Alcohols like ethanol CH3CH2OH, have a proton attached to the oxygen atom that can be acidic. Can a base like CH3NH2 deprotonate the -OH group? (pka CH3CH2OH=15.7 pKa CH3NH3=10.8) justify your answer

Answers

Answer 1

Answer:

No. CH₃NH₂ cannot deprotonate CH₃CH₂OH.

Explanation:

Organic species like ethanol and methylamine can behave as acid or bases. The pKa value is necessary to indentify in a reaction if the species can act as an acid or a base.

In this case, despite alcohols being recognised as acids and amines as bases, the amine  CH₃NH₂ has an smaller pKa than the alcohol CH₃CH₂OH , which means that  CH₃NH₂  is a stronger acid than  CH₃CH₂OH . Therefore, CH₃NH₂ will not deprotonate the -OH group.

Answer 2
Final answer:

The likelihood of methylanime (CH3NH2) deprotonating the -OH group of ethanol (CH3CH2OH) is low as the conjugate acid CH3NH3+ (formed after CH3NH2 accepts a proton) is a stronger acid than ethanol, demonstrated by its lower pKa value (10.8 compared to 15.7).

Explanation:

In the realm of chemistry, it's known that the strength of an acid is determined by its pKa value. The lower the pKa, the stronger the acid. As given, ethanol (CH3CH2OH) has a pKa of 15.7 and CH3NH3+ (ammonium ion) has a pKa of 10.8. As such, ethanol is a weaker acid than the ammonium ion.

For a base to deprotonate an acid, the conjugate acid of that base (formed after it accepts a proton) has to be weaker than the initial acid. In your case, the conjugate acid of CH3NH2 (methylamine) would be CH3NH3+. As CH3NH3+ is a significantly stronger acid than ethanol, it is therefore unlikely that methylamine could effectively deprotonate the -OH group in ethanol.

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Related Questions

What is the effect on the outlet temperatures in both the hot and cold streams, when the surface area (length and diameter) is reduced?

a- The outlet temperature of the cold stream decreases and the outlet temperature of the hot stream increases.

b- Both the outlet temperature increases

c- The outlet temperature of cold fluid increases and the outlet temperature of hot fluid decreases

Answers

Answer:

The right answer is option A: The outlet temperature of the cold stream decreases and the outlet temperature of the hot stream increases.

Explanation:

In a heat exchanger, when the exchange area is reduced, and at the same condition for the rest of the variables, the amount of heat exchanged by the two fluids is reduced.

In the case of the hot fluid, it will rise at a higher temperature, compared to the temperature at which it would come out with a larger exchange area.

In the case of the hot fluid, it will come out at a lower temperature, compared to the temperature at which it would come out with a larger exchange area.

The correct answer is option A.

Answer:

a- The outlet temperature of the cold stream decreases and the outlet temperature of the hot stream increases.

Explanation:

Hello,

In heat transfer processes, specific units such as heat exchangers are useful to either heat up a cold stream with a hot one or vise-versa. However, the heat transfer efficiency increases as surface area increases, it means that the higher the surface area the higher the heat exchanger's efficiency, therefore, if the surface area is reduced, the efficiency is reduced as well, due to the lack of more space for the heat transfer.

In such a way, the outer temperature of the cold fluid will be lower (less heating up) and the outer one of the hot fluid will be higher (less cooling down), thus, the outlet temperature of the cold stream decreases and the outlet temperature of the hot stream increases (a option).

Best regards.

A mixture of methanol and methyl acetate contains 7.0 weight percent methanol. Determine the number of gmols of methanol in 300.0 kilograms of the mixture.

Answers

Answer:

656,25 moles of CH3OH

Explanation:

If the mixture has 7% of weight in methanol (CH3OH), it means that for every 100 kg of the mixture there are 7 kg of methanol.

To solve the problem we just use this relation and convert the kilograms of methanol to gmoles of methanol using the molecuar weight of the methanol (32g/mol):

[tex]300 kgmixture*\frac{7kgCH3OH}{100kg}*\frac{1000g}{1kg}*\frac{1molCH3OH}{32g}=656,25molesCH3OH[/tex]

In the soap testing experiment, what is the purpose of adding some MgCl2 to the soap solution at the end?

Answers

Explanation:

The water is hard due to the presence of ions of Mg²⁺. Now in hard water soaps are ineffective . Since in hard water Mg²⁺ ion forms precipitate , which concludes less number of soap molecules are present in the solution and less amount of  froth  which disables the cleansing property of the soap .

Now the experiment , the very last step is the hard water test where some amount of MgCl₂ is added ,

Now this magnesium salt act as a source of Mg²⁺ ion and now the soap action can be determined , whether it is able to form froth or not .

MgCl2 is added to the soap solution during a soap testing experiment to simulate the conditions of hard water. This allows for assessment of the soap's effectiveness as a cleansing agent in hard water, as soaps often form insoluble precipitates when reacting with minerals like magnesium.

In a soap testing experiment, MgCl2 (magnesium chloride) is added to the soap solution as a way to simulate hard water conditions because hard water contains high levels of magnesium and other minerals. One property of soap is its ability to react with ions present in water. In hard water, soaps react with calcium or magnesium ions to form insoluble precipitates, which makes the soap less effective as a cleaning agent. This is notable because soaps form insoluble calcium and magnesium compounds in hard water, while detergents form water-soluble products.

The behavior of soap in this soap-magnesium chloride solution, therefore, would give an indication of its prospective behaviour in hard water conditions.

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Draw the Lewis Structure for CH4.

Answers

Explanation:

Methane (CH₄)

Valence electrons of carbon = 4

Valence electrons of hydrogen = 1  

The total number of the valence electrons  = 4 + 4(1) = 8

The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete. So,  

The Lewis structure is:

                H

                 :

     H    :    C   :    H

                 :

                H

Final answer:

To draw the Lewis Structure for CH4, follow the steps to determine the total number of valence electrons, place the atoms and lone pairs, and draw the final structure. Carbon in CH4 forms 4 single bonds with Hydrogen, resulting in a tetrahedral arrangement of Hydrogen atoms around the Carbon atom.

Explanation:

The Lewis Structure for CH4, which represents the arrangement of electrons in the molecule, can be drawn using the following steps:

Determine the total number of valence electronsDetermine the central atom, which is Carbon (C) in this case, and surround it with the Hydrogen (H) atomsPlace the remaining electrons as lone pairs on the central atom and the outer atomsCheck if the central atom has an octet of electrons. If not, move a lone pair from an outer atom to form a double bond with the central atomDraw the final Lewis structure

For CH4, Carbon has 4 valence electrons and Hydrogen has 1 valence electron each. The total number of valence electrons is 4 + (4 x 1) = 8. Following the steps, Carbon is surrounded by 4 Hydrogen atoms, and each Hydrogen atom shares its electron with Carbon. Carbon ends up with a complete octet by sharing electrons with the Hydrogen atoms, forming 4 single bonds. The final Lewis structure for CH4 is a tetrahedral arrangement of the Hydrogen atoms around the Carbon atom, represented by a molecular formula, structural formula, or other models.

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What is the appropriate chemical formula for Iron(III) Sulfate? Sulfate is SO_4^2 Fe3SO4 O a . Fe2(SO4)3 O b. O c. Fe2SO4 O d. Fe2(SO4)2 O e. Fe3(SO4)2

Answers

Answer: The chemical formula of Iron (III) sulfate is [tex]Fe_2(SO_4)_3[/tex]

Explanation:

Iron is the 26th element of periodic table having electronic configuration of [tex][Ar]3d^64s^2[/tex].

To form [tex]Fe^{3+}[/tex] ion, this element will loose 3 electrons.

Sulfate ion is a polyatomic ion having chemical formula of [tex]SO_4^{2-}[/tex]

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for Iron (III) sulfate is [tex]Fe_2(SO_4)_3[/tex]

What are the major causes of environmental problems?

Answers

Answer: This world is rapidly moving towards to some serious enviornmental problems, the cause of this problems are mentioned below;

(1) The increase in average temperature of Earth.

(2) Deforestation

(3) Over Population

(4) Mining

Because of these reasons mentioned above there are many enviornmental issue occurring, and many problems we will face in future for example, water crisis, lack of resources.

Define "Hydrogen Bond"

Answers

Answer:

Hydrogen bond is a special type of dipole-dipole interaction between the hydrogen atom in a polar bond, such as N-H, O-H, or F-H, and electronegative O,N or F atom. The interaction is written:

                                        A-B . . . B        or           A-H . . . .A

A and B represent O,N or F; A-H is one molecule or part of a molecule and B is a part of another molecule, and the dotted represents the hydrogen bond. So, hydrogen bond is a type of intermolecular forces (attractive forces between molecules, NOT between atoms of the same molecule), and only a few elements can participate on hydrogen bond formation.

Note: dipole-dipole interactions are forces between polar molecules, that is, between molecules that possess dipole moments.  

Final answer:

A hydrogen bond is a weak intermolecular attraction where a partially positive hydrogen atom, bonded to an electronegative atom, is attracted to a lone pair of electrons on a nearby electronegative atom.

Explanation:

A hydrogen bond is a type of intermolecular or intramolecular attractive force. It occurs when a hydrogen atom, covalently bonded to a highly electronegative atom such as nitrogen, oxygen, or fluorine, develops a partial positive charge. This partially positive hydrogen atom is then attracted to the lone pair of electrons on a neighboring electronegative atom, often in a different molecule. This attraction is relatively weak compared to covalent bonds but is stronger than other dipole-dipole interactions. In water, hydrogen bonds are a crucial part of the molecule's properties, including its high boiling point and surface tension.

For light with a frequency of 5.371 x 10^14 s^-1, what is the energy (in J)?

Answers

Answer:

The energy is 3.559 * 10⁻¹⁹ J

Explanation:

For this problem we need to use Planck's equation, this equation describes the relationship between energy radiated by light (a photon, to be precise) and its frequency. The equation is:

E = h * v

Where E is energy (in Joules), h is Planck's constant (6.626 * 10⁻³⁴ J·s), and v is the frequency (in s⁻¹).

Now we solve the equation, using the data given in the problem:

E = 6.626 * 10⁻³⁴ J·s * 5.371 * 10¹⁴ s⁻¹ = 3.559 * 10⁻⁴⁷ J

Thus the energy is 3.559 * 10⁻¹⁹ J

Determine the temperature, volume, and quality of 1kg of H2O under the following conditions

A) U=2900kj/kg, P= 1.7MPa
B) U= 350 kj/kg P= 0.3 Mpa

Answers

the answer is the second one (B)

Write 0.000000093425 in Scientific Notation with 4 significant figures.

Answers

Answer: The given number in scientific notation is [tex]9.342\times 10^{-8}[/tex]

Explanation:

Scientific notation is the notation where a number is expressed in the decimal form. This means that the number is always written in the power of 10 form. The numerical digit lies between 0.1.... to 9.9.....

If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.

We are given:

A number having value = 0.000000093425

Converting this into scientific notation, we get:

[tex]\Rightarrow 0.000000093425=9.342\times 10^{-8}[/tex]

Hence, the given number in scientific notation is [tex]9.342\times 10^{-8}[/tex]

Answer:

[tex]9.343\times 10^{-8}[/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.

The given number:

0.000000093425 can be written as [tex]9.3425\times 10^{-8}[/tex]

Answer upto 4 significant digits = [tex]9.343\times 10^{-8}[/tex]

Which compound has the bigger lattice energy? K Br CaBr2 o O Na, NaF Rb20 0 Rb, S 0

Answers

Answer:

The answer is NaF.

Explanation:

Lattice energy:

It can be define as "the amount of energy released when ions combine to from the ionic solids or energy required to break the compound into its ions"

Example:

Consider the example of ionic solid NaCl. The sodium chloride is formed when ions like Na+ and Cl - are combine. During the formation of NaCl crystals the energy is released which is called lattice energy of sodium chloride.

Dependence of lattice energy:

Lattice energy depend upon following factors:

1. Charges of the ions

2. Sizes of the ions

As the charge of ion increase the lattice energy also increase. There is a direst relationship between them.

The size of the ions and lattice energy have inverse relation. As the size decreases, lattice energy increases. Therefor, the compound NaF consist of smallest cation (Na+) and the smallest anion (F-) have largest lattice energy.

                F= q1 × q2/ r²

As the size of ions decreases the distance between the oppositely charged ions in a ionic compound decreases. They will more tightly packed in a crystal and their center will more closer to each others hence increase the lattice energy.

Express each of the following values in the given base unit. 1st attempt Part 1 (2.5 points) m = 2.54 cm Part 2 (2.5 points) S = 316 MS

Answers

Answer:

1) 2.54 cm = 0.0254 m

2) 316 Ms = [tex]3.16\times 10^{8} s[/tex]

Explanation:

1) 2.54 cm

In 1 centimeter is there are 0.01 meters.

1 cm = 0.01 m

[tex]2.54 cm = 2.54\times 0.01 m = 0.0254 m[/tex]

2) 316 Ms

In 1 mega second  is there are 1 million seconds.

[tex]1 Ms = 10^{6} s[/tex]

[tex]316 Ms= 316\times 10^{6} s= 3.16\times 10^{8} s[/tex]

How does the energy of the activated complex compare with the energies of reactants an products? a. It is lower than the energy of both reactants and products. b. It is lower than the energy of reactants but higher than the energy of products. c. It is higher than the energy of reactants but lower than the energy of products. d. It is higher than the energy of both reactants and products.

Answers

Answer:

Option d, It is higher than the energy of both reactants and products.

Explanation:

When reactant molecules collide with each other during the course of the reaction, an intermediate state is reached before the formation of product. This intermediate state is also called activated complex.

Activated complex consist energy higher than that of both the reactants and products. Because of possessing higher energy, it is unstable and temporary.

Hence, among given, option d is correct.

Answer:

d. It is higher than the energy of both reactants and products.

Explanation:

The activated complex is a transient state, or intermediate phase, between reagents (weak or not), in which the final product has not yet been formed. This effective shock, like any chemical bond, needs energy. To form this transient state requires an activation energy, a minimum energy for this intermediate phase to occur.

Within this phenomenon, the activation energy must occur by some criteria, which characterizes a chemical bond in an activated complex:

Binding reagent molecules need to be involved; This collision in the activation energy must be as close as possible to the geometric formation of the complex in order to activate it; The collision cannot have a lower energy than the activation energy. Must be equal to or greater than propagated during reaction.

Within these three rules, it is clearer to define this complex as the moment when a molecule of a given atom collides with another molecule of any other atom and breaks the bond established between them.

For it to happen, the energy needs to be high. When this potential energy is present at a high level during the reaction, it is also necessary to have a high energy charge to complete the complex and also the collision of the reactant molecules to form the binding products.

How many moles of solute particles are present in 7.94 mL of 0.887 M NaNO3? When you have your answer, take the LOG (base 10) of your answer and enter it with 2 decimal places.

Answers

Final answer:

To calculate the number of moles of solute particles in a solution, multiply the molarity by the volume of the solution. In this case, the result is 0.0070 mol. Taking the LOG (base 10) of the result gives -2.15.

Explanation:

To determine the number of moles of solute particles in a solution, you need to consider the molarity and volume of the solution. In this case, the molarity is 0.887 M and the volume is 7.94 mL. To calculate the number of moles, you can use the formula:

Moles = Molarity x Volume

Converting mL to L:

7.94 mL = 7.94/1000 L = 0.00794 L

Now, substitute the values into the formula:

Moles = 0.887 M x 0.00794 L = 0.0070 mol

Taking the LOG (base 10) of 0.0070, we get -2.15 (rounded to 2 decimal places).

Hydrogen gas and oxygen gas react to form water. The balanced equation for the reaction is the following. 2 H2(g) + O2(g) → 2 H2O(g) Write equations to describe the rate of disappearance of H2(g) and the rate of disappearance of O2(g). Then write a third equation to show how the rates are related. Rate of disappearance of H2(g): rate of disappearance of H2 = − Δ[H2] Δt rate of disappearance of H2 = Δ[H2] Δt rate of disappearance of H2 = Δt Δ[H2] rate of disappearance of H2 = − Δt Δ[H2] Rate of disappearance of O2(g): rate of disappearance of O2 = Δ[O2] Δt rate of disappearance of O2 = Δt Δ[O2] rate of disappearance of O2 = − Δ[O2] Δt rate of disappearance of O2 = − Δt Δ[O2] Relationship between the rates: rate of disappearance of H2 = − 2 mol H2 1 mol O2 × rate of disappearance of O2 rate of disappearance of H2 = 2 mol H2 1 mol O2 × rate of disappearance of O2 rate of disappearance of H2 = 1 mol H2 2 mol O2 × rate of disappearance of O2 rate of disappearance of H2 = − 1 mol H2 2 mol O2 × rate of disappearance of O2

Answers

Answer: Thus rate of disappearance of [tex]H_2=2\times {\text {rate of disappearance of}O_2}[/tex]

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex]

Given: Order with respect to [tex]H_2[/tex] = 2

Order with respect to [tex]O_2[/tex] = 1

Thus rate law is:

[tex]Rate=k[H_2]^2[O_2]^1[/tex]

k= rate constant

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

[tex]Rate=-\frac{1d[H_2]}{2dt}=-\frac{1d[O_2]}{dt}[/tex]

[tex]Rate=-\frac{1d[H_2]}{dt}=2\times -\frac{1d[O_2]}{dt}[/tex]

Thus rate of disappearance of [tex]H_2=2\times {\text {rate of disappearance of}O_2}[/tex]

You're a newly appointed Engineer in APE Chemical Sdn Bhd. and your first task is to design a 0.35m vessel to be used to store saturated vapor ethanol at 480°C and 6000 kPa. For these conditions, analyze (6) The molar volume of the saturated vapor ethanol. . (15 marks) (6) The mass of the saturated vapor ethanol in the vessel

Answers

Answer:

molar volume of vapor ethanol is [tex]V = 1.043 \times 10^{-3} m^3/mol[/tex]

mass of ethanol is 15430.22 g

Explanation:

By using ideal gas equation

PV = nRT

Where P is pressure , R is gas constant

so, volume is

[tex]V = \frac{RT}{P} = \frac{8.314\times (480+273)}{6000\times 10^3}[/tex]

[tex]V = 1.043 \times 10^{-3} m^3/mol[/tex]

molar volume of vapor ethanol is

[tex]V = 1.043 \times 10^{-3} m^3/mol[/tex]

b)

volume of vessel is given [tex]0.35 m^3[/tex]

therefore total moles of ethanol in given vessel will be

[tex]n =\frac{V_{vessel}}{V}[/tex]

[tex]n =\frac{0.35}{1.043\times10^{-3}}[/tex]

n = 335.44 mol

we know that

mole is given as [/tex]n = \frac{mass}{moleculae weight}[/tex]

weight of ethanol is 46 g/mol

n\times 46 = mass

[tex]335.44\times 46 = 15430.22 g[/tex]

mass of ethanol is 15430.22 g

Which of the following non-hydrogen atom transitions does the photon have at its long wavelength?

Move from Level n=6 to Level n=1
Move from Level n=4 to Level n=1
Move from Level n=3 to Level n=2
Move from Level n=5 to Level n=4

Please explain thae way, how I can solve such questions and how to calculate (If needed) in future?

Answers

Answer:

The move from Level n=3 to Level n=2 has the long wavelength.

Explanation:

First, due to the selection rules, only transitions between adjacent levels are allowed, thus, only a transition between Level n=3 to Level n=2 or Level n=5 to Level n=4 are allowed. The two first options are wrong.

Second, analyzing the transition between Level n=3 to Level n=2 and the transition between Level n=5 to Level n=4 it is necessary to think in terms of the equation of the difference of energy for these type of transitions:

Δ[tex]E = \frac{h^{2}}{8.m.L}(n_{LUMO}^{2} -n_{HOMO}^{2} )[/tex] (1)

The difference in energy (ΔE) is directly proportional to the quadratic difference between the 'n' levels of transition. Therefore, If the transition occurs between smaller 'n' levels the difference of energy will be smaller too.

Also, the energy (ΔE) is inversely proportional to the wavelength (λ) so a smaller energy means a larger wavelength.

ΔE = c / λ (2)

Hence, the move from Level n=3 to Level n=2 has a long wavelength.

In order to calculate this wavelength is necessary to replace the data on equation (1) and (2).

An antibacterial ointment with 8% muciprocin as active ingredient contains _ mupirocin per 25 g of the ointment? g

Answers

Answer:

2 grams

Explanation:

The percentage in weight/weight is equivalent to the solute mass divided by the solution mass and multiplied by 100. For example, if you have 25g/100g then the percentage will be 25%. In this case you just need to multiply 8*25 and divided it by 100, so the answer is 2 grams of mupirocin per 25 g of the ointment.

The specific gravity of gasoline is approximately 0.70.

(a) Estimate the mass (kg) of 50.0 liters of gasoline.

(b) The mass flow rate of gasoline exiting a refinery tank is 1150 kg/min. Estimate the volumetric flow rate in liters/s.
(c) Estimate the average mass flow rate lbm/min delivered by a gasoline pump when the gas pump has a pump rate of 10 gallons per 2 minutes.

Answers

Answer:

a) mass=35 kg

b) volumetric flow rate= 27.37 [tex]\frac{liters}{s}[/tex]

c) average mass flow rate=29.21 [tex]\frac{lbm}{min}[/tex]

Explanation:

Specific gravity is defined as the relation between the density of one substance and the density of another reference substance (it is usual that water is used in this case). In this case the specific gravity is the relationship between the density of gasoline and the density of water. So:

[tex]0.7=\frac{densitygasoline}{densitywater}[/tex]

Now you can know the density of gasoline  with a simple mathematical operation and knowing that densitywater≅ 1000 [tex]\frac{kg}{m^{3} }[/tex]:

densitygasoline=0.7*densitywater

densitygasoline=0.7*1 [tex]\frac{kg}{liters}[/tex]

densitygasoline=0.7 [tex]\frac{kg}{liters}[/tex]

a)

Now that the density is known, you can calculate the mass in 50 liters of gasoline.

By definition of density, you know that in 1 liter there are 700 kg of gasoline. So using The Rule of Three( tool that allows you to solve problems based on proportions), you can calculate the mass in 50 liters:

[tex]\frac{700 kg}{1 liter} =\frac{mass}{50 liters}[/tex]

[tex]mass=\frac{0.7 kg* 50 liters}{1 liter}[/tex]

mass=35 kg

b)

First, it is convenient to change units: from kg / min to kg / seconds. For that you know that 1 min= 60 seconds. So:

[tex]\frac{1150 kg}{min} =\frac{1150 kg}{60 s}[/tex]

[tex]\frac{1150 kg}{min} =\frac{19.16 kg}{s}[/tex]

By definition, density is the relationship between the mass and volume of a substance: This is: [tex]density=\frac{mass}{volume}[/tex]

Applied to this case, it would be [tex]density=\frac{mass flow rate}{volumetric flow rate}[/tex]

The mass flow rate and the density of the gasoline are known, so you can calculate the volumetric flow rate:

[tex]volumetric flow rate= \frac{mass flow rate}{density gasoline}[/tex]

volumetric flow rate= 27.37 [tex]\frac{liters}{s}[/tex]

c)

You know that the pump rate is 10 gallons per 2 minute. This is:

[tex]\frac{10 gallons}{2 minutes} = 5 \frac{gallons}{minute}[/tex]

To calculate the average mass flow, you must relate this data to density and do the necessary unit conversions.  Conversions are done similarly to what was previously applied in the previous exercises.

You have to know that:

1 gallon = 3.785 liters1 kg= 2.205 kg

So:

[tex]\frac{5 gallons*3.785 liters}{1gallons*min} =\frac{18.925 liters}{min}[/tex]

You know that the density is [tex]density=\frac{mass flow rate}{volumetric flow rate}[/tex]

The volumetric flow rate (pump rate) and the density of the gasoline are known, so you can calculate the average mass flow rate:

average mass flow rate=density*volumetric flow rate

[tex]average mass flow rate=\frac{0.7 kg}{liters} *\frac{18.925 liters}{min}[/tex]

average mass flow rate=13.2475 [tex]\frac{kg}{min}[/tex]

Converting the units:

[tex]average mass flow rate=\frac{13.2475 kg*2.205 lbm}{1 kg*min}[/tex]

average mass flow rate=29.21 [tex]\frac{lbm}{min}[/tex]

A turbine operating adiabatically is fed with steam at 400 °C and 8.0 MPa at the rate of 1000 kg/h. Process steam saturated at 0.5 MPa is withdrawn from an intermediate location in the turbine at a rate of 300 kg/h, and the remaining steam leaves the turbine saturated at 0.1 MPa. The kinetic energies and differences in potential energies of all streams are negligible. What is the power output of the turbine?

Answers

Explanation:

We select the enthalpy of steam at state 1 at [tex]800^{o}C[/tex] and 8.0 MPa from the steam tables as follows.

                     [tex]h_{1}[/tex] = 3138 kJ/kg

Also, we select the enthalpy of steam at state 2 at 0.5 MPa from the steam tables as follows.

                     [tex]h_{2}[/tex] = 2748.6 kJ/kg

At state 3 also, from the steam tables at state 3 at 0.1 MPa.

                     [tex]h_{3}[/tex] = 2675.1 kJ/kg

Hence, calculate the mass flow rate at state 3 as follows.

                  [tex]m_{3} = m_{1} - m_{2}[/tex]

                              = 1000 kg/h - 300 kg/h

                              = 700 kg/h

Now, we will calculate the power output of the turbine as follows.

                 [tex]W_{r} = m_{1}(h_{1} - h_{2}) + m_{3}(h_{2} - h_{3})[/tex]

                             = 1000 kg/h (3138 kJ/kg - 2748.6 kJ/kg) + 700 kg/h (2784.6 kJ/kg - 2675.1 kJ/kg)

                            = 440850 kJ/h

It is known that 1 kJ/h = 0.000278 kW.

Therefore,        [tex]440850 kJ/h \times \frac{0.000278 kW}{1 kJ/hr}[/tex]    

                          = 122.56 kW

Thus, we can conclude that the power output of the turbine is 122.56 kW.

Suppose the gas above the soda in a bottle of soft drink
ispure CO2 at a pressure of 2atm. Calculate [CO2] at 25
degreeC.
Henry's law at 25C of CO2 is 0.033363 (mol/L*atm).

Answers

Answer: The concentration of carbon dioxide in bottle of soft drink is [tex]6.6\times 10^{-2}mol/L[/tex]

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{CO_2}=K_H\times p_{liquid}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]0.033363mol/L.atm[/tex]

[tex]p_{CO_2}[/tex] = partial pressure of gas in a bottle of soft drink = 2 atm

Putting values in above equation, we get:

[tex]C_{CO_2}=0.033363mol/L.atm\times 2atm\\\\C_{CO_2}=0.066726mol/L=6.6\times 10^{-2}[/tex]

Hence, the concentration of carbon dioxide in bottle of soft drink is [tex]6.6\times 10^{-2}mol/L[/tex]

A mixture of Fe2O3 and FeO was found to contain 72.00% Fe
bymass. What is the mass of Fe2O3 in 0.500 g of this mixture...how
doi work this..i dont even know what to look for as the given
asneeded?

Answers

Answer:

The mass of  Fe₂O₃ in 0.500 g of mixture is 0.367 g.

Explanation:

First off, we know that 72% of the mass of the mixture is iron. The information also tells us that the remaining 28% of the mass is oxygen.

Now we calculate the total mass of iron and the total mass of oxygen in the mixture:

0.500 g * 0.72 = 0.360 g of Fe0.500 g * 0.28 = 0.140 g of O

With the mass of each element we can calculate the number of moles of each atom, using the atomic weight:

0.360 g Fe * 1 mol / 55.845 g = 0.00645 moles of Fe

0.140 g O * 1 mol / 16 g = 0.00875 moles of O

The number of moles of Fe in the mixture is equal to the number of moles of FeO plus two times the number of moles of Fe₂O₃:

0.00645 = [tex]2*n_{Fe2O3} +n_{FeO}[/tex]             eq A

The number of moles of O in the mixture is equal to the number of moles of FeO plus three times the number of moles of Fe₂O₃:

0.00875 = [tex]3*n_{Fe2O3} +n_{FeO}[/tex]               eq B

So now we have a system of two equations and two unknowns, we solve for [tex]n_{Fe2O3}[/tex]:

From eq A:

[tex]n_{FeO3}=0.00645-2*n_{Fe2O3}[/tex]

Replacing in  eq B:

[tex]0.00875=3*n_{Fe2O3} + (0.00645-2*n_{Fe2O3})\\0.00230=n_{Fe2O3}[/tex]

Now we just need to convert moles of Fe₂O₃ into grams, using the molecular weight:

0.00230 moles * 159.66 g/mol = 0.367 g Fe₂O₃

How many atoms of Phosphorus (symbol P) are in 45.67g of P? (Your answer should just be a number. Do not include atoms as the unit. For scientific notation use the "^" symbol before the exponent. For example 2.54 x 10^5)

Answers

Find the Ar of P, then apply the moles = mass divided by Mr. Then do mr x 45.67 x 6.02 x 10^23 and that should give you the answer.

Answer:

8.882 × 10^23

Explanation:

The molar mass of phosphorus is 30.97 g/mol. The moles of P corresponding to 45.67 g are:

45.67 g × (1 mol/30.97 g) = 1.475 mol

In 1 mole of P, there are 6.022 × 10²³ atoms of P (Avogadro's number). The number of atoms of P in 1.475 moles are:

1.475 mol × (6.022 × 10²³ atoms/1 mol) = 8.882 × 10²³ atoms

Which of the following statements is not true regarding molecular orbital theory? O Bonding molecular orbitals have lower energy than the atomic orbitals from which they are formed. O Molecular orbitals are formed by the combination of atomic orbitals. Two atomic orbitals combine to form one molecular orbital. O Molecular orbitals are regions of a molecule where the electrons are most likely to be found.

Answers

Answer:

Option c, Two atomic orbitals combine to form one molecular orbital

Explanation:

Molecular orbitals are formed by linear combination of atomic orbitals.

Some of the important facts of molecular orbital theories are as follows:

No. of the molecular orbitals formed are equal to the no. of atomic orbitals participated.Half of the molecular orbitals are bonding molecular orbitals and half of the molecular orbitals are anti bonding molecular orbitals.Anti bonding molecular orbitals have energy higher than participating atomic orbitals. Bonding molecular orbitals have energy lower than participating atomic orbitals.Molecular orbitals are that region in the molecule where electrons are most likely to found.

So, among given, option c which is 'atomic orbitals combine to form one molecular orbital' is incorrect.

Final answer:

All the statements provided are true with regards to Molecular Orbital Theory, except for 'Two atomic orbitals combine to form one molecular orbital.' In fact, two atomic orbitals combine to form two molecular orbitals, one being a lower-energy bonding molecular orbital and the other one being a higher-energy antibonding orbital.

Explanation:

In the context of Molecular Orbital Theory, three out of the four statements made in the question are correct. Namely, molecular orbitals are indeed formed by the combination of atomic orbitals, and they represent regions of a molecule where electrons are most likely to be found. Additionally, bonding molecular orbitals do have lower energy than the atomic orbitals from which they are formed.

However, the statement suggesting that 'Two atomic orbitals combine to form one molecular orbital' is not accurate. In reality, when two atomic orbitals combine, they form two molecular orbitals: one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital.

This nuanced concept is a cornerstone of MO theory and helps explain a range of phenomena, from the paramagnetism of the oxygen molecule to violations of the octet rule. Understanding this allows us to further appreciate the differences between bonding and antibonding orbitals and their roles in the structure and stability of molecules.

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A gas mixture of 20 moles nitrogen and 70 moles hydrogen is fed to a reactor in which ammonia is produced. Assuming that the reaction proceeds to completion, determine the composition of the exit gas stream both in mole and mass %. If 75% of nitrogen gets converted, what will be the corresponding fractions?

Answers

Answer:

Complete reaction

mole %: 0% N₂; 20% H₂; 80% NH₃

Gram %: 0% N₂; 2,9% H₂; 97,1% NH₃

75% reaction:

mole%: 8,3% N₂; 41,7% H₂; 50% NH₃

Gram: 20% N₂; 7,2% H₂; 72,8% NH₃

Explanation:

The global reaction in the reactor is:

N₂(g) + 3 H₂ (g) → 2 NH₃ (g)

For a complete reaction of 20 moles of N₂(g) you need:

20 moles N₂ ₓ [tex]\frac{3 mol H_2}{1 mol N_2}[/tex] = 60 moles of H₂(g)

So, 10 moles of H₂(g) will stay in the end.

The moles produced of NH₃ are:

20 moles N₂ ₓ [tex]\frac{2 mol NH_3}{1 mol N_2}[/tex] = 40 moles of NH₃(g)

Thus, the final composition in moles is:

0 moles N₂; 10 moles H₂; 40 moles of NH₃

The mole % is:

0% N₂; ¹⁰/₅₀ ₓ 100 = 20% H₂; ⁴⁰/₅₀ ₓ100 = 80% NH₃

In grams you have:

10 moles H₂ [tex]\frac{2,02 g}{1 mol}[/tex] = 20,2 g of H₂

40 moles NH₃ [tex]\frac{17,03 g}{1 mol}[/tex] = 681,2 g of NH₃

0% N₂; [tex]\frac{20,2 g}{701,4}[/tex] ₓ 100 = 2,9% H₂; [tex]\frac{681,2 g}{701,4}[/tex] ₓ100 = 97,1% NH₃

With a 75% of conversion you have that the moles produced of NH₃ are:

40 moles of NH₃(g) × 75% = 30 moles of NH₃

The necessary moles to produce these moles of NH₃ are:

30 moles NH₃ ₓ [tex]\frac{3 mol H_2}{2 mol NH_3}[/tex] = 45 moles of H₂(g)

So, 25 moles of H₂(g) will stay in the end.

30 moles NH₃ ₓ [tex]\frac{1 mol N_2}{2 mol NH_3}[/tex] = 15 moles of N₂(g)

So, 5 moles of H₂(g) will stay in the end.

Thus, the final composition in moles is:

5 moles N₂; 25 moles H₂; 30 moles of NH₃

The mole % is:

⁵/₆₀ ₓ 100 =8,3% N₂; ²⁵/₆₀ ₓ 100 = 41,7% H₂; ³⁰/₆₀ ₓ100 = 50% NH₃

In grams you have:

5 moles N₂ [tex]\frac{28 g}{1 mol}[/tex] = 140 g of N₂

25 moles H₂ [tex]\frac{2,02 g}{1 mol}[/tex] = 50,5 g of H₂

30 moles NH₃ [tex]\frac{17,03 g}{1 mol}[/tex] = 510,9 g of NH₃

[tex]\frac{140 g}{701,4}[/tex] ₓ 100 =20% N₂; [tex]\frac{ 50,5g}{701,4}[/tex] ₓ 100 = 7,2% H₂; [tex]\frac{510,9 g}{701,4}[/tex] ₓ100 = 72,8% NH₃

I hope it helps!

If 37.1 mL AgNO3 solution reacts with excess potassium chloride solution to yield 1.56 g of AgCl precipitate, what is the molarity of silver ion in the original solution? AgCl = 143.3 g/mol Enter your answer in decimal format with three decimal places and no units.

Answers

Answer:

Molarity of silver ion: 0.296 M

Explanation:

Reaction:

AgNO₃ + KCl → AgCl↓ + K⁺ + NO₃⁻

From the reaction, we know that the moles of AgCl produced will be the same as the moles of initial silver.

First, let´s calculate the number of moles of AgCl produced:

1.56 g AgCl was produced, that is, (1.56 g AgCl * 1 mol AgCl/143.3 g AgCl) 0.011 moles AgCl.

The moles of silver ion present in the original solution was 0.011 mol. Since this number of moles was present in a 37.1 ml solution, then, in 1000 ml:

moles of silver ion per liter = 1000 ml * 0.011 mol / 37.1 ml = 0.296 mol

Molarity of silver ion = 0.296 M

Final answer:

The molarity of silver ion in the original AgNO3 solution is 0.293 M.

Explanation:

To determine the molarity of silver ion in the original AgNO3 solution when 37.1 mL reacts to yield 1.56 g of AgCl, we first need to calculate the moles of AgCl produced using the molar mass of AgCl.

The molar mass of AgCl is 143.3 g/mol. So, the moles of AgCl formed are calculated as follows:

(1.56 g AgCl) / (143.3 g/mol) = 0.010878 mol AgCl

Since AgNO3 reacts with KCl in a 1:1 mole ratio to produce AgCl, the moles of AgNO3 that reacted is also 0.010878 mol. To find the molarity of AgNO3, we divide the moles by the volume of solution in liters:

(0.010878 mol AgNO3) / (0.0371 L) = 0.293 M

explain why sodium (metal) is soft and can be bend, whereas NaCl is hard and brittle.

Answers

Answer:

Sodium chloride has a crystalline face-centered  cubic structure whereas metallic sodium body-centered  cubic structure.

Explanation:

Hello, atomic arrangements provide the molecules different features and behaviors, since the sodium metal has a body-centered cubic structure (https://en.wikipedia.org/wiki/Cubic_crystal_system#/media/File:Cubic-body-centered.svg) the lack of inner atoms, allows the material to be soft and bendable. On the other hand the compacted sodium chloride's face-centered  cubic structure (https://en.wikipedia.org/wiki/Cubic_crystal_system#/media/File:Cubic-face-centered.svg), provides it a crystalline structure which is hard and brittle since the atoms are closer.

Best regards!

The Prandtl number, Pr, is a dimensionless group important in heat transfer. It is defined as Pr = Cp*mu/k where Cp is the heat capacity of a fluid, mu is the fluid viscosity, and k is the fluid thermal conductivity. For a given fluid, Cp = 0.58 J/(g* deg C), k = 0.28 W/(m * deg C), and mu = 1934 Ibm / (ft * h). Determine the value of the Prandtl number for this fluid.

Answers

Answer:

The value of the Prandlt number for this fluid is: 1656.04

Explanation:

As it is stated in the problem : Pr = Cp*mu/k

where:

cp: heat capacity of the fluid

mu: viscosity of the fluid

k: thermal conductivity of the fluid

Now for a given fluid we have

cp= 0.58 J/(g* deg C)

mu=1934 Ibm / (ft * h)

k = 0.28 W/(m * deg C)

If we put these values in the ecuation of the Prandlt number we have:

Pr = (0.58 J/(g* deg C)) × (1934 Ibm / (ft * h) / 0.28 W/(m * deg C)) =

As we can see we have to convert the units so we can operate all the values in the same units of measurement and then cancel them so as to obtain a dimensionless result.

Converting the value of: mu = 1934 Ibm / (ft * h)

1 ft= 0.3048 m

1 h= 3600 s

1 lbm= 453.59 g

mu= 1934 Ibm / (ft * h) × (453,59 g/ lbm) × (1h/3600 s) × (1 ft/0.3048 m) = 799.47 g/ (m *s).

Pr = (0.58 J/(g* deg C)) × (799.47 g/ (m *s) / 0.28 W/(m * deg C)) = 1656.04

A chemist prepares a solution of nickel(II) chloride (NiCly) by measuring out 61.3 umol of nickel(II) chloride into a 200 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in umol/L of the chemist's nickel(II) chloride solution. Round your answer to 3 significant digits. x 5 ? Explanation Check 2009 Mew Econ Aires Reserved. Terms of Use

Answers

Final answer:

The concentration of the nickel(II) chloride solution is determined by dividing the amount of substance (in umol) by the volume of the solution (in liters). This results in a concentration of 306.5 umol/L.

Explanation:

The concentration of a solution is determined by the formula c = n/V, where 'c' is the concentration, 'n' is the amount of substance (in moles), and 'V' is the volume of the solution (in liters).

In this case, the chemist has used 61.3 umol of nickel(II) chloride (NiCl2), and the total volume of the solution is 200 ml. We need to convert this volume to liters, giving us 0.2 L. Now, using the formula, we calculate: c = 61.3 umol / 0.2 L, which equals 306.5 umol/L.

Make sure to always convert the units appropriately before you substitute the values into the formula.

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Give the ΔH value for the combustion of butane as shown in the reaction 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)+5315 kJ.

Express your answer using four significant figures. If the value is positive, do not include the + sign in your answer.

Answers

Final answer:

The ΔH value for the combustion of butane is given as 5315 kJ for the reaction as written, which is for 2 moles of butane. To find the enthalpy of combustion per mole, divide this number by 2, yielding -2658 kJ/mol.

Explanation:

The student asked to give the ΔH value for the combustion of butane as shown in the reaction 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) + 5315 kJ. The ΔH value for the reaction is given as +5315 kJ, indicating the amount of energy released during the combustion of butane.

Now, the enthalpy of combustion is typically expressed on a per mole basis, referring to the amount of heat released when one mole of a substance is burned. Since the reaction above shows the combustion for 2 moles of butane, we can calculate the enthalpy of combustion per mole by dividing the total energy by 2. So, the enthalpy of combustion for one mole of butane would be 5315 kJ / 2 = 2657.5 kJ/mol, which is typically reported in kilojoules per mole (kJ/mol).

Expressed with four significant figures, the enthalpy of combustion per mole of butane would be -2658 kJ/mol (the negative sign indicates that energy is released).

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