Explanation:
Fischer Projections allow to represent the three dimensional molecular structures in two dimensional environment without the change in the properties or the structural integrity of the compound. It consists of horizontal as well as vertical lines both, where horizontal lines represent atoms which are pointed toward viewer while vertical line represents atoms which are pointed away from viewer. The point of the intersection between horizontal and vertical lines represents central carbon.
A slab of dry wood of 4-in thickness and sealed edges is exposed to air of 40% relative humidity. Assuming that the two unsealed surfaces of the wood immediately jump to an equilibrium moisture content of 8 lb H2O per 100 lb of dry wood, determine the time for the moisture content at the center of the slab to reach 1% of the equilibrium value using the semi-infinite slab solution. Assume a diffusivity of water of 6.00e-006 cm2/s.
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.
A freezer compartment is covered with a 2-mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient air at 20 °C and a coefficient of h 2 W/(m2 K) characterizes heat transfer by natural convection from the exposed surface of the layer, estimate the time required to completely melt the frost. The frost may be assumed to have a mass density of 700 kg/m3 and a latent heat of 334 kJ/kg.
The time required to completely melt the frost can be calculated by first determining the mass of the frost using its thickness and the assumed surface area of the freezer compartment. Once we know the mass, we can calculate the amount of heat required to melt the frost. We can then find the time by dividing the heat required by the rate at which heat is transferred.
Explanation:To solve this problem, we first need to calculate the volume of frost since we know its thickness (2 mm) and lets assume average surface area of the freezer. Lets say the surface area of the freezer compartment is A. Hence volume of frost would be 0.002m * A. Subsequently, using the volume and the density of the frost, we can find the mass of the frost. Mass = Density * Volume = 700 kg/m3 * 0.002m * A.
Once we have the mass of the frost, we can then calculate the amount of heat needed to completely melt the frost, Q. The formula for this is the mass of the frost multiplied by the latent heat of fusion (Q= mass * Lf), where Lf here is 334 kJ/kg. The time required to complete the process can finally be calculated by dividing the heat required to do so (Q) by the heat transferred per unit time (P = h*A*ΔT). Here, A is the surface area, h is the coefficient of heat transfer and ΔT is the temperature difference.
Note: Here we have used the area A which is lesser when the frost starts to melt and greater when the frost has been almost been melted.
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For this device the liquid to be tested is drawn into the tube to a level above the top etched line. The time is then obtained for the liquid to drain to the bottom etched line. The kinematic viscosity, ν, in m2/s is then obtained from the equation ν = KR4t where K is a constant, R is the radius of the capillary tube in mm, and t is the drain time in seconds. When glycerin at 20 ˚C (υ=1.19·10-3 m2/s) is used as a calibration fluid in a particular viscometer the drain time is 1430 s. When a liquid having a density of 970 kg/m3 is tested in the same viscometer the drain time is 900 s. What is the dynamic viscosity of this liquid?
Answer:
The dynamic viscosity of the liquid is 0.727 kg/m*s
Explanation:
In the equation for that viscosimeter, ν = KR⁴t, the terms K and R are not dependent on the liquid that is being tested, unlike ν and t.
Using that equation and the data given in the problem, we can calculate the product of K and R⁴.
1.19*10⁻³m²/s = (KR⁴)* 1430 s
KR⁴=8,32*10⁻⁷m²/s²
We can now calculate the kinematic viscosity of the unknown liquid.
ν=8,32*10⁻⁷m²/s²*900s
ν=7.49*10⁻⁴m²/s
The relationship between the kinematic viscosity and the dynamic viscosity is given by the equation μ=ν * ρ, where μ is the dynamic viscosity and ρ is the density. Thus:
μ=7.49*10⁻⁴m²/s * 970 kg/m³
μ=0.727 kg/m*s
Answer:
the dynamic viscosity is 0.72653 N s/m²
Explanation:
the solution is in the attached Word file
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?
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
[tex]T_2=62.9^o[/tex]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?
temperature of benzeneGenerally 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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Calculate the pH of the following simple solutions:
53.1 mM HCl
0.223 M KOH
53.1 mM HCl + 0.223 M KOH
Is the solution "c" a buffer? Why or why not?
Answer :
(1) pH = 1.27
(2) pH = 13.35
(3) The given solution is not a buffer.
Explanation :
(1) 53.1 mM HCl
Concentration of HCl = [tex]53.1mM=53.1\times 10^{-3}M[/tex]
As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.
So, Concentration of hydrogen ion= [tex]53.1\times 10^{-3}M[/tex]
pH : It is defined as the negative logarithm of hydrogen ion concentration.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (53.1\times 10^{-3})[/tex]
[tex]pH=1.27[/tex]
(2) 0.223 M KOH
Concentration of KOH = 0.223 M
As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.
So, Concentration of hydroxide ion= 0.223 M
Now we have to calculate the pOH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (0.223)[/tex]
[tex]pOH=0.65[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35[/tex]
(3) 53.1 mM HCl + 0.223 M KOH
Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.
It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.
As we know that the pH of strong acid and strong base solution is always 7.
So, the given solution is not a buffer.
Tania's mom boils water in a pot on the stove to make soup. Which of these ingrediente will dissolve in the water?
Answer:
The only one which will dissolve is salt try other ones can't dissolve.
Answer:
Salt
Explanation:
Salt can be dissolved in water.
A computer depreciates at a rate of 15% per year. If the computer is worth $324.50 when it is 5 years old, how much was it worth when it was new? Round your answer to two decimal places
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]
Which observation allows you to tell that two cubes are made of different metals? Select the two correct answers.
A. Their volumes are the same, and their masses are the same.
B. Their volumes are the same, and their masses are different.
C. Their volumes are different, and their masses are the same.
D. Their volumes are unknown, and their masses are the same.
E. Their volumes are the same, and their masses are unknown.
Answer:
B. Their volumes are the same, and their masses are different.
C. Their volumes are different, and their masses are the same.
Explanation:
In order to ascertain that two metal cubes are made of different materials, their densities would be very helpful in our experiment.
Every metal has its own unique density and we can use it to differentiate it from other metals.
Density is defined as the mass per unit volume of a substance. Therefore, if the mass and volume are the same, their densities would be the same. Density is expressed a:
Density = [tex]\frac{mass}{volume}[/tex]
If the volumes are the same but given different masses, we can differentiate the metals using their quite different densities.
If the volumes are different but the masses the same, we would still obtain different values of density and the metals can be differentiated.
Which of the following describes carbon dioxide? O Has polar covalent bonds and is a polar molecule O Has non-polar covalent bonds and is a polar molecule O Has ionic bonds and is an polar molecules O Has ionic bonds and is a non-polar molecule O Has polar bonds and is a non-polar molecule O Has non-polar bonds and is a non-polar molecules
Answer: Option "Has polar bonds and is a non-polar molecule" is the correct answer.
Explanation:
A covalent compound is defined as the compound in which there will be sharing of electrons between the combining atoms.
An ionic compound is defined as the compound in which there will be transfer of electrons from one atom to another.
A non-polar covalent compound is defined as the compound in which there will be equal sharing of electrons between the combining atoms.For example, a chlorine atom has 7 valence electrons.
So, when a chlorine atom chemically combines with another chlorine atom then there occurs equal sharing of electrons.
Hence, molecule [tex]Cl_{2}[/tex] is formed which is non-polar covalent compound in nature with non-polar covalent bond.
A polar covalent bond is defined as the bond in which there will be unequal sharing of electrons between the combining atoms.For example, a carbon atom has 4 valence electrons and an oxygen atom has 6 valence electrons.
To complete their octet a carbon atom needs 4 more electrons and an oxygen atom needs 2 more electrons.
And, when one carbon and two oxygen atoms chemically combine together then there occurs unequal sharing of electrons.
Hence, a polar covalent compound [tex]CO_{2}[/tex] is formed.
Since, oxygen is more electronegative than carbon atom. Hence, dipoles in the linear carbon dioxide molecules will cancel out each other.
Therefore, a carbon dioxide molecule is non-polar compound that has polar covalent bonds.
Thus, we can conclude that the statement, has polar bonds and is a non-polar molecule, correctly describes carbon dioxide.
Find the volume in m3 of 52.6 lbm of iron:
Answer: The volume of iron is [tex]3.031\times 10^{-3}m^3[/tex]
Explanation:
To calculate volume of a substance, we use the equation:
[tex]\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}[/tex]
We are given:
Mass of iron = 52.6 lbm = 23.86 kg (Conversion factor: 1 kg = 2.205 lbm)
Density of iron = [tex]7873kg/m^3[/tex]
Putting values in above equation, we get:
[tex]7873kg/m^3=\frac{23.86kg}{\text{Volume of iron}}\\\\\text{Volume of iron}=3.031\times 10^{-3}m^3[/tex]
Hence, the volume of iron is [tex]3.031\times 10^{-3}m^3[/tex]
Spread Spectrum (SS) encoding: a) Name four spread spectrum techniques. b) Which one of them was originally devised against eavesdropping in the military environment?
Explanation:
( a )
The four types of spread spectrum techniques are as follows -
1. Direct sequence spread spectrum .
2. frequency hopping spread spectrum .
3. chirp spread spectrum .
4. time hopping spread spectrum .
( b )
The Direct sequence spread spectrum was devised for eavesdropping in the military .
In the field of telecommunications , the Direct sequence spread spectrum , it is the technique of spread spectrum modulation which is used to reduce the overall inference of the signal .
One kilogram is slightly more than A) 2 B) 5 U.S. pounds. C) 0.5 D) 1 11) The mass of a proton is 1.67 A) 1.67. 10-30 G B 10-27 kg.
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.
Calculate how many g of ammonia will be produced from 9.35 g of nitrogen gas and excess hydrogen using the following equation: 2 NH(g) 3 H2(g) N(g) 4 answer
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.
The molecular mass of magnesium sulfate heptahydrate is 246.46 g/mol. What weight is needed to obtain 0.150 moles?
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.
The molar mass of MgSO₄·7H₂O is given as 246.46 g/mol.The number of moles needed is 0.150 moles.To find the mass needed, use the formula: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.
In AG calculations, temperature is expressed in a. degrees Celsius. b. kelvins. c. degrees Fahrenheit, d. kilojoules.
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).
Arrange the following cellular components in order of decreasing size: hexokinase ribosome CO2 glucose
Answer:
Ribosome, Hexokinase, Glucose, CO2.
Ribosomes are proteins that sintetize the proteins in the cell, depending of the organism, the can size up to 30 nm. Hexokinase is an enzyme that measures approximately 50 kDa, and in its spatial conformation it sizes about 5 to 6 nm in diameter. Glucose is a molecule that sizes about 1 nm, and CO2 is another molecule that sizes 0.232 nm.
Compare Newton's law of viscosity and Hooke's law of elasticity. What is the origin of these "laws"?
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.
Comparison between Newton's law of viscosity and Hooke's law of elasticityNewton'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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Which situation is an indication that a chemical change has occurred? No color change occurs. A gas is formed. No heat is released. The temperature does not change.
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.
Why is a cis-1,3-disubstituted cyclohexane more stable thanits
trans isomer?
Answer:
Explanation has been given below
Explanation:
In diaxial conformation of cis-1,3-disubstituted cyclohexane, 4 gauche-butane interactions along with syn-diaxial interaction are present. Hence it readily gets converted to diequitorial conformation where no such gauche-butane interaction is presentIn two possible conformations of trans-1,3-disubstituted cyclohexane, 2 gauche-butane interactions are present in each of them.Hence cis-1,3-disubstituted cyclohexane exists almost exclusively in diequitorial form. But trans-1,3-disubstituted cyclohexane has no such option. Trans-1,3-disubstituted cyclohexane experiences gauche butane interaction in each of the two conformations.Therefore cis-1,3-disubstituted cyclohexane is more stable than trans conformationA 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.
Explanation: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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A water treatment plant has a total flow of 45 MGD (mega gallons per day). The plant was designed with five sedimentation basins, each with a total volume of 2500 m3.
Originally, the plant was operating the basins in series: the entire flow goes through one tank, then the second, and so on. What is the retention time in each tank?
The engineer realizes that the sedimentation would improve if they changed the operation to parallel flow. Under this new configuration, the flow is evenly split between the four tanks. What is the retention time in each tank?
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
A water solution contains 1.704 [kg] of HNO3 per [kg] of water, and has a specific gravity of 1.382 at 20 [°C]. Please, express the composition of this as: (a) weight percent HNO3; (b) [lb HNO3/ft3]; HNO3 molarity; [mol HNO3/m3).
Answer:
(a) The weight percent HNO3 is 63%.
(b) Density of HNO3 = 111.2 lb/ft3
(c) Molarity = 13792 mol HNO3/m3
Explanation:
(a) Weight percent HNO3
To calculate a weight percent of a component of a solution we can express:
[tex]wt = \frac{mass \, of \, solute}{mass \, of \, solution}=\frac{mass\,HNO_3}{mass \, HNO_3+mass \, H_2O}\\ \\wt=\frac{1.704}{1.704+1} =\frac{1.704}{2.704}= 0.63[/tex]
(b) Density of HNO3, in lb/ft3
In this calculation, we use the specific gravity of the solution (1.382). We can start with the volume balance:
[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}+\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{2.956/ \rho_w}= 0.576*\rho_w[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}-\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{0.956/\rho_w}= 1.782*\rho_w[/tex]
The density of HNO3 is 1.782 times the density of water (Sp Gr of 1.782). If the density of water is 62.4 lbs/ft3,
[tex]\rho_{HNO3}= 1.782*\rho_w=1.782*62.4 \, lbs/ft3=111.2\, lbs/ft3[/tex]
(c) HNO3 molarity (mol HNO3/m3)
If we use the molar mass of HNO3: 63.012 g/mol, we can say that in 1,704 kg (or 1704 g) of HNO3 there are 1704/63.012=27.04 mol HNO3.
When there are 1.704 kg of NHO3 in solution, the total mass of the solution is (1.704+1)=2.704 kg.
If the specific gravity of the solution is 1.382 and the density of water at 20 degC is 998 kg/m3, the volume of the solution is
[tex]Vol=\frac{M_{sol}}{\rho_{sol}}=\frac{2.704\, kg}{1.382*998 \, kg/m3} = 0.00196m3[/tex]
We can now calculate the molarity as
[tex]Molarity HNO_3=\frac{MolHNO3}{Vol}=\frac{27.04mol}{0.00196m3} =13792 \frac{molHNO3}{m3}[/tex]
The composition of an HNO3 solution can be expressed in various ways. The weight percent is calculated by dividing the mass of HNO3 by the total mass of the solution and multiplying by 100. The quantity in [lb HNO3/ft3] can be obtained by unit conversion. The molarity is calculated by dividing the moles of HNO3 by the volume of the solution in liters. Lastly, [mol HNO3/m3] is obtained by multiplying the molarity by 1000.
Explanation:To tackle this problem, we need to make a variety of calculations using the information given and our knowledge of chemical molar masses and concentration calculations.
Firstly, (a) The weight percent HNO3 can be calculated by dividing the mass of HNO3 by the total mass of the solution (HNO3 + water), and then multiplying by 100 to get a percentage. Since the solution has 1.704 kg of HNO3 for every 1 kg of water, the total mass of the solution is 2.704 kg. Thus, the weight percent of HNO3 is (1.704 kg / 2.704 kg) x 100 = ~63%.
Secondly, (b) To express the composition as [lb HNO3/ft3], we have to convert kg of HNO3 to lb, and then use the specific gravity to determine the volume of the solution in ft3.
Next, we calculate the HNO3 molarity by dividing the moles of HNO3 by the liters of solution. The moles of HNO3 can be calculated from the mass using the molar mass of HNO3, which is about 63.01 g/mol.
Finally, [mol HNO3/m3] can be calculated by multiplying the molarity by 1000, since 1 m3 is equal to 1000 L.
Learn more about Solution Composition here:https://brainly.com/question/8598915
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A balloon has a volume of 3.50 liters at 20.0°C. What will be the volume of the balloon at 100.0°C? Assume the pressure doesn't change and the amount of gas in the balloon doesn't change O a. 0.224 Ob.2.75L O c. 17.5L O d. 65.45 Oe. None of the above
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
Prepare benzoic acid from from benzyl alcohol. What is the purpose of the addition of the sulfuric acid?Include chemical reaction.
Answer:
Benzoic acid can be prepared from benzyl alcohol using potassium dichromate and sulfuric acid.
Explanation:
Benzyl alcohol is a primary alcohol that can be transformed into benzoic acid by an oxidation reaction.
The reactants needed are potassium dichromate and sulfuric acid.
The sulfuric acid is needed to react with the potassium dichromate and form cromic acid, which is a strong oxidizing agent that will convert the alcohol to its equivalent carboxylic acid.
The chemical reaction is
3ФCH₂OH + 2Cr₂O⁷⁻ + 16 H⁺ 3ФCOOH + 4Cr³⁺ + 11H₂O
Measurement of heat conductivity A metal panel of area A = 100 cm² and thickness Y = 0.5 cm was found to conduct heat at a rate of 3 W at steady state with temperatures To = 25 °C and T, = 30 °C imposed on the two main surfaces. What is the average thermal conductivity of the material at this temperature range?
Answer:
The average thermal conductivity of the material at this temperature range is 3*10^(-9) W/mK.
Explanation:
The heat flow through this area of the metal panel can be expressed as:
[tex]q/A=-k\frac{(T_2-T_1)}{Y}[/tex]
being A the surface area, k the thermal conductivity, T2 and T1 the temperatures in each side and Y the thickness of the metal panel.
We have to rearrange the equation to clear k:
[tex]k=\frac{-(q/A)*Y}{\Delta T} = \frac{-(3 W/100 cm2)*0.5cm}{5 ^\circ K}*(\frac{1m}{100cm} )^{3} \\\\k=\frac{0.015}{5*100^{3} }W/mK= 3*10^{-9}W/mK[/tex]
The average thermal conductivity of the material at this temperature range is 3*10^(-9) W/mK.
Chlorine is used to disinfect swimming pools. The accepted concentration for this purpose is 1.00 ppm chlorine, or 1.00 g of chlorine per million grams of water. Calculate the volume of a chlorine solution (in milliliters) a homeowner should add to her swimming pool if the solution contains 5.50 percent chlorine by mass and there are 6.52 × 104 gallons (gal) of water in the pool (1 gal = 3.79 L; density of liquids = 1.00 g/mL). Enter your answer in scientific notation.
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
Define "aldose" and "ketone" and recognize them structurally
Explanation:
Aldose is monosaccharide sugar in which the carbon backbone chain has carbonyl group on endmost carbon atom which corresponds to an aldehyde, and the hydroxyl groups are connected to all other carbon atoms.
Example - Glucose
Ketone is functional group with structure RC(=O)R', where the groups, R and R' can be variety of the carbon-containing substituents.
Example, Ethylmethylketone
Structurally, Aldose can be recognized by observing the carbonyl group on endmost carbon atom which corresponds to an aldehyde in a sugar.
Structurally, Ketone can be recognized by observing the functional group with structure RC(=O)R'.
Define "Dielectric" and list the Dielectric Constant values for vacuum and water
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:
For vacuum is always 1 for water depending on the temperature can be 87.9 at 0º C, 80.2 at 20º C and 55.5 at 100º C.
You have a red ballooe that has a volume of 20 liters, a pressure of 1.5 atmospheres and a temperature of 28 C. What is the volume of the balloon if you change the pressure to 5 atmospheres and change the temperature to S0°C Oa.1.07 L Ob3.0 L 0.64 L O d.o.42 L
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.
What is the heat of vaporization of ethanol, given that ethanol has a normal boiling point of 63.5°C and the vapor pressure of ethanol is 1.75 atm at 780 g.
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.
Assume that you have developed a tracer with first-order kinetics and k = 0.1 s − 1 . For any given mass, how long will it take the tracer concentration to drop by (a) 50% and (b) by 75%?
Answer: It will take 6.93 sec for tracer concentration to drop by 50% and 13.9 sec for tracer concentration to drop by 75%
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant =[tex]0.1s^{-1}[/tex]
t = age of sample
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process
a) for tracer concentration to drop by 50%
a - x = amount left after decay process = 50
[tex]t=\frac{2.303}{k}\log\frac{100}{50}[/tex]
[tex]t=\frac{0.693}{0.1}[/tex]
[tex]t=6.93sec[/tex]
It will take 6.93 sec for tracer concentration to drop by 50%
b) for tracer concentration to drop by 75 %
a - x = amount left after decay process = 25
[tex]t=\frac{2.303}{k}\log\frac{100}{100-75}[/tex]
[tex]t=\frac{2.303}{0.1}\times \log\frac{100}{25}[/tex]
[tex]t=13.9sec[/tex]
It will take 13.9 sec for tracer concentration to drop by 75%.