7: A piston cylinder device initially contains 3.6L of air at 137 kPa and 28 C. Air is now compressed to a final state of 830 kPa and 231 C. The useful work input is 1.9 kJ. Assume the surroundings are at 112 kPa and 15 C. The gas constant of air is R = 0.287 kPa.m3 /kg.K. The specific heats of air at the average temperature of 360 K are cp = 1.009 kJ/kg·K and cv = 0.722 kJ/kg·K. a. Determine the entropy change per unit mass, (s2 - s0). b. Determine the volume change per unit mass, (v2 – v0). c. Determine the exergy of the air at the initial and the final states. d. Determine the minimum work that must be supplied to accomplish this compression process. e. Determine the second law efficiency of this process.
Answers
Answer:
See explaination
Explanation:
To calculate the second law of efficiency, temperatures here should be in Kelvin → K = ºC + 273.15 or Rankin = 460 + ºF. 2. Second Law Efficiency Second Law efficiency is a measure of how much of the theoretical maximum (Carnot) you achieve, or in other words, a comparison of the system's thermal efficiency to the maximum possible efficiency.
See attachment for the step by step solution of the given problem.
To solve this problem, we'll follow these steps:
a. Determine the entropy change per unit mass, (s2 - s0): The entropy change per unit mass can be calculated using the following equation:
[tex]\[ \Delta s = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) \][/tex]
where:
- [tex]\( c_p \)[/tex] is the specific heat at constant pressure,
- [tex]\( R \)[/tex]is the gas constant,
- [tex]\( T_1 \) and \( T_2 \)[/tex] are the initial and final temperatures in Kelvin,
-[tex]\( P_1 \) and \( P_2 \)[/tex]are the initial and final pressures.
Given values:
- [tex]\( c_p = 1.009 \) kJ/kg·K[/tex]
- [tex]\( R = 0.287 \) kPa·m³/kg·K[/tex]
-[tex]\( T_1 = 28 + 273.15 \) K[/tex]
- [tex]\( T_2 = 231 + 273.15 \) K[/tex]
- [tex]\( P_1 = 137 \) kPa[/tex]
-[tex]\( P_2 = 830 \) kPa[/tex]
Substitute the values into the entropy change equation to calculate[tex]\( \Delta s \) per unit mass.[/tex]
[tex]\[ \Delta s = 1.009 \ln\left(\frac{231 + 273.15}{28 + 273.15}\right) - 0.287 \ln\left(\frac{830}{137}\right) \][/tex]
b. Determine the volume change per unit mass, (v2 – v0): The volume change per unit mass can be calculated using the equation:
[tex]\[ \Delta v = R \ln\left(\frac{V_2}{V_1}\right) \][/tex]
where:
-[tex]\( V_1 \) and \( V_2 \)[/tex] are the initial and final volumes.
Given values:
- [tex]\( V_1 = 3.6 \) L[/tex]
- [tex]\( V_2 \)[/tex] can be calculated using the ideal gas law:
[tex]\( V_2 = \frac{nRT_2}{P_2} \),[/tex] where \( n \) is the number of moles, [tex]\( R \)[/tex] is the gas constant,[tex]\( T_2 \)[/tex] is the final temperature in Kelvin, and [tex]\( P_2 \)[/tex] is the final pressure.
c. Determine the exergy of the air at the initial and the final states: Exergy [tex](\( X \))[/tex] is given by the equation:
[tex]\[ X = h - h_0 - T_0(s - s_0) \][/tex]
where:
- [tex]\( h \)[/tex] is the enthalpy per unit mass,
-[tex]\( h_0 \)[/tex] is the enthalpy at a reference state,
- [tex]\( T_0 \)[/tex] is the temperature at the reference state,
- [tex]\( s \)[/tex] is the entropy per unit mass,
- [tex]\( s_0 \)[/tex] is the entropy at the reference state.
Given values:
-[tex]\( T_0 = 15 + 273.15 \)[/tex]K (temperature of the surroundings)
- [tex]\( P_0 = 112 \) kPa[/tex] (pressure of the surroundings)
We'll need to calculate [tex]\( h \), \( h_0 \), \( s \), and \( s_0 \)[/tex] for both initial and final states.
d. Determine the minimum work that must be supplied to accomplish this compression process: The minimum work required [tex](\( W_{\text{min}} \))[/tex]for an isentropic process can be calculated using the following equation:
[tex]\[ W_{\text{min}} = h_1 - h_2 \][/tex]
where:
-[tex]\( h_1 \)[/tex] is the enthalpy at the initial state,
- \( h_2 \) is the enthalpy at the final state.
e. Determine the second law efficiency of this process: The second law efficiency[tex](\( \eta \))[/tex] is given by the ratio of the actual work done[tex](\( W_{\text{actual}} \)) to the minimum work (\( W_{\text{min}} \)[/tex]):
[tex]\[ \eta = \frac{W_{\text{actual}}}{W_{\text{min}}} \][/tex]
Given values:
- [tex]\( W_{\text{actual}} = 1.9 \) kJ[/tex]
Now, let's go step by step and calculate each part of the problem.
a. Entropy Change per Unit Mass[tex](\( \Delta s = s_2 - s_0 \))[/tex]:
Using the given specific heats [tex]\( c_p \), \( R \)[/tex], temperatures [tex]\( T_1 \), \( T_2 \)[/tex], pressures [tex]\( P_1 \), and \( P_2 \)[/tex]:
[tex]\[ \Delta s = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) \][/tex]
Substitute the values:
[tex]\[ \Delta s = 1.009 \ln\left(\frac{231 + 273.15}{28 + 273.15}\right) - 0.287 \ln\left(\frac{830}{137}\right) \][/tex]
[tex]\[ \Delta s \approx 1.009 \ln(1.842) - 0.287 \ln(6.0657) \][/tex]
[tex]\[ \Delta s \approx 0.6087 \, \text{kJ/kg·K} \][/tex]
b. Volume Change per Unit Mass [tex](\( \Delta v = v_2 - v_0 \))[/tex]:
First, calculate [tex]\( V_2 \)[/tex] using the ideal gas law:
[tex]\[ V_2 = \frac{nRT_2}{P_2} \][/tex]
where [tex]\( n \)[/tex] is the number of moles of air, which can be calculated using the initial conditions:
[tex]\[ n = \frac{P_1V_1}{RT_1} \][/tex]
Substitute the values and calculate[tex]\( V_2 \)[/tex]:
[tex]\[ n = \frac{137 \times 3.6}{0.287 \times (28 + 273.15)} \][/tex]
[tex]\[ n \approx 1.932 \, \text{kg} \][/tex]
[tex]\[ V_2 = \frac{1.932 \times 0.287 \times (231 + 273.15)}{830} \][/tex]
[tex]\[ V_2 \approx 1.476 \, \text{L/kg} \][/tex]
Now, calculate [tex]\( \Delta v \)[/tex] :
[tex]\[ \Delta v = R \ln\left(\frac{V_2}{V_1}\right) \][/tex]
[tex]\[ \Delta v \approx 0.287 \ln\left(\frac{1.476}{3.6}\right) \][/tex]
[tex]\[ \Delta v \approx -0.5305 \, \text{m}^3/\text{kg} \][/tex]
The negative sign indicates compression.
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Related Questions
Radiation detectors use which of the following properties to detect radioactivity?
○ charge
○ ionizing
○ mass
○ energy
Answers
Answer:
ionizing
I hope I helped :)
Ethanol has a 38% lower energy density by volume than gasoline. Partially offsetting this disadvantage is the higher octane rating of ethanol, which allows it to be used in engines having a higher compression ratio. In fact, a standard gasoline-powered engine typically runs at a compression ratio r = 10, while an ethanol-powered one can run at r = 16. Internal combustion engines can be approximated by the ideal Otto cycle, for which the efficiency is given by eo = 1 − r −0.4. Assume that a real engine has one-third the ideal Otto efficiency, and calculate how much improvement this would make to the efficiency of the ethanol-fueled engine over a standard gasoline engine. Is it enough to offset the lower fuel energy density of ethanol?
Answers
Answer:
Find the given attachment
Calculate the expected pH of the solution at the equivalence point using YOUR AVERAGE VALUES for the concentrations of NaOH and acetic acid and the volumes of each that you used. (Just like in Prelab Q3, you will only have the conjugate base and spectator ions present at the equivalence point in a volume that is the sum of the volumes of acid, water, and base you combined.) The Ka for acetic acid is 1.8 x 10-5.'
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The pH is [tex]pH = 4.94[/tex]
Explanation:
From the question we are told that
The average concentration of NaOH is [tex][NaOH] = 0.101 M[/tex]
The volume of NaOH is [tex]V__{NaOH}} = 15.00 mL[/tex]
The average concentration of Acetic acid is [tex][Acetic \ Acid] =0.497 \ M[/tex]
The volume of Acetic acid is [tex]V__{Acetic \ Acid}} = 5.00 \ mL[/tex]
The chemical equation for this reaction is
[tex]NaOH + CH_3COOH ---> CH_3 COONa + H_2 O[/tex]
The total volume of the solution is
[tex]V__{Total}} = V__{NaOH}} + V__{Acetic \ Acid}}[/tex]
Substituting values
[tex]V__{Total}} = 15 + 5[/tex]
[tex]V__{Total}} = 20mL = 20 *10^{-3} L[/tex]
The number of moles of NaOH is mathematically represented as
[tex]n__{NaOH}} = [NaOH] * V__{NaOH}}[/tex]
substituting values
[tex]n__{NaOH}} = 0.101 * 15*10^{-3}[/tex]
[tex]n__{NaOH}} = 0.001515 \ moles[/tex]
The number of moles of Acetic acid is mathematically represented as
[tex]n__{Acetic acid}} = [Acetic \ acid] * V__{Acetic acid}}[/tex]
substituting values
[tex]n__{Acetic acid}} = 0.497 * 5*10^{-3}[/tex]
[tex]n__{Acetic acid}} = 0.002485\ moles[/tex]
From the chemical equation
1 mole of NaOH reacts with 1 mole of Acetic acid to produce 1 mole of [tex]CH_3 COONa[/tex] salt and 1 mole of [tex]H_2 O[/tex]
So
0.001515 moles of NaOH reacts with 0.001515 moles of Acetic acid to produce 0.001515 moles of [tex]CH_3 COONa[/tex] salt and 0.001515 moles of [tex]H_2 O[/tex]
This implies the number of moles of NaOH remaining after the react would be
[tex]\Delta n__{NaOH}} = 0.001515 - 0.001515[/tex]
[tex]\Delta n__{NaOH}} = 0 \ mole[/tex]
the number of moles of Acetic acid remaining after the react would be
[tex]\Delta n__{Acetic acid}} = 0.002485 - 0.001515[/tex]
[tex]\Delta n__{Acetic acid}} = 0.00097 \ moles[/tex]
the number of moles of [tex]CH_3 COONa \ salt[/tex] remaining after the react would be
[tex]\Delta n__{CH_3 COONa \ salt}} = 0 + 0.001515[/tex]
[tex]\Delta n__{CH_3 COONa \ salt}} = 0.001515 \ moles[/tex]
the number of moles of [tex]H_2 O[/tex] remaining after the react would be
[tex]\Delta n__{H_2O}} = 0 + 0.001515[/tex]
[tex]\Delta n__{H_2O}} = 0.001515 \ moles[/tex]
The expected pH is mathematically evaluated as
[tex]pH = pK_a + log [\frac{[CH_3 COONa]}{[Acetic \ acid]} ][/tex]
Where [tex]pKa[/tex] is mathematically evaluated as
[tex]pK_a = - log (K_a)[/tex]
The concentration of [tex]CH_3 COONa \ salt[/tex] is mathematically evaluated a s
[tex][CH_3 COONa] = \frac{\Delta n_CH_3 COONa \ salt }{V__{Total}}}[/tex]
substituting values
[tex][CH_3 COONa] = \frac{0.001515}{20 *10^{-3}}[/tex]
[tex][CH_3 COONa] = 0.07575M[/tex]
The concentration of Acetic acid is mathematically evaluated as
[tex][Acetic acid] = \frac{\Delta n__Acetic acid}{V__{Total}}}[/tex]
substituting values
[tex][CH_3 COONa] = \frac{0.00097}{20 *10^{-3}}[/tex]
[tex][CH_3 COONa] = 0.0485 M[/tex]
Substituting values into the equation for pH
[tex]pH = - log (1 .8 *10^{-5}) + log [\frac{0.07575}{ 0.0485} ][/tex]
[tex]pH = log [\frac{0.07575}{ 0.0485} ] - log (1 .8 *10^{-5})[/tex]
[tex]pH = log [\frac{1.561856}{1.8*10^{-5}} ][/tex]
[tex]pH = 4.94[/tex]
Which of these substances is the least acidic? solution A, pOH = 1.5 solution B, pOH = 7.0 solution C, pOH = 13.5
Answers
The elementary gas-phase reaction takes place isobarically and isothermally in a PFR where 63.2% conversion is achieved. The feed is pure A. It is proposed to put a CSTR of equal volume upstream of the PFR. Based on the entering molar flow rate to A to the first reactor, what will be the intermediate from the CSTR, X 1 , and exit conversion from the PFR, X 2 , based on the feed to first reactor?
Answers
Answer:
Explanation:
check below for explanation in the attached files.
The molecular formula of butane is C4H10. It is obtained from petroleum and is used commonly in LPG (Liquefied Petroleum Gas) cylinders (a common source of cooking gas). It has two arrangements of carbon atoms: a straight chain and a branched chain. Using this information, draw the structure of the tertiary butyl radical that will form upon removal of a hydrogen atom. Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. Include all free radicals by right-clicking on an atom on the canvas and then using the Atom properties to select the monovalent radical.
Answers
Answer:
See explaination
Explanation:
Please see the attached file for the structural diagram of the molecular formation.
It is detailed and properly presented at the attachment.
The reaction between sucrose and water to produce fructose and glucose is first order // overall. The data below shows the change in concentration of sucrose over time at 298 K. C12H22011 (aq)sucrose + H2O(l)艹C6H1206(aq)fructose + C6H1206(aq)oucose C12H2201l M Time (minutes) 1.002 0.808 0.630 0.0 60.0 130.0 a. Find the average rate ofdisappearance of C12H22011 from t = 0 min to t-60 min. b. Find the average rate of appearance of fructose from t 0 min to t 60 min. c. Calculate the rate constant, k, for the decomposition of sucrose at 25°C. Include units. d. How long will it take for the concentration of sucrose to drop from 1.002 to 0.212 M? e. What is the half-life for the decomposition of sucrose at 25°C?
Answers
Answer:
(a)
Rate of appearance of sucrose = - d[C12H22O11] / dt = - ( 0.808 - 1.002 ) / ( 60.0 - 0.0) = 0.00323 M/s
(b)
Rate of appearance of fructose = d[C6H12O6] / dt = (1.002 - 0.808) / (60.0 - 0.0) = 0.00323 M/s
(c)
k = (1 / t ) * ln[A]/[A]t
k = ( 1 / 60.0 ) * ln[1.002 / 0.808]
k = 0.00359 min-1
(d)
0.00359 = ( 1 / t ) * ln[1.002 / 0.212]
t = 432.6 min
(e)
Half life time = 0.693 / k = 0.693 / 0.00359 = 193 min
Explanation:
First-order reactions are defined as the chemical reactions in which rate of the reaction is linearly dependent on the concentration of only one reactant.
The answers can be explained as:
(a) Rate of appearance of the sucrose from the chemical reaction is:
Rate = [tex]\dfrac{\text d [\text C_{12}\text H_{22}\text O_{11}]}{\text {dt}}[/tex] = [tex]\dfrac{0.808 - 1.002}{60.0 -0.0}[/tex]
Rate = 0.00323 m/s
(b) Rate of appearance of Fructose from the given chemical reaction is:
Rate = [tex]\dfrac{\text d [\text C_{6}\text H_{12}\text O_{6}]}{\text {dt}}[/tex] = [tex]\dfrac{1.002 - 0.808 }{60.0 -0.0}[/tex]
Rate = 0.00323 m/s
(C) Rate constant for the reaction is:
[tex]\text k &= \dfrac{1}{\text t}\times \dfrac {\text{ln [A]}}{\text {[A]} \text t}[/tex]
[tex]\text k &= \dfrac{1}{60}\times \dfrac {\text{ln} (1.002)}{(0.808)}[/tex]
k = 0.00359 minute⁻¹
(d) Time required for the concentration of sucrose to drop from 1.002 to 0.212 M is:
[tex]0.00359 &= \dfrac{1}{t} \times {\text{ln}\dfrac{[1.002]}{[0.212]}[/tex]
t = 432.6 minutes
(e) The half-life of the decomposition of sucrose at 25°C is:
Half-life = [tex]\dfrac{0.693}{\text k} = \dfrac{0.693}{0.00359}[/tex]
Half-life = 193 minutes.
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True False The colors emitted depends on the number of free electrons passing through the lamp. True False When a free electron hits an atom, the atom is always excited to the highest energy level possible. True False The kinetic energy of the free electron at the point of collision increases as the voltage of the battery increases. True False The kinetic energy of the free electron at the point of collision is higher if the atom is closer to the source of electrons. True False The only way to emit IR photons is if there are empty electronic energy levels really close to the ground state (lowest energy level). True False When atomic electrons are excited to a higher level, they always return to their lowest energy level by jumping down one level at a time.
Answers
Answer:
False
False
True
False
False
False
Explanation:
The colour emitted by an excited atom depends on the emission Spectra of the atom rather than on the number of free electrons passing through the lamp.
When a free electrons hits an atom, the atom can be excited to various intermediate states other than the highest energy level depending on the amount of energy it absorbed.
Increase in voltage of the battery also increases the kinetic energy of the electron
The kinetic energy of the electron depends on the voltage difference not on the distance from the atom.
The Paschen, Brackett, and Pfund series of lines are due to transitions from higher-energy orbits to orbits with n = 3, 4, and 5, respectively; these transitions release substantially less energy, corresponding to infrared radiation.
Excited atoms can jump from a higher level to the ground state in a series of steps or directly to the ground state
1. A cylinder of gas has a pressure of 4.40 atm at 25°C. At what temperature in °C
will it reach a pressure of 6.50 atm? Answer 167°C.
Answers
Answer: The final temperature in °C is 167
Explanation:
To calculate the final temperature of the system, we use the equation given by Gay lussac's Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant volume.
Mathematically,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
[tex]P_1=4.40atm\\T_1=25^oC=(25+273)K=298K\\P_2=6.50atm\\T_2=?[/tex]
Putting values in above equation, we get:
[tex]\frac{4.40}{298}=\frac{6.50}{T_2}\\\\T_2=440K=(440-273)^0C=167^0C[/tex]
Thus the final temperature in °C is 167
To determine at what temperature a gas cylinder will reach 6.50 atm pressure from 4.40 atm at 25°C, we use Gay-Lussac's Law and find that the temperature is 167°C.
We can solve this problem using Gay-Lussac's Law, which states that for a given mass and constant volume of an ideal gas, the pressure is directly proportional to its temperature.
The formula for Gay-Lussac's Law is:
P1/T1 = P2/T2
Where:
P1 is the initial pressureT1 is the initial temperature in KelvinP2 is the final pressureT2 is the final temperature in KelvinFirst, we need to convert the initial temperature into Kelvin:
T1 (in Kelvin) = 25°C + 273 = 298 K
Now, we rearrange the equation to solve for T2:
T2 = (P2 × T1) / P1
Substitute the known values into the formula:
T2 = (6.50 atm × 298 K) / 4.40 atm
T2 ≈ 440 K
Finally, convert the temperature back to Celsius:
T2 (in °C) = 440 K - 273 = 167°C
Therefore, the temperature at which the gas will reach a pressure of 6.50 atm is 167°C.
On January 2, 2020, Nash Company purchases a call option for $290 on Merchant common stock. The call option gives Nash the option to buy 1,050 shares of Merchant at a strike price of $51 per share. The market price of a Merchant share is $51 on January 2, 2020 (the intrinsic value is therefore $0). On March 31, 2020, the market price for Merchant stock is $54 per share, and the time value of the option is $210.
Answers
Final answer:
A call option is a financial contract that gives the buyer the right to purchase shares of a stock at a predetermined price. The net profit from this call option transaction can be calculated by determining the difference between the market price and the strike price, multiplied by the number of shares.
Explanation:
A call option is a type of financial contract that gives the buyer the right, but not the obligation, to purchase a specified number of shares of a stock at a predetermined price within a certain time frame. In this case, Nash Company purchased a call option to buy 1,050 shares of Merchant stock at a strike price of $51 per share. On March 31, 2020, the market price for Merchant stock is $54 per share, and the time value of the option is $210.
The net profit from this stock transaction can be calculated by determining the difference between the market price of the stock and the strike price, multiplied by the number of shares:
Net Profit = (Market Price - Strike Price) x Number of Shares
Using the given information, the net profit from this call option transaction would be:
Net Profit = ($54 - $51) x 1,050 = $3 x 1,050 = $3,150
A 3-column table with 4 rows. The first column labeled Scenario has entries A, B, C, D. The second column labeled Object 1 has entries lotion at 26 degrees Celsius, lasagna at 170 degrees Celsius, ocean water at 25 degrees Celsius, eggs at 2 degrees Celsius. The third column labeled Object 2 has entries skin at 37 degrees Celsius, plate at 20 degrees Celsius, lava 950 degrees Celsius, sidewalk at 80 degrees Celsius.
Indicate the direction of heat flow in each scenario.
Scenario A:
Scenario B:
Scenario C:
Scenario D:
Answers
The direction of heat flow in each scenario will be:
Scenario A = toward object 1Scenario B = toward object 2Scenario C = toward object 1Scenario D = toward object 1What is heat transfer?It should be noted that heat transfer means the transfer of heat from a location to another.
It is the process in which the molecules are moved from a region of higher temperature to that of a lower temperature.
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Answer:
It Is Scenario D.
Explanation:
Determine the amount of energy required to boil 50 g of
ethanol.
Answers
Answer:
42050 J.
Explanation:
Data obtained from the question:
Mass (M) of ethanol = 50g
Heat of vaporisation (ΔHv) of ethanol = 841 J/g
Heat (Q) =.?
The heat required to boil 50g of ethanol can be obtained as follow:
Q = MΔHv
Q = 50 x 841
Q = 42050 J.
Therefore, the heat required to boil 50g of ethanol is 42050 J.
Calculate the percent oxygen in sodium oxide.
Answers
Answer:
% is 51.6% of Oxygen
Explanation:
According to the working in the photo
51.6% is the percent of oxygen in sodium oxide. It represents the amount of solute in a given amount of solution.
What is mass percentage?A concentration is expressed as a mass percent. Furthermore, it defines the component of a particular blend. The solution component can be expressed in terms of mass percentage. It represents the amount of solute in a given amount of solution.
The amount of solute is measured in mass or moles. We shall study the mass percent equation using numerous solved numerical examples in this post. Mass is often represented in grams, although any unit of measurement is acceptable as long as an equivalent unit is used both for the component or solute mass and thus the total or solution mass.
mass percentage= (mass of oxygen / mass of sodium oxide)×100
=(32/62)×100
= 51.6%
Therefore, 51.6% is the percent of oxygen in sodium oxide.
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A geochemist in the field takes a sample of water from a rock pool lined with crystals of a certain mineral compound . He notes the temperature of the pool, , and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to . Then he filters it and evaporates all the water under vacuum. Crystals of are left behind. The researcher washes, dries and weighs the crystals. They weigh 3.00 g.
a. Using only the information above, can you calculate the solubility of X in water at 26 degrees Celsius.?
b. If yes than calculate the solubility of X. Round your answer to 3 significant digits
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Which of these statements indicates an experiment is systematic?
A. The experiment is repeated many times.
B. The experiment is about a testable question.
C. The experiment is very complicated and uses lots of chemicals.
D. The experiment includes a good plan for what measurements to take.
Answers
Answer:
D. The experiment includes a good plan for what measurements to take.
Explanation:
A balloon at 30.0°C has a volume of 222 mL. If the temperature is
increased to 53.1°C and the pressure remains constant, what will the new volume be, in ml?
Answers
Answer : The new volume will be, 238.9 mL
Explanation :
Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles of gas.
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=222mL\\T_1=30.0^oC=(30.0+273)K=303K\\V_2=?\\T_2=53.1^oC=(53.1+273)K=326.1K[/tex]
Putting values in above equation, we get:
[tex]\frac{222mL}{303K}=\frac{V_2}{326.1K}\\\\V_2=238.9mL[/tex]
Therefore, the new volume will be, 238.9 mL
The isothermal, first-order reaction of gaseous A occurs within the pores of a spherical catalyst pellet. The reactant concentration halfway between the external surface and the center of the pellet is equal to one-fourth the concentration at the external surface. (a) What is the relative concentration of A near the center of the pellet? (b) By what fraction should the pellet diameter be reduced to give an effectiveness factor of 0.77
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
17
¿Cuál de las siguientes
propiedades distingue una solución
sobresaturada de una diluida?
Answers
Answer:
Which of the following
properties distinguishes a solution
oversaturated with a dilute?
The supersaturated solution is one in which the solvent has dissolved more solute than it can dissolve in the saturation equilibrium. The solute can be a solid, or a gas. The molecules of the solvent surround those of the solute and seek to open space between themselves to be able to harbor more amount of solute.A dilute solution is a solution that has not reached the maximum concentration of solute dissolved in a solvent. The additional solute will dissolve when added in a dilute solution and will not appear in the aqueous phase. It is considered a state of dynamic equilibrium where the speeds in which the solvent dissolves the solute are greater than the recrystallization rate.At pH0 the more thermodynamically stable species is ______ and the most strongly oxidizing species is _______ . At pH14 the more thermodynamically stable species is _______ and the most strongly oxidizing species is ______ . At pH0 ______ will tend to disproportionate and pH14 ________will tend to disproportionate. Under acidic conditions mixing Mn3+(aq) and MnO42- (aq) will result in the formation of______.
Answers
Answer:
At pH0 the more thermodynamically stable species is lower and the most strongly oxidizing species is higher. At pH14 the more thermodynamically stable species is higher and the most strongly oxidizing species is lower . At pH0 convex curve will tend to disproportionate and pH14 concave curce will tend to disproportionate. Under acidic conditions mixing Mn3+(aq) and MnO42- (aq) will result in the formation of redox oxidation-reduction reaction.
A 100g sample of Carbon-14 has a half-life of 5 years. How much Carbon-14 is left after 10 years?
Answers
Answer: The carbon-14 left after 10 years is 25 g
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
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for rate constant
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{5years}=0.139years^{-1}[/tex]
b) for amount left after 10 years
[tex]t=\frac{2.303}{k}\log\frac{a}{(a-x)}[/tex]
[tex]10=\frac{2.303}{0.139}\log\frac{100}{(a-x)}[/tex]
[tex](a-x)=25g[/tex]
Thus carbon-14 left after 10 years is 25 g
The magnesium and calcium ions present in seawater ([Mg2+] = 0.059 M and [Ca2+] = 0.011 M) can be separated by selective precipitation with KOH.
If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×10−13.)
Answers
Answer:
the minimum [OH−] triggers precipitation = 2.3x10^-6 M
Explanation:
Mg2+ + 2OH- --> Mg(OH)2(s)
Ksp = [Mg2+][OH-]^2
(2.06x10^-13) = (0.039*x^2)
x = [OH-] = 2.3*10^-6 M
the minimum [OH−] triggers precipitation = 2.3x10^-6 M
Final answer:
The minimum [OH-] that triggers the precipitation of Mg2+ ion can be calculated using the Ksp value.
Explanation:
The minimum [OH-] that triggers the precipitation of Mg2+ ion can be calculated using the Ksp value. Given that the concentration of Mg2+ in the solution is 0.039 M, and the Ksp value is 2.06×10⁻¹³, the minimum [OH-] can be determined to initiate precipitation.
when 1564 J of heat energy is added to a sample of gold at 25.0°C, the temperature of the gold increases to 449°C. what is the means of the gold sample? (specific heat of gold= 0.129 J/g-°C
Answers
Answer:
The mass of the gold sample is 28.59 g
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system. The amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without changing its physical state (solid, liquid or gaseous) is calculated by:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
Q= 1564 Jc= 0.129 [tex]\frac{J}{g*C}[/tex]m=?ΔT= Tfinal - Tinitial= 449°C - 25°C= 424 °CReplacing:
[tex]1564 J=0.129\frac{J}{g*C}*m*424 C[/tex]
Solving:
[tex]m=\frac{1564 J}{0.129\frac{J}{g*C}*424 C}[/tex]
m=28.59 g
The mass of the gold sample is 28.59 g
what kind of solution would have a Ka value that is much less than 1
Answers
Answer:
A weakly acidic solution.
Explanation:
Final answer:
A solution with a Ka value much less than 1 refers to a weak acid, where only a small portion of the acid is ionized, resulting in relatively low concentrations of hydronium ions and the conjugate base.
Explanation:
The type of solution that would have a Ka value much less than 1 is a weak acid solution. This is because the Ka (acid dissociation constant) value indicates the strength of the acid in solution; the lower the Ka, the weaker the acid. For a Ka value significantly less than 1, this means that only a small fraction of the acid molecules donate protons (H+) to the water, resulting in few hydronium ions (H₋₃O⁺) and the conjugate base (A-).
In contrast, strong acids have a Ka value that approaches infinity since they are almost completely ionized in solution. Acetic acid, with a Ka value of 1.75 x 10⁻⁵, is an example of a weak acid, indicating that in equilibrium, the concentration of the undissociated acetic acid is much greater than the concentrations of the acetate and hydronium ions.
On occasion, it has been found that the oxidation of borneol doesn't go to completion (possibly because of poor stirring or insufficient Oxone). This reaction would probably be easy to monitor via TLC, however. Which component should have a lower Rf, and why
Answers
Answer:
Check the explanation
Explanation:
Here, Nitrogen (N) undergoes oxidation and Chlorine (Cl) undergoes reduction.
To answer your question:
N is oxidized from an oxidation number of -3 to an oxidation number of -1.
Cl is reduced from oxidation number of +1 to an oxidation number of -1.
Now,
Borneol should have a lower Rf because of boiling point.
The ph of an aqueous solution at 25.0°c is 10.66. what is the molarity of h+ in this solution? the ph of an aqueous solution at 25.0°c is 10.66. what is the molarity of h+ in this solution? 4.6 à 10-4 3.3 2.2 à 10-11 1.1 à 10-13 4.6 à 1010
Answers
Answer: The molarity of [tex]H^+[/tex] in this solution is [tex]2.2\times 10^{-11}M[/tex]
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
[tex]pH=-\log [H^+][/tex]
Given : pH = 10.66
Putting in the values:
[tex]10.66=-\log[H^+][/tex]
[tex][H^+]=10^{-10.66}[/tex]
[tex][H^+]=2.2\times 10^{-11}[/tex]
Thus the molarity of [tex]H^+[/tex] in this solution is [tex]2.2\times 10^{-11}M[/tex]
The molarity of hydrogen ions, H⁺ in this solution is 2.2 * 10⁻¹¹
The molarity of a solution is a measure of the concentration in moles of a substance present in a liter of solution of that substance.
The pH of a solution is the negative logarithm in base 10 of the hydrogen ion concentration of the solution.
pH = -log[H⁺]
-pH = log[H⁺]
pH of solution = 10.66
-10.66 = log[H⁺]
taking antilogarithm of both sides
[H⁺] = 10⁻¹⁰°⁶⁶
[H⁺] = 2.2 * 10⁻¹¹
Therefore, the molarity of hydrogen ions, H⁺ in this solution is 2.2 * 10⁻¹¹
Learn more at: https://brainly.com/question/22624846
a gas occupies 34.5 liters at a pressure of 345 torr what will be the volume of this gas if the pressure becomes 760 torr
Answers
Answer:
THE NEW PRESSURE IS 15.5 L.
Explanation:
Using Boyle's law which states that the pressure of a given mass of gas is inversely proportional to the volume of the gas at constant temperature.
Mathematically,
P1V1 =P2V2
P1 = 345 torr = 345/760 atm = 0.45 atm
P2 = 760 torr = 760/760 atm = 1 atm
V1 = 34.5 L
V2 = ?
Rearranging the variables, making V2 the subject of the formula, we obtain:
V2 = P2 V1 / P1
V2 = 1 * 34.5 / 0.45
V2 = 15.525 Litres
The volume of the gas when the pressure becomes 760 torr or 1 atm is 15.5 Litres
Given the following reaction:
\ce{2KClO3 -> 2KCl + 3O2}2KClOX
3
2KCl+3OX
2
How many moles of \ce{KCl}KClK, C, l will be produced from 15.0 \text{ g}15.0 g15, point, 0, start text, space, g, end text of \ce{KClO3}KClOX
3
?
moles (round to three significant figures)
Answers
Answer:
0.122 moles
Explanation:
Final answer:
0.122 moles of KCl will be produced from the reaction of 15.0 g of KClO3, using stoichiometry and the molar mass of KClO3 which is 122.55 g/mol.
Explanation:
To determine the amount of KCl produced from 15.0 g of KClO3, we first need to know the molar mass of KClO3, which is 122.55 g/mol according to LibreTexts. By using stoichiometry, we can find out how many moles of KClO3 we have and, subsequently, how many moles of KCl will be produced according to the balanced chemical equation.
First, calculate the number of moles of KClO3:
15.0 g KClO3 × (1 mol KClO3 / 122.55 g) = 0.122 mol KClO3
From the balanced equation Î{2KClO3 -> 2KCl + 3O2}, we can see that 2 moles of KClO3 produce 2 moles of KCl. Therefore, the number of moles of KCl produced is equal to the number of moles of KClO3 available:
0.122 mol KClO3 × (2 mol KCl / 2 mol KClO3) = 0.122 mol KCl
Thus, 0.122 mol of KCl are produced from the reaction of 15.0 g of KClO3.
Push against a door that does not move. Describe the mechanical energy
Answers
Answer:
Work done on the door is = 0
Mechanical energy of the person is 100%
Mechanical energy of the door is 50%.
Explanation:
Mechanical energy, M.E., is the energy an object possesses due to its position and motion. When we talk of position, the body is at rest = potential energy. In terms of motion, the body is moving = kinetic energy.
This means that M.E = potential + kinetic
Also, Work done = force × distance
For the person pushing the door, the potential of the energy stored in the person and the kinetics of pushing the door handle are all present. Therefore, M.E of person is 100%.
But despite the doors potential energy present, there is no motion from it, which means the M.E is only half or 50%.
Also, since distance moved by the door is 0, no work is done on the door.
Measurements show that the enthalpy of a mixture of gaseous reactants decreases by 162. kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -194. kJ of work is done on the mixture during the reaction. a.Calculate the change in energy of the gas mixture during the reaction. Round your answer to 3 significant digits b. Is the reaction exothermic or endothermic?
Answers
Answer:
= -356KJ
therefore, the reaction where heat is released is exothermic reaction since theΔH is negative
Explanation:
given that enthalpy of gaseous reactants decreases by 162KJ and workdone is -194KJ
then,
change in enthalpy (ΔH) = -162( released energy)
work(w) = -194KJ
change in enthalpy is said to be negative if the heat is evolved during the reaction while heat change(ΔH) is said to be positive if the heat required for the reaction occurs.
At constant pressure the change in enthalpy is given as
ΔH = ΔU + PΔV
ΔU = change in energy
ΔV = change in volume
P = pressure
w = -pΔV
therefore,
ΔH = ΔU -W
to evaluate energy change we have,
ΔU =ΔH + W
ΔU = -162+ (-194KJ)
= -356KJ
therefore, the reaction where heat is released is exothermic reaction since theΔH is negative
Final answer:
The change in energy of the gas mixture during the reaction is 32 kJ. The reaction is exothermic because it releases heat to the surroundings as indicated by the negative enthalpy change.
Explanation:
To calculate the change in energy of the gas mixture during the reaction, we use the formula for change in internal energy (ΔU) at constant pressure:
ΔU = ΔH - PΔV
where ΔH is the change in enthalpy and PΔV is the work done on or by the system. Given that the enthalpy of the system decreases by 162 kJ and the work done on the system is -194 kJ, we can substitute these values into the equation:
ΔU = -162 kJ - (-194 kJ)
ΔU = -162 kJ + 194 kJ
ΔU = 32 kJ
Therefore, the change in energy of the gas mixture during the reaction is 32 kJ. The reaction can be classified as exothermic since the enthalpy decreases (ΔH is negative), which means heat is released by the system to the surroundings.
A 0.0200 gram piece of unknown alkaline earth metal, M, is reacted with excess 0.500 M H 2 SO 4 , and the hydrogen gas produced is collected over water. The total gas pressure inside the collecting tube is 1.01 atm., the temperature 24.0 o C. The volume of gas collected is 19.6 mL. The gas in the tube contains water vapor, at a pressure of 0.029 atm. M(s) + H 2 SO 4 (aq) → MSO 4 (aq) + H 2 (g). What is the partial pressure of the dry hydrogen gas collected in the tube?
Answers
Answer:
0.981atm
Explanation:
According ot Dalton's law total pressure of a mixture of non-reactive gas is equal to sum of partial pressures of individual gases.
total pressure= 1.01at
Number of gases=2
Gases: water vapor and hydrogen
partial pressure of water vapor= 0.029atm
1.01= partial pressure of water vapor+ partial pressure of hydrogen
1.01= 0.029 + partial pressure of hydrogen
partial pressure of hydrogen = 0.981atm