A solution contains 0.45 M hydrofluoric acid (HF; KA = 6.8 X 10−4). Write the dissociation reaction. Determine the degree of ionization and the pH of the solution

Answer: The mass of solid NaOH required is 80 g

Explanation:

Equivalent weight is calculated by dividing the molecular weight by n factor. The equation used is:

[tex]\text{Equivalent weight}=\frac{\text{Molecular weight}}{n}[/tex]

where,

n = acidity for bases = 1 (For NaOH)

Molar mass of NaOH = 40 g/mol

Putting values in above equation, we get:

[tex]\text{Equivalent weight}=\frac{40g/mol}{1eq/mol}=40g/eq[/tex]

Normality is defined as the umber of gram equivalents dissolved per liter of the solution.

Mathematically,

[tex]\text{Normality of solution}=\frac{\text{Number of gram equivalents} \times 1000}{\text{Volume of solution (in mL)}}[/tex]

Or,

[tex]\text{Normality of solution}=\frac{\text{Given mass}\times 1000}{\text{Equivalent mass}\times \text{Volume of solution (in mL)}}[/tex]         ......(1)

We are given:

Given mass of NaOH = ?

Equivalent mass of NaOH = 40 g/eq

Volume of solution = 400 mL

Normality of solution = 5 eq/L

Putting values in equation 1, we get:

[tex]5eq/L=\frac{\text{Mass of NaOH}\times 1000}{40g/eq\times 400mL}\\\\\text{Mass of NaOH}=80g[/tex]

Hence, the mass of solid NaOH required is 80 g

Answer:

Significant hydrogen bonding is possible in [tex]NH_{2}Cl[/tex]

Explanation:

The only substance from above in which significant hydrogen bonding possible is chloramine ( NH2Cl )

Organic compounds are substances which contain carbon and hydrogen. Some few groups of organic compounds include:

So therefore, the only substance from above in which significant hydrogen bonding possible is chloramine ( NH2Cl )

Learn more about organic compounds:

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Answer:

4057.85 g/mol

Explanation:

Hello, the numerical procedure is shown in the attached file.

- In this case, since we don't have the density of the protein, we must assume that the volume of the solution is solely given by the benzene's volume, in order to obtain the moles of the solute (protein).

-Van't Hoff factor is assumed to be one.

Best regards.

To calculate the molar mass of the protein, use the formula M = (RT) / (V * P), where M is the molar mass, R is the ideal gas constant, T is the temperature, V is the volume, and P is the osmotic pressure.

To calculate the molar mass of the protein, we can use the formula:

M = (RT) / (V * P)

Where M is the molar mass, R is the ideal gas constant (0.0821 L.atm/mol.K), T is the temperature in Kelvin (24.4 + 273 = 297.4 K), V is the volume in liters (44 mL / 1000 = 0.044 L), and P is the osmotic pressure in atm (50.9 torr / 760 = 0.067 atm).

Substituting the values into the formula:

M = (0.0821 * 297.4) / (0.044 * 0.067) = 25567.14 g/mol

Therefore, the molar mass of the protein is approximately 25567.14 g/mol.

The energy of the photons emitted by the AM and FM radio stations can be calculated using Planck's equation (E = hf). The calculated energies show that the FM radio station's photons have higher energy than the AM radio station's photons.

To calculate the energy of the photons emitted by radio stations, we use the formula:

E = hf, where E is the energy, h is Planck's constant (6.62607015 × 10⁻³⁴ Js) and f is the frequency.

(For the sake of this calculation, we are going to convert the frequency from kHz and MHz to Hz.)

Comparing these energy values, it's clear that the FM radio station's emitted photons have a higher energy than the AM radio station's photons, due to its higher frequency.

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Answer:

Exit velocity of air is 96.43 m/s.

Explanation:

Given that

[tex]V_1=400\ m/s[/tex]

[tex]T_1=25C[/tex]

[tex]T_2=100C[/tex]

For air

[tex]C_p=1.005\ KJ/kg.K[/tex]

Now from first law of thermodynamics for open system at steady state

[tex]h_1+\dfrac{V_1^2}{2000}+Q=h_2+\dfrac{V_2^2}{2000}+w[/tex]

For diffuser

[tex]h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}[/tex]

We know that

[tex]h=C_pT[/tex]

[tex]h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2000}[/tex]

[tex]1.005\times 25+\dfrac{400^2}{2000}=1.005\times 100+\dfrac{V_2^2}{2}[/tex]

[tex]V_2=96.43\ m/s[/tex]

So the exit velocity of air is 96.43 m/s.

Answer:

pH of acetic acid solution is 2.88

Explanation:

[tex]pK_{a}=4.75[/tex]

or, [tex]-log(K_{a})=4.75[/tex]

or, [tex]K_{a}=10^{-4.75}=1.78\times 10^{-5}[/tex]

We have to construct an ICE table to determine concentration of [tex]H^{+}[/tex] and corresponding pH. Initial concentration of acetic acid is 0.1 M.

[tex]CH_{3}COOH\rightleftharpoons CH_{3}COO^{-}+H^{+}[/tex]

I(M):  0.1                            0                    0

C(M): -x                              +x                 +x

E(M): 0.1-x                         x                    x

So, [tex]\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}=K_{a}[/tex]

or, [tex]\frac{x^{2}}{0.1-x}=1.78\times 10^{-5}[/tex]

or, [tex]x^{2}+(1.78\times 10^{-5}\times x)-(1.78\times 10^{-6})=0[/tex]

So, [tex]x=\frac{-(1.78\times 10^{-5})+\sqrt{(1.78\times 10^{-5})^{2}+(4\times 1\times 1.78\times 10^{-6})}}{2\times 1}[/tex](M)

so, [tex]x=1.33\times 10^{-3}M[/tex]

Hence [tex]pH=-log[H^{+}]=-log(1.33\times 10^{-3})=2.88[/tex]

Answer:

The intermolecular forces between the solute and solvent.

Explanation:

When you are heating a solvent, the intermolecular forces are reduced because the distances between molecules are large. Thus, in a solution where solvent is hot the intermolecular forces between solute and solvent are lower than those solutions where solvent is in room temperature.

The covalent bonds do not change because this mean a chemical reaction that doesn't occur in a solution.

Usually solid solutes melts in a higher temperature than boiling point in solvents. Thus, a compound normally doesn't melt in a hot solvent.

I hope it helps!

When a compound is dissolved in hot ethanol during recrystallization, the intermolecular forces between the solute and solvent are changing on a molecular level. The process of solution formation involves the formation of new intermolecular forces between the solute and solvent molecules, allowing the compound to dissolve in the solvent.

Intermolecular forces are attractive or repulsive forces between molecules. They include London dispersion forces, dipole-dipole interactions, and hydrogen bonding. These forces influence physical properties like boiling points and solubility in molecular substances. The intermolecular forces between the solute and solvent are changing on a molecular level when a compound is dissolved in hot ethanol during recrystallization.

When the compound is dissolved in the hot solvent, the solute-solvent interactions which is also known as solvatio then naturally takes place. In addition to this, these interactions involve the formation of new intermolecular forces between the solute and solvent molecules, which are nearly as strong as the intermolecular forces within the solute and solvent alone. This favorable solution formation process allows the compound to dissolve in the solvent.

Answer: The weight fraction of ethanol in the mixture is

Explanation:

We are given:

Mass of water = 1 kg = 2.205 lb   (Conversion factor:  1 kg = 2.205 lb)

Mass of ethanol = 1.9 lb

Mass of ethyl acetate = 4.6 lb

Mass of mixture = [2.205 + 1.9 + 4.6] = 8.705 lb

To calculate the percentage composition of ethanol in mixture, we use the equation:

[tex]\%\text{ composition of ethanol}=\frac{\text{Mass of ethanol}}{\text{Mass of mixture}}\times 100[/tex]

Mass of mixture = 8.705 lb

Mass of ethanol = 1.9 lb

Putting values in above equation, we get:

[tex]\%\text{ composition of ethanol}=\frac{1.9lb}{8.705lb}\times 100=21.8\%[/tex]

Hence, the weight fraction of ethanol in the mixture is 21.8 %

Answer:

Due to the low difference of electronegativity in carbon and nitrogen, these will form a covalent bond, and this covalent union of C-N is one of the most common bonds in the organic chemistry and biological systems. This type of bonds can be found in amines, amides, imines, etc. Also, a single atom of C and a single atom of N can form a cyanide, that is a triple covalent bond between the atoms.

Answer:

The expression to calculate the mass of the reactant is [tex]m = \frac{1.080kJ}{31.2kJ/g}[/tex]

Explanation:

The amount of heat released is equal to the amount of heat released per gram of reactant times the mass of the reactant. To keep to coherence between units we need to transform 1,080 J to kJ. We do so with proportions:

[tex]1,080J.\frac{1kJ}{10^{3}J } =1.080kJ[/tex]

Then,

[tex]1.080kJ=31.2\frac{kJ}{g} .m\\m = \frac{1.080kJ}{31.2kJ/g}[/tex]

Answer:

1. The expression is: [tex]m=\frac{E}{\Delta _rH}[/tex]

2. The computed mass is: [tex]m=0.0346g[/tex]

Explanation:

Hello,

In this case, we know the so called enthalpy of reaction whose symbol and value is shown below:

[tex]\Delta _rH=31.2\frac{kJ}{g}[/tex]

In addition, we know that the energy released by the involved reactant is:

[tex]E=1080 J[/tex]

Therefore, the expression to compute the required mass, based on the given units is:

[tex]m=\frac{E}{\Delta _rH}[/tex]

Finally, the computed mass turns out:

[tex]m=\frac{1080J*\frac{1kJ}{1000J} }{31.2\frac{kJ}{g}} \\m=0.0346g[/tex]

Best regards.

Answer:

[tex]5.5*10x^{-9}m[/tex]

Explanation:

As you have the diameter of the sphere in nanometers (nm), you need to use de conversion factor to find the diameter in meters (m):

First you should put the quantity that you want to convert with its respective units:

[tex]Diameter=5.5nm[/tex]

Then you put the conversion factor, always you should put the same unit that you want to convert in the denominator:

Diameter = [tex]5.5nm*\frac{1*10^{-9}m}{1nm}[/tex]

And finally, you should multiply and/or divide the quantities:

Diameter = [tex]5.5*10^{-9}m[/tex]

Answer:

Dispersion system is a system in which certain particles are scattered in a continuous liquid or solid medium. The two phases present in this system are the dispersed particles and dispersion medium. These phases may or may not be present in the same state.

In a dispersion system, the particles that are dispersed are known as the dispersed particles and the medium in which the particles are dispersed is known as the dispersion medium.

Answer:

There are three major sources of errors in numerical method, namely human error, truncation error and round off error, but here we have to discuss only two sources-

Answer:

sdfsdfsdf

Explanation:dsf

Answer: The correct answer is Option B.

Explanation:

The chemical equation for the reaction of lead (II) nitrate and ammonium chloride follows:

[tex]Pb(NO_3)_2(aq.)+2NH_4Cl(aq.)\rightarrow PbCl_2(s)+2NH_4NO_3(aq.)[/tex]

Ionic form of the above equation follows:

[tex]Pb^{2+}(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)+2NH_4^+(aq.)+2NO_3^-(aq.)[/tex]

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

[tex]Pb^{2+}(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)[/tex]

Solubility rules:

Hence, the correct answer is Option B.

Answer:

49.3% 81Br and 50.7% 79Br or [tex]\frac{493}{1000}[/tex] 81Br and [tex]\frac{507}{1000}[/tex] 79BR

Step-by-step explanation:

To get the fraction-of-occurrences of the isotopes we must write the following equation. x is the isotopic abundance of 81Br, we can use 1 - x to get the isotopic abundance of 79Br.

(78.918)(1 - x) + (80.916)(x) = 79.903

78.918 - 78.918x + 80.916x = 79.903

1.998x = 0.985

x = 0.493

0.493 × 100 = 49.3% 81Br

100 - 49.3 = 50.7% 79Br

49.3% 81Br and 50.7% 79Br or [tex]\frac{493}{1000}[/tex] 81Br and [tex]\frac{507}{1000}[/tex] 79BR

Answer:

There are 1.041×10²³ atoms in 4.20g of Magnesium.

Explanation:

To find the amount of atoms in 4.20 g of Magnesium we need de molar mass of Mg: 24.305 g/mol

According to Avogadro number there are 6.022×10²³ particles in 1 mol, so the number of atoms of Mg is:

[tex]4.20 g Mg*\frac{1 molMg}{24.305gMg} *\frac{6.022*10^{23}atoms }{1 mol Mg} = 1.041*10^{23}atoms Mg[/tex]

Answer: The number of atoms found in given amount of magnesium is [tex]1.042\times 10^{23}[/tex]

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of magnesium = 4.20 g

Molar mass of magnesium = 24.31 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of magnesium}=\frac{4.20g}{24.31g/mol}=0.173mol[/tex]

According to mole concept:

1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms

So, 0.173 moles of an element contains = [tex]0.173\times 6.022\times 10^{23}=1.042\times 10^{23}[/tex] number of atoms

Hence, the number of atoms found in given amount of magnesium is [tex]1.042\times 10^{23}[/tex]

Answer:

The final dilution is 1:400

Explanation:

Let's analyze what we are told: we have an initial 1:5 dilution of protein lysate. This means that the initial solution (stock solution) was diluted 5 times. Then, from this dilution the student prepared another dilution taking 2 mL of the first dilution in 8 mL of water. This is the same as saying we took 1 mL of first dilution in 4 mL of water (the ratio is the same), so we now have a second 1:4 dilution of the first dilution (1:5). Finally, the student made a third 1:20 dilution, this means that the second dilution was further diluted 20 times.

So, to calculate the final dilution of protein lysate, we have to multiply all the dilution factors of every dilution prepared: in this case we have a final dilution of 1:20, this means we have a factor dilution of 20. But it was previously diluted 4 times, so we have a factor dilution of 20×4 = 80. However, this dilution was also previously diluted 5 times, so the new dilution factor is 80 × 5 = 400

This means that the final dilution of the compound was diluted a total of 400 times compared to the initial concentration of stock solution.

Answer:

a. 0.14042

b. 0.221

c. equal to a.

5) 0.579 L

3) 8.1e6 ms

Explanation:

Hello,

In the attached photo you'll find the numerical procedure to get the results.

Best regards.

Answer:

It is necessary 1 meter of thickness.

Explanation:

You need to know the equation (P/A)=(k*T)/(L), where P/A is the heat flux, k is the thermal conductivity, T the temperature drop, and L the unknown thickness. So if rearrange the variables you will have that L = (k*T)/(P/A), and substituting the terms L = (0.2*400)/(200) = 1 meter

Final answer:

The necessary thickness of insulation to achieve a temperature drop of 400°C with a heat flux of 200 W/m^2, given a thermal conductivity of 0.2 W/m°C, is 0.4 meters.

Explanation:

The necessary thickness of insulation for a specific temperature drop and heat flux. By using the equation of thermal conduction Q = (kAΔT)/d, where Q is the heat flux, k is the thermal conductivity, A is the area, ΔT is the temperature difference, and d is the thickness, we can solve for the thickness d.

To start, we know that k (thermal conductivity) is 0.2 W/m°C, the temperature drop ΔT is 400°C, and the heat flux Q is 200 W/m^2. Plugging these values into the equation gives us d = (kAΔT)/Q. Since the area A is not specified, it will cancel out in this case, allowing us to solve for d directly. Rearranging the equation yields d = (kΔT)/Q, which equals (0.2 W/m°C * 400°C)/200 W/m^2, resulting in d = 0.4 m. Therefore, the thickness required for this insulation is 0.4 meters.

Answer:

% of mass of electrons in C = 0.0272

Explanation:

Atomic no. of carbon = 6

So, no. of electron = 6

Mass of an electron = [tex]\frac{1}{1836} amu[/tex]

Mass of 6 electrons = [tex]6 \times \frac{1}{1836} = 0.003268 amu[/tex]

Mass of C = 12.0107 u

% of mass of electrons in C = [tex]\frac{Mass\ of\ all\ electrons}{Mass\ of\ C}\times 100[/tex]

% of mass of electrons in C = [tex]\frac{0.003268}{12.0107}\times 100=0.0272 \%[/tex]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression [tex]d=\frac{m}{V}[/tex] that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

[tex]d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL[/tex]

[tex]d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL[/tex]

[tex]d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL[/tex]

[tex]d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL[/tex]

The problem says that all the samples have the same mass, so:

[tex]m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m[/tex]

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

[tex]V_{lithium}=\frac{m}{d_{lithium}}[/tex]

[tex]V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m[/tex]

[tex]V_{lithium}=1.88\frac{mL}{g}*m[/tex]

[tex]V_{gold}=\frac{m}{d_{gold}}[/tex]

[tex]V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m[/tex]

[tex]V_{gold}=5.18*10^{-2}\frac{mL}{g}*m[/tex]

[tex]V_{aluminum}=\frac{m}{d_{aluminum}}[/tex]

[tex]V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m[/tex]

[tex]V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m[/tex]

[tex]V_{lead}=\frac{m}{d_{lead}}[/tex]

[tex]V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m[/tex]

[tex]V_{lead}=8.85*10^{-2}\frac{mL}{g}*m[/tex]

If we assume m = 1g, we find that:

[tex]V_{lithium}=1.88mL[/tex]

[tex]V_{gold}=5.18*10^{-2}mL[/tex]

[tex]V_{aluminum}=3.70*10^{-1}mL[/tex]

[tex]V_{lead}=8.85*10^{-2}mL[/tex]

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

Explanation:

Orifice meters require to compensate for pressure and temperature when one uses these meters to measure the steam or the gas flow in the pipes with the variable operating pressure as well as temperature conditions.

Normally chemists do not have online density measurement tool and thus to avoid complications, the chemists consider density as constant parameter.

In the steam or the gas flow measurement, density of steam or gas changes as the pressure and the temperature change. This significant  change in the density can affect accuracy of measured flow rate if the change is uncompensated and thus, this has to be fixed in order to avoid errors. Therefore, it is important to compensate for the pressure and the temperature when orifice is used to measure the steam or the gas flow.

Answer:

Few important points related to [tex]S_N2[/tex] reaction:

1. [tex]S_N2[/tex] is a one-step reaction that follows second order kinetics.

2. In [tex]S_N2[/tex] reaction, a transition state is formed in situ.

3. Strong nucleophiles like [tex]OH^- \ or\  CN^-[/tex] are used in case of bi-molecular nucleophilic substitution reaction.

Ethyl acetate can be prepared by a second-order nucleophilic substitution reaction between acetic acid and ethyl bromide.  

The reaction between acetic acid and ethyl bromide is drawn below:

In the SN2 reaction to create ethyl acetate, the nucleophile acetate ion attacks the alkyl bromide, leading to the formation of ethyl acetate and bromide ions.

Ethyl acetate can be prepared by an SN2 reaction. In this context, an alkylbromide and nucleophile initiate the reaction. Ethyl bromide (CH3CH2Br) would be the alkylbromide used in the reaction. The nucleophile, in this case, would be acetate ion (CH3COO-).

Ethyl bromide is a good leaving group and acetate ion is a strong nucleophile, which comes from acetic acid (ethanoic acid). In SN2 reactions, the nucleophile attacks the substrate, and the leaving group (Br- in this case) leaves, leading to a reverse configuration. In this reaction, the nucleophile (acetate ion) will attack the alkyl bromide, resulting in the formation of ethyl acetate, CH3COOCH2CH3, and bromide ions, which is a good leaving group.

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Answer:

There are 0.1125 g of O₂ less in 1 L of air at 14,000 ft than in 1 L of air at sea level.

Explanation:

To solve this problem we use the ideal gas law:

PV=nRT

Where P is pressure (in atm), V is volume (in L), n is the number of moles, T is temperature (in K), and R is a constant (0.082 atm·L·mol⁻¹·K⁻¹)

Now we calculate the number of moles of air in 1 L at sea level (this means with P=1atm):

1 atm * 1 L = n₁ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₁=0.04092 moles

Now we calculate n₂, the number of moles of air in L at an 14,000 ft elevation, this means with P = 0.59 atm:

0.59 atm * 1 L = n₂ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₂=0.02414 moles

In order to calculate the difference in O₂, we substract n₂ from n₁:

0.04092 mol - 0.02414 mol = 0.01678 mol

Keep in mind that these 0.01678 moles are of air, which means that we have to look up in literature the content of O₂ in air (20.95%), and then use the molecular weight to calculate the grams of O₂ in 20.95% of 0.01678 moles:

[tex]0.01678mol*\frac{20.95}{100} *32\frac{g}{mol} =0.1125 gO_{2}[/tex]

Answer:

The required volume is 1.6 x 10³mL.

Explanation:

When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:

C₁ . V₁ = C₂ . V₂

where,

C₁ and V₁ are the concentration and volume of the concentrated solution

C₂ and V₂ are the concentration and volume of the dilute solution

In this case, we want to find out V₁:

C₁ . V₁ = C₂ . V₂

[tex]V_{1} = \frac{C_{2}.V_{2}}{C_{1}} = \frac{1.5M \times1.11.10^{4}mL }{10.2M} =1.6\times10^{3} mL[/tex]

Answer:

12.4 × 10∧3 atoms

Explanation:

Given data:

moles of oxygen molecule= 1.0000 x 10-20 mol

atoms =?

Solution:

32 g O2 = 1 mol = 6.02 × 10∧23

1.0000 x 10∧-20 mol × 6.02 × 10∧23 × 2 = 12.4 × 10∧3 atoms

Answer:

a) m  = 2398256.64 Ib/hr

b) Q = 38433.6 ft^3/hr

c) J = = 898560 lb/h-ft2

Explanation:

a) To get the mass flow rate in lb/h we are going to use this formula:

m = ρνA

note: we have to change some units to reach to the final units

when m is the mass flow rate which we need to calculate

ρ is the density of water in lb/ft3 = 62.4 lbs/ft3 it is a constant numbers you can get it from anywhere

ν is the velocity of the water = 4 ft/s we need to change it to ft/h unit so  

when 1 ft/s = 3600 ft/hr (note : 3600 is a result of converting hr to sec 60min*60sec)

so  the velocity of the water = 4 ft/s * 3600 ft/hr= 14400 ft/hr

now , we need to get the area of the pipe in feet also when:

A = π*diameter * length

diameter = 3.068 inches

when 1 inch = 0.0833333333 feet so,

diameter in feet = 3.068 * 0.0833333333 =.300 0.2557ft

and length = 40 * 0.0833333333  = 3.333 ft

so the area of the pipe in feet^2 = 3.14*0.255*3.33 =2.669 ft^2

by substitution in the mass flow rate:

m = 62.4 lbs/ft3 * 14400 ft/hr *2.669 ft^2 = 2398256.64 Ib/hr

b) now to calculate volumetric flow rate in ft /h  we are going to use this formula:

Q = AV

when Q is  the volume flow rate

A is the cross-sectional area filled by water =2.669 ft^2

V is the average velocity of the water = 14400 ft/hr

so , by substitution:

Q = 2.669  ft^2* 14400 ft/hr= 38433.6 ft^3/hr

C) To calculate mass velocity (mass flux) in lb/h-ft  we are going to use this formula:

J = m / A

when J is the mass flux

and m is the mass flow rate which we calculated above = 2398256.64 Ib/hr

and A is the cross section area of the pipe which we calculated above and = 2.669 ft^2

so, by substitution:

J = 2398256.64 Ib/hr /  2.669 ft^2

 = 898560 lb/h-ft2

As the volume of confined gas decreases at the constant temperature, the pressure exerted by the gas increases .  The correct option is C.

Boyle–Mariotte law is a gas law, shows the relation between pressure and volume. With the increase in the volume, the pressure decreases.

The pressure exerted by the mass, is inversely proportional to the volume of the gas.

Thus, the correct option is C, increases.

Learn more about the Boyle's law

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Answer:

ghdtgfgrdvreeeegrwwggegteefewqrwefrwerrftrtdsfgyuytfgererdf

Explanation:

The given data is as follows.

     [tex]V_{1}[/tex] = 1.50 L,    [tex]T_{1}[/tex] = 159 K,

      [tex]P_{1}[/tex] = 5.00 atm,   [tex]V_{2}[/tex] = 0.2 L,

       [tex]T_{2}[/tex] = ?,       [tex]P_{2}[/tex] = 50.0 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{5.0 atm \times 1.50 L}{159 K} = \frac{5 atm \times 0.2 L}{T_{2}}[/tex]

            [tex]T_{2}[/tex] = 21.27 K

Thus, we can conclude that the final temperature is 21.27 K .

The correct chemical formula for Iron(III) hydroxide is Fe(OH)3, which is formed by combining one iron ion with a +3 charge (Fe3+) with three hydroxide ions (OH-).

The chemical formula for Iron(III) hydroxide is Fe(OH)3. This compound consists of an iron ion with a +3 charge (Fe3+) and three hydroxide ions (OH-). The number in parentheses after iron indicates the oxidation state of iron, which in this case is +3. The formula reflects the stoichiometry required to balance the charges, resulting in one Fe3+ for every three OH- ions to form the neutral compound. Therefore, the correct answer is Fe(OH)3.

The correct answer is d. Fe(OH)₃.