integrated rate law for second order unimolecular irreversible

CH3OH (l)

Methanol.

This an alcohol, in ambient temperature is liquid, and in contact with water, they dissolve.

HC2H3O2 (l)

Acetic Acid.

This is a liquid acid, common in vinegar, giving them it's taste and sour flavor.

With water they do not repel, they can create a solution.

C12H22O11 (s)

Sucralose.

This is what we know as a common sugar.

It is a solid that dissolves in water.

.

Answer:

A) The hydroxide ion concentration in the solution would be greater than that in pure water.

Explanation:

Spermine has four amino groups. In water, these amino groups will be protonate -In NH₂⁺ or NH₃⁺- decreasing the H⁺ concentration and, by water equilibrium, increasing OH⁻ concentration.

Thus, in an aqueous solution of spermine the hydroxide ion concentration in the solution would be greater than that in pure water.

I hope it helps!

Answer:

The correct answer is option A)  The hydroxide ion concentration in the solution would be greater than that in pure water.

Explanation:

Spermine is basic in nature as it contains di-amine group (NH2), which is basic in nature that is, it easily gives off electrons.

Thus being basic spermine solution will have more concentration of Hydroxide ion [OH-].

Answer:

c. None of the above

Explanation:

The equation is:

H₃O⁺ + CH₃COO⁻ ⇒ CH₃COOH + H₂O

An acid is a species that donates a proton. Thus, the acid is the species on the reactants' side that has one proton more than it's conjugate base species on the products's side.

The available options are:

a. CH₃COOH

b. H₂O

c. None of the above.

Both CH₃COOH and H₂O are on the right-hand side of the equation, so the answer must be c. The answer to the problem is H₃O⁺, since it has lost a proton and become H₂O.

Answer:

d. calorimeter

Explanation:

Calorimeter -

It is the instrument use for the measurement of the heat that is absorbed or released during a physical change or in chemical change .

The instrument or devise is used for the study thermodynamics , biochemistry and chemistry .

Hence , From the question information , the instrument which can be used is a Calorimeter .

Energy absorbed or released as heat during chemical or physical changes is measured by a calorimeter, which calculates this based on temperature changes and the specific heat and mass of the substances involved.

The correct answer is d. Calorimeter. Energy absorbed or released as heat in a chemical or physical change is measured by a calorimeter. A calorimeter is a device designed specifically for measuring the amount of heat involved in chemical or physical processes. For instance, during an exothermic reaction, where heat is produced, the temperature of the solution within the calorimeter increases.

Conversely, for an endothermic reaction, where heat is required, the temperature of the solution decreases as it absorbs heat from the thermal energy of the solution. These temperature changes, in conjunction with specific heat and mass calculations, allow the calorimeter to accurately measure the heat absorbed or released. Calorimetry relies on capturing the temperature changes that occur as a result of heat transfer during these processes, using them to calculate the total amount of heat involved.

Answer:

c. when the raction quotient is greater than the Ksp

Explanation:

∴ Ksp = [ A+ ]∧x  *  [ B- ]∧y.....solubility product constant

∴ I.P = xA+ * yB-............ionic product

⇒ the relationship between Ksp and I.P, will give us the precipitation conditions.

∴ I.P > Ksp ⇒ precipitation

∴ I.P < Ksp ⇒ no precipitation (disolution)

∴ I.P = Ksp ⇒ equilibrium

⇒ the ionic product is directly related to the reaction quotient Q, where Q = [A+] * [B-] / [AxBy] = PI / [AB], then the higher this product the higher the reaction quotient will be and therefore will be greater than the value of the constant and will have precipitated, the correct answer is the c

Answer: Mass of gas is 0.001 pounds.

Explanation:

Density is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{Volume}[/tex]

Given : Mass of gas = ?

Density of gas = [tex]1.29g/L[/tex]

Volume of gas = 389 ml = 0.389 L     (1L=1000ml)

Putting in the values we get:

[tex]1.29g/L=\frac{mass}{0.389L}[/tex]

[tex]mass=0.5grams[/tex]  

[tex]mass=0.5\times 0.002lb=0.001lb[/tex]       (1g =0.002 lb)

Thus the mass of gas is 0.001 pounds.

Final answer:

To find the mass in pounds of 389 mL of a gas with a density of 1.29 g/L, you convert the density to g/mL, calculate the mass in grams, and then convert that mass to pounds, resulting in approximately 0.001106 pounds.

Explanation:

To calculate the mass of the gas in pounds, we first need to convert the density from grams per liter (g/L) to grams per milliliter (g/mL) since the volume of gas given is in milliliters (mL). With the given density of 1.29 g/L, we can use dimensional analysis to calculate the mass of 389 mL of the gas:

Convert density to g/mL: 1.29 g/L = 0.00129 g/mL.

Calculate the mass: mass (in grams) = density (in g/mL)  imes volume (in mL) = 0.00129 g/mL  imes 389 mL = 0.50181 g.

Convert the mass to pounds using the conversion factor 1 lb = 453.59 g: mass (in pounds) = mass (in grams) / conversion factor = 0.50181 g / 453.59 g/lb

Carrying out the division, the mass is approximately 0.001106 pounds.

Answer:

[tex]578.2\times 10^{-6}[/tex]

Explanation:

Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10. Engineering notation is the same version of the scientific notation but the number can be between 1 and 1000 and in this exponent of the ten is divisible by three.

For example, [tex]1000^2[/tex] is to be written as [tex]10^6[/tex] in engineering notation.

The given number:

0.00057824 can be written as [tex]578.24\times 10^{-6}[/tex]

Answer upto 4 significant digits = [tex]578.2\times 10^{-6}[/tex]

Answer:

[S₂] = 1.27×10⁻⁷ M

Explanation:

2 H₂S(g) ⇄ 2 H₂(g) + S₂(g), Kc=1,625x10⁻⁷

The equation of this reaction is:

1,625x10⁻⁷ = [tex]\frac{[H_2]^2[S_2]}{[H_{2}S]^2}[/tex]

The equilibrium concentrations are:

[H₂S] = 0,162 - 2x

[H₂] = 0,184 + 2x

[S₂] = x

Replacing:

1,625x10⁻⁷ = [tex]\frac{[0,184+2x]^2[x]}{[0,162-2x]^2}[/tex]

Solving:

4x³ + 0,736x² + 0,033856x - 4,3x10⁻⁹

x = 1.27×10⁻⁷

Thus, concentration of S₂ is:

[S₂] = 1.27×10⁻⁷ M

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = 20144 mol air/h

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = 289 kg O₂

43058 mol air×29g/mol 1249 kg air

Percent of oxygen is: [tex]\frac{289kg}{1249 kg}[/tex] =0,231 kg O₂/ kg air

I hope it helps!

Answer:

A) Mass flow rate of air = [tex]22.982[/tex] kmol/hr

B) percentage by mass of oxygen in the product gas = [tex]$22.52 \%$[/tex]

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have [tex]0.09[/tex] mole of methane(CH4) and the remaining will be the air which is [tex]$(100 \%-9 \%)=91 \%$=0.91[/tex]

We are given the average molecular weight of air = [tex]29[/tex] g/mol Thus; Average molar mass of air and methane mixture is;

M-air- [tex]$=(0.09 \times 16)+(0.91 \times 29)$[/tex]

[tex]=$27.83 \mathrm{~g} / \mathrm{mol}$[/tex]

Mass fraction of oxygen in the product gas= [tex]=43.016 \mathrm{kmol} / \mathrm{h} \times(0.21 \mathrm{molO} 2 / 1 \mathrm{~mol}$ air $) \times(32 \mathrm{~kg}$ oxygen/1kmol oxygen $) \times(1 / 1283.596 \mathrm{~kg} / \mathrm{h})=0.2252$[/tex]

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Answer:

Volumetric flow:

Mixture = 373.55 l/h; Benzene = 310.05 l/h; n-hexane = 63.5 l/h

Mass flow

Mixture = 317.52 kg/h; Benzene = 275.66 kg/h; n-hexane = 41.85 kg/h

Explanation:

First, we can express the mixture mass flow as

[tex]M=700\frac{lbm}{h}*\frac{453.6g}{1lbm} = 317,520g/h=317,52kg/h[/tex]

The volumetric flow of the mixture is

[tex]V=M/\rho = \frac{317520g/h}{0.85g/ml} =373,553ml/h=373.55l/h[/tex]

The mass balance can be written as

[tex]M_1+M_2=M\\[/tex]

And the volumetric flow as

[tex]V_1+V_2=V\\\\\rho_1*M_1+\rho_2*M_2=\rho*M\\\\\rho_1(M-M_2)+\rho_2*M_2=\rho*M\\\\(\rho_2-\rho_1)*M_2=(\rho-\rho_1)*M\\\\M_2=\frac{(\rho-\rho_1)}{(\rho_2-\rho_1)} *M[/tex]

If fluid 2 is n-hexane, we have

[tex]M_2=\frac{(\rho-\rho_1)}{(\rho_2-\rho_1)} *M\\\\M_2=\frac{(0.85-0.879)}{(0.659-0.879)} *317.52kg/h\\\\M_2=0.1318*317.52kg/h=41.85kg/h[/tex]

The mass flow of n-hexane is 41.85 kg/h.

Its volumetric flow is 63.5 litre/h

[tex]V=M/\rho=41.85\frac{kg}{h}*\frac{1}{0.659g/ml} *\frac{1000g}{1kg}\\\\V = 63505 ml/h=63.5 l/h[/tex]

The mass flow of benzene can be deducted by difference:

[tex]M_1= M-M_2= 317.52-41.85=275.66 kg/h\\\\V_1=V-V_2= 373.55 - 63.5=310.05 l/h[/tex]

Answer:

Re=309926.13

Explanation:

density=92.8lbm/ft3*(0.45kg/1lbm)*(1ft3/0.028m3)=1491.43kg/m3

viscosity=4.1cP*((1*10-3kg/m*s)/1cP)=0.0041kg/m*s

velocity=237ft/min*(1min/60s)*(0.3048m/1ft)=1.2m/s

diameter=28inch*(0.0254m/1inch)=0.71m

Re=(density*velocity*diameter)/viscosity=(1491.43kg/m3*1.2m/s*0.71m)/0.0041kg/m*s

Re=309926.13

Answer:

Explanation:

Using the law of mass conservation, we have that the moles of any component in the inlet air stream equals the moles of this compenent in the two outlets stream. Being A the mole flowrate for the permeate stream and B the mole flowrate for the reject stream (both in mol/min), we have equation (1) and (2):

2 mol/min * 0,21 = A mol/min * 0,50 + B mol/min * 0,13    ......................(1)

0,42 = 0,5A + 0,13B...........................(1)

2 mol/min * 0,79 = A mol/min * 0,50 + B mol/min * 0,87    ......................(2)

1,58 = 0,5A + 0,87B...........................(2)

Solving this system:

1,58 = 0,5A + 0,87B...........................(2)

0,42 = 0,5A + 0,13B...........................(1)

1,16 = 0A + 0,74B...............................(2) - (1)

1.16=0,74B

1,16/0,74=B

B=1,57 mol/min

And now replacing B in equation (1):              

1,58 = 0,5A + 0,87*1,57

1,58-1,37=0,5A

0,21/0,5=A

A=0,43 mol/min

To calculate the volumetric flowrate of the permeate stream (M) and the reject stream (J), we just replace the given and calculate data into the ideal gas equation:

PV=nRT,.................... V = nRT/P

[tex]M= \frac{0,43 \frac{mol}{min}*8,314\frac{m3Pa}{molK}*293K  }{0,1 MPa*\frac{10^{6}Pa }{1 MPa} }=0,01047\frac{m^{3} }{min} * \frac{1000 L}{1 m^{3} }=10,47 L/min[/tex]

[tex]J= \frac{1,57 \frac{mol}{min}*8,314\frac{m3Pa}{molK}*293K  }{0,1 MPa*\frac{10^{6}Pa }{1 MPa} }=0,0382\frac{m^{3} }{min} * \frac{1000 L}{1 m^{3} }=38,24 L/min[/tex]

Explanation:

A chemical reaction equation that contains equal number of atoms on both reactant and product side is known as a balanced chemical equation.

For example, when the products are given as [tex]Ca(ClO_{4})_{2}(aq) + H_{2}O(l)[/tex]

So, reactants for this equation will be as follows.

        [tex]HClO_{4}(aq) + Ca(OH)_{2}(aq) \rightarrow Ca(ClO_{4})_{2}(aq) + H_{2}O(l)[/tex]

Number of atoms present on reactant side are as follows.

H = 3

Cl = 1

O = 6

Ca = 1

Number of atoms present on product side are as follows.

H = 2

Cl = 2

O = 9

Ca = 1

Therefore, to balance this equation we multiply [tex]HClO_{4}[/tex] by 2 on reactant side and multiply [tex]H_{2}O[/tex] by 2 on product side.

Hence, the balanced chemical reaction equation will be as follows.

           [tex]2HClO_{4}(aq) + Ca(OH)_{2}(aq) \rightarrow Ca(ClO_{4})_{2}(aq) + 2H_{2}O(l)[/tex]

The reactants for the given chemical reaction are likely to be calcium carbonate (CaCO₃) and hydrochloric acid (2HCl). These react to form the given products: calcium chloride (CaCl₂) and water (2H₂O), with the balanced equation being: CaCO₃ + 2HCl --> CaCl₂ + H₂O + CO₂.

Predicting the reactants of the chemical reaction requires understanding of chemistry concepts. First, let's rearrange the given product side which is CaCl₂(aq) + 2H₂O(l). From this, we can find that the reactants are likely CaCO₃ + 2HCl, as calcium carbonate generally reacts with hydrochloric acid to form calcium chloride and water.

In terms of steps, the HCl will dissolve the limestone (CaCO₃), and the resulting calcium ions (Ca₂+) will combine with the chloride ions (Cl-) to form calcium chloride (CaCl₂). Meanwhile, the carbonate ions (CO₃₂-) will combine with the hydrogen ions (H+) from the HCl to form water (H₂O) and carbon dioxide (CO₂).

The balanced molecular equation would therefore be: CaCO₃ + 2HCl --> CaCl₂+ H₂O + CO₂.

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Answer :

(a) The change in internal energy of the gas is 22.86 kJ.

(b) The change in enthalpy of the gas is 34.29 kJ.

Explanation :

(a) The formula used for change in internal energy of the gas is:

[tex]\Delta U=nC_v\Delta T\\\\\Delta U=nC_v(T_2-T_1)[/tex]

where,

[tex]\Delta U[/tex] = change in internal energy = ?

n = number of moles of gas = 5 moles

[tex]C_v[/tex] = heat capacity at constant volume = 2R

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]

[tex]T_2[/tex] = final temperature = [tex]300^oC=273+300=573K[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta U=nC_v(T_2-T_1)[/tex]

[tex]\Delta U=(5moles)\times (2R)\times (573-298)[/tex]

[tex]\Delta U=(5moles)\times 2(8.314J/mole.K)\times (573-298)[/tex]

[tex]\Delta U=22863.5J=22.86kJ[/tex]

The change in internal energy of the gas is 22.86 kJ.

(b) The formula used for change in enthalpy of the gas is:

[tex]\Delta H=nC_p\Delta T\\\\\Delta H=nC_p(T_2-T_1)[/tex]

where,

[tex]\Delta H[/tex] = change in enthalpy = ?

n = number of moles of gas = 5 moles

[tex]C_p[/tex] = heat capacity at constant pressure = 3R

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]

[tex]T_2[/tex] = final temperature = [tex]300^oC=273+300=573K[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta H=nC_p(T_2-T_1)[/tex]

[tex]\Delta H=(5moles)\times (3R)\times (573-298)[/tex]

[tex]\Delta H=(5moles)\times 3(8.314J/mole.K)\times (573-298)[/tex]

[tex]\Delta H=34295.25J=34.29kJ[/tex]

The change in enthalpy of the gas is 34.29 kJ.

Final answer:

The balanced equation for the combustion of octane is 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O. From this equation, we can determine the moles of carbon dioxide produced from a given amount of octane.

Explanation:

The balanced equation for the combustion of octane is:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

From the equation, we can see that for every 2 moles of octane burned, 16 moles of carbon dioxide are produced. Therefore, if we know the number of moles of octane burned, we can calculate the moles of carbon dioxide produced.

Final answer:

The frequency of light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level is approximately -2.06 x 10^14 Hz.

Explanation:

The frequency of light emitted by a hydrogen atom during a transition of its electron can be calculated using the formula:

f = R(1/n1^2 - 1/n2^2)

where f is the frequency, R is a constant (R = 3.29 x 10^15 Hz), and n1 and n2 are the initial and final principal energy levels, respectively.

In this case, the electron transitions from n = 4 to n = 1. Plugging the values into the formula, we get:

f = 3.29 x 10^15 (1/4^2 - 1/1^2)

f = 3.29 x 10^15 (1/16 - 1/1)

f = 3.29 x 10^15 (0.9375 - 1)

f = 3.29 x 10^15 (-0.0625)

f = -2.06 x 10^14 Hz

The frequency of the light emitted by the hydrogen atom during this transition is approximately -2.06 x 10^14 Hz.

Answer: The mass and number of moles of sulfur hexafluoride gas is 2.8 g and 0.019 moles respectively.

Explanation:

To calculate the moles of sulfur hexafluoride gas, we use the equation given by ideal gas:

PV = nRT

where,

P = Pressure of sulfur hexafluoride = 0.090 atm

V = Volume of the sulfur hexafluoride gas = 5.0 L

n = number of moles of gas = ?

R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

T = Temperature of sulfur hexafluoride gas = [tex]14^oC=[273+14]K=287K[/tex]

Putting values in above equation, we get:

[tex]0.090atm\times 5.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 287K\\\\n=0.019mol[/tex]

To calculate the mass of sulfur hexafluoride, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Molar mass of sulfur hexafluoride = 146.0 g/mol

Moles of sulfur hexafluoride = 0.0191 moles

Putting values in equation 1, we get:

[tex]0.019mol=\frac{\text{Mass of sulfur hexafluoride}}{146.0g/mol}\\\\\text{Mass of sulfur hexafluoride}=2.8g[/tex]

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits. Here, the least precise number of significant digits are 2.

Hence, the mass and number of moles of sulfur hexafluoride gas is 2.8 g and 0.019 moles respectively.

To calculate the mass and number of moles of sulfur hexafluoride gas, we use the ideal gas law equation PV = nRT. By substituting the given values and solving for n, we find that 0.1903 moles of sulfur hexafluoride were collected. The mass of the gas is determined by multiplying the moles by the molar mass of sulfur hexafluoride, resulting in approximately 27.8 grams.

To calculate the mass and number of moles of sulfur hexafluoride gas, we can use the ideal gas law equation: PV = nRT. Rearranging the equation to solve for moles (n), we have n = PV/RT. We are given the pressure (0.090 atm), volume (5.0 L), temperature (14.0°C which needs to be converted to Kelvin), and the ideal gas constant (R = 0.0821 L·atm/(mol·K)). After converting the temperature to Kelvin (287 K), we substitute the values and solve for n. n = (0.090 atm * 5.0 L) / (0.0821 L·atm/(mol·K) * 287 K). This gives us n = 0.1903 moles of sulfur hexafluoride gas.

To calculate the mass, we use the formula mass = moles * molar mass. The molar mass of sulfur hexafluoride (SF6) is  SF6 = (32.06 g/mol * 6) + (19.00 g/mol * 1). This gives us a total molar mass of 146.06 g/mol. We can now calculate the mass by multiplying the moles by the molar mass: mass = 0.1903 moles * 146.06 g/mol. The mass of sulfur hexafluoride gas that were collected is approximately 27.8 grams.

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Answer:

Considering the half-life of 10,000 years, after 20,000 years we will have a fourth of the remaining amount.

Explanation:

The half-time is the time a radioisotope takes to decay and lose half of its mass. Therefore, we can make the following scheme to know the amount remaining after a period of time:

Time_________________ Amount

t=0_____________________x

t=10,000 years____________x/2

t=20,000 years___________x/4

During the first 10,000 years the radioisotope lost half of its mass. After 10,000 years more (which means 2 half-lives), the remaining amount also lost half of its mass. Therefore, after 20,000 years, the we will have a fourth of the initial amount.

Final answer:

After 20,000 years, or two half-lives, one-fourth of the original amount of a radioisotope with a half-life of 10,000 years would remain.

Explanation:

If the half-life of a radioisotope is 10,000 years, the amount of the isotope remaining after 20,000 years would be one-fourth of the original amount. This can be understood by realizing that with each half-life period, the quantity of the isotope is reduced by half. After the first 10,000 years, one half-life, half of the isotope will have decayed leaving 50%. After another 10,000 years, which makes two half-lives in total, half of the remaining 50% will have decayed, leaving 25% of the original amount. Hence, the correct answer to the question is a fourth, or 25%.

Answer : The concentration of [tex]H_2,I_2[/tex] and [tex]HI[/tex] at equilibrium 0.033 M, 0.033 M and 0.234 M respectively.

Solution :  Given,

[tex]K_c=50.3[/tex]

Concentration = 0.100 M

The given equilibrium reaction is,

                              [tex]H_2(g)+I_2(g)\rightleftharpoons 2HI(g)[/tex]

Initially conc.       0.100     0.100       0.100

At equilibrium  (0.100-x)   (0.100-x)  (0.100+2x)

The expression of [tex]K_c[/tex] will be,

[tex]K=\frac{[HI]^2}{[H_2][I_2]}[/tex]

Now put all the given values in this expression, we get:

[tex]50.3=\frac{(0.100+2x)^2}{(0.100-x)\times (0.100-x)}[/tex]

By solving the term x, we get

[tex]x=0.067\text{ and }0.158[/tex]

From the values of 'x' we conclude that, x = 0.158 can not more than initial concentration. So, the value of 'x' which is equal to 0.158 is not consider.

Thus, the concentration of [tex]H_2[/tex] and [tex]I_2[/tex] at equilibrium = (0.100-x) = 0.100 - 0.067 = 0.033 M

The concentration of [tex]HI[/tex] at equilibrium = (0.100+2x) = 0.100 + 2(0.067) = 0.234 M

Explanation:

The given data is as follows.

         Thickness (dx) = 0.87 m,       thermal conductivity (k) = 13 W/m-K

        Surface area (A) = 5 [tex]m^{2}[/tex],       [tex]T_{1} = 14^{o}C[/tex]

        [tex]T_{2} = 93^{o}C[/tex]

According to Fourier's law,

                    Q = [tex]-kA \frac{dT}{dx}[/tex]

Hence, putting the given values into the above formula as follows.

                      Q = [tex]-kA \frac{dT}{dx}[/tex]

                          = [tex]-13 W/m-k \times 5 m^{2} \times \frac{(14 - 93)}{0.87 m}[/tex]

                          = 5902.298 W

Therefore, we can conclude that the rate of heat transfer is 5902.298 W.

Answer:

Required compressor work W=8560.44  KJ/Kmol

Explanation:

Given that

Initial pressure = 1 bar

[tex]P_1=1\ bar[/tex]

Final pressure = 40 bar

[tex]P_2=40\ bar[/tex]

Process is isothermal and T=280 K

We know that ,work done in isothermal process given as

[tex]W=P_1V_1\ln \dfrac{P_1}{P_2}[/tex]

given taht gas is ideal so

P V =m R T

[tex]W=mRT\ln \dfrac{P_1}{P_2}[/tex]

R for ethylene

R=0.296 KJ/kg.K

Now by putting the values

[tex]W=mRT\ln \dfrac{P_1}{P_2}[/tex]

[tex]W=1\times 0.296\times 280 \ln \dfrac{1}{40}[/tex]

W= -305.73 KJ/kg

Negative sign indicates that work done on the system.

Required compressor work W=305.73 KJ/kg

Molar mass of ethylene M= 28 Kg/Kmol

So W= 305.73 x 28  KJ/Kmol

W=8560.44  KJ/Kmol

Required compressor work W=8560.44  KJ/Kmol

Answer: The mass of carbon dioxide produced is 352 grams

Explanation:

We are given:

Moles of ethanol = 4 mol

For the given chemical equation:

[tex]C_2H_5OH(g)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)[/tex]

By Stoichiometry of the reaction:

1 mole of ethanol produces 2 moles of carbon dioxide

So, 4 moles of ethanol will produce = [tex]\frac{2}{1}\times 4=8mol[/tex] of carbon dioxide

To calculate the mass of carbon dioxide, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of carbon dioxide = 8 moles

Molar mass of carbon dioxide = 44 g/mol

Putting values in above equation:

[tex]8mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(8mol\times 44g/mol)=352g[/tex]

Hence, the mass of carbon dioxide produced is 352 grams

Final answer:

The combustion of 4 moles of ethanol produces 352 g of CO₂, based on the stoichiometry of the balanced chemical equation.

Explanation:

The question involves a stoichiometry calculation based on the balanced chemical equation for the combustion of ethanol, C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g). Given that 4 moles of ethanol are combusted, we need to determine the amount of CO2 produced.

From the equation, it is clear that 1 mole of ethanol produces 2 moles of CO₂. Therefore, 4 moles of ethanol will produce 4 * 2 = 8 moles of CO₂.

To calculate the weight of these CO₂ moles, we use the molar mass of CO₂, which is approximately 44 g/mol. Hence, the weight of CO₂ produced is 8 moles * 44 g/mol = 352 g.

Answer:

0.152 moles

Explanation:

Given that:

The mass of the oxygen gas produced = 4.87 g

Also, The molar mass of oxygen gas, [tex]O_2[/tex] = [tex]2\times 16\ g/mol[/tex] = 32 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{4.87\ g}{32\ g/mol}[/tex]

[tex]Moles=0.152\ mol[/tex]

Both given values and the answer is in 3 significant digits.

Answer:

B

Explanation:

Limiting reagent is the one which is present in small molar amount. It got exhausted at the end of the reaction and the formation of the products is governed by it.

The given reaction:

2A + 3B → D+E

2 moles of A react with 3 moles of B

1 mole of A react with 3 / 2 moles of B

Given moles of A = 3 moles

Moles of B = 4 moles

So,

3 moles of A will react with 1.5 * 3 moles of B

Moles of B = 4.5

Since, only we have 4 moles of B. So, B is the limiting reagent.

To determine the molar mass of an unknown protein, first calculate the molar concentration using the osmotic pressure formula. Next, calculate the moles of solute from this molarity. Finally, determine the molar mass by dividing the mass of the protein by the moles of solute.

First, we need to calculate the molarity (M) of the solution using the osmotic pressure formula II = MRT. Here, 'II' represents the osmotic pressure in atm, which is 0.0861 atm, 'R' is the gas constant (0.08206 L atm/mol K), and 'T' is the temperature in Kelvin. Convert the given temperature 25.0 °C to Kelvin by adding 273.15, which results in 298.15 K.

Substituting the known values into the formula, we get M = II / RT. Now, compute the molar concentration (M) from the osmotic pressure.

Next, the molar mass of the protein can be determined. Divide the mass of the solute by the number of moles in that mass. In this case, the mass of the solute is 400mg, which equals 0.4 g (as 1 g = 1000 mg). By using the calculated molarity, you can get the moles of solute, from which the molar mass can be determined by dividing the mass of solute by moles of solute.

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To determine the molar mass of the protein, use the osmotic pressure equation. After solving for molarity and then moles, the molar mass is found to be approximately 22,700 g/mol. This process involves converting units and applying the ideal gas constant.

To find the molar mass of the protein, we use the formula for osmotic pressure (π):

π = MRT

π (osmotic pressure) = 0.0861 atm

M (molarity) = n/V (number of moles/volume)

R (gas constant) = 0.0821 L·atm/(K·mol)

T (temperature) = 25 + 273 = 298 K

First, convert the mass of the protein to grams:

400 mg = 0.400 g

The volume of the solution is 5.00 mL, which is 0.00500 L.

We can rearrange the formula to find the molarity (M):

M = π / (RT)

M = 0.0861 atm / (0.0821 L·atm/(K·mol) × 298 K)

M = 0.00352 mol/L

Now, the number of moles of the protein (n) can be related to its molarity and the volume of the solution:

n = M × V

n = 0.00352 mol/L × 0.00500 L = 1.76 x 10⁻⁵ mol

Finally, the molar mass (M) is given by the mass of the protein divided by the number of moles:

Molar mass = mass / n

Molar mass = 0.400 g / 1.76 x 10⁻⁵ mol ≈ 22727 g/mol

Rounded to three significant digits, the molar mass of the protein is 22,700 g/mol.

Answer:

1.0489 M is the concentration (in M) of the NaOH solution that was added

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCl[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=1\\M_1=0.0500 M\\V_1=0.450 L\\n_2=1\\M_2=?\\V_2=21.45 mL=0.02145 L[/tex]

Putting values in above equation, we get:

[tex]1\times 0.0500 M\times 0.450 L=1\times M_2\times 0.02145 L\\\\M_2=1.0489 M[/tex]

1.0489 M is the concentration (in M) of the NaOH solution that was added

Answer:

Every substance, whether an element or a compound that contains Avogadro's number, is said to form a mole of it. Therefore for the fluorine atoms that form the Avogadro number we say that we have one mole of fluorine atoms and this weighs 19 g, so 19 g/mol

Explanation:

You can search the F in the Periodic Table where you have the molar mass, 19 g/mol but if you don't have a Periodic Table you can calculate like this:

1 atom of F weighs 19 Da (unified atomic mass unit) and 1 Da is 1,661 x10*-24 g, so try to apply a rule of three, twice.

1 Da ........... 1,661x10*-24 g

19 Da .......... x

x= (19 Da . 1,661x10*-24 g ) / 1 Da = 3,1559x10*-23 g

1 atom F ..............  3,1559x10*-23 g

6,022x10*23 atoms F ............. x

X= (6,022x10*23 atoms F . 3,1559x10*-23 g) / 1 atom F = 19.0048 g

Explanation:

It is given that molarity is 0.15 M and volume is 250.0 ml. As 1 ml equals 0.001 l.

Hence, 250.0 ml = 0.25 l.

Also, molarity is equal to the number of moles present in liter of solution. Therefore, calculate number of moles as follows.

                   Molarity = [tex]\frac{\text{no. of moles}}{volume}[/tex]

                      0.15 M = [tex]\frac{\text{no. of moles}}{0.25 l}[/tex]

             No. of moles = 0.0375 mol

Thus, we can conclude that moles present in given solution are 0.0375.

Final answer:

To find the number of moles in a 250.0 mL solution of 0.15 M nitric acid, you convert the volume to liters and multiply by the molarity, resulting in 0.0375 moles of HNO3.

Explanation:

To calculate the number of moles in a 0.15 M nitric acid (HNO3) solution with a volume of 250.0 mL, you would first convert the volume from milliliters (mL) to liters (L) because concentration (Molarity, M) is typically expressed in moles per liter. Therefore, 250.0 mL is equivalent to 0.250 L (since 1 L = 1000 mL).

Using the formula for molarity:

Molarity (M) = moles of solute / liters of solution

We can rearrange the formula to solve for the number of moles:

moles of solute = Molarity (M) × liters of solution

Thus, the number of moles of nitric acid in the solution is:

0.15 moles/L × 0.250 L = 0.0375 moles of HNO3

Answer: The average atomic mass of element X is 96.6 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]    .....(1)

Mass of isotope 1 = 95.0 amu

Percentage abundance of isotope 1 = 10.68 %

Fractional abundance of isotope 1 = 0.1068

Mass of isotope 2 = 96.0 amu

Percentage abundance of isotope 2 = 16.90 %

Fractional abundance of isotope 2 = 0.1690

Mass of isotope 3 = 97.0 amu

Percentage abundance of isotope 3 = 72.42 %

Fractional abundance of isotope 3 = 0.7242

Putting values in equation 1, we get:

[tex]\text{Average atomic mass of X}=[(95\times 0.1068)+(96\times 0.1690)+(97\times 0.7242)][/tex]

[tex]\text{Average atomic mass of X}=96.6amu[/tex]

Hence, the average atomic mass of element X is 96.6 amu.

To calculate the atomic mass of the element with three isotopes, you multiply the percentage abundance by the mass of each isotope and then add the values.

To calculate the atomic mass of the element, we need to multiply the percentage abundance of each isotope by its respective mass and then sum up the products.

For isotope X-95: (10.68/100) x 95.0 amu = 10.1376 amu

For isotope X-96: (16.90/100) x 96.0 amu = 16.224 amu

For isotope X-97: (72.42/100) x 97.0 amu = 70.2774 amu

Adding the values: 10.1376 amu + 16.224 amu + 70.2774 amu = 96.639 amu

Therefore, the atomic mass of the element is 96.639 amu (rounded to 3 significant figures).

Explanation:

The given data is as follows.

     [tex]V_{1}[/tex] = 2.36 L,    [tex]T_{1}[/tex] = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K,

      [tex]P_{1}[/tex] = 10.0 atm,   [tex]V_{2}[/tex] = 7.79 L,

       [tex]T_{2}[/tex] = ?,       [tex]P_{2}[/tex] = 5.56 atm  

And, according to ideal gas equation,  

               [tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]

Hence, putting the given values into the above formula to calculate the value of final temperature as follows.          

 [tex]\frac{10.0 atm \times 2.36 L}{298 K} = \frac{5.56 atm \times 7.79 L}{T_{2}}[/tex]

            [tex]T_{2}[/tex] = [tex]\frac{43.3124}{0.0792}[/tex] K

                     = 546.87 K

Thus, we can conclude that the final temperature is 546.87 K.

Final answer:

The resulting temperature when transferring cyclopropane from one container to another is approximately 12.4 °C.

Explanation:

To find the resulting temperature when transferring cyclopropane from one container to another, we can use the ideal gas law: PV = nRT. We are given the initial and final pressures and volumes, and we need to find the final temperature.

First, we can calculate the initial number of moles using the ideal gas law: n1 = (P1 * V1) / (R * T1). Then, we can use this number of moles to calculate the final temperature using the ideal gas law: T2 = (P2 * V2) / (n1 * R).

Substituting the given values, we have: T2 = (5.56 atm * 7.79 L) / ((10.0 atm * 2.36 L) / (0.0821 atm L/mol K)). Solving this equation, we find that the resulting temperature is approximately 12.4 °C.