Given,Initial volume, Vi = 10-3 m3Pressure, P = 10^4 PaTemperature, T1 = 25°CAvogadro's Number, NA = 6.022×1023 atoms/molAtomic weight of Argon gas, m = 0.040 kg/mole
Explanation: 1) What is the number density of atoms in this chamber?Number density is given by:N/V = PNAT1V1 = 10^4×6.022×1023/8.314×298×10-3N/V = 2.4299 × 1024 atoms/m3Therefore, the number density of atoms in the chamber is N/V = 2.4299 × 1024 atoms/m3
2) What is the new temperature, T?Volume of the container is changed from V1 to V2Pressure remains constantTemperature of the gas changes from T1 to T2Since the expansion is free expansion, the internal energy of the gas remains constantFor an ideal gas,U = (3/2)Nk(T2 - T1)Where k is the Boltzmann constant or the gas constant divided by the Avogadro number k = R/NA = 8.314/6.022×1023 = 1.381×10-23 JK-1Therefore, U = (3/2)PV(T2 - T1)/kV1 = (3/2)(P/NA)(T2 - T1)V1/kV2 = V1 × 3 = 3×10-3m3T2 = T1 × V1/V2T2 = 25 × 10-3/3 = 8.33°CThus, the new temperature T is T = 8.33°C
3) What is the new pressure, P?According to Boyle's Law, P1V1 = P2V2P2 = P1V1/V2P2 = 10^4×10-3/(3×10-3)P2 = 3333 PaTherefore, the new pressure is P2 = 3333 Pa
4) How much work is needed to do this?In the compression process, work is done on the system.W = -∫PdVWhere, P = P(V) is the pressure as a function of the volume V.The compression is done slowly and isothermal, which means that the temperature remains constant at T1 = 25°CSo the ideal gas law,PV = NkTTemperature remains constant during the compression,So, P = NkT/V = nRT/VWhere n is the number of moles of gas and R is the molar gas constantWe have seen before thatN/V = P/kTRearranging this expression gives us N = (PV/kT)Therefore,W = -∫PdV = -∫(nRT/V)dV = -nRT ln(Vf/Vi)The amount of gas remains constant, so n is constant.The final volume is Vf = Vi = 10-3 m3W = -nRT ln(Vf/Vi)W = -PV ln(Vf/Vi)Since Vf/Vi = 1/3,W = -PV ln(1/3)W = PV ln(3)W = 10^4 × 10-3 × 0.040 × 8.31 × ln(3)W = -106.6 JThus, the amount of work needed to compress the gas back to its initial volume is W = -106.6 J.
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in 400 bce, the greek philosopher democritus first proposed the idea that all matter was composed of atoms. since that time, scientists have learned that, far from resembling tiny marbles, atoms actually have very complex structures. since it has been changed so many times, why is it referred to as the atomic theory rather than the atomic hypothesis?
The term "atomic theory" is used instead of "atomic hypothesis" because it signifies the evolution and acceptance of the concept over time.
While Democritus initially proposed the idea of atoms in 400 BCE, it was merely a hypothesis without substantial experimental evidence. Over centuries, scientific investigations and advancements led to a deeper understanding of atomic structure and behavior.
The term "atomic theory" acknowledges that the concept of atoms has undergone refinement and modification based on experimental evidence and theoretical developments.
It recognizes that the understanding of atoms has evolved from a speculative hypothesis to a well-established scientific theory supported by extensive experimental observations, mathematical models, and empirical data.
The term "theory" conveys the comprehensive and validated nature of our understanding of atoms, encompassing their complex structures and behavior.
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The diffusion coefficient for aluminum in silicon is D_Al in Si = 3 times 10^- 16 cm^2/s at 300 K. What is a reasonable value for D_Al in Si at 600 K? 1.5 times 10^-16 cm^2/s 3 times 10^-16 cm^2/s 6 times 10^-16 cm^2/s 1.5 times 10^-16 cm^2/s > 6 times 10^-16 cm^2/s
The comparison between an electrical circuit and a water circuit can be helpful in understanding the concepts and principles of electricity by drawing parallels with a familiar system like the flow of water.
In both circuits, the potential energy or pressure that drives the flow is represented by voltage or PSI. Just as pipes provide a path for water to flow, conductors in an electrical circuit provide a path for electricity. The pump in a water circuit acts as the source of energy, similar to a battery in an electrical circuit. Both valves and switches control or regulate the flow by either opening or closing the circuit or pathway. Restrictions in a water circuit and resistance in an electrical circuit impede the flow and reduce the overall current or flow rate. The water meter and ammeter measure the flow rate or current passing through the circuit. Water itself in a water circuit and electrons in an electrical circuit act as carriers of energy. The high-pressure output and positive voltage represent the part of the circuit with higher potential energy, while the low-pressure intake and negative voltage represent the part with lower potential energy. When a valve is closed, it corresponds to an open circuit, interrupting the flow or current. Conversely, when a valve is open, it can be compared to a closed circuit, allowing the flow or current to pass through. The flow rate in a water circuit, measured in liters/second, is similar to the current in an electrical circuit, measured in amps.
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How might the idea of continental drift explain 300-million-old glacial grooves on four separate southern continents?
The idea of Continental drift explains the presence of 300-million-old glacial grooves on four separate southern continents as they were once joined together and subject to the same climate conditions.
The theory of Continental drift states that the continents were once joined together as a supercontinent called Pangaea and have since drifted apart. This movement has caused the formation of geological features and altered the climate of the earth. 300-million-old glacial grooves have been found on four separate southern continents. This suggests that the continents were once joined together and were subject to the same climate conditions at the time of the formation of these grooves. These continents could have been connected through land bridges or narrow passages, which allowed for the migration of flora and fauna. The movement of these landmasses could have been caused by tectonic activity, which can be linked to the theory of Continental drift.
In conclusion, the idea of Continental drift explains the presence of 300-million-old glacial grooves on four separate southern continents as they were once joined together and subject to the same climate conditions.
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Give the expected hybridization of the central atom for the following molecules or ions.
(a) NO3−
(b) CCl4
(c) NCl3
(d) NO2−
(e) OCN− (carbon is the central atom)
(f) SeCl2
The expected hybridization of the central atom varies depending on the molecular geometry of the molecule or ion.
(a) NO3− - sp2 hybridization
(b) CCl4 - sp3 hybridization
(c) NCl3 - sp3 hybridization
(d) NO2− - sp2 hybridization
(e) OCN− (carbon is the central atom) - sp hybridization
(f) SeCl2 - sp3 hybridization
(a) NO3−:
The central atom in NO3− is nitrogen (N). Nitrogen is bonded to three oxygen (O) atoms. The nitrogen atom forms three sigma bonds with the three oxygen atoms, resulting in a trigonal planar molecular geometry. In a trigonal planar geometry, the nitrogen atom is sp2 hybridized.
(b) CCl4:
The central atom in CCl4 is carbon (C). Carbon is bonded to four chlorine (Cl) atoms. The carbon atom forms four sigma bonds with the four chlorine atoms, resulting in a tetrahedral molecular geometry. In a tetrahedral geometry, the carbon atom is sp3 hybridized.
(c) NCl3:
The central atom in NCl3 is nitrogen (N). Nitrogen is bonded to three chlorine (Cl) atoms. The nitrogen atom forms three sigma bonds with the three chlorine atoms, resulting in a trigonal pyramidal molecular geometry. In a trigonal pyramidal geometry, the nitrogen atom is sp3 hybridized.
(d) NO2−:
The central atom in NO2− is nitrogen (N). Nitrogen is bonded to two oxygen (O) atoms and has one lone pair of electrons. The nitrogen atom forms two sigma bonds with the two oxygen atoms, resulting in a bent molecular geometry. In a bent geometry, the nitrogen atom is sp2 hybridized.
(e) OCN− (carbon is the central atom):
The central atom in OCN− is carbon (C). Carbon is bonded to an oxygen (O) atom, a carbon (C) atom, and has one lone pair of electrons. The carbon atom forms two sigma bonds with the oxygen and carbon atoms, resulting in a linear molecular geometry. In a linear geometry, the carbon atom is sp hybridized.
(f) SeCl2:
The central atom in SeCl2 is selenium (Se). Selenium is bonded to two chlorine (Cl) atoms and has two lone pairs of electrons. The selenium atom forms two sigma bonds with the two chlorine atoms, resulting in a bent molecular geometry. In a bent geometry, the selenium atom is sp3 hybridized.
The expected hybridization of the central atom varies depending on the molecular geometry of the molecule or ion. The hybridization determines the arrangement of the atomic orbitals and is related to the geometry of the molecule.
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Find [H+] of a 0.056 M hydrofluoric acid solution. Ka = 1.45 x 10-7
Answer: [tex][H^+][/tex] of 0.056 M HF solution is [tex]8.96\times 10^{-5}[/tex]
Explanation:
[tex]HF\rightarrow H^+F^-[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Give c= 0.056 M and [tex]\alpha[/tex] = ?
[tex]K_a=1.45\times 10^{-7}[/tex]
Putting in the values we get:
[tex]1.45\times 10^{-7}=\frac{(0.056\times \alpha)^2}{(0.056-0.056\times \alpha)}[/tex]
[tex](\alpha)=0.0016[/tex]
[tex][H^+]=c\times \alpha[/tex]
[tex][H^+]=0.056\times 0.0016=8.96\times 10^{-5}[/tex]
Thus [tex][H^+][/tex] of 0.056 M HF solution is [tex]8.96\times 10^{-5}[/tex]
The combustion of acetylene in the presence of excess oxygen yields carbon dioxide and water:
2C2H2 (g) + 5O2 (g) ----> 4CO2 (g) + 2H2O (l) The value of delta So for this reaction is __________ J/K (Answer: +122.3)
To determine the value of ΔSo (change in entropy) for the given reaction, we need to consider the difference in the number of moles of gas between the reactants and the products.
Reactants:
2 moles of C2H2 (g)
5 moles of O2 (g)
Products:
4 moles of CO2 (g)
2 moles of H2O (l)
The change in the number of moles of gas is given by Δn = (moles of gas in products) - (moles of gas in reactants).
Δn = (4 moles of CO2 + 2 moles of H2O) - (2 moles of C2H2 + 5 moles of O2)
= 4 - 2 + 2 - 5
= -1
The ΔSo value can be calculated using the equation ΔSo = ΣnΔSo(products) - ΣnΔSo(reactants).
Since Δn is -1, we have:
ΔSo = (4 mol x ΔSo(CO2) + 2 mol x ΔSo(H2O)) - (2 mol x ΔSo(C2H2) + 5 mol x ΔSo(O2))
Assuming the standard entropy values, we have:
ΔSo = (4 mol x 213.7 J/(mol·K) + 2 mol x 69.9 J/(mol·K)) - (2 mol x 200.8 J/(mol·K) + 5 mol x 205.0 J/(mol·K))
= 122.3 J/K
Therefore, the value of ΔSo for the given reaction is +122.3 J/K.
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Which of the following is the incorrect IUPAC name of a compound?
A. Pent-3-ene
B. Prop-1-en-2-yne
C. 1-methylpropane
D. All are incorrect.
The IUPAC names of the following organic compounds are correct with the given naming conventions:A. Pent-3-eneB. Prop-1-en-2-yneC. 1-methylpropaneD. All are incorrect - This is the incorrect option because all the given options have correct IUPAC names of the organic compounds. Hence, option D is incorrect.
The International Union of Pure and Applied Chemistry (IUPAC) is an organization that establishes a set of rules for the naming of chemical compounds. This is done to make sure that all scientists in the world use the same names for the same compounds. Therefore, the names should be unique and unambiguous. The IUPAC name of a compound provides information about its molecular structure, functional groups, and substituents. Some of the examples are given below:A. Pent-3-ene - It is a five-carbon molecule with a double bond between the third and fourth carbons. Hence, the name of the compound is pent-3-ene.B. Prop-1-en-2-yne - It is a three-carbon molecule with a triple bond between the first and second carbons and a double bond between the second and third carbons. Hence, the name of the compound is prop-1-en-2-yne.C. 1-methylpropane - It is a three-carbon molecule with one methyl group attached to the first carbon. Hence, the name of the compound is 1-methylpropane.
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Listen and select the one of two statements that corresponds to each drawing. October 28 11:59 PM 1 attempt remaining Grade settings External referencesVocabulary list Grammar explanation 136-139 Questions Modelo You see:Illustration of a girl in a record store with headphones on. You hear: a. Ella sale a bailar. / b. Ella oye música. You select: b
The statement that corresponds to the drawing of a girl in a record store with headphones on is " Ella oye música." The correct statement is B.
This means "She is listening to music" in English. The drawing depicts a girl wearing headphones while standing in a record store. This indicates that she is most likely listening to music rather than going out to dance, which is what statement a suggests.The correct statement is b, Ella oye música, which correctly describes what the girl in the drawing is doing. She is listening to music.To sum up, when presented with the drawing of a girl in a record store with headphones on, the statement that corresponds to it is "b. Ella oye música." This means "She is listening to music" in English.For more questions on music
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There is a series of nitrogen oxides with the general formula N?O?. What is the empirical formula of one that contains 63.66% nitrogen?
Answer:
N₂O
Explanation:
Empirical formula is defined as the simplest whole number ratio of atoms presents in a molecule
For a nitrogen oxide that contains 63.66% of nitrogen, the percent of oxygen must be:
100-63.66 = 36.34% Oxygen
This percent is the percent in mass. To solve this question we must convert the mass of each atom to moles in order to find the simplest whole number ratio:
Moles of the atoms:
N = 63.66g * (1mol / 14g) = 4.547 moles N
O = 36.34g * (1mol / 16g) = 2.271 moles O
The ratio N:O is:
4.547 moles N / 2.271 moles O = 2
That means there are 2 atoms of N per atom of O and the empirical formula is:
N₂OWhich elements are
considered "Noble Metals"?
Answer:
ruthenium (Ru), rhodium (Rh), palladium (Pd), osmium (Os), iridium (Ir), platinum (Pt), gold (Au), silver (Ag).
Explanation:
The term "Noble Metals" traditionally refers to a group of metals that are resistant to corrosion and oxidation in moist or chemically aggressive environments. The elements commonly considered noble metals are Gold, Platinum, Palladium, Palladium, etc.
Gold is perhaps the most well-known noble metal. It is highly resistant to corrosion and oxidation. Platinum is another widely recognized noble metal. It is extremely resistant to corrosion and has a high melting point.
Palladium is a noble metal that exhibits excellent chemical stability and resistance to corrosion.
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.Calculate ΔS°rxn for the following. 4 NH3 (g) + 5 O2 (g) à 4 NO (g) + 6 H2O (g)
Substance
S° (J/mol × K)
NH3 (g)
192.8
O2 (g)
205.2
NO (g)
210.8
H2O (g)
188.8
can this be explained step by step?
The standard entropy change for the given reaction is -1052.0 J/K. The entropy change of the reaction is negative, which indicates that the reaction is not spontaneous.
Given, Reaction: 4 NH3 (g) + 5 O2 (g) à 4 NO (g) + 6 H2O (g)SubstanceS° (J/mol × K)NH3 (g)192.8O2 (g)205.2NO (g)210.8H2O (g)188.8The formula for finding the standard entropy of reaction is as follows:ΔS°rxn = ΣS°products – ΣS°reactantsWhere, ΔS°rxn = Standard entropy changeS°products = Sum of the standard entropies of the productsS°reactants = Sum of the standard entropies of the reactants. For the given reaction, the standard entropy change is calculated by finding the difference between the sum of standard entropies of products and the sum of standard entropies of reactants.ΔS°rxn = ΣS°products – ΣS°reactants= [4(S°NO) + 6(S°H2O)] – [4(S°NH3) + 5(S°O2)]= [4(210.8) + 6(188.8)] – [4(192.8) + 5(205.2)]= 843.6 – 1895.6= -1052.0 J/K. Thus, the standard entropy change for the given reaction is -1052.0 J/K. The entropy change of the reaction is negative, which indicates that the reaction is not spontaneous.
In the given question, we are asked to calculate the standard entropy change, ΔS°rxn. The formula for finding the standard entropy of reaction is:ΔS°rxn = ΣS°products – ΣS°reactantsWhere, ΔS°rxn = Standard entropy changeS°products = Sum of the standard entropies of the productsS°reactants = Sum of the standard entropies of the reactants. By using this formula and substituting the values from the table, we get:ΔS°rxn = ΣS°products – ΣS°reactants= [4(S°NO) + 6(S°H2O)] – [4(S°NH3) + 5(S°O2)]= [4(210.8) + 6(188.8)] – [4(192.8) + 5(205.2)]= 843.6 – 1895.6= -1052.0 J/K. Thus, the standard entropy change for the given reaction is -1052.0 J/K.
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1. A sample of oxygen gas has a volume of 150 ml when its pressure is 440 mmHg. If the pressure is
Increased to standard pressure and the temperature remains constant, what will the new volume be?
Answer:86.84
Explanation:
2. What property of a gas molecule affect (increase/decrease) the speed at which it diffuses? A Temperature B Kinetic energy C Vibration D Brownian motion
The property of a gas molecule that affects the speed at which it diffuses is the kinetic energy (Option B).
Diffusion refers to the process of gas molecules spreading out and mixing with other molecules in a medium, such as air. The speed at which diffusion occurs is influenced by the kinetic energy of the gas molecules.
As temperature increases (Option A), the kinetic energy of gas molecules also increases. Higher kinetic energy means that the gas molecules move faster and collide more frequently with each other and with other molecules in the medium. These collisions promote the mixing and spreading out of the gas molecules, leading to faster diffusion.
Vibration (Option C) and Brownian motion (Option D) are related to the movement and behavior of particles but are not directly associated with the speed of diffusion. Vibration refers to the oscillation of particles around a fixed position, while Brownian motion refers to the random motion of particles due to collisions with surrounding molecules.
In summary, it is the kinetic energy, influenced by temperature, that primarily determines the speed at which gas molecules diffuse. Higher kinetic energy leads to faster molecular motion and more rapid diffusion.
Option B is correct.
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b) If 35 grams of copper (II) chloride react with 20 grams of sodium nitrate, how much sodium chloride can be formed? what is the limiting reactant ? what is the excessive reactant ?
Answer:
CuCl₂ is the excessive reactant and NaNO₃ the limiting reactant, 13.7g of NaCl can be formed.
Explanation:
Based on the reaction:
CuCl₂ + 2NaNO₃ → 2NaCl + Cu(NO₃)₂
To solve the problem we must convert the mass of each reactant in order to find limiting reactant as follows:
Moles CuCl₂ -Molar mass: 134.45g/mol-:
35g * (1mol / 134.45g) = 0.26 moles
Moles NaNO₃ -Molar mass: 84.99g/mol-:
20g * (1mol / 84.99g) = 0.235 moles
For a complete reaction of 0.235 moles of NaNO₃ there are required:
0.235 moles NaNO₃ * (1 mol CuCl₂ / 2 mol NaNO₃) = 0.118 moles CuCl₂
As there are 0.26 moles CuCl₂,
CuCl₂ is the excessive reactant and NaNO₃ the limiting reactantAs 2 moles of NaNO₃ produce 2 moles of NaCl, he moles of NaCl produced are: 0.235 moles NaCl. The mass is:
Mass NaCl -Molar mass: 58.44g/mol-:
0.235 moles NaCl * (58.44g / mol) =
13.7g of NaCl can be formedThank you it means alot if you help :)!
Answer:
10.Semi-Solid
11.Liquid
12.Solid
13:Semi-Solid
sub-atomic particles like negatively charged electrons, positively charged
protons and electrically ________ neutrons
Answer:
neutral
Explanation:
There are three basic subatomic particles. These are;
Protons (positively charged)Electrons (negatively charged)Neutrons (neutral)A neutron has no charge unlike the proton and the electron. It is present in the nucleus and contributes to the mass of the atom.
Write an essay about the harm ful effects of sunlight to people. Minimum of ten sentences cite some expale of possible
Answer: See explanation
Explanation:
The sun is a star that's made up of several gases. It is known to be the most vital source of energy to living things.
Despite the importance of the sun in our lives, it's still harmful when one is exposed to it. They rays of light that's given our by the sun can be harmful to us.
Being exposed to the sun can cause sunburn and this can result in the damage of skin cells and development of cancer. Also, sunlight can lead to early aging as the skin ages faster when one is exposed to the sun.
Furthermore, exposure to sunlight can lead to eye injuries the tissue in our eyes can be damaged by the UV rays. Lastly, it also brings about lowered immune system.
How many molecules of N204 are in 85.0 g of N2O4?
Answer:
5.56 × 10^23 molecules
Explanation:
The number of molecules in a molecule can be calculated by multiplying the number of moles in that molecule by Avagadro's number (6.02 × 10^23)
Using mole = mass/molar mass
Molar mass of N2O4 = 14(2) + 16(4)
= 28 + 64
= 92g/mol
mole = 85.0/92
= 0.9239
= 0.924mol
number of molecules of N2O4 (nA) = 0.924 × 6.02 × 10^23
= 5.56 × 10^23 molecules
Sooner or later your new school won't feel so strange.get.
I'm hoping the same for my new coaching classes
This is my first time going out to study ngl-
What was the purpose of using water/soap solution for one of the trials?
In one of the trials, the purpose of using water/soap solution was to compare the cleanliness of the hand with washing by water alone.
Hand washing is one of the simplest, most effective ways to avoid getting sick and prevent the spread of germs. Washing your hands with soap and water is still one of the most important steps you can take to avoid getting sick and to avoid spreading germs to others. The purpose of using water/soap solution for one of the trials was to compare the cleanliness of the hand with washing by water alone.The experiment involves two trials to investigate the effectiveness of soap and water in removing bacteria from hands. In one trial, the participant washed their hands with soap and water. While in the other trial, the participant washed their hands with water alone. After washing, their hands were pressed on a petri dish with culture medium to grow the bacteria. Then, the plates were placed in an incubator at 37°C for two days to grow bacteria. The soap and water solution are effective in removing bacteria from hands because the soap helps to lift dirt, grease, and microbes off skin and onto the surfaces of the lather, so that it can be rinsed away by water.
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C.
Calculate the number of moles in 62g of CO2
Answer:
32÷5
I'm just tryna get points I'm sorry
goodluck tho❤
What statement best explains how life functions a unicellular organism are carried out?
Answer:
The structures in the cell work together to perform its life functions
Explanation:
Answer:
What statement best explains how life functions a unicellular organism are carried out?
Explanation:
in a certain viscous (glycerine), incompressible flow field with zero body forces the velocity components are
u = ay - b(cy - y^2)
v = w = 0
where a, b, and c are constants. (a) Use the Navier-Stokes equations to determine an expression for the pressure gradient in the x direction, (b) For what combination of the constants a, b, and c (if any) will the shearing stress, r xyz, be zero at y = 0 where the velocity is zero?
(a) The expression for the pressure gradient in the x direction is given by -∂p/∂x = ρ(u∂u/∂x) - μ(∂[tex]^2u[/tex]/∂x[tex]^2)[/tex].
(b) The combination of constants a, b, and c
To determine the expression for the pressure gradient in the x-direction using the Navier-Stokes equations, we start by writing the x-component of the Navier-Stokes equation for an incompressible flow with zero body forces:
ρ(∂u/∂t + u∂u/∂x + v∂u/∂y + w∂u/∂z) = -∂p/∂x + μ(∂[tex]^2u[/tex]/∂[tex]x^2[/tex] + ∂[tex]^2u[/tex]/∂[tex]y^2[/tex] + ∂[tex]^2u[/tex]/∂[tex]z^2[/tex])
where ρ is the fluid density,
u, v, and w are the velocity components in the x, y, and z directions respectively,
t is time, p is pressure,
μ is the dynamic viscosity,
and (∂/∂x) and (∂[tex]^2/[/tex]∂[tex]x^2[/tex]) denote partial derivatives with respect to x.
Given the velocity components u and v provided, we can see that v and w are both zero. Therefore, the x-component of the Navier-Stokes equation simplifies to:
ρ(∂u/∂t + u∂u/∂x) = -∂p/∂x + μ(∂[tex]^2u[/tex]/∂[tex]x^2)[/tex]
Since the flow is steady (there is no time dependence) and we are interested in the expression for the pressure gradient (∂p/∂x), we can neglect the (∂u/∂t) term.
The equation then becomes:
ρ(u∂u/∂x) = -∂p/∂x + μ(∂[tex]^2u[/tex]/∂[tex]x^2[/tex])
Rearranging the terms, we get:
-∂p/∂x = ρ(u∂u/∂x) - μ(∂[tex]^2u[/tex]/∂[tex]x^2[/tex])
Now, let's address part (b) of the question. We are looking for a combination of constants a, b, and c that will make the shearing stress, τ[tex]_{xyz},[/tex] zero at y = 0 where the velocity is zero.
The shearing stress τ[tex]_{xyz}[/tex] is given by:
τ[tex]_{xyz}[/tex]= μ(∂u/∂y + ∂v/∂x)
Since v is zero and we are interested in the point y = 0, the expression simplifies to:
τ[tex]_{xyz}[/tex] = μ(∂u/∂y)
Evaluating this expression at y = 0, we have:
τ_xyz|y=0
= μ(∂u/∂y)|y
=0
Given u = ay - b(cy - [tex]y^2[/tex]), we can find (∂u/∂y)|y=0 by taking the partial derivative with respect to y and evaluating it at y = 0:
(∂u/∂y)|y=0
= a - b(c - 2y)|y
=0
= a - b(c - 0)
= a - bc
Therefore, for the shearing stress to be zero at y = 0, we need:
τ[tex]_{xyz}[/tex]|y=0
= μ(∂u/∂y)|y
=0
This implies that a - bc = 0, or a = bc.
In summary:
(a) The expression for the pressure gradient in the x direction is given by
-∂p/∂x = ρ(u∂u/∂x) - μ(∂[tex]^2u[/tex]/∂x[tex]^2)[/tex].
(b) The combination of constants a, b, and c
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Which indicator would show a pH change from 6 to 7?
A. Red litmus indicator
B. Methyl red indicator
C. Phenol red indicator
D. Blue litmus indicator
Answer:
c
Explanation:
1. litmus paper is used when showing a change between a greater range in ph levels - so A and D are automatically a no.
2. methyl red is used to show a range in ph levels between 4.8-6
3. Option C is the only one left so im going to assume its C because its definitely not A, B, or D
In this experiment it takes about 10 microliters of solution to produce a spot 1 cm in diameter. If the Co (NO3)2 solution contains about 6 g Co2+ per liter, how many micrograms of Co2+ ion are there in one spot? 1 microliter - 1E-6 L 1 microgram = 1E-6 g
If the Co (NO₃)₂ solution contains about 6 g Co₂+ per liter, then there are 60 micrograms of Co₂+ ions in one spot.
To calculate the number of micrograms of Co₂+ ions in one spot, we need to convert the given concentration from grams per liter (g/L) to micrograms per microliter (µg/µL) and then multiply it by the volume of the spot.
Given:
Volume of solution for one spot = 10 µL
Concentration of Co₂+ ions in the solution = 6 g/L
First, we need to convert the concentration from grams per liter to micrograms per microliter:
6 g/L = 6 × (1E+6) µg/L (since 1 g = 1E+6 µg)
= 6 × (1E+6) µg / (1E+6) µL (since 1 L = 1E+6 µL)
= 6 µg/µL
Now, we can calculate the number of micrograms of Co₂+ ions in one spot:
Number of micrograms of Co₂+ ions = Concentration of Co₂+ ions × Volume of solution for one spot
= 6 µg/µL × 10 µL
= 60 µg
Therefore, there are 60 micrograms of Co₂+ ions in one spot.
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Which claim about the universality of gravity is not supported by evidence?
Answer:
b because i said so
Explanation:
Why isn’t there a lunar eclipse every time Earth is in between the sun and the Moon?
Answer: Exploratorium Senior Scientist Paul Doherty explains why not—the orbit of the moon is tilted relative to the orbit of the Earth around the sun, so the moon often passes below or above Earth. At those times, it does not cross the line between the sun and the Earth, and therefore does not create a solar eclipse.
HOPE THIS HELPS
Explain why has a higher boiling point than NH3
Explanation:
Melting point = -33.34°cboiling point=77.73°csomeone help me on these two 2
Answer:
Question 4 is- Solubility
Question 5 is- Suspension
Hopes this helps >:D
if the halo of our falazy is sypherically symmetric what is the mass density
if the halo of our galaxy is spherically symmetric and has a constant mass density, the mass density remains the same at all radii within the halo. This assumption simplifies the calculation, allowing us to consider a uniform mass density throughout the spherically symmetric halo.
If the halo of our galaxy is spherically symmetric, we can make certain assumptions about its mass distribution. Let's consider a simplified model where the halo has a constant mass density throughout its volume.
In this case, the mass density (ρ) represents the amount of mass per unit volume. To calculate the mass density, we divide the total mass of the halo by its volume. However, since we do not have the specific values for the total mass or the volume, we will express the mass density in terms of an equation.
Let's denote the total mass of the halo as M and the volume of the halo as V. The mass density (ρ) is then given by:
ρ = M / V
Since we assume the halo is spherically symmetric, we can use the formula for the volume of a sphere:
V = (4/3)πr³
where r represents the radius of the sphere.
To obtain the mass density as a function of the radius (ρ(r)), we need to find an expression for the total mass (M) in terms of the radius.
Assuming a constant mass density throughout the halo, the mass (M) enclosed within a sphere of radius r is given by:
M = ρ * V = ρ * (4/3)πr³
Substituting this expression for M into the equation for mass density, we have:
ρ(r) = (ρ * (4/3)πr³) / ((4/3)πr³)
Simplifying the equation, we find that the mass density is constant and independent of the radius:
ρ(r) = ρ
Therefore, if the halo of our galaxy is spherically symmetric and has a constant mass density, the mass density remains the same at all radii within the halo. This assumption simplifies the calculation, allowing us to consider a uniform mass density throughout the spherically symmetric halo.
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