We now consider three moles of ideal gas at the same initial state (3.0 L at 273 K). This time, we will first perform an isobaric compression, then an isothermal expansion to bring the gas to the final state with the same volume of 3.0 L, and at temperature 110 K. Hint (a) How much work (in J) is done on the gas during the isobaric compression? Wcompression = J (b) How much work (in J) is done on the gas during the isothermal expansion? Wexpansion = J
Answers
Answer:
Explanation:
check the image
a)
The workdone on the gas during during the isobaric compression
W = pΔV
= nRΔT
= 2 * 8.314 * (273-110)
=2710J
b) The workdone on the gas during during the isothermal expansion
[tex]W= nRT_fIn(\frac{V_f}{V_i} )\\\\2*8.314*110In\frac{110}{273} \\\\=-1662.6J[/tex]
Answer:
a) The work done during the isobaric compression is 4060.91 J.
b) The work done during the isothermal expansion is 2489.58 J.
Explanation:
a) Here we have two different processes. We calculate the work done during the isobaric compression using the following formula:
[tex] W= -P. (V₂ - V₁) [/tex]
Since the process is isobaric, the pressure is constant. First we need to find this value. In order to do that, we use the ideal gas equation:
[tex] p.V= n.R.T [/tex]
[tex] p= [n.R.T] ÷ V = [3 mol × 0.082 (L.atm÷mol.K) × 273 K] ÷ 3 L [/tex]
p = 22.39 atm
Now, we need to find V₂. We do this using Charles' law:
[tex] V₁/T₁ = V₂/T₂ [/tex]
[tex] V₂ = V₁/T₁ × T₂ = 3 L ÷ 273 K × 110 K [/tex]
V₂ = 1.21 L
Now we are in condition to calculate the work done during the isobaric compression:
[tex] W= -P. (V₂ - V₁) [/tex]
[tex] W= -22.39 atm. (1.21 - 3) L [/tex]
W= 40. 07 L.atm
We multiply this value by 101.325 to convert it to joules: W= 4060.91 J
b) To calculate the work done during the isothermal expansion, we use the following equation:
[tex] W = n.R.T. ln (V₂/V₁) [/tex]
[tex] W = 3 mol ×0.082 (L.atm/mol.K) × 110 K × ln (3L / 1.21 L) [/tex]
W= 24.57 L.atm
We convert this value to joules: W= 2489.58 J
Related Questions
Predict what will be observed in experiment below.
Experiment:
Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in 500mL of hot water (70°C). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature.
He prepares batch B by dissolving sugar in 500mL of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature.
Predicted observation (Choose one).
O It is likely that more rock candy will be formed in batch A.
O It is likely that less rock candy will be formed in batch A.
O It is likely that no rock candy will be formed in either batch.
O I need more information to predict which batch is more likely to form rock candy.
Answers
Answer:
Option A: It is likely that more rock candy will be formed in batch A.
Explanation:
The difference between batch A and batch B is that batch A uses temperature to dissolve the sugar, while the dissolution of the sugar in batch B is produced at room temperature.
When he use temperature (hot water) to dissolve the sugar he is increasing the solubility of the sugar in the water, so in batch A we will have more quantity of sugar dissolved than in the batch B. The cooling of the solution at room temperature favors the formation of bigger sugar crystals in the process of crystallization.
From all of the above, the correct predicted observation is the option A: It is likely that more rock candy will be formed in batch A, for the increase of the solubility by the use of hot water. Also, you say that Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization.
I hope it helps you!
easy question, need help
Answers
Answer:
210 amps
Explanation:
B
The Colstrip Power Plant
in Montana burns _______
releasing _______
into the air.
A. Iron; HCI
B.Carbon; base(soap)
C. Coal; H2SO4(sulfuric acid)
D. None of the above
Answers
Answer:
Coal; H2SO4(sulfuric acid)
Explanation:
Coal has been very helpful in power generation for decades. It has also become a major contributor to global warming, and has major negative effects on human health and the environment.
Coal is formed when dead plant matter submerged in swamp environments is subjected to the geological forces of heat and pressure over hundreds of millions of years. Over time, the plant matter transforms from moist, low-carbon peat, to coal, an energy- and carbon-dense black or brownish-black sedimentary rock.
There are many types of coal. Each type of coal must contain sulfur, which, when burned, releases toxic air pollution. The extent Sulfur content in a given coal sample is largely decided by the conditions under which the coal is formed. Low-sulfur coal usually develop in freshwater environments; high-sulfur coals are formed in brackish swamps.
The burning of coal releases its sulphur content as oxides of sulphur (SOx). These oxides of sulphur dissolve in rain water to form acid rain. Hence rain falling around the Colstrip power plant will contain H2SO4 resulting from the dissolution of SOx in rain droplets.
The statements in the tables below are about two different chemical equilibria. The symbols have their usual meaning, for example AG Gibbs free energy of reaction and stands for the equilibrium constant. stands for the standard In each table, there may be one statement that is false because it contradicts the other three statements. If you find a false statement, check the box next to It. Otherwise, check the "no false statements box under the table. statement false? statement false? Ink>0
AH°
R<1
AG'>0
AG'>0
In > AH">TAS
no false statements
Answers
Answer:
see explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
Gneiss rock forms from a great amount of pressure and heat underground. What type of rock is it?
Answers
Answer:
Gneiss is a foliated metamorphic rock
Explanation:
Gneiss is a high grade metamorphic rock, meaning that it has been subjected to higher temperatures and pressures .
What determines the average kinetic energy of the particles in a gas?
A.
the number of collisions
B
the number of particles
C
the size of the particles
D.
the temperature
Answers
Answer:
Explanation:
D
The total measure
of angles in a
triangle is 180° the triangle is?
Answers
What is the formula of the ion formed when tin achieves a stable electron configuration?
A. Sn2-
B. Sn3+
C. Sn4+
D. Sn4-
Answers
Answer:
Sn2–
Explanation:
The formula of the ion when tin achieves a stable electron configuration is Sn⁴⁺. Therefore, option (C) is correct.
What are the oxidation states of tin?Tin is a element with the atomic number 50 and the chemical symbol Sn. Tin is present in group 14 as a post-transition metal. Tin has two common oxidation states, +2 and +4. The oxidation state +4 of tin metal is somewhat more stable.
Sn⁴⁺ ion is chemically comparable to both of its neighbor germanium and lead in group 14. The electronic configuration of the tin metal is [Kr] 4d¹⁰5s²5p². It has four electrons in its valence shell so when it losses four electrons it gets configuration with fully filled subshells.
Sn²⁺ is called the stannous ion, while its compound SnCl₂ is known as stannous chloride. Sn⁴⁺ is called the stannic ion, while its compound SnCl⁴ is stannic chloride which is a volatile liquid.
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How do you draw structural formulas 2,4-dimethylhexane; 4-methyl-2-pentene; 4-chloro-7-methyl-2-nonyne
Answers
Answer:
Structural formulas are the graphical representation of chemical compunds and shows the chemical bonds between the atoms of a molecule.
structural formulas of following compunds is attached below:
2,4-dimethylhexane - C8H18, it will have single bonds between carbon atom and will have methyl group at position 2 and 4 carbon.
4-methyl-2-pentene - C6H12, it will have double bond at 2nd carbon and methyl group at 4th carbon.
4-chloro-7-methyl-2-nonyne - C10H17Cl, it will have triple bond at 2nd position, chloride group at 4th carbon and a methyl group at 7th carbon.
When 177. g of alanine (C3H7NO2) are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is 5.9 °C lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is 7.2 °C lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X. Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits.
Answers
Answer:
Van't Hoff factor of KBr is 1.63
Explanation:
Freezing point depression due the addition of a solute follows the formula:
ΔT = Kf×m×i
Where ΔT is change in freezing point, Kf is freezing point depression constant of the solvent X, m is molality of the solution (mol / kg) and i is Van't Hoff factor.
Moles of 177g of alanine (Molar mass: 89.09g/mol) are:
177g × (1mol / 89.09g) = 1.99 moles. In 0.800kg:
1.99mol / 0.800kg = 2.49m
Replacing in freezing point depression formula:
5.9°C = Kf×2.49m×1
Alinine has a Van't Hoff factor of 1
The Kf of the solvent is:
2.37 °C/m
Molality of the 177.0g of KBr solution (Molar mass: 119g/mol) is:
177.0g × (1mol / 119g) = 1.487 moles / 0.800kg = 1.859m
And the freezing point depression formula is:
7.2°C = 2.37°C/m×1.859m×i
1.63 = i
Van't Hoff factor of KBr is 1.63
A solution contains one or more of the following ions: Ag+, Ca2+, and Fe2+. When sodium chloride is added to the solution, no precipitate forms. When sodium sulfate is added to the solution, a white precipitate forms. Then, the precipitate is filtered off and sodium carbonate is added to the remaining solution, and a precipitate forms.
Which of the ions were present in the original solution?
Write net ionic equations for the formation of each of the precipitates observed.
Answers
Explanation:
when NaSO4 added, Ca2+ + SO42- = CaSO4
and Calcium sulphate is a white insoluble precipitate
when NaCO3 added, Fe2+ + CO32- = FeCO3
and it is a green precipitate
What distinguishes effusion from diffision? How are these processes similar?
Answers
Answer:
In effusion, a substance escapes through a tiny pinhole, or other hole whereas diffusion is the spreading out of a substance within a dispersing medium.
Effusion and diffusion are similar in terms of rates and speeds.
Explanation:
According to effusion, if there is a small hole in a container, a gas will leak from it. So, in this process a substance escapes through a tiny pinhole, or other hole.
According to diffusion, the two solutions will reach the lowest possible concentration of both, if they are combined and not mixed. So, diffusion is the spreading out of a substance within a dispersing medium.
Effusion and diffusion are similar in terms of rates and speeds.
A student runs two experiments with a constant-volume "bomb" calorimeter containing 1500.g of water.
First, a 7.500g tablet of benzoic acid C6H5CO2H is put into the "bomb" and burned completely in an excess of oxygen. (Benzoic acid is known to have a heat of combustion of 26.454 kJ/g.) The temperature of the water is observed to rise from 10.00°C to 36.99°C over a time of 13.0 minutes.
Next, 5.260g of ethanol C2H5OH are put into the "bomb" and similarly completely burned in an excess of oxygen. This time the temperature of the water rises from 10.00°C to 28.03°C.
Use this information, to answer the questions below about this reaction:
C2H5OH(l)+ 3O2(g)→ 2CO2(g)+ 3H2O(g)
a. Is this reaction exothermic, endothermic, or neither?
b. If you said the reaction was exothermic or endothermic, calculate the amount of heat that was released or absorbed by the reaction in the second experiment.
c. Calculate the reaction enthalpy ΔHrxn per mole of CO2
Answers
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
Since there has been a rise in the reaction temperature, there has been an exothermic reaction.
The amount of heat energy released in the second step has been -132.54 kJ.
The reaction enthalpy per mole has been -1160.85 kJ/mol.
(a) To determine whether the reaction has been exothermic or endothermic, the heat absorbed or released has been calculated.
Since there has been a rise in the temperature of the solution with the combustion, the reaction has been termed as the exothermic reaction.
(b) Amount of heat released in second experiment:
In the bomb calorimeter:
[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]
[tex]\rm q_w_a_t_e_r[/tex] has been given a:q = mc[tex]\Delta[/tex]T
q = 1500 g × 4.184 J/g.[tex]\rm ^\circ C[/tex] × (36.99 - 10[tex]\rm ^\circ C[/tex] )
[tex]\rm q_w_a_t_e_r[/tex] = 169389.24 J.
[tex]\rm q_b_o_m_b\;[/tex] can be given as:q = C [tex]\Delta[/tex]T
q = c (36.99 - 10[tex]\rm ^\circ C[/tex] )
[tex]\rm q_b_o_m_b\;[/tex] = 26.99 [tex]\rm ^\circ C[/tex] × c
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] can be given by:q = mass × heat of combustion of benzoic acid
q = 7.5 g × 26.454 kJ/g
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = 198405 J
[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]
[tex]\rm q_b_o_m_b\;[/tex] = - ([tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_w_a_t_e_r[/tex])
[tex]\rm q_b_o_m_b\;[/tex] = - (198405 J + 169389.24 J )
[tex]\rm q_b_o_m_b\;[/tex] = 29015.76 J.
[tex]\rm q_b_o_m_b\;[/tex] = 26.99 [tex]\rm ^\circ C[/tex] × c
29015.76 J = 26.99 [tex]\rm ^\circ C[/tex] × c
c of bomb = 1075.05 J/[tex]\rm ^\circ C[/tex].
For the second reaction of combustion of ethanol:
[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]
[tex]\rm q_w_a_t_e_r[/tex] has been given as:q = mc[tex]\Delta[/tex]T
q = 1500 g × 4.184 J/g.[tex]\rm ^\circ C[/tex] × (28.03 - 10[tex]\rm ^\circ C[/tex] )
[tex]\rm q_w_a_t_e_r[/tex] = 113156.28 J.
[tex]\rm q_b_o_m_b\;[/tex] can be given as:q = C [tex]\Delta[/tex]T
q = 1075.05 J/[tex]\rm ^\circ C[/tex] × (28.03 - 10[tex]\rm ^\circ C[/tex] )
[tex]\rm q_b_o_m_b\;[/tex] = 19383.15 J
Moles of ethanol = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles of ethanol = [tex]\rm \dfrac{5.26\;g}{46.07\;g/mol}[/tex]
Moles of ethanol = 0.11417 mol.
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] for ethanol combustion can be given by:q = moles of ethanol × [tex]\Delta[/tex]H of reaction
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = 0.11417 × [tex]\Delta[/tex]H of reaction
[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = - ([tex]\rm q_b_o_m_b\;[/tex] + [tex]\rm q_w_a_t_e_r[/tex])
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = - (19383.15 J + 113156.28 J)
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = -132539.43 J
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = -132.54 kJ.
The amount of heat energy released in the second step has been -132.54 kJ.
(c) The reaction enthalpy per mole can be given as:
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = 0.11417 mol × [tex]\Delta[/tex]H of reaction
-132.54 kJ = 0.11417 mol × [tex]\Delta[/tex]H of reaction
[tex]\Delta[/tex]H of reaction = -1160.85 kJ/mol.
The reaction enthalpy per mole has been -1160.85 kJ/mol.
Since there has been a rise in the reaction temperature, there has been an exothermic reaction.
The amount of heat energy released in the second step has been -132.54 kJ.
The reaction enthalpy per mole has been -1160.85 kJ/mol.
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If I have a 200 L container filled with nitrogen at a pressure of 1.0 atm, how many moles of nitrogen are present at 25 C?
0
Select one:
O a. 0,085 moles
O b. 81.8 moles
O C. 19.3 moles
O d. 8.18 moles
Answers
Hydrogen,H2,and nitrogen,N2(g), combined to form ammonia,NH3(g):3H2+N2–>2NH3(g) what amount, in moles, of nitrogen will react 18 mols of hydrogen
Answers
Answer:
In this reaction, [tex]6\; \rm mol[/tex] of [tex]\rm N_2[/tex] will react with [tex]18\; \rm mol[/tex] of [tex]\rm H_2[/tex].
Explanation:
In the balanced equation for this reaction, the ratio between the coefficient of [tex]\rm H_2[/tex] and [tex]\rm N_2[/tex] is [tex]3 : 1[/tex]. That is: [tex]n(\mathrm{H_2}) : n(\mathrm{N_2}) = 3:1[/tex]. That's the same as saying that for every one mole of [tex]\rm N_2[/tex] consumed, three moles of [tex]\rm H_2[/tex] will be consumed.
Rewrite this ratio as a fraction:
[tex]\displaystyle \frac{n(\mathrm{H_2})}{n(\mathrm{N_2})} = \frac{3}{1}[/tex].
Take the reciprocal of both sides to obtain:
[tex]\displaystyle \frac{n(\mathrm{N_2})}{n(\mathrm{H_2})} = \frac{1}{3}[/tex].
It is given that [tex]18\; \rm mol[/tex] of [tex]\rm H_2[/tex] was consumed; in other words, [tex]n(\mathrm{H_2}) = 18\; \rm mol[/tex]. The question is asking for [tex]n(\mathrm{N_2})[/tex], the number of moles of [tex]\rm N_2[/tex] required. Apply this ratio:
[tex]\begin{aligned}n(\mathrm{N_2}) &= n(\mathrm{H_2}) \cdot \frac{n(\mathrm{N_2})}{n(\mathrm{H_2})}\\ &= 18\; \rm mol \times \frac{1}{3} = 6\; \rm mol\end{aligned}[/tex].
Hence the conclusion: in this reaction, it will take (at least) [tex]6\; \rm mol[/tex] of [tex]\rm N_2[/tex] to react with [tex]18\; \rm mol[/tex] of [tex]\rm H_2[/tex].
A 2.40 kg block of ice is heated with 5820 J of heat. The specific heat of water is 4.18 J•g^-1•C^-1. By how much will it’s temperature rise, assuming it does not melt?
Answers
Answer: The temperature rise is [tex]0.53^0C[/tex]
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed by ice = 5280 J
m = mass of ice = 2.40 kg = 2400 g (1kg=1000g)
c = heat capacity of water = [tex]4.18J/g^0C[/tex]
Initial temperature = [tex]T_i[/tex]
Final temperature = [tex]T_f[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i=?[/tex]
Putting in the values, we get:
[tex]5280J=2400g\times 4.18J/g^0C\times \Delta T[/tex]
[tex]\Delta T=0.53^0C[/tex]
Thus the temperature rise is [tex]0.53^0C[/tex]
Answer: 0.580 C
Explanation: On Ck-12 I got it right sooo...
what kind of air pressure would you find in Bridgeport, Connecticut?
Answers
Answer:
Warm
Explanation: because it would be less hot air population
A balloon with a volume of 5.0 L is filled with a gas at 760 tore. If the pressure is reduced to 389 torr without a change in temperature. What will be the volume of the balloon?
Answers
Answer:
[tex]V_2=9.97L[/tex]
Explanation:
Hello,
In this case, we can apply the Boyle's law in order to understand the pressure-volume relationship as an inversely proportional relationship:
[tex]P_1V_1=P_2V_2[/tex]
Thus, we can solve for the resulting volume at the second state as follows:
[tex]V_2=\frac{P_1V_1}{P_2}=\frac{5.0L*760torr}{389tor} \\\\V_2=9.97L[/tex]
Best regards.
A gas absorbs 21.39 kJ of energy while expanding against 0.276 atm from a volume of 0.0432 L to 1.876 L. What is the energy change of the gas?
Answers
Answer:
Q = 21.896kJ
Explanation:
Q = ?
∇U = 21.39kJ
W = ?
W = P∇V
W = P (V2 - V1)
W = 0.276 × (1.876 - 0.04232)
W = 0.276 × 1.83368
W = 0.5060J
Q = ∇U + W
Q = 21.39 + 0.5060
Q = 21.896kJ
The energy change corresponds to the work done by the system
Answer:
[tex]\Delta E=21.34kJ[/tex]
Explanation:
Hello,
In this case, we should apply the first law of thermodynamics to compute the energy change:
[tex]\Delta E=Q-W[/tex]
Thus, with the given volume change we compute the corresponding work in kJ:
[tex]W=P\Delta V=0.276atm*(1.876L-0.0432L)*\frac{101.325kPa}{1atm}*\frac{1m^3}{1000L}=0.0513kJ[/tex]
Then, we compute the energy change:
[tex]\Delta E=21.39kJ-0.0512kJ\\\\\Delta E=21.34kJ[/tex]
Best regards.
Which of these represents the correctly balanced equation?
HCI + Na2S - H2S + NaCl
4HCI +2Na2S - 2H2S + 4NaCl
2HCI + Na2S - H2S + 2NaCl
2HCI + Na2S - H2S + NaCl
Answers
Answer:
2HCI + Na2S ----> H2S + 2NaCl
Explanation:
For a balanced chemical equation, the number of moles of atoms on the reaction side must equal the number of moles of atoms on the product side, in accordance with the law of conservation of mass.
The correct answer is:
2HCI + Na2S ----> H2S + 2NaCl
Since it has equal number of moles of each atom on both sides of the equation: 2 atoms each of hydrogen, chlorine, and sodium on both sides of the equation as well as 1 atom of sulphur on both sides of the equation.
If the volume of a spherical ball is 113.04 cubic inches, what is the radius?
3 inches
9 inches
18 inches
27 inches
Answers
Answer:
r= 3 inches
Explanation:
V= (4/3)*pi* r^3
113.04= (4/3)*3.14*r^3
113.04*(3/4)= 3.14*r^3
84.78 = 3.14*r^3
84.78/3.14 = r^3
27 = r^3 Take the cubed root of both sides.
r = 3 inches
Based on Table I, which compound dissolves in water by an exothermic process? *
NH4NO3
NH4Cl
NaOH
NaCl
Answers
The compound whose dissolution is exothermic in the list is NaOH.
What is an exothermic reaction?An exothermic reaction is one in which heat is gven out in the process. In an exothermic reaction, the reaction vessel feels hot after the reaction is complete.
The table is not shown here. However, we know that the dissolution of NaOH in water is an exothermic process hence the answer is option C.
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The outer surface of a steel gear is to be hardened by increasing its carbon content; the carbon is to be supplied from an external carbon-rich atmosphere, which is maintained at an elevated temperature. A diffusion heat treatment at 850 °C (1123 K) for 10 min increases the carbon concentration to 0.90 wt% at a position 1.0 mm below the surface. Estimate the diffusion time required at 650 °C (923 K) to achieve this same concentration also at a 1.0-mm position. Assume that the surface carbon content is the same for both heat treatments, which is maintained constant. D = 1.1 x 10-6 m2 /s and Qd = 87,400 J/mol C diffusion in a-Fe. (8 points)
Answers
Answer:
[tex]\mathbf{t_2 = 75.696 \ min}[/tex]
Explanation:
From the question:
The outer surface of a steel gear is to be hardened by increasing its carbon content
Given that :
Diffusion of heat temperature at [tex]T_1[/tex] 850 °C = 1123 K
Diffusion time [tex]t_1[/tex] = 10 min
diffusion after the carbon concentration at a position [tex]x_1[/tex] ( 1.0 mm) below the surface = 0.90 wt%
Preexponential = 1.1 × 10⁻⁶ m²/s
Activation Energy [tex]Q_d[/tex] = 87400 J/mol
We are to determine the time [tex]t_2[/tex] at 650 °C (923 K) to achieve the same diffusion result as at 850 °C (1123 K) for [tex]t_1[/tex] = 10 min
Considering Fick's second law for the condition of Constant surface concentration; we have:
[tex]\frac{Cx-C_0}{C_s-C_0} = 1-erf(\frac{x}{2\sqrt{Dt} } )[/tex] ------ equation (1)
where;
[tex]C_0 =[/tex] concentration of the diffusing solute atom before diffusion
[tex]C_s[/tex] = Constant surface concentration
[tex]C_x[/tex] = Concentration at depth x after time t
[tex]erf(\frac{x}{2\sqrt{Dt} } )[/tex] = Gaussian error function
At some desired specific concentration of solute [tex]C_1[/tex] in an alloy ; the left side of the above equation (1) thus becomes constant ;
i.e [tex]\frac{Cx-C_0}{C_s-C_0} = \mathbf{ constant}[/tex]
Then ; [tex]\frac{x}{2\sqrt{Dt} }[/tex] = constant
[tex]\frac{x^2}{Dt}[/tex] = constant
Dt = constant
Thus; [tex]D_1t_1 = D_2t_2[/tex]
Therefore, the time [tex]t_2[/tex] at 650°C([tex]T_2[/tex] = 923 K) required to produce the same diffusion on result as at 850°C ([tex]T_1[/tex] = 1123 K) for [tex]t_1[/tex] = 10 min is [tex]t_2 = \frac{D_1t_1}{D_2}[/tex]
We need to first determine the Diffusion coefficient at 1123 K and 923 K ( i.e [tex]D_1[/tex] and [tex]D_2[/tex])
At [tex]T_1[/tex] = 1123 K , Diffusion coefficient [tex]D_1[/tex] is calculated by the equation [tex]D_1 = D_0 exp ( - \frac{Q_d}{RT_1})[/tex] (equation from temperature dependence of the diffusion coefficient)
[tex]D_1 = 1.1 * 10^{-6} \ exp ( - \frac{87,400}{8.314*1123} )[/tex]
[tex]D_1 = 9.462*10^{-11} \ m^2/s[/tex]
[tex]D_2 = D_0 exp ( - \frac{Q_d}{RT_2})[/tex]
[tex]D_2 = 1.1 * 10^{-6} \ exp ( - \frac{87,400}{8.314*923} )[/tex]
[tex]D_2 = 1.25*10^{-11} m^2/s[/tex]
[tex]t_2 = \frac{ 9.462*10^{-11}*10}{ 1.25*10^{-11} }[/tex]
[tex]\mathbf{t_2 = 75.696 \ min}[/tex]
What system of units is used by only a small number of countries in the world, including the U.S.?
Answers
Answer:
The correct answer is the imperial system.
Explanation:
Only three countries in the world, that is, the United States, Myanmar, and Liberia use the imperial system. These include measurements in the form of inches, ounces, Fahrenheit, and feet. In the imperial system, the distances, height, weight, or area measurements are used eventually that traced back to everyday items or parts of the body.
In comparison to other metric systems, the units used in the imperial system are not further differentiated easily into parts of hundreds or thousands, and are thus, regarded of less use in comparison to other metric systems by some. The real follower of the imperial system at present in the world is the United States.
When making a fire, without gas, matches, or a lighter, what are two important things about the wood?
Answers
Answer:
that is a solid and that there are other ways to make a fire?
A 1 liter solution contains 0.247 M nitrous acid and 0.329 M sodium nitrite. Addition of 0.271 moles of calcium hydroxide will: (Assume that the volume does not change upon the addition of calcium hydroxide.)
a. Raise the pH slightly
b. Lower the pH slightly
c. Raise the pH by several units
d. Lower the pH by several units
e. Not change the pH
f. Exceed the buffer capacity
Answers
Answer:
a. Raise the pH slightly
Explanation:
We know that
Pka of HNO2/KNO2 =3.39
Moles of HNO2 in the buffer=0.247 mol/L×1L=0.247 moles
Moles of NO2-=0.329mol/L×1L=0.329 moles
If 0.271 moles of Ca(OH)2 is added it will neutralise 0.136 moles of acid ,HNO2,remaining HNO2=0.247-0.136=0.111 moles
Moles of NO2- will increase as 0.0333 moles Ca(NO)2 will be formed =0.0333+0.036=0.0693 moles
pH=pka+log [base]/[acid] {henderson -hasselbach equation}
=3.39+log (0.0693/0.0317)=3.39+0.34=3.73
pH=3.73
As a system becomes less random, its entropy *
decreases
remains the same
increases
Answers
Answer:
decreases
Explanation:
The higher the entropy, the more unpredictable and the more random a value is, this is because entropy describes disorder of a system. If things are less random, that means they are predictable, which means they are more ordered.
How human health is expected to be affected by climate change?
“IN YOUR OWN WORDS”
Answers
A measurement gave a mass of .34kg. This is the same as 340g
Answers
Answer:
Yes it is. 0.34 kg = 340 g (Kilograms and grams)
Explanation:
When we convert from kg to grams we multiply by 1000 because kilo mean 1000.
0.34*1000= 340
If the half life of a first order reaction is 24 days. Calculate the rate constant for the reaction
Answers
Answer:
Half-lives of first order reactions
[A]1/2[A]o=12=e−kt1/2.
ln0.5=−kt.
t1/2=ln2k≈0.693k.