, the stepwise oxidation take place as above firstly to the corresponding aldehyde which undergoes further oxidation to the corresponding carboxylic acid.

You can find that the formed aldehyde after first oxidation is butanal and the only organic product, due to the strong oxidizing reagent is butanoic acid.

Thus, the major product formed is butanoic acid.

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N-Octanol and water are chosen because the connection between a substance's hydrophilicity and lipophilicity is measured by [tex]K_{OW}[/tex] (n-Octanol/Water partition coefficient). When a chemical is more dissolves in fat-like solvents like n-octanol, the value is more significant than one,  when it's more dissolved in water, the value is lower.

What is the partition coefficient?

So, N-Octanol is chosen because it has a carbon/oxygen ratio that is comparable to that of lipids and because it shows both hydrophobic and hydrophilic properties. N-octanol, therefore, resembles the makeup and characteristics of cells and other living things.

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Root mean square speed is the estimation of the effect of temperature and weight on kinetic energy. The root means square speed for a mixture of helium and methane is 2. Thus, option d is correct.

Root mean square speed is the estimation of the square root of the mean value of the squared speeds of the individual gas particles in the mixture.

As it is known that,

Root mean square speed = √{3RT/M}

Where T is constant temperature and R is gas constant. The molecular mass of helium gas is 4 g/mol and of methane is 16 g/mol.

Vrms = √{3RT/M}

For helium, Vrms = √{3RT/4}

And, for methane = Vrms = √{3RT/16}

So, (Vrms,helium) / (Vrms,methane) = 4 / 2 = 2

Therefore, option d. 2 is the root-mean-square speed for the mixture.

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Your question is incomplete, but most probably your full question was, Helium and methane gases are mixed together in a container. what is the ratio of their root-mean-square speeds, i. e. (Vrms, helium) / (Vrms, methane)?

If a person wearing such a mask exhales 0. 65g of [tex]CO_{2}[/tex] every minute then the  number of grams of [tex]O_{2}[/tex]  that will be produced in 15 minutes is  7.09 grams .

Calculation ,

Since number of moles of carbon dioxide exhale is equal to the number of moles of oxygen produce ,

Mass of carbon dioxide exhale per minute = 0.65 g

Mass of carbon dioxide exhale 15 minute = 0.65 g × 15 = 9.75 g

Number of moles of carbon dioxide = given mass / molar mass = 9.75 g/44

Number of moles of carbon dioxide = 0.221 moles

Number of moles of oxygen  = 0.221 moles = given mass / molar mass

Number of moles of oxygen  = 0.221 moles  = given mass / 32

Mass of oxygen  = 0.221 moles × 32 = 7.09 grams

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[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

B. How much of the reactants are needed and how much product will be made.

Chemists use mole in calculations to calculate the amount of product that will be formed when certain known amount of reactants are used at the end of reaction.

The moles are used to determine the atoms and molecules in a substance. It is used by chemists to determine the amount of reactants needed and products produced. Thus, option B is correct.

A mole is said to be defined as the estimation of the small entities like the atoms as that of the Avagadro's number 6.022 × 10²³. It defines the number of particles contained in a substance.

The moles of the substances are determined by the mass of the substance and its molar mass. The moles are given as,

Moles = mass ÷ molar mass

The moles give the estimation of the amount of the reactants needed to produce the products in a chemical reaction.

Therefore, option B. the moles tell the amount of the reactants required to produce the product.

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Answer:

1. Carbon is very special because it can form so many compounds

2. Many carbon-based compounds are not attracted to water and so in general do not dissolve in water. As a result, water alone cannot be used to remove grease or oil from a surface, nor will water dissolve our skin, because all of these things are carbon compounds.

3. Carbon can also link together in long chains or rings, carbon to carbon to carbon to carbon and so on.

Explanation:

The rate of reaction of the value of KIS 3 and [A] and [B] are each 2M is: 24 (mol/L)/s (Option B)

In chemistry, the rate of reaction describes how quickly a chemical reaction develops.

It is frequently described in terms of either the concentration of a reactant that is spent in a unit of time or the concentration of a product that is generated in a unit of time (amount per unit volume).

Recall that the rate equation is:
Rate = K[A]²[B]

Hence:

Concentration of A = 2M

Concentration of B = 2M

K = 3

Taking the values and substituting them, we have:

Rate = 3 (mol/L)⁻³/s×[2]² ×[2]

Rate = 24 (mol/L)⁻³/s

The rate of reaction can be used as a valuable diagnostic tool. We may devise strategies to increase production by learning how quickly things are created and what slows down reactions.

This knowledge is necessary for the industrial production of various chemicals, such as:

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Solution is the term which best describes mixture A which was made by mixing different components together but has only one phase.

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Answer:

136 g Al₂O₃

Explanation:

Assuming you do not need to find the limiting reactant, to find the mass of Al₂O₃, you need to (1) convert grams O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles Al₂O₃ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles Al₂O₃ to grams Al₂O₃ (via molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value (64.0 g).

Molar Mass (O₂): 32 g/mol

Molar Mass (Al₂O₃): 102 g/mol

4 Al + 3 O₂ -----> 2 Al₂O₃

64.0 g O₂           1 mole           2 moles Al₂O₃            102 g
-----------------  x  --------------  x  ------------------------  x   -------------  =  136 g Al₂O₃
                             32 g               3 moles O₂             1 mole

Answer:

Explanation:

Based of the fact that you were given 2 masses I would assume this to be a limiting reagent question. However amount on the left side both equal 2. Ignoring limiting reagents and focusing on just O2 the steps would be:

1. Make sure the equation is balanced ( already given)

2- Use given values to find the mols of O2 (mass/molar mass)

3. Mols are conserved but due to the coefficients the molar value from O2 must be divided by three and multiplied by 2 to ensure proper ratios

4. Using that amount the mass can derived using amount/molar mass

5. Use proper significant digits and units(3 in this case)

The half-life of the substance is 3.106 years.

Here,

a = the initial amount of substance

1-r is the decay rate

x = time span

The equation is given in its correct form as follows:

a = [tex]a_{0}[/tex]×[tex](0.8)^{t}[/tex]

As this is an exponential decay of a first order reaction, t is an exponent of 0.8.

Now let's figure out the half life. Since the amount left is half of the initial amount at time t, that is when:

a = 0.5 a0

0.5[tex]a_{0}[/tex] = [tex]a_{0}[/tex]×[tex](0.8)^{t}[/tex]

0.5 = [tex](0.8)^{t}[/tex]

taking log on both sides

t log 0.8 = log 0.5

t = log 0.5/log 0.8

t = 3.106 years

The half-life of the substance is 3.106 years.

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The molecular weight of this gas will be 45 g/mol .

The state equilibrium equation for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it represents a decent approximation of the activity of many gases under various conditions.

Ideal gas law can be expressed as:

PV =nRT

Calculation of molecular weight by using ideal as law.

Given data:

P = 3 atm

T = 127 °c

Density =  4. 20 grams per liter

PV =nRT

where p is pressure , T is temperature and R is gas constant.

PV = gram / molecular weight RT

Molecular weight = (g/v)( 1/P) RT

Putting the given data in above equation.

Molecular weight =4.20 × 1/ 3× 400 × 0.0831

Molecular weight = 45 g/mol.

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Number of oxygen atoms present in 7.15×10²³ molecules of mannose are 1.92≈2 a m u .

D-mannose is a simple sugar found in many fruits. It is related to glucose. In some cells it occurs naturally in the body.

Mannose is a six-carbon sugar found in a variety of fruits and vegetables. This sugar is not found free in foods. It is a part of polysaccharide chains attached to a variety of proteins.

To calculate number of oxygen atoms present in 7.15×10²³  molecules of mannose-

Since,

1 molecule of oxygen=2.303×10²³ no. of atoms

Here,  total number of molecules of oxygen is 6.

Therefore, 6 molecule of oxygen =6×2.303×10²³÷7.15×10²³

=1.92≈2 a m u .

Hence, the total number of oxygen atoms in 7.15×10²³ molecules of mannose is 1.92 ≈2 a m u.

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Answer:

3.3 x 10⁻¹⁷ Hz    

Explanation:

To find the frequency, you can use the following equation:

E = h / f

In this equation,

-----> E = energy (J)

-----> h = Planck's Constant (6.626 x 10⁻³⁴ J*s)

-----> f = frequency (Hz)

You can plug the given values into the equation and simplify to find the frequency. This equation will require a little bit of rearranging.

E = h / f                                                        <----- Given equation

(2.0 x 10⁻¹⁷ J) = (6.626 x 10⁻³⁴ J*s) / f         <----- Insert values

(2.0 x 10⁻¹⁷ J) x f = (6.626 x 10⁻³⁴ J*s)        <----- Multiply both sides by f

f = 3.3 x 10⁻¹⁷ Hz                                        <----- Divide both sides by 2.0 x 10⁻¹⁷

The frequency of an x-ray wave with an energy of 2.0 x 10^-17 J is 3.3 x 10⁻¹⁷ Hz.

The term frequency is defined as the number of waves that pass a fixed point in unit time. Frequency is measured in hertz which is equal to one event per second.

Frequency also describes the number of cycles undergoes during one unit of time by a body in periodic motion.

Calculating the frequency, you can use the following equation:

E = h / f

Where,

E = energy (J)

h = Planck's Constant (6.626 x 10⁻³⁴ J*s)

f = frequency (Hz)

Insert his values in the given equation

(2.0 x 10⁻¹⁷ J) = (6.626 x 10⁻³⁴ J × s) / f        

(2.0 x 10⁻¹⁷ J) x f = (6.626 x 10⁻³⁴ J × s)        

f = 3.3 x 10⁻¹⁷ Hz                

Thus, The frequency of an x-ray wave with an energy of 2.0 x 10^-17 J is 3.3 x 10⁻¹⁷ Hz.

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Answer: B: Your ears send a message through the nerves to the brain.

Answer:

B!!!!

Explanation:

hope this helps! :D

The hybridization of the central atom in SCl₂ is sp³.

Lewis structure is also known as Lewis dot structure which simply represent the valence electron of an atom.

Hybridization is defined as the mixing of two atomic orbitals belongs to same atom but having entirely different shapes , energies which produces a new hybridized orbitals.

Now first write the electronic configuration of central atom.

The central atom in SCl₂ is S (Sulphur).

Electronic configuration of S: [Ne] 3s² 3p⁴

One 3s and three 3p undergo hybridization to form sp³ orbitals. Here each atom forms bond with one sp³ hybrid orbital.

Thus from the above conclusion we can say that The hybridization of the central atom in SCl₂ is sp³.

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The pH at the equivalence point for the titration of 0. 22 m HCN with 0. 22 m NaOH is 11.17

Calculation,

Concentration of NaCN = 0. 22 m/ 2 = 0.11 M ( at equal volumes of acid and base will be used).

The equilibrium is ,

HCN +[tex]H_{2} O[/tex]  → [tex]H^{+} + CN^{-}[/tex]

C(1-x)              Cx      Cx

Where x , is the degree of hydrolysis and

[tex]K_{h}[/tex] = C[tex]x^{2}[/tex]/(1-x)

We know that [tex]K_{h}[/tex] = [tex]K_{w}/K_{a}[/tex] = 1 ×[tex]10^{-14}[/tex]/4. 9 ×[tex]10^{-10}[/tex] =  2.04×[tex]10^{-5}[/tex]

[tex]K_{h}[/tex] =  C[tex]x^{2}[/tex] =   2.04×[tex]10^{-5}[/tex]  = 0.11 M×[tex]x^{2}[/tex]

[tex]x^{2}[/tex] =  2.04×[tex]10^{-5}[/tex]/0.11 M

x = 1.36×[tex]10^{-2}[/tex]

[tex][OH^{-} ][/tex] = Cx =  1.36×[tex]10^{-2}[/tex] × 0.11 M = 0.15×[tex]10^{-2}[/tex]

[tex][H^{+} ][/tex] =  1 ×[tex]10^{-14}[/tex]/ 0.15×[tex]10^{-2}[/tex] = 6.66×[tex]10^{-12}[/tex]

pH = -㏒[tex][H^{+} ][/tex] = -㏒6.66×[tex]10^{-12}[/tex] = 11.17

The pH at the equivalence point for the titration is  11.17.

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The option that is most likely to be true regarding L76, L77, and L79  is that option B: The assembly would incur an entropic penalty if they occupied a solvent-exposed site.

The entropic penalty in regards to ordered water is known to be one that tells or account  for any form of  weaker binding of the antibiotic novobiocin to what we call resistant mutant of DNA gyrase.

Note that in regards to the scenario above, the Entropic penalty is seen as the thermodynamically disfavored needs that is required in forming a cage of polar solvent molecules that is known to be seen around surface that has exposed hydrophobic potion of a molecule.

Hence, The option that is most likely to be true regarding L76, L77, and L79  is that option B: The assembly would incur an entropic penalty if they occupied a solvent-exposed site.

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See full question below

When Bs1A is assembled as an octamer, what is most likely to be true regarding L76, L77, and L79?

A. They are oriented toward the solvent-exposed exterior of the protein assembly.

B. The assembly would incur an entropic penalty if they occupied a solvent-exposed site.

C. Unlike in a monomer, they are not situated within the hydrophobic cap.

D. Their physiochemical properties are not substantially dependent on their hydrophobicity

3.74×[tex]10^{21}[/tex]

3.74 ×[tex]10^{21}[/tex] molecules of propane were in the erlenmeyer flask.

number of moles of propane can be calculated as moles of propane.

mass of propane =  0.274 g

molar mass of propane = 44.1

So this gives us the value of 6.21×[tex]10^{-3}[/tex] moles of propane

No one mole of propane As a 6.0-2 × [tex]10^{23}[/tex]

so, 6.21 ×[tex]10^{-3}[/tex] × 6. 022 × 10^23

= 3.74 ×[tex]10^{21}[/tex]

Therefore, molecules of propane were in the erlenmeyer flask is found to be 3.74 ×[tex]10^{21}[/tex]

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