An ideal gas with MW = 30 is flowing at a steady rate through a circular pipe (cross-section area of 57.8 cm^2) at a temperature of 1.60 × 10^2 °C and pressure of 49.3 psig. The volumetric flow rate is 940 L/min, what is the molar flow rate (mol/s)?

Answer:

The energy transferred between samples of matter because of a difference in their temperatures is called a. heat.

Explanation:

The first law of thermodynamics establishes that when two bodies with different temperatures are put in contact they will find thermic equilibrium to a final temperature by transferring heat. Thus the correct answer is (a).

Thermochemistry is the study of the transformations of heat energy on the chemical reactions. Chemical kinetics is the study of the rate of chemical reactions. And temperature is the measure of the heat.

Answer: b. accepts [tex]H^+[/tex] ions from acids

Explanation:

According to Arrhenius concept, a base is defined as a substance which donates hydroxide ions [tex](OH^-)[/tex] when dissolved in water and an acid is defined as a substance which donates hydronium ions [tex](H_3O^+)[/tex] in water.

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates [tex]H^+[/tex] ions and a base is defined as a substance which accepts [tex]H^+[/tex] ions.

[tex]NH_3+H_2O\rightarrow NH_4^++OH^-[/tex]

Here water is donating [tex]H^+[/tex] ions, and thus act as acid and ammoia [tex]NH_3[/tex] is accepting [tex]H^+[/tex] ions from water and thus is a base.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

Thus base is a substance which accepts  [tex]H^+[/tex] ions from acids.

Explanation:

According to the Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the direction of equilibrium where there is less stress or disturbance.

For example, [tex]A + B \rightleftharpoons C + D[/tex]

So, when we increase the concentration of D in this reaction then disturbance will increase on the product side.

And, frequency of successful collisions between the reactant atoms will decrease. Hence, there will be less stress on reactant side.

As a result, equilibrium will shift in the backward direction that is, on the reactant side it will shift.

Final answer:

One photon of microwave radiation with a wavelength of 0.122m has an energy of approximately 1.63 x 10⁻²⁴ joules, calculated using Planck's equation (E=hc/λ).

Explanation:

The energy of a photon can be calculated using the equation E = hc/λ, where E is the energy of the photon in joules, h is Planck's constant (6.626 x 10⁻³⁴ J·s), c is the speed of light in a vacuum (3.00 x 10⁸ m/s), and λ is the wavelength of the photon in meters. Using the given wavelength of 0.122 meters for the microwave radiation, we can calculate the energy of one photon by substituting the values into the equation:

E = (6.626 x 10⁻³⁴ J·s)(3.00 x 10⁸ m/s) / 0.122 m

E = 1.63 x 10⁻²⁴ Joules per photon. Therefore, one photon of microwave radiation with a wavelength of 0.122m has an energy of approximately 1.63 x 10⁻²⁴ joules.

Explanation:

The given data is as follows.

           [tex]T_{1}[/tex] = [tex]80.1^{o}C[/tex] = (80 + 273) = 353 K

  Atmospheric pressure = 445 torr

   Heat of vaporization = 30.72 kJ/mol = [tex]30.72 kJ/mol \times \frac{1000 J/}{1 kJ}[/tex]

                                      = 30720 J/mol

Now, according to Clausius-Clapereryon equation,

              [tex]ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times \frac{1}{T_{2}} - \frac{1}{T_{1}}[/tex]

Putting the given values into the above equation as follows.

                   [tex]ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times \frac{1}{T_{2}} - \frac{1}{T_{1}}[/tex]

               [tex]ln (\frac{445}{760}) = \frac{-30720 J/mol}}{8.314 J/mol K} \times \frac{1}{T_{2}} - \frac{1}{353 K}[/tex]    

                   -0.5352 = [tex]-3694.9723 \times \frac{1}{T_{2}} - \frac{1}{353 K}[/tex]                  

                   [tex]\frac{1}{T_{2}}[/tex] = 0.002977

                      [tex]T_{2}[/tex] = 335.909 K

or,                             = [tex](335.909 - 273)^{o}C[/tex]

                                = [tex]62.909^{o}C[/tex]

Thus, we can conclude that benzene boils at [tex]62.909^{o}C[/tex] when the external pressure is 445 torr.

We have that for the Question "At what temperature does benzene boil when the external pressure is 445 torr" it can be said to be at

From the question we are told

Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1°C. At what temperature does benzene boil when the external pressure is 445 torr?

Generally the equation for the Pressure ratio  is mathematically given as

[tex]ln(p2/p1)=-(heat of vapour/R)*(1/t2-1/t1}\\\\Therefore\\\\ln(445/760)=-(30720/8.3k)*(\frac{1}{1/t2-1/353.1}\\\\T2=335.919k\\\\[/tex]

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Explanation:

The given data is as follows.

 [tex]P_{1}[/tex] = 1.5 atm,     [tex]V_{1}[/tex] = 20 L,  [tex]T_{1}[/tex] = (28 + 273) K = 301 K

   [tex]P_{2}[/tex] = 5 atm,     [tex]V_{2}[/tex] = ?,  [tex]T_{2}[/tex] = (50 + 273) K = 323 K

Formula to calculate the volume will be as follows.

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

Putting the given values into the above formula as follows.

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

        [tex]\frac{1.5 atm \times 20 L}{301 K}[/tex] = [tex]\frac{5 atm \times V_{2}}{323 K}[/tex]  

                 [tex]V_{2}[/tex] = 0.64 L

Thus, we can conclude that the change in volume of the balloon will be 0.64 L.

Answer:

The number of moles of calcium phosphate is 0.0485 mol. The mass is 15.1 g.

Explanation:

The balanced equation is:

[tex]3 Ca(OH)_{2} + 2 H_{3}PO_{4} => Ca_{3}(PO_{4})_{2} + 6H_{2}O[/tex]

As we can see, 3 moles of calcium hidroxide produced 1 mol of calcium phosphate. So the quantity of moles of Calcium posphate is:

[tex]10.8g Ca(OH)_{2}*\frac{1 molCa(OH)_{2}}{74.093gCa(OH)_{2}} *\frac{1mol Ca_{3}(PO_{4})_{2}}{3molCa(OH)_{2}} =0.0485 molCa_{3}(PO_{4})_{2} [/tex]

The mass of calcium phosphate in grams is:

[tex]0.0485 molCa_{3}(PO_{4})_{2}*\frac{310.176g}{mol} =15.1 g[/tex]

Answer: The volume of balloon at 100°C is 4.46 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where,

[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.

[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.

We are given:

[tex]V_1=3.50L\\T_1=20^oC=(20+273)K=293K\\V_2=?L\\T_2=100^oC=(100+273)K=373K[/tex]

Putting values in above equation, we get:

[tex]\frac{3.5L}{293K}=\frac{V_2}{373K}\\\\V_2=4.46L[/tex]

Hence, the volume of balloon at 100°C is 4.46 L

Answer: In the reaction rate law the rate is expressed in terms of concentrations of species. It is important to know how much time a reaction will take to complete itself. It depends on some factors. Temperature, concentration of component, catalyst and pressure. On increasing these factors the rate of reaction of a respective reaction increases. It doesn't depends upon reactor type.

Explanation:

The chemical reaction is as follows.

            [tex]CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O[/tex]

It is given that 2094 [tex]ft^{3}/hr[/tex]. And, it is known that 1 [tex]m^{3}/s[/tex] = 127133 [tex]ft^{3}/hr[/tex]

Hence, convert 2094 [tex]ft^{3}/hr[/tex] into [tex]m^{3}/s[/tex] as follows.

                [tex]\frac{2094 ft^{3}/hr}{127133 ft^{3}/hr} \times 1 m^{3}/s[/tex]

                  = [tex]0.0165 m^{3}/s[/tex]

As ideal gas equation is PV = nRT. So, calculate the number of moles as follows.

                   n = [tex]\frac{PV}{RT}[/tex]

                      = [tex]\frac{1 atm \times 0.0165 m^{3}}{0.0821 \times 298 K}[/tex]

                       = 0.673 mol/sec

According to the stoichiometry of the given reaction, 1 mol of methane reacts with 2 mol of oxygen.

So, 1 mol [tex]CH_{4}[/tex] = 2 mol [tex]O_{2}[\tex]

Hence, [tex]O_{2}[\tex] required theoretically = [tex]2 \times 0.673 mol/s[/tex] = 1.346 mol/s

Hence, air required theoretically = [tex]\frac{1.346}{0.21}[/tex] = 6.4095 mol/s.

Since, 6% of excess air is being supplied. Therefore, total air supplied will be calculated as follows.

                Total air supplied = [tex]6.4095 mol/s [1 + \frac{6}{100}][/tex]

                                              = 6.794 mol/s

Now, calculate the volume using ideal gas law equation as follows.

                            PV = nRT

           [tex]1 atm \times V = 6.794 mol/s \times 8.21 \times 10^{-5} Latm/K mol \6 times 298 K[/tex]

                           V = 0.166229 [tex]m^{3}/s[/tex]

Converting calculated volume into [tex]ft^{3}/hr[/tex] as follows.

                  1 [tex]m^{3}/s[/tex] = 127133 [tex]ft^{3}/hr[/tex]

So,        0.166229 [tex]m^{3}/s[/tex] = [tex]0.166229 m^{3}/s \times 127133 \frac{ft^{3}/hr}{1 m^{3}/s}[/tex]  

                                        = 21133.191 [tex]ft^{3}/hr[/tex]

Thus, we can conclude that 21133.191 [tex]ft^{3}/hr[/tex] of air are drawn from outside  per hour by the fan that supplies the air.

Explanation:

According to Clausius-Claperyon equation,

       [tex]ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]

The given data is as follows.

         [tex]T_{1} = 63.5^{o}C[/tex] = (63.5 + 273) K

                         = 336.6 K

        [tex]T_{2} = 78^{o}C[/tex] = (78 + 273) K

                         = 351 K

         [tex]P_{1}[/tex] = 1 atm,             [tex]P_{2}[/tex] = ?

Putting the given values into the above equation as follows.    

        [tex]ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]

       [tex]ln (\frac{1.75 atm}{1 atm}) = \frac{-\text{heat of vaporization}}{8.314 J/mol K} \times [\frac{1}{351 K} - \frac{1}{336.6 K}][/tex]

                      [tex]\Delta H[/tex] = [tex]\frac{0.559}{1.466 \times 10^{-4}} J/mol[/tex]

                                  = [tex]0.38131 \times 10^{4} J/mol[/tex]

                                  = 3813.1 J/mol

Thus, we can conclude that the heat of vaporization of ethanol is 3813.1 J/mol.

Answer:

The term dielectric makes reference to materials that are insulators (the charges in the material does not flow freely and for this reason an insulator material cannot conduce electricity). BUT if we applied this material in an electric field they can be polarized.

Dielectric Constant: It is also defined as a relative permittivity and it is the amount of charge required to produce one unit of electric flux( electric field in a surface)  in a given medium.

it is typically denoted as Epsilon (ε)

and its formula is: ε = ε(ω) /ε(0)

where:

ε(ω) = the frequency-dependent permittivity of the material

ε(0) = the dielectric constant value in vacuum

Dielectric Constant values:

Answer:

Explanation has been given below

Explanation:

A cis-1,3-disubstituted cyclohexane is more stable than its trans isomer due to lower steric hindrance in the cis configuration as compared to the trans.

A cis-1,3-disubstituted cyclohexane is more stable than its trans isomer because of a property known as steric hindrance. In organic chemistry, steric hindrance is a phenomenon where the size of groups within a molecule prevents chemical reactions from taking place. In a cis isomer, the substituents are on the same side of the ring, specifically on the ring's axial position. This arrangement minimizes the steric hindrance, which makes it more stable. On the other hand, the trans isomer has the substituents on opposite sides of the ring, causing more steric hindrance and thus it is less stable.

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Answer:

a) 21.133 minutes for series

b) 84.54 minutes when split into 4

Explanation:

Data provided:

Total flow, V₀ = 45 MGD

total volume of each basin, V = 2500 m³

Now,

1 MGD = 3785.4118 m³/day

also,

1 day = 1440 minutes

thus,

45 MGD =  45 × 3785.4118 m³/day

or

45 MGD = 170343.5305 m³/day

and,

170343.5305 m³/day  in 'm³/min'

= 170343.5305 / 1440

= 118.2941 m³/min.

Therefore,

The retention time, t = [tex]\frac{\textup{Volume of tank}}{\textup{Volumetric Flowrate }}[/tex]

or

The retention time, t = [tex]\frac{\textup{2500}}{\textup{118.2941 }}[/tex]

or

The retention time, t = 21.13 min

Hence,

for series of tanks the retention time is 21.133 min.

Now,

on splitting the tanks in 4

the volumetric flow rate will be

= [tex]\frac{\textup{118.2941}}{\textup{4}}[/tex]

= 29.57 m³/min.

Therefore,

The retention time = [tex]\frac{\textup{Total Volume of tank}}{\textup{Volumetric Flowrate }}[/tex]

or

The retention time, t = [tex]\frac{\textup{2500}}{\textup{29.57 }}[/tex]

or

The retention time, t = 84.54 min

Answer:

The correct option is: a) 2 U.S. pounds;

B) 1.672622×10⁻²⁷ kg

Explanation

Pound is a United States customary unit for mass and is abbreviated as lb. One pound is defined as being equal to 0.45359237 kilograms (kg), exactly.

Kilogram is the SI unit of mass and is abbreviated as kg.

Since, 1 lb = 0.45359237 kg  

Therefore, 1 kg = 1 ÷ 0.45359237 lb ≈ 2.20462 lb

Therefore, we can say that 1 kilogram is slightly more than 2 pounds.

A proton is a positively charged subatomic particle, that possesses +1e electric charge and has mass approximately equal to 1.672622×10⁻²⁷ kg.

Answer:

Its A

Explanation:

Its A

The correct answer is B. A gas is formed

Explanation:

In chemistry, changes in substances or elements occur as a product of chemical reactions in which a substance reacts with others or decomposes. In any of these cases, the original substance or element changes into new or new substances, which does not occur in physical changes. This change in composition can be identified through different changes and signals that include a change in odor, color or temperature; the production of bubbles or gas; and the formation of a precipitate. According to this, the situation that is an indication of a chemical change is the formation of a gas, because only if the original substance has changed there would be a new substance such as a gas.

Answer:

The volume of a chlorine solution a homeowner should add to her swimming pool is [tex]4.480\times 10^3 mL[/tex].

Explanation:

The solution contains 5.50 percent chlorine by mass.

Mass of the chlorine required to added in the pool= x

Mass of the water in the pool = y

Volume of the water ,V= [tex]6.52\times 10^4 gal=2.464\times 10^5 L[/tex]

(1 gal = 3.78 L)

V = [tex]2.464\times 10^5 L=2.464\times 10^8 mL[/tex]

( 1 L = 1000 mL)

Density of the water ,d= 1.00 g/mL

[tex]y=d\times V=1.00 g/mL\times 2.464\times 10^8 mL=2.464\times 10^8 g[/tex]

y = [tex]2.464\times 10^8 g[/tex]

1.00 g of chlorine per million grams of water .

Then for  [tex]2.464\times 10^8 g[/tex] of water, we required x amount of chlorine:

[tex]z=\frac{2.464\times 10^8 g}{10^6 g}=246.4 g[/tex]

x = 246.4 grams of chlorine

Mass of the chlorine solution = M'

[tex](w/w)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100[/tex]

[tex]5.5\%=\frac{z}{M'}\times 100[/tex]

[tex]M'=\frac{246.4}{5.5}\times 100=4,480 g[/tex]

Density of the chlorine solution = D = 1 g/mL

Volume of the chlorine solution to be added in the pool : v

[tex]v=D\times M'=1 g/mL\times 4,480 g=4480 mL[/tex]

[tex]v=4.480\times 10^3 mL[/tex]

The volume of a chlorine solution a homeowner should add to her swimming pool is [tex]4.480\times 10^3 mL[/tex].

Answer:

4.45 L

Explanation:

The concentration of chlorine in the pool must be 1.00g/10⁶g, and the solution has 5.50g/100g, so it will be diluted in the pool.

The product of the concentration by the volume is always constant in a solution, thus, we can use the equation:

C1V1 = C2V2

Where C is the concentration, V is the volume, 1 represents the solution, and 2 the pool. So, C1 = 5.50g/100g, C2 = 1.00g/10⁶g, V2 = 6.52x10⁴ gal = 247108 L. Thus,

(5.50/100)*V1 = (1.00/10⁶)*247108

0.055V1 = 0.247108

V1 = 4.5 L

Answer:

The correct option is: a. degrees Celsius

Explanation:

The anion gap is the difference in the cations and anions in plasma, serum or urine, calculated from medical lab test results. It can be calculated by measuring the concentration of the anions or cations, which are expressed in millimoles/litre (mmol/L) or milliequivalents/liter (mEq/L).

The temperature in this test is expressed in degrees Celsius (°C).

Answer: The mass of ammonia produced in the reaction is 11.36 g

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of nitrogen gas = 9.35 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of nitrogen gas}=\frac{9.35g}{28g/mol}=0.334mol[/tex]

The chemical reaction for the formation of ammonia from hydrogen and nitrogen follows:

[tex]3H_2+N_2\rightarrow 2NH_3[/tex]

As, hydrogen gas is present in excess. So, it is considered as an excess reagent.

Nitrogen gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of nitrogen gas is producing 2 moles of ammonia gas

So, 0.334 moles of nitrogen gas will produce = [tex]\frac{2}{1}\times 0.334=0.668mol[/tex] of ammonia gas.

Now, calculating the mass of ammonia gas by using equation 1, we get:

Moles of ammonia gas = 0.668 mol

Molar mass of ammonia gas = 17 g/mol

Putting values in equation 1, we get:

[tex]0.668mol=\frac{\text{Mass of ammonia gas}}{17g/mol}\\\\\text{Mass of ammonia gas}=11.36g[/tex]

Hence, the mass of ammonia produced in the reaction is 11.36 g

Final answer:

To calculate the mass of ammonia produced from 9.35 g of nitrogen gas, convert the mass of nitrogen to moles, use the stoichiometric relationship from the balanced reaction to find moles of ammonia, and then convert back to grams. The result is 11.37 g of NH3.

Explanation:

To calculate how many grams of ammonia (NH3) will be produced from 9.35 g of nitrogen gas (N2) using the chemical equation N₂(g) + 3H₂(g) → 2NH3(g), we need to perform stoichiometric calculations. First, we determine the molar mass of nitrogen gas (N2) which is 28.02 g/mol. Using this, we find that 9.35 g of N2 is equivalent to 9.35 g / 28.02 g/mol = 0.3338 mol of N2.

From the balanced chemical equation, we know that 1 mole of nitrogen gas produces 2 moles of ammonia. Therefore, 0.3338 mole N2 × (2 moles NH3 / 1 mole N2) = 0.6676 mole NH3. Finally, using the molar mass of ammonia, which is 17.03 g/mol, we can find the mass of ammonia produced: 0.6676 mol × 17.03 g/mol = 11.37 g of NH3.

Answer:

12 mL

Explanation:

Density is expressed as mass divided by volume, so the expression can be rearranged to solve for volume:

D = m/V ⇒ V = m/D

V = m/D = (8.410 g)/(0.71g/mL) = 12 mL

Final answer:

The volume of a substance with a mass of 8,410 g and a density of 0.71 g/mL is 12,000 mL, rounded to two significant figures to match the precision of the given density.

Explanation:

To calculate the volume of a substance, you can use the density formula: Density = Mass / Volume. Given the density of the substance is 0.71 g/mL and the mass is 8,410 g, the volume can be found by rearranging the formula to Volume = Mass / Density.

Volume = 8,410 g / 0.71 g/mL = 11,845.07042 mL. This raw calculation has more significant figures than is justified by the precision of the given data. Since the density value has two significant figures, the volume should also be reported with two significant figures, resulting in 12,000 mL.

Answer:

c) 3.41 atm

Explanation:

We can calculate the final pressure using Boyles Law, P₁V₁ = P₂V₂.

P₁V₁ = P₂V₂

(0.983 atm)(10.0 L) = P₂(2.88 L)

9.83 ÷ 2.88 = P₂

P₂ = 3.41 atm

The final pressure can be determined by using Boyles law and will be 3.41 atm.

The force delivered perpendicularly to a surface of the structure per unit area throughout whom that force would be dispersed is known as pressure.

According to Boyle's law, the relationship between a gas's pressure as well as volume seems to be inverse.

Given data:

[tex]P_{1} = 0.983 atm\\V_{1} = 10 L\\V_{2} = 2.88 L[/tex]

Calculation of pressure.

The formula of Boyles law:

[tex]P_{1} V_{1} = P_{2} V_{2}[/tex]

Put the value of given data:

[tex](0.983) (10 ) = P_{2} (2.88)\\P_{2} = 3.41 atm[/tex]

Therefore, the final pressure will be 3.41 atm.

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Answer:

Newton's law of viscosity

It relates shear stress in a fluid flow to velocity gradient in the direction perpendicular to the flow of fluid.

[tex]\tau\ \alpha \ \frac{\mathrm{d} u}{\mathrm{d} y}[/tex]

Or

[tex]\tau\ =\mu \times \frac{\mathrm{d} u}{\mathrm{d} y}[/tex]

[tex]\tau\ =Shear\ stress[/tex]

[tex]\frac{du}{dy} =[/tex] Rate of shear deformation

[tex]\mu\ =Viscosity[/tex]

Hooke's Law

It states that within the limit of elasticity, the stress-induced (σ ) in the solid due to some external force is always in proportion with the strain (ε ). In other words, the force causing stress in a solid is directly proportional to the solid's deformation.

[tex]\sigma\ \alpha\ \epsilon[/tex]

[tex]\sigma=E\ \epsilon[/tex]

where E is constant of proportionality known as Young's Modulus and it represents the stiffness of the material.

The type of matter is different in both Newton's and Hook's laws.

Newton's law of viscous deformation deals with deformation of fluids that is subjected to a load. This law states that shears stress is proportional to shear strain.

While on the other hand, Hooke's law of elasticity deals with deformation in solids which are subjected to a load so we can conclude that the type of matter is different in both Newton's and Hook's laws.

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Answer : The correct answer is, [tex]0.274\times 10^{12}[/tex]

Explanation :

Engineering notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form times 10 raise to the power.  It is similar to the scientific notation but in engineering notation, the powers of ten are always multiples of 3.

The engineering notation written in the form:

[tex]a\times 10^b[/tex]

where,

a = the number which is greater than 0 and less than 999

b = an integer multiple of 3

For example : [tex]45.89\times 10^3[/tex]  or [tex]56.45\times 10^{-6}[/tex]

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.

As we are given the 274,541,005,000 in standard notation.

Now converting this into engineering notation, we get:

[tex]\Rightarrow 274,541,005,000=0.274\times 10^{12}[/tex]

As, the decimal point is shifting to left side, thus the power of 10 is positive.

Hence, the correct answer is, [tex]0.274\times 10^{12}[/tex]

Final answer:

To write 274,541,005,000 in Engineering Notation with 3 significant figures, we need to move the decimal point so that we have a number between 1 and 10 multiplied by a power of 10. The number becomes 2.75 × 10¹¹.

Explanation:

Engineering Notation is a way of expressing numbers in scientific notation but with the power of 10 always being a multiple of 3. To write 274,541,005,000 in Engineering Notation with 3 significant figures, we need to move the decimal point so that we have a number between 1 and 10 multiplied by a power of 10. The number becomes 2.75 × 10¹¹.

Answer:

the price of the new computer was [tex]\$ 731.34[/tex]

Explanation:

let's assume that the price of the computer was [tex]\$ x[/tex] when it was new.

As price depreciates at a rate of 15% therefore we can mathematically express this situation when the computer is 5 yrs old as-

              [tex]x\times (1-0.15)^{5}=324.50[/tex]

            or,   [tex]x\times (0.85)^{5}=324.50[/tex]

            or, [tex]x=\frac{324.50}{(0.85)^{5}}[/tex]

             or, [tex]x=731.34[/tex]

So, the price of the new computer was [tex]\$ 731.34[/tex]

Explanation:

Molar mass of [tex]CH_{3}SSCH_{3}[/tex] is 94 g/mol. As it is known that number of moles is equal to mass of a substance divided by its molar mass.

Then, calculate the number of moles as follows.

     No. of moles = [tex]\frac{86.0 g}{94 g/mol}[/tex] in 1 s

                           = 0.914 mol

So, in 60 sec number of moles will be equal to 0.914 x 60 = 54.89 mol/min.

Hence, the molar flow rate = 54.89 mol/min

Also, density is equal to mass of a substance divided by its volume.

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

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

                                     = [tex]\frac{86.0 g}{1.0625 g/cm^{3}}[/tex]

                                     = 80.941 [tex]cm^{3}[/tex]

As, 80.941 [tex]cm^{3}[/tex] of volume flows in 1 s . Therefore, flow of volume in 1 hour will be calculated as follows.

                 In 1 hr = 80.941 [tex]cm^{3} \times 3600[/tex]

                            = 291388.24 [tex]cm^{3}/hr[/tex]

Since, 1 [tex]cm^{3}[/tex] = 0.001 L.

So,              291388.24 [tex]cm^{3}/hr \times 0.001 L/cm^{3}[/tex]

                            = 291.38824 L/hr

Thus, we can conclude that molar flow rate in mol/min is 54.89 mol/min and the volumetric flow rate in L/hr is 291.38824 L/hr.

Answer:

pH = 5,22

Explanation:

A buffer consist in a solution with both conjugate acid (HA) and conjugate base (A⁻). To know the concentration of both A⁻ and HA you use Henderson-Hasselbalch equation:

pH = pka + log₁₀ [tex]\frac{A^{-} }{HA}[/tex]

In the problem:

5,23 = 5,14 + log₁₀ [tex]\frac{A^{-} }{HA}[/tex]

1,23 = [tex]\frac{A^{-} }{HA}[/tex] (1)

Also, you know buffer concentration is 1,0 M:

1,0 M = A⁻ + HA (2)

Replacing (2) in (1):

HA = 0,45 M, thus:

A⁻ = 0,55 M.

The addition of 1.50 mL of 5.00 M HCl represents a concentration of:

1,50x10⁻³ L×[tex]\frac{5,00 mol}{L}[/tex] ÷ 0,5 L = 0,015 M of HCl

The buffer equilibrium is:

HA ⇄ A⁻ + H⁺ ka = [tex]10^{-5,14}[/tex]

The concentrations in equilibrium are:

[HA] = 0,45 M + x

[A⁻] = 0,55 M - x

[H⁺] = 0,015 M -x

Because the equilibrium is displaced to the left.

The equation of equilibrium is:

[tex]10^{5,14}[/tex] = [tex]\frac{[0,015M -x][0,55M-x]}{[0,45M-x]}[/tex]

You will obtain:

x² - 0,565x + 8,24674x10⁻³ = 0

Solving:

x = 0,5500061 ⇒ No physical sense

x = 0,01499391

Thus, [H⁺] = 0,015 - 0,01499391 = 6,09x10⁻⁶

As pH = -log₁₀ [H⁺]

pH = 5,22

I hope it helps!

Answer:

(a) Yes, he should be worried. The Fatal accident rate (FAR) is too high according to standars of the industry. This chemical plant has a FAR of 167, where in average chemical plants the FAR is about 4.

(b) FAR=167 and Death poer person per year = 0.0033 deaths/year.

(c) The expected number of fatalities on a average chemical plant are one in 12500 years.

Explanation:

Asumming 50 weeks of work, with 40 hours/week, we have 2000 work hours a year.

In 300 years we have 600,000 hours.

With these estimations, we have (1/600,000)=1.67*10^(-6) deaths/hour.

If we have 2000 work hours a year, it is expected 0.0033 deaths/year.

[tex]1.67*10^{-6} \frac{deaths}{hour}*2000 \frac{hours}{year}=0.0033 deaths/year[/tex]

The Fatal accident rate (FAR) can be expressed as the expected number of fatalities in 100 millions hours (10^(8) hours).

In these case we have calculated 1.67*10^(-6) deaths/hour, so we can estimate FAR as:

[tex]FAR=1.67*10^{-6} \frac{deaths}{hour}*10^{8}  hours=1.67*10^{2} =167[/tex]

A FAR of 167 is very high compared to the typical chemical plants (FAR=4), so the worker has reasons to be worried.

If we assume FAR=4, as in an average chemical plant, we expect

[tex]4\frac{deaths}{10^{8} hour} *2000\frac{hours}{year}=8*10^{-5} \frac{deaths}{year}[/tex]

This is equivalent to say

[tex]\frac{1}{8*10^{-5} } \frac{years}{death}=1.25*10^{4} \frac{years}{death} =12500 \, \frac{years}{death}[/tex]

The expected number of fatalities on a average chemical plant are one in 12500 years.

Answer:

The only one which will dissolve is salt try other ones can't dissolve.

Answer:

Salt

Explanation:

Salt can be dissolved in water.

Explanation:

It is given here that a slab is made up of dry wood.

Thickness of slab = 4 inch

It is given condition that the edges of the slab are sealed and exposed to air.

Given the relative humidity of air = 40%

The surface of slab is unsealed and reached to the equilibrium moisture content = [tex]\frac{8 lb H_{2}O}{100 lb dry wood}[/tex].

Given, the diffusivity of water in the slab = [tex]6 \times 10^{-6} cm^{2}/s[/tex].

As per the given condition 1% moisture content will be taken by multiplying the moisture content by 1%, i.e. 0.01.

It is given that the thickness of slab is in inch, so converting the thickness of slab in cm as follows.

        Thickness of slab = [tex]4 inch \times 2.54 inch/cm[/tex]

                                     = 10.16 cm             (As, 1 inch = 2.54 cm)

Since, the equilibrium moisture content is required to calculate at the center of the slab.

The center of the slab, from the semi-infinite slab solution will be taken from thickness t = 0.

The thickness at center = 10.16 cm

Hence, calculate total thickness at the center of slab as follows.

          [tex]\frac{\text{total thickness of slab}}{2}[/tex]

                = 5.08 cm

Now, by semi-infinite solution, and by 40% relative humidity of air, time for moisture content to reach at the center of the slab, to 1% of its equilibrium value will be calculated as follows.

       Time(Seconds) = [tex]\frac{thickness^{2}(cm^{2}) \times \text{relative humidity} \times 0.01 \times \text{equilibrium content} \times \text{equilibrium moisture}}{\text{diffusivity(cm/s)}}[/tex]

       Time (sec) = [tex]\frac{5.08 cm \times 5.08 cm \times 0.4 \times \text{relative humidity} \times 0.01 \text{equilibrium content} \times 8 lb H_{2}O}{100 lb dry.wood \times 6 \times 10{-6} cm^{2}/s \text{diffusivity}}[/tex]

               Time (Seconds) = 1376.34 seconds

Thus, we can conclude that the time required for the moisture content at the center of the slab to reach to 1 % of equilibrium value is 1376.34 seconds.

Answer : The mass of magnesium sulfate heptahydrate needed are, 36.969 grams.

Explanation : Given,

Moles of magnesium sulfate heptahydrate = 0.150 mole

Molar mass of magnesium sulfate heptahydrate = 246.46 g/mole

Formula used :

[tex]\text{Mass of magnesium sulfate heptahydrate}=\text{Moles of magnesium sulfate heptahydrate}\times {\text{Molar mass of magnesium sulfate heptahydrate}[/tex]

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

[tex]\text{Mass of magnesium sulfate heptahydrate}=0.150mole\times 246.46g/mole[/tex]

[tex]\text{Mass of magnesium sulfate heptahydrate}=36.969g[/tex]

Therefore, the mass of magnesium sulfate heptahydrate needed are, 36.969 grams.

To obtain 0.150 moles of magnesium sulfate heptahydrate, you will need 36.969 grams.

To determine the weight needed to obtain 0.150 moles of magnesium sulfate heptahydrate (MgSO₄·7H₂O), you can use the molar mass and the number of moles given.

Mass = Number of Moles x Molar Mass

Substitute the values into the formula:

Mass = 0.150 moles x 246.46 g/mol

Mass = 36.969 g

Therefore, to obtain 0.150 moles of magnesium sulfate heptahydrate, you need 36.969 grams.