Answer: -2.373 x 10^-24J/K(particles
Explanation: Entropy is defined as the degree of randomness of a system which is a function of the state of a system and depends on the number of the random microstates present.
The entropy change for a particle in a system depends on the initial and final states of a system and is given by Boltzmann equation as
S = k ln(W) .
where S =Entropy
K IS Boltzmann constant ==1.38 x 10 ^-23J/K
W is the number of microstates available to the system.
The change in entropy is given as
S2 -S1 = kln W2 - klnW1
dS = k ln (W2/W1)
where w1 and w2 are initial and final microstates
from the question, W2(final) = 0.842 x W1(initial), so:
= 1.38*10-23 ln (0.842)
=1.38*10-23 x -0.1719
= -2.373 x 10^-24J/K(particles)
A rule of thumb is that a reaction rate roughly doubles for every 10 °C increase in temperature. What is the activation energy of a reaction whose rate exactly doubles between 25.0 °C and 35.0 °C
Answer:
FOR EVERY 10 DEGREE CELSIUS INCREASE IN TEMPERATURE, THE ACTIVATION ENERGY THAT SHOWS THIS IS 52.4 KJ/MOL
Explanation:
From Arrhenius equation, the relationship between the rate constant and the temperature is as shown below:
k = Ae^ -Ea/RT
At initial temperature T1, the initial rate constant is (k1)
At final temperature T2, the final rate constant is k2
For the reaction rate to be doubled, we must double the rate constant which shows that the ratio of k2 / k1 must be equal to 2.
That is, k2 / k1 = 2 (rate is doubled)
Equating this into the Arrhenius equation, we have:
k2 / k1 = Ae^ (-Ea / R ) (1/ T2 - 1/T1)
2 = e^ (-Ea / R) (1 / T2 = 1 / T1)
Taking the natural logarithm of both sides:
ln 2 = - (Ea / R) (1 / T2 - 1 / T1)
Making Ea the subject of the formula, we obtain:
Ea = - (ln 2 R / (1 / T2- 1 / T1))
Let T1 = 25 C = 25 + 273 K = 298 K
T2 = 35 C = 35 + 273 K = 308 K
R = 8.314
So,
Ea = - (ln 2 * 8.314 / ( 1/308 - 1 / 298))
Ea = - (0.693 * 8.314 / 0.00324 - 0.00335)
Ea = - 5.7616 / -0.00011
Ea = 52 378,18 J / mol
So therefore, the activation energy Ea is 52.4 kJ/mol.
A 11.0 mLmL sample of 0.30 MHBrMHBr solution is titrated with 0.16 MNaOHMNaOH. Part A What volume of NaOHNaOH is required to reach the equivalence point? Express the volume to two significant figures and include the appropriate units. nothingnothing
Answer:
21 mL of NaOH is required.
Explanation:
Balanced reaction: [tex]HBr+NaOH\rightarrow NaBr+H_{2}O[/tex]
Number of moles of HBr in 11.0 mL of 0.30 M HBr solution
= [tex](\frac{0.30}{1000}\times 11.0)[/tex] moles = 0.0033 moles
Let's say V mL of 0.16 M NaOH solution is required to reach equivalence point.
So, number of moles of NaOH in V mL of 0.16 M NaOH solution
= [tex](\frac{0.16}{1000}\times V)[/tex] moles = 0.00016V moles
According to balanced equation-
1 mol of HBr is neutralized by 1 mol of NaOH
So, 0.0033 moles of HBr are neutralized by 0.0033 moles of NaOH
Hence, [tex]0.00016V=0.0033[/tex]
[tex]\Rightarrow V=\frac{0.0033}{0.00016}=21[/tex]
So, 21 mL of NaOH is required.
12.39 g sample of phosphorus (30.97 g/mol) reacts with 52.54 g of chlorine gas, Cl2
(70.91 g/mol) to form only phosphorus trichloride, PC13 (137.33 g/mol). Which is the
limiting reactant?
Answer:
P is the limiting reagent
Explanation:
P = phosphorus = 30.97g/mol
Cl2 = Chlorine = 70.91g/mol
PCl3 = Phosphorus Trichloride = 137.33g/mol
P + Cl2 = PCl3
Left Side
P = 1
Cl = 2
Right Side
P = 1
Cl = 3
So equation needs to be balanced first
2P + 3Cl = 2PCl3
Left Side
P = 2
Cl = 6
Right Side
P = 2
Cl = 6
That's better.
Ok so we have 12.39g of P so we have 0.4 moles of it
We then have 52.54g of Cl so we have 0.74 moles of it
For every P we need 1.5 Cl so we have an excess of Cl
The compound ClF contains Group of answer choices polar covalent bonds with partial negative charges on the Cl atoms. ionic bonds. nonpolar covalent bonds. polar covalent bonds with partial negative charges on the F atoms.
Answer:
polar covalent bonds with partial negative charges on the F atoms.
Explanation:
A covalent bond could be polar or nonpolar depending on the relative electro negativity difference between the two bonding atoms. In this case, the bonding atoms are chlorine and fluorine.
In the Pauling's scale, fluorine has an electro negativity value of 3.98 while chlorine has an electro negativity value of 3.16. The difference in electro negativity between the two atoms is about 0.82. This magnitude of electro negativity difference between the two bonding atoms correspond to the existence of a polar covalent bond in the molecule.
The direction of the dipole depends on the relative electro negativity values of the two bonding atoms. Since fluorine is more electronegative than chlorine, the fluorine atom will be partially negative and the chlorine atom will be partially positive accordingly.
The compound ClF (chlorine monofluoride) contains polar covalent bonds with partial negative charges on the F atoms. Therefore, option D is correct.
In ClF, chlorine (Cl) is more electronegative than fluorine. As a result, the shared electrons in the Cl-F bond are pulled closer to the chlorine atom, creating a partial negative charge on the fluorine atoms and a partial positive charge on the chlorine atom.
This polarity in the Cl-F bond gives the molecule an overall polarity, making it a polar molecule. Thank you for pointing out the error, and I apologize for any confusion caused.
Thus, option D is correct.
To learn more about the polar covalent bonds, follow the link:
https://brainly.com/question/28295508
#SPJ6
What is the temperature at which the substance can be both in the solid and the liquid phase?
Answer: Gas–liquid–solid triple point
The single combination of pressure and temperature at which liquid water, solid ice, and water vapor can coexist in a stable equilibrium occurs at approximately 273.1575 K (0.0075 °C; 32.0135 °F) and a partial vapor pressure of 611.657 pascals (6.11657 mbar; 0.00603659 atm).
Explanation:
It represents the equilibrium between the liquid and gas phases. The point on this curve where the vapor pressure is 1 atm is the normal boiling point of the substance. The vapor-pressure curve ends at the critical point (B), which is at the critical temperature and critical pressure of the substance.
Covalent bonds can be best described as
Answer:
neutral atoms coming together to share electrons
Answer:
a
Explanation:
neutral atoms coming together to share electrons
A gaseous hydride of Nitrogen
contains its own volume of Nitrogen
and twice its volume of Hydrogen
and has vapour density 16. The
formula of the hydride is.
Select one:
a. NH2
b. NH3
c. N3H
• d. N2H4
Answer:
N2H4
Explanation:
A hydride is a binary compound of hydrogen and another element. Binary compounds contain only two atoms. We have to x-ray the hydrides of nitrogen given in the question in order to make our choice. Remember that we were told that that the hydride contains its own volume of nitrogen and twice its volume of hydrogen.
Now consider the hydride N2H4.
N2H4(g) -----> N2(g) + 2H2(g)
The volume ration of nitrogen gas to hydrogen gas in N2H4 is 1:2.
The molecular mass of the compound is;
N2H4= 2(14) + 4(1)= 28+4= 32
Since
molecular mass= 2 vapour density
Vapour density= molecular mass/2
Vapour density= 32/2
Vapour density = 16
Therefore the hydride of nitrogen referred to in the question is N2H4
A gas has volume of 800.0mL at -23.0°c and 300.0torr. What would the volume of the gas be at 227.0°c and 600.0torr of pressure
Answer:
Explanation:
use gas law eqation
P1 * V1 / T1 = P2 * V2 /T2
600*V1/227 = 300*800/23
V1 = 300*800*227 / 23*600 = ............ can you solve this and get the answer?
30
Drag each number to the correct location on the equation. Each number can be used more than once, but not all numbers will be used.
Balance the equation with the correct coefficients.
2
3
4
5
SIO2 +
HF → SiF4 +
H2O
Reset
Next
Answer:
The balanced equation is given below:
SiO2 + 4HF —> SiF4 + 2H2O
The coefficients are 1, 4, 1, 2
Explanation:
The equation for the reaction is given below:
SiO2 + HF —> SiF4 + H2O
The above equation can be balance by as follow:
There are 4 atoms of F on the right side and 1 atom on the left side. It can be balance by putting 4 in front of HF as shown below:
SiO2 + 4HF —> SiF4 + H2O
Therefore are 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:
SiO2 + 4HF —> SiF4 + 2H2O
Now the equation is balanced.
Name the advantages of coronavirus
Answer: Positive environmental changes.
Explanation: Without many humans around, the environment has been getting better as more sea life have been spotted in places they haven't been for decades, as well as clearer waters and less rubbish about. Pollution levels have dropped as there are barley any planes in the sky and not many cars about.
Answer:
honestly,i can say that socially being away from people reduces stress
Explanation:
1.Draw the born-Haber lattice energy cycle for sodium chloride. Explain the concept of resonance using the nitrate ion structure.
Answer:
1. Born Haber cycle is used to calculate enthalpy of formation of an ionic solid
2. Resonance structures are used to represent the bonding in some chemical species.
Explanation:
The Born–Haber cycle is a method popularly known in chemistry used in computing enthalpy. The enthalpy of formation of an ionic solid cannot be measured directly. The lattice enthalpy refers to the enthalpy change involved in the formation of an ionic compound from gaseous ions the process is exothermic process. A Born–Haber cycle works on the principle of Hess's law. It can be used to calculate the lattice enthalpy by comparing the standard enthalpy change of formation of the ionic compound from the elements to the enthalpy required to make gaseous ions from the elements.
Resonance is an idea introduced by Linus Pauling to explain chemical bonding from the valence bond perspective. The idea of resonance affords us the opportunity to describe the bonding in certain molecules by combining several structures called chemical or canonical structures. The real structure of the specie lie somewhere between the structures indicated by the resonance structures. The resonance structures of the nitrate ion are shown in the image attached.
A certain metal forms a soluble nitrate salt M(NO3)3. Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0mM solution of M(NO3)3 and the right half cell with a 3.0M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0 C.
Required:
a. Which electrode will be positive?
b. What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode.
Answer:
1.The electrode on the right is positive
2. 0.058V
Explanation:
The above cell is a concentration cell.
A concentration cell is an electrolytic cell that is made of two half-cells with the same electrodes, but differs in concentrations of the solutions. A concentration cell functions by diluting the more concentrated solution and concentrating the more dilute solution, creating a voltage as the cell reaches an equilibrium thereby transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.
In the above cell, electrons flow from the left electrode (less concentrated) to the right electrode (more concentrated). Therefore, the right electrode is the positive electrode (cathode).
Part 2: Please, see the attachment below for the calculations.
Gold can be separated from sand by panning or by a sluice-box. In panning, water is mixed with the sand and the resulting slurry is swirled in a shallow, saucer-shaped metal pan. In the sluice-box technique, running water is passed over an agitated sand- gold mixture. What physical property and what technique make this separation possible?
Answer:
Decantation by means of difference in relative densities.
Explanation:
The specific gravity (relative density) of the gold to the soil/sand is the physical property exploited in panning gold. The particles with lower density would float whilst the heavier gold sinks lower to the bottom of the pan by gravity and is decanted off.
I hope this explanation is easy to comprehend.
Identify the particle represented by each symbol as an alpha particle, a beta particle, a gamma ray, a positron, a neutron, or a proton.
a. 11P
b. 42He
c. +10e
Answer:
[tex]_1^{1} {P}[/tex] is symbol for proton emission in the nucleus.
[tex]_2^{4} {He}[/tex] symbolises alpha emission, equivalent to helium atom emission of a radioactive particle
[tex]+_1^{0} {e}[/tex] is the radiation symbol for positiron particle. which occurs when beta + radioactive decay occurs
If the sign for delta G is negative (spontaneous process) and the sign for delta S is positive (more disorder) for both dissolving processes, how could one be endothermic (positive delta H) and one be exothermic (negative delta H)
Answer: From your question,
One could be exothermic which means that the final enthalpy will be less than the initial enthalpy. H= Hf-Hi(Hf<Hi).
In Endothermic reaction, the entropy is lowered by absorbing energy in the surronding. By so doing, the surronding losses energy and the reaction is not spontaneous.
H is positive and S (entropy) is positive.
Explanation:
Exothermic reaction is the reaction where heat is released In the surronding which lead to increase in the surronding Temperature.
Endothermic reaction is the reaction that absorb heat from the surronding and decrease the surronding Temperature.
A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction:
2MnO4^-(aq)+16H+(aq)+5Pb(s)-->2Mn^2+(aq)+8H2O(l)+5Pb^2+(aq)
Suppose the cell is prepared with 1.87 M MnO−4 and 1.37 M H+ in one half-cell and 3.23 M Mn+2 and 6.62 M Pb+2 in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.
Answer:
1.63 V
Explanation:
Let us state the reaction equation again for the purpose of clarity;
2MnO4^-(aq)+16H+(aq)+5Pb(s)-->2Mn^2+(aq)+8H2O(l)+5Pb^2+(aq)
The reduction potentials for the two half reaction equations are;
MnO 4 - (aq) + 8H + (aq) + 5e - → Mn2+(aq) + 4H2O(l) Eo=1.51 V
Pb2+(aq) + 2e - → Pb(s) Eo= -0.13 V
E°cell = E°red – E°Ox
E°cell = 1.51 - (-0.13)
E°cell = 1.51 + 0.13
E°cell = 1.64 V
But Q= [Mn^2+]^2 [Pb^2+]^5/[MnO4^-]^2 [H^+]^16
Q= [3.23]^2 [6.62]^5/[1.87]^2 [1.37]^16
Q= 10.43 × 12714.22/3.4969 × 154
Q= 132609.3/538.5226
Q= 246.25
From Nernst equation
E= E° - 0.0592/n log Q
Where n=10
E= 1.64- 0.0592/10 log 246.25
E= 1.64-0.0142
E= 1.63 V
Which of the following obervations would be classified as a physical change? A) Fireworks releasing light B) Antacid fizzing in water C) Steam condensing on a mirror D) Apple turning brown
Answer:
C) Steam condensing on a mirror
Explanation:
This was just a change in the physical state.
A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.
Answer: The molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]
Explanation:
Formula used for Elevation in boiling point :
[tex]\Delta T_b=k_b\times m[/tex]
or,
[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_b-T^o_b =(125.2-120.7)^0C=4.5^0C[/tex]
[tex]k_b[/tex] = boiling point constant = ?
m = molality
[tex]w_2[/tex] = mass of solute (urea) = 55.4 g
[tex]w_1[/tex] = mass of solvent X = 500 g
[tex]M_2[/tex] = molar mass of solute (urea) = 60 g/mol
Now put all the given values in the above formula, we get:
[tex]4.5^oC=k_b\times \frac{55.4g\times 1000}{60\times 500g}[/tex]
[tex]k_b=2.4^0C/m[/tex]
Thus the molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]
Chemical formula for copper gluconate I have 1.4g of Copper gluconate. There is .2g of copper within the copper gluconate. Determine the chemical formula for Copper gluconate with the given information: Copper Gluconate: Cu(C6H11O?)? Cu = 63.55 g/mol H = 12.01 g/mol O = 1.008 g/mol Cu = 63.55 g/mol
Answer:
The simplest chemical formula of the compound is Cu(C₆H₁₁O₇)₂
Explanation:
Given mass of sample = 1.4 g
mass of copper in the sample = 0.2 g
mass of the gluconate =1.4 - 0.2 = 1.2 g
The mole ratio is determined first using the formula;
mole ratio = reacting mass / atomic mass
atomic mass of copper = 63.55
mass of gluconate, C₆H₁₁O₇ = 12*6 + 1*11 + 16*7 = 195 g/mol
mole ratio ( copper : gluconate) = 0.2/63.55 : 1.4/195
mole ratio ( copper : gluconate) = 0.003 : 0.007
convert to whole number ratios by dividing with the smallest ratio
mole ratio ( copper : gluconate) = 0.003/0.003 : 0.007/0.003
mole ratio ( copper : gluconate) = 1 : 2
Therefore, the simplest chemical formula of the compound is Cu(C₆H₁₁O₇)₂
Which of the following is most likely to make scientific knowledge stronger?
A.) Creating new hypotheses
B.) Lack of observational evidence
C.) Few scientists working towards it
D.) Evaluating it with experimentation and argument
Answer:
D) Evaluating it with experimentation and argument.
Explanation:
If you don't have enough observational evidence, then you can't really strengthen your scientific knowledge, because you haven't collected enough information to answer the questions you're asking.
If you only have a few scientists working on a question, it will take longer to find an answer. With more people working on a concept, it can be solved in more time.
It can be good to create new hypotheses, but only if you've already tested your original hypothesis (with experimentation and argument), and it turned out to be wrong. Once you've rejected your first hypothesis, only then should you make a new one.
Answer:
D
Explanation:
Which accurately labels the lysosome?
Answer:
One of the organelles in eukaryotic cells that carry out digestion and waste removal.
Answer:
It's X
Explanation:
Indicate whether each of the following indicates that a physical or chemical change has taken place when a piece of magnesium metal is studied: (a) Can be cut into tiny pieces (b) Fizzling occurs when placed water (c) Light is emitted when burned (d) Turns to ash
Answer:
a) Can be cut into tiny pieces - Physical Change
b) Fizzling occurs when placed water -Chemical Change
c) Light is emitted when burned -Chemical Change
d) Turns to ash -Chemical Change
Explanation:
Which compound would you expect to be least soluble in water? Explain.
a. CCl4
b. CH3Cl
c. NH3
d. KF
Answer: a.CCl4 aka carbon tetrachloride
Explanation:
ionic compounds and polar molecules can be dissolved in water which is a polar solvent.
choice d (KF) is a salt (an ionic compound) and can be dissolved in water /(K+ and F- ions would be formed in water).
choice c (NH3 or ammonia) is a very polar molecule and thus can be dissolved in water(Hydrogen bonding).
choice b (CH3Cl) is slightly polar because the atoms surrounding the central carbon atom are different(3 H atoms and 1 chlorine atom) and can be dissolved in water(Dipole-dipole interaction).
choice a is nonpolar and cannot be dissolved in water.
Best example of potential energy?
Answer:
water stored in a dam
Explanation:
when the water is in dam it is ready to move bit is not moving
The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 50 mg. Find a formula for the mass of the sample that remains after t years. (b) Find the mass after 500 years correct to the nearest milligram. (c) When will the mass be reduced to 40 mg
Answer:
Explanation:
a )
m = m₀ [tex]e^{-\lambda t[/tex]
m is mass after time t . original mass is m₀ , λ is disintegration constant
λ = .693 / half life
= .693 / 1590
= .0004358
m = m₀ [tex]e^{- 0.0004358 t}[/tex]
b )
m = 50 x [tex]e^{-.0004358\times 500}[/tex]
= 40.21 mg .
c )
40 = 50 [tex]e^{-.0004358t[/tex]
.8 = [tex]e^{-.0004358t[/tex]
[tex]e^{.0004358t[/tex] = 1.25
.0004358 t = .22314
t = 512 years .
RUIGA GIRLS
CHEMISTRY FORM 3. 23/06/2020
MR. GICHURU
IZ
1
Narne the elements present in
Common salt
(2 miks)
Hydrated copper (11) Sulphate.
(2 ks)
Sulphuric (VI) acid,
2 Why is a reaction between zinc metal and Nitric acid not suitable for preparing
hydrogen gae in the laboratory
(2 mi)
(1 m)
3.
What is relative atomic mass?
b)
Define 'isotopes
c)Determine the relative atomic mass of element K whose isotople misure occur in
the proportione:
(2 marks)
Calculate the percent saturated fat in the total fat in butter
A 1.00 liter solution contains 0.42 moles nitrous acid and 0.32 moles sodium nitrite .
If 0.16 moles of nitric acid are added to this system, indicate whether the following statements are true or false.
(Assume that the volume does not change upon the addition of nitric acid.)
A. The number of moles of HNO2 will decrease.
B. The number of moles of NO2- will remain the same.
C. The equilibrium concentration of H3O+ will increase.
D. The pH will decrease.
E. The ratio of [HNO2] / [NO2-] will increase
Answer:
E. The ratio of [HNO2] / [NO2-] will increase
D. The pH will decrease.
Explanation:
Nitrous acid ( HNO₂ ) is a weak acid and NaNO₂ is its salt . The mixture makes a buffer solution .
pH = pka + log [ salt] / [ Acid ]
= 3.4 + log .32 / .42
= 3.4 - .118
= 3.282 .
Now .16 moles of nitric acid is added which will react with salt to form acid
HNO₃ + NaNO₂ = HNO₂ + NaNO₃
concentration of nitrous acid will be increased and concentration of sodium nitrite ( salt will decrease )
concentration of nitrous acid = .42 + .16 = .58 M
concentration of salt = .32 - .16 = .16 M
ratio of [HNO₂ ] / NO₂⁻]
= .42 / .32 = 1.3125
ratio of [HNO₂ ] / NO₂⁻] after reaction
= .42 + .16 / .32 - .16
= 58 / 16
= 3.625 .
ratio will increase.
Option E is the answer .
pH after reaction
= 3.4 + log .16 / .58
= 2.84
pH will decrease.
most reactions give off energy in the form of heat and are called what
Answer:
Exothermic reaction.
Explanation:
A filtration system continuously removes water from a swimming pool, passes the water through filters, and then returns it to the pool. Both pipes are located near the surface of the water. The flow rate is 15 gallons per minute. The water entering the pump is at 0 psig, and the water leaving the pump is at 10 psig.
A. The diameter of the pipe that leaves the pump is 1 inch. How much flow work is done by the water as it leaves the pump and enters the pipe?
B. The water returns to the pool through an opening that is 1.5 inches in diameter, located at the surface of the water, where the pressure is 1 atm. How much work is done by the water as it leaves the pipe and enters the pool?
C. "The system" consists of the water in the pump and in the pipes that transport water between the pump and the pool. Is the system at steady state, equilibrium, both, or neither?
Answer:
A . [tex]\mathbf{W = 7133.2 \dfrac{ft. lb_f}{min} }[/tex]
B. [tex]\mathbf{W = 4245.24 \dfrac{ft. lb_f}{min} }[/tex]
C. The system is at steady state but not at equilibrium
Explanation:
Given that:
The volumetric flow rate of the water = 15 gallons per minute
The diameter of the pipe that leaves the pump is 1 inch.
A. The objective here is to determine how much work flow is done by the water as it leaves the pump and enters the pipe
The work flow that is said to be done can be expressed by the relation :
W = P × V
where;
P = pressure
V = volume
Also the given outlet pressure is the gauge pressure
The pressure in the pump P is can now be expressed by the relation:
[tex]P_{absolute} = P_{guage} + P_{atmospheric}[/tex]
[tex]P_{absolute}[/tex] = 10 psig + 14.7 psig
[tex]P_{absolute}[/tex] = 24.7 psig
W = P × V
W = 24.7 psig × 15 gal/min
[tex]W = (24.7 \ psig * \dfrac{\frac{lb_f}{in^2}}{psig}) * ( 15 \frac{gal}{min}* \dfrac{0.1337 \ ft^3}{1 \ gal }* \dfrac{144 \ in^2}{1 \ ft^2})[/tex]
[tex]\mathbf{W = 7133.2 \dfrac{ft. lb_f}{min} }[/tex]
Thus ; the rate of flow of work is said to be done by the water at [tex]\mathbf{W = 7133.2 \dfrac{ft. lb_f}{min} }[/tex]
B.
Given that :
The water returns to the pool through an opening that is 1.5 inches in diameter.
where the pressure is 1 atm.
Then ; the rate of work done by the water as it leaves the pipe and enter the pool is as follows:
W = P × V
W = 1 atm × 15 gal/min
[tex]W = 1 \ atm * ( 15 \frac{gal}{min}* \dfrac{0.1337 \ ft^3}{1 \ gal }* \dfrac{144 \ in^2}{1 \ ft^2})[/tex]
[tex]\mathbf{W = 4245.24 \dfrac{ft. lb_f}{min} }[/tex]
Thus ; the rate of flow of work done by the water leaving the pipe and enters into the pool is at [tex]\mathbf{W = 4245.24 \dfrac{ft. lb_f}{min} }[/tex]
C.
We can consider the system to be at steady state due to the fact that; the data given for the flow rate and pressure doesn't reflect upon the change in time in the space between the pump and the pool.
On the other-hand the integral factor why the system is not at equilibrium is that :
the pressure leaving the pipe is different from that of the water at the surface of the pool as stated in the question.