Answer:
691.84g
Explanation:
I'm assuming that by CzHg, you mean C3H2
First, use the mole ratio in the equation to find the moles of CO2
n (CO2)= n ( C3H2) × 3
= 5.24 × 3
= 15.72
To find the mass of CO2 produced in grams, complete the following calculation
m= n × MM
where
m = mass
n= moles
MM= molecular mass
m= 15.72 × (12.01 +( 16×2))
m =691.8372
m= 691.84g
You have a 0.5 L of a 1.0 M buffer solution with pka = 5.14 and current pH = 5.23. Calculate the new pH when 1.50 mL of 5.00 M HCl is added. Be mindful of units. (How can you tell if your answer makes sense?) pH = pka + 109 U base 5.23 - 5.14+ log ( )
Answer:
pH = 5,22
Explanation:
A buffer consist in a solution with both conjugate acid (HA) and conjugate base (A⁻). To know the concentration of both A⁻ and HA you use Henderson-Hasselbalch equation:
pH = pka + log₁₀ [tex]\frac{A^{-} }{HA}[/tex]
In the problem:
5,23 = 5,14 + log₁₀ [tex]\frac{A^{-} }{HA}[/tex]
1,23 = [tex]\frac{A^{-} }{HA}[/tex] (1)
Also, you know buffer concentration is 1,0 M:
1,0 M = A⁻ + HA (2)
Replacing (2) in (1):
HA = 0,45 M, thus:
A⁻ = 0,55 M.
The addition of 1.50 mL of 5.00 M HCl represents a concentration of:
1,50x10⁻³ L×[tex]\frac{5,00 mol}{L}[/tex] ÷ 0,5 L = 0,015 M of HCl
The buffer equilibrium is:
HA ⇄ A⁻ + H⁺ ka = [tex]10^{-5,14}[/tex]
The concentrations in equilibrium are:
[HA] = 0,45 M + x
[A⁻] = 0,55 M - x
[H⁺] = 0,015 M -x
Because the equilibrium is displaced to the left.
The equation of equilibrium is:
[tex]10^{5,14}[/tex] = [tex]\frac{[0,015M -x][0,55M-x]}{[0,45M-x]}[/tex]
You will obtain:
x² - 0,565x + 8,24674x10⁻³ = 0
Solving:
x = 0,5500061 ⇒ No physical sense
x = 0,01499391
Thus, [H⁺] = 0,015 - 0,01499391 = 6,09x10⁻⁶
As pH = -log₁₀ [H⁺]
pH = 5,22
I hope it helps!
Chlorine is used to disinfect swimming pools. The accepted concentration for this purpose is 1.00 ppm chlorine, or 1.00 g of chlorine per million grams of water. Calculate the volume of a chlorine solution (in milliliters) a homeowner should add to her swimming pool if the solution contains 5.50 percent chlorine by mass and there are 6.52 × 104 gallons (gal) of water in the pool (1 gal = 3.79 L; density of liquids = 1.00 g/mL). Enter your answer in scientific notation.
Answer:
The volume of a chlorine solution a homeowner should add to her swimming pool is [tex]4.480\times 10^3 mL[/tex].
Explanation:
The solution contains 5.50 percent chlorine by mass.
Mass of the chlorine required to added in the pool= x
Mass of the water in the pool = y
Volume of the water ,V= [tex]6.52\times 10^4 gal=2.464\times 10^5 L[/tex]
(1 gal = 3.78 L)
V = [tex]2.464\times 10^5 L=2.464\times 10^8 mL[/tex]
( 1 L = 1000 mL)
Density of the water ,d= 1.00 g/mL
[tex]y=d\times V=1.00 g/mL\times 2.464\times 10^8 mL=2.464\times 10^8 g[/tex]
y = [tex]2.464\times 10^8 g[/tex]
1.00 g of chlorine per million grams of water .
Then for [tex]2.464\times 10^8 g[/tex] of water, we required x amount of chlorine:
[tex]z=\frac{2.464\times 10^8 g}{10^6 g}=246.4 g[/tex]
x = 246.4 grams of chlorine
Mass of the chlorine solution = M'
[tex](w/w)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100[/tex]
[tex]5.5\%=\frac{z}{M'}\times 100[/tex]
[tex]M'=\frac{246.4}{5.5}\times 100=4,480 g[/tex]
Density of the chlorine solution = D = 1 g/mL
Volume of the chlorine solution to be added in the pool : v
[tex]v=D\times M'=1 g/mL\times 4,480 g=4480 mL[/tex]
[tex]v=4.480\times 10^3 mL[/tex]
The volume of a chlorine solution a homeowner should add to her swimming pool is [tex]4.480\times 10^3 mL[/tex].
Answer:
4.45 L
Explanation:
The concentration of chlorine in the pool must be 1.00g/10⁶g, and the solution has 5.50g/100g, so it will be diluted in the pool.
The product of the concentration by the volume is always constant in a solution, thus, we can use the equation:
C1V1 = C2V2
Where C is the concentration, V is the volume, 1 represents the solution, and 2 the pool. So, C1 = 5.50g/100g, C2 = 1.00g/10⁶g, V2 = 6.52x10⁴ gal = 247108 L. Thus,
(5.50/100)*V1 = (1.00/10⁶)*247108
0.055V1 = 0.247108
V1 = 4.5 L
In a combustion furnace, 2094 standard ft3 per hour
of natural gas (Methane) is burned with 6% excess air.
How many standard ft3 of air are drawn from outside
per hour by the fan that supplies the air?
Explanation:
The chemical reaction is as follows.
[tex]CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O[/tex]
It is given that 2094 [tex]ft^{3}/hr[/tex]. And, it is known that 1 [tex]m^{3}/s[/tex] = 127133 [tex]ft^{3}/hr[/tex]
Hence, convert 2094 [tex]ft^{3}/hr[/tex] into [tex]m^{3}/s[/tex] as follows.
[tex]\frac{2094 ft^{3}/hr}{127133 ft^{3}/hr} \times 1 m^{3}/s[/tex]
= [tex]0.0165 m^{3}/s[/tex]
As ideal gas equation is PV = nRT. So, calculate the number of moles as follows.
n = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{1 atm \times 0.0165 m^{3}}{0.0821 \times 298 K}[/tex]
= 0.673 mol/sec
According to the stoichiometry of the given reaction, 1 mol of methane reacts with 2 mol of oxygen.
So, 1 mol [tex]CH_{4}[/tex] = 2 mol [tex]O_{2}[\tex]
Hence, [tex]O_{2}[\tex] required theoretically = [tex]2 \times 0.673 mol/s[/tex] = 1.346 mol/s
Hence, air required theoretically = [tex]\frac{1.346}{0.21}[/tex] = 6.4095 mol/s.
Since, 6% of excess air is being supplied. Therefore, total air supplied will be calculated as follows.
Total air supplied = [tex]6.4095 mol/s [1 + \frac{6}{100}][/tex]
= 6.794 mol/s
Now, calculate the volume using ideal gas law equation as follows.
PV = nRT
[tex]1 atm \times V = 6.794 mol/s \times 8.21 \times 10^{-5} Latm/K mol \6 times 298 K[/tex]
V = 0.166229 [tex]m^{3}/s[/tex]
Converting calculated volume into [tex]ft^{3}/hr[/tex] as follows.
1 [tex]m^{3}/s[/tex] = 127133 [tex]ft^{3}/hr[/tex]
So, 0.166229 [tex]m^{3}/s[/tex] = [tex]0.166229 m^{3}/s \times 127133 \frac{ft^{3}/hr}{1 m^{3}/s}[/tex]
= 21133.191 [tex]ft^{3}/hr[/tex]
Thus, we can conclude that 21133.191 [tex]ft^{3}/hr[/tex] of air are drawn from outside per hour by the fan that supplies the air.
The energy transferred between samples of matter because of a difference in their temperatures is called a. heat. b. thermochemistry c. chemical kinetics. d. temperature.
Answer:
The energy transferred between samples of matter because of a difference in their temperatures is called a. heat.
Explanation:
The first law of thermodynamics establishes that when two bodies with different temperatures are put in contact they will find thermic equilibrium to a final temperature by transferring heat. Thus the correct answer is (a).
Thermochemistry is the study of the transformations of heat energy on the chemical reactions. Chemical kinetics is the study of the rate of chemical reactions. And temperature is the measure of the heat.
Tania's mom boils water in a pot on the stove to make soup. Which of these ingrediente will dissolve in the water?
Answer:
The only one which will dissolve is salt try other ones can't dissolve.
Answer:
Salt
Explanation:
Salt can be dissolved in water.
What is the energy in joules of one photon of
microwaveradiation with a wavelength 0.122m?
Final answer:
One photon of microwave radiation with a wavelength of 0.122m has an energy of approximately 1.63 x 10⁻²⁴ joules, calculated using Planck's equation (E=hc/λ).
Explanation:
The energy of a photon can be calculated using the equation E = hc/λ, where E is the energy of the photon in joules, h is Planck's constant (6.626 x 10⁻³⁴ J·s), c is the speed of light in a vacuum (3.00 x 10⁸ m/s), and λ is the wavelength of the photon in meters. Using the given wavelength of 0.122 meters for the microwave radiation, we can calculate the energy of one photon by substituting the values into the equation:
E = (6.626 x 10⁻³⁴ J·s)(3.00 x 10⁸ m/s) / 0.122 m
E = 1.63 x 10⁻²⁴ Joules per photon. Therefore, one photon of microwave radiation with a wavelength of 0.122m has an energy of approximately 1.63 x 10⁻²⁴ joules.
What is the heat of vaporization of ethanol, given that ethanol has a normal boiling point of 63.5°C and the vapor pressure of ethanol is 1.75 atm at 780 g.
Explanation:
According to Clausius-Claperyon equation,
[tex]ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]
The given data is as follows.
[tex]T_{1} = 63.5^{o}C[/tex] = (63.5 + 273) K
= 336.6 K
[tex]T_{2} = 78^{o}C[/tex] = (78 + 273) K
= 351 K
[tex]P_{1}[/tex] = 1 atm, [tex]P_{2}[/tex] = ?
Putting the given values into the above equation as follows.
[tex]ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]
[tex]ln (\frac{1.75 atm}{1 atm}) = \frac{-\text{heat of vaporization}}{8.314 J/mol K} \times [\frac{1}{351 K} - \frac{1}{336.6 K}][/tex]
[tex]\Delta H[/tex] = [tex]\frac{0.559}{1.466 \times 10^{-4}} J/mol[/tex]
= [tex]0.38131 \times 10^{4} J/mol[/tex]
= 3813.1 J/mol
Thus, we can conclude that the heat of vaporization of ethanol is 3813.1 J/mol.
Which of the following is a definition of a base? O a. behaves as a catalyst. O b. accepts H ions from acids. Oc. turns cloudy when dissolved in water O d. donates H ions when dissolved in water O e. releases heat when dissolved in water
Answer: b. accepts [tex]H^+[/tex] ions from acids
Explanation:
According to Arrhenius concept, a base is defined as a substance which donates hydroxide ions [tex](OH^-)[/tex] when dissolved in water and an acid is defined as a substance which donates hydronium ions [tex](H_3O^+)[/tex] in water.
According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates [tex]H^+[/tex] ions and a base is defined as a substance which accepts [tex]H^+[/tex] ions.
[tex]NH_3+H_2O\rightarrow NH_4^++OH^-[/tex]
Here water is donating [tex]H^+[/tex] ions, and thus act as acid and ammoia [tex]NH_3[/tex] is accepting [tex]H^+[/tex] ions from water and thus is a base.
According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.
Thus base is a substance which accepts [tex]H^+[/tex] ions from acids.
The molecular mass of magnesium sulfate heptahydrate is 246.46 g/mol. What weight is needed to obtain 0.150 moles?
Answer : The mass of magnesium sulfate heptahydrate needed are, 36.969 grams.
Explanation : Given,
Moles of magnesium sulfate heptahydrate = 0.150 mole
Molar mass of magnesium sulfate heptahydrate = 246.46 g/mole
Formula used :
[tex]\text{Mass of magnesium sulfate heptahydrate}=\text{Moles of magnesium sulfate heptahydrate}\times {\text{Molar mass of magnesium sulfate heptahydrate}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Mass of magnesium sulfate heptahydrate}=0.150mole\times 246.46g/mole[/tex]
[tex]\text{Mass of magnesium sulfate heptahydrate}=36.969g[/tex]
Therefore, the mass of magnesium sulfate heptahydrate needed are, 36.969 grams.
To obtain 0.150 moles of magnesium sulfate heptahydrate, you will need 36.969 grams.
To determine the weight needed to obtain 0.150 moles of magnesium sulfate heptahydrate (MgSO₄·7H₂O), you can use the molar mass and the number of moles given.
The molar mass of MgSO₄·7H₂O is given as 246.46 g/mol.The number of moles needed is 0.150 moles.To find the mass needed, use the formula:Mass = Number of Moles x Molar Mass
Substitute the values into the formula:
Mass = 0.150 moles x 246.46 g/mol
Mass = 36.969 g
Therefore, to obtain 0.150 moles of magnesium sulfate heptahydrate, you need 36.969 grams.
Consider the following reaction: A+B C+D According to Le Chatelier's Principle what would happen if you added more D to the reaction? From the point of view of the molecules and atoms why would the reaction respond this way? Your answer may use a maximum of 15 words. Using more than 15 words will result in a 0 score
Explanation:
According to the Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the direction of equilibrium where there is less stress or disturbance.
For example, [tex]A + B \rightleftharpoons C + D[/tex]
So, when we increase the concentration of D in this reaction then disturbance will increase on the product side.
And, frequency of successful collisions between the reactant atoms will decrease. Hence, there will be less stress on reactant side.
As a result, equilibrium will shift in the backward direction that is, on the reactant side it will shift.
In AG calculations, temperature is expressed in a. degrees Celsius. b. kelvins. c. degrees Fahrenheit, d. kilojoules.
Answer:
The correct option is: a. degrees Celsius
Explanation:
The anion gap is the difference in the cations and anions in plasma, serum or urine, calculated from medical lab test results. It can be calculated by measuring the concentration of the anions or cations, which are expressed in millimoles/litre (mmol/L) or milliequivalents/liter (mEq/L).
The temperature in this test is expressed in degrees Celsius (°C).
A worker is told her chances of being killed by a particular process are 1 in every 300 years. Should the worker be satisfied or alarmed? What is the FAR (assuming normal working hours) and the deaths per person per year? What should her chances be, assuming an average chemical plant?
Answer:
(a) Yes, he should be worried. The Fatal accident rate (FAR) is too high according to standars of the industry. This chemical plant has a FAR of 167, where in average chemical plants the FAR is about 4.
(b) FAR=167 and Death poer person per year = 0.0033 deaths/year.
(c) The expected number of fatalities on a average chemical plant are one in 12500 years.
Explanation:
Asumming 50 weeks of work, with 40 hours/week, we have 2000 work hours a year.
In 300 years we have 600,000 hours.
With these estimations, we have (1/600,000)=1.67*10^(-6) deaths/hour.
If we have 2000 work hours a year, it is expected 0.0033 deaths/year.
[tex]1.67*10^{-6} \frac{deaths}{hour}*2000 \frac{hours}{year}=0.0033 deaths/year[/tex]
The Fatal accident rate (FAR) can be expressed as the expected number of fatalities in 100 millions hours (10^(8) hours).
In these case we have calculated 1.67*10^(-6) deaths/hour, so we can estimate FAR as:
[tex]FAR=1.67*10^{-6} \frac{deaths}{hour}*10^{8} hours=1.67*10^{2} =167[/tex]
A FAR of 167 is very high compared to the typical chemical plants (FAR=4), so the worker has reasons to be worried.
If we assume FAR=4, as in an average chemical plant, we expect
[tex]4\frac{deaths}{10^{8} hour} *2000\frac{hours}{year}=8*10^{-5} \frac{deaths}{year}[/tex]
This is equivalent to say
[tex]\frac{1}{8*10^{-5} } \frac{years}{death}=1.25*10^{4} \frac{years}{death} =12500 \, \frac{years}{death}[/tex]
The expected number of fatalities on a average chemical plant are one in 12500 years.
Which situation is an indication that a chemical change has occurred? No color change occurs. A gas is formed. No heat is released. The temperature does not change.
Answer:
Its A
Explanation:
Its A
The correct answer is B. A gas is formed
Explanation:
In chemistry, changes in substances or elements occur as a product of chemical reactions in which a substance reacts with others or decomposes. In any of these cases, the original substance or element changes into new or new substances, which does not occur in physical changes. This change in composition can be identified through different changes and signals that include a change in odor, color or temperature; the production of bubbles or gas; and the formation of a precipitate. According to this, the situation that is an indication of a chemical change is the formation of a gas, because only if the original substance has changed there would be a new substance such as a gas.
calcualte pressure at STP in 10.0 L vessel after reaction of 1.0 L hydrochloride acid (concentration 35% and density 1.28 g/cub.cm) with 1.0 kg of limestone. yield of reaction 97%
Answer:
The pressure in the vessel is 13,3 atm.
Explanation:
The reaction that occurs in vessel (where limestone is 96% of CaCO₃) is:
2 HCl (aq)+ CaCO₃ (s) → CaCl₂(aq)+ H₂O(l)+ CO₂(g)
The increase in the pressure of the vessel after the reaction is by formation of a gas (CO₂). So we have to find the produced moles of this gas and apply the gas ideal law to find the pressure.
We have to find the limit reactant, to do so, we have to calculate the moles of each reactant in the reaction, the one that have the less moles will be the limit reactant:
HCl:
1,0L × (35/100) × (1000 cm³/1L) × (1,28 g/ 1cm³) × (1mol HCl/ 36,46 g) ÷ 2mol
(Concentration) (L to cm³) (cm³ to g) (g to mol) (moles of reaction)
moles of HCl= 6,14 mol
CaCo₃:
1,0 kg × (96/100) × (1000 g/1kg) × (1 mol/100,09g)
(Limestone) (CaCo₃ in limestone) (kg to g) (g to mol)
moles of CaCo₃= 9,59 mol
So, reactant limit is HCl
This reaction have a yield of 97%. So, the CO₂ moles are:
6,14 mol × 97÷ = 5,96 mol CO₂
The ideal gas formula to obtain pressure is:
P = nRT/V
Where: n = 5,96mol; R= 0,082 atm×L/mol×K; T = 273,15 (until STP conditions) and V= 10,0 L
Replacing this values in the equation the pressure is
P = 13,3 atm
I hope it helps!
Balance the equation and show the calculation of the number of moles and grams of Ca3(PO42 formed from 10.8 grams of Ca(OH)2. Show your answers to 3 significant figures Ca(OH)2 H3PO4-- Ca3(PO4)2 + H2O]
Answer:
The number of moles of calcium phosphate is 0.0485 mol. The mass is 15.1 g.
Explanation:
The balanced equation is:
[tex]3 Ca(OH)_{2} + 2 H_{3}PO_{4} => Ca_{3}(PO_{4})_{2} + 6H_{2}O[/tex]
As we can see, 3 moles of calcium hidroxide produced 1 mol of calcium phosphate. So the quantity of moles of Calcium posphate is:
[tex]10.8g Ca(OH)_{2}*\frac{1 molCa(OH)_{2}}{74.093gCa(OH)_{2}} *\frac{1mol Ca_{3}(PO_{4})_{2}}{3molCa(OH)_{2}} =0.0485 molCa_{3}(PO_{4})_{2} [/tex]
The mass of calcium phosphate in grams is:
[tex]0.0485 molCa_{3}(PO_{4})_{2}*\frac{310.176g}{mol} =15.1 g[/tex]
One kilogram is slightly more than A) 2 B) 5 U.S. pounds. C) 0.5 D) 1 11) The mass of a proton is 1.67 A) 1.67. 10-30 G B 10-27 kg.
Answer:
The correct option is: a) 2 U.S. pounds;
B) 1.672622×10⁻²⁷ kg
Explanation
Pound is a United States customary unit for mass and is abbreviated as lb. One pound is defined as being equal to 0.45359237 kilograms (kg), exactly.
Kilogram is the SI unit of mass and is abbreviated as kg.
Since, 1 lb = 0.45359237 kg
Therefore, 1 kg = 1 ÷ 0.45359237 lb ≈ 2.20462 lb
Therefore, we can say that 1 kilogram is slightly more than 2 pounds.
A proton is a positively charged subatomic particle, that possesses +1e electric charge and has mass approximately equal to 1.672622×10⁻²⁷ kg.
Compare Newton's law of viscosity and Hooke's law of elasticity. What is the origin of these "laws"?
Answer:
Newton's law of viscosity
It relates shear stress in a fluid flow to velocity gradient in the direction perpendicular to the flow of fluid.
[tex]\tau\ \alpha \ \frac{\mathrm{d} u}{\mathrm{d} y}[/tex]
Or
[tex]\tau\ =\mu \times \frac{\mathrm{d} u}{\mathrm{d} y}[/tex]
[tex]\tau\ =Shear\ stress[/tex]
[tex]\frac{du}{dy} =[/tex] Rate of shear deformation
[tex]\mu\ =Viscosity[/tex]
Hooke's Law
It states that within the limit of elasticity, the stress-induced (σ ) in the solid due to some external force is always in proportion with the strain (ε ). In other words, the force causing stress in a solid is directly proportional to the solid's deformation.
[tex]\sigma\ \alpha\ \epsilon[/tex]
[tex]\sigma=E\ \epsilon[/tex]
where E is constant of proportionality known as Young's Modulus and it represents the stiffness of the material.
The type of matter is different in both Newton's and Hook's laws.
Comparison between Newton's law of viscosity and Hooke's law of elasticityNewton's law of viscous deformation deals with deformation of fluids that is subjected to a load. This law states that shears stress is proportional to shear strain.
While on the other hand, Hooke's law of elasticity deals with deformation in solids which are subjected to a load so we can conclude that the type of matter is different in both Newton's and Hook's laws.
Learn more about law here: https://brainly.com/question/25545050
Liquid dimethyl disulfide (CH3SSCH3) flows through a pipe with a mass flow rate of 86.0 g's. Given that the density of dimethyl disulfide is 1.0625 g/cm, find: The molar flow rate in molimin: Number mol/min The volumetric flow rate in L/hr.
Explanation:
Molar mass of [tex]CH_{3}SSCH_{3}[/tex] is 94 g/mol. As it is known that number of moles is equal to mass of a substance divided by its molar mass.
Then, calculate the number of moles as follows.
No. of moles = [tex]\frac{86.0 g}{94 g/mol}[/tex] in 1 s
= 0.914 mol
So, in 60 sec number of moles will be equal to 0.914 x 60 = 54.89 mol/min.
Hence, the molar flow rate = 54.89 mol/min
Also, density is equal to mass of a substance divided by its volume.
Density = [tex]\frac{mass}{volume}[/tex]
Volume = [tex]\frac{mass}{Density}[/tex]
= [tex]\frac{86.0 g}{1.0625 g/cm^{3}}[/tex]
= 80.941 [tex]cm^{3}[/tex]
As, 80.941 [tex]cm^{3}[/tex] of volume flows in 1 s . Therefore, flow of volume in 1 hour will be calculated as follows.
In 1 hr = 80.941 [tex]cm^{3} \times 3600[/tex]
= 291388.24 [tex]cm^{3}/hr[/tex]
Since, 1 [tex]cm^{3}[/tex] = 0.001 L.
So, 291388.24 [tex]cm^{3}/hr \times 0.001 L/cm^{3}[/tex]
= 291.38824 L/hr
Thus, we can conclude that molar flow rate in mol/min is 54.89 mol/min and the volumetric flow rate in L/hr is 291.38824 L/hr.
The reaction rate low is independent of which of the following charactetics • Reactor type] • Temperature • Concentration of components • Pressure
Answer: In the reaction rate law the rate is expressed in terms of concentrations of species. It is important to know how much time a reaction will take to complete itself. It depends on some factors. Temperature, concentration of component, catalyst and pressure. On increasing these factors the rate of reaction of a respective reaction increases. It doesn't depends upon reactor type.
A slab of dry wood of 4-in thickness and sealed edges is exposed to air of 40% relative humidity. Assuming that the two unsealed surfaces of the wood immediately jump to an equilibrium moisture content of 8 lb H2O per 100 lb of dry wood, determine the time for the moisture content at the center of the slab to reach 1% of the equilibrium value using the semi-infinite slab solution. Assume a diffusivity of water of 6.00e-006 cm2/s.
Explanation:
It is given here that a slab is made up of dry wood.
Thickness of slab = 4 inch
It is given condition that the edges of the slab are sealed and exposed to air.
Given the relative humidity of air = 40%
The surface of slab is unsealed and reached to the equilibrium moisture content = [tex]\frac{8 lb H_{2}O}{100 lb dry wood}[/tex].
Given, the diffusivity of water in the slab = [tex]6 \times 10^{-6} cm^{2}/s[/tex].
As per the given condition 1% moisture content will be taken by multiplying the moisture content by 1%, i.e. 0.01.
It is given that the thickness of slab is in inch, so converting the thickness of slab in cm as follows.
Thickness of slab = [tex]4 inch \times 2.54 inch/cm[/tex]
= 10.16 cm (As, 1 inch = 2.54 cm)
Since, the equilibrium moisture content is required to calculate at the center of the slab.
The center of the slab, from the semi-infinite slab solution will be taken from thickness t = 0.
The thickness at center = 10.16 cm
Hence, calculate total thickness at the center of slab as follows.
[tex]\frac{\text{total thickness of slab}}{2}[/tex]
= 5.08 cm
Now, by semi-infinite solution, and by 40% relative humidity of air, time for moisture content to reach at the center of the slab, to 1% of its equilibrium value will be calculated as follows.
Time(Seconds) = [tex]\frac{thickness^{2}(cm^{2}) \times \text{relative humidity} \times 0.01 \times \text{equilibrium content} \times \text{equilibrium moisture}}{\text{diffusivity(cm/s)}}[/tex]
Time (sec) = [tex]\frac{5.08 cm \times 5.08 cm \times 0.4 \times \text{relative humidity} \times 0.01 \text{equilibrium content} \times 8 lb H_{2}O}{100 lb dry.wood \times 6 \times 10{-6} cm^{2}/s \text{diffusivity}}[/tex]
Time (Seconds) = 1376.34 seconds
Thus, we can conclude that the time required for the moisture content at the center of the slab to reach to 1 % of equilibrium value is 1376.34 seconds.
A water treatment plant has a total flow of 45 MGD (mega gallons per day). The plant was designed with five sedimentation basins, each with a total volume of 2500 m3.
Originally, the plant was operating the basins in series: the entire flow goes through one tank, then the second, and so on. What is the retention time in each tank?
The engineer realizes that the sedimentation would improve if they changed the operation to parallel flow. Under this new configuration, the flow is evenly split between the four tanks. What is the retention time in each tank?
Answer:
a) 21.133 minutes for series
b) 84.54 minutes when split into 4
Explanation:
Data provided:
Total flow, V₀ = 45 MGD
total volume of each basin, V = 2500 m³
Now,
1 MGD = 3785.4118 m³/day
also,
1 day = 1440 minutes
thus,
45 MGD = 45 × 3785.4118 m³/day
or
45 MGD = 170343.5305 m³/day
and,
170343.5305 m³/day in 'm³/min'
= 170343.5305 / 1440
= 118.2941 m³/min.
Therefore,
The retention time, t = [tex]\frac{\textup{Volume of tank}}{\textup{Volumetric Flowrate }}[/tex]
or
The retention time, t = [tex]\frac{\textup{2500}}{\textup{118.2941 }}[/tex]
or
The retention time, t = 21.13 min
Hence,
for series of tanks the retention time is 21.133 min.
Now,
on splitting the tanks in 4
the volumetric flow rate will be
= [tex]\frac{\textup{118.2941}}{\textup{4}}[/tex]
= 29.57 m³/min.
Therefore,
The retention time = [tex]\frac{\textup{Total Volume of tank}}{\textup{Volumetric Flowrate }}[/tex]
or
The retention time, t = [tex]\frac{\textup{2500}}{\textup{29.57 }}[/tex]
or
The retention time, t = 84.54 min
Why is a cis-1,3-disubstituted cyclohexane more stable thanits
trans isomer?
Answer:
Explanation has been given below
Explanation:
In diaxial conformation of cis-1,3-disubstituted cyclohexane, 4 gauche-butane interactions along with syn-diaxial interaction are present. Hence it readily gets converted to diequitorial conformation where no such gauche-butane interaction is presentIn two possible conformations of trans-1,3-disubstituted cyclohexane, 2 gauche-butane interactions are present in each of them.Hence cis-1,3-disubstituted cyclohexane exists almost exclusively in diequitorial form. But trans-1,3-disubstituted cyclohexane has no such option. Trans-1,3-disubstituted cyclohexane experiences gauche butane interaction in each of the two conformations.Therefore cis-1,3-disubstituted cyclohexane is more stable than trans conformationA cis-1,3-disubstituted cyclohexane is more stable than its trans isomer due to lower steric hindrance in the cis configuration as compared to the trans.
Explanation:A cis-1,3-disubstituted cyclohexane is more stable than its trans isomer because of a property known as steric hindrance. In organic chemistry, steric hindrance is a phenomenon where the size of groups within a molecule prevents chemical reactions from taking place. In a cis isomer, the substituents are on the same side of the ring, specifically on the ring's axial position. This arrangement minimizes the steric hindrance, which makes it more stable. On the other hand, the trans isomer has the substituents on opposite sides of the ring, causing more steric hindrance and thus it is less stable.
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Calculate the volume of 8,410 g of a substance whose density is 0.71 g/mL. Express your answer in milliliters using the correct number of significant figures. Do not enter your answer using scientific notation.
Answer:
12 mL
Explanation:
Density is expressed as mass divided by volume, so the expression can be rearranged to solve for volume:
D = m/V ⇒ V = m/D
V = m/D = (8.410 g)/(0.71g/mL) = 12 mL
Final answer:
The volume of a substance with a mass of 8,410 g and a density of 0.71 g/mL is 12,000 mL, rounded to two significant figures to match the precision of the given density.
Explanation:
To calculate the volume of a substance, you can use the density formula: Density = Mass / Volume. Given the density of the substance is 0.71 g/mL and the mass is 8,410 g, the volume can be found by rearranging the formula to Volume = Mass / Density.
Volume = 8,410 g / 0.71 g/mL = 11,845.07042 mL. This raw calculation has more significant figures than is justified by the precision of the given data. Since the density value has two significant figures, the volume should also be reported with two significant figures, resulting in 12,000 mL.
A computer depreciates at a rate of 15% per year. If the computer is worth $324.50 when it is 5 years old, how much was it worth when it was new? Round your answer to two decimal places
Answer:
the price of the new computer was [tex]\$ 731.34[/tex]
Explanation:
let's assume that the price of the computer was [tex]\$ x[/tex] when it was new.
As price depreciates at a rate of 15% therefore we can mathematically express this situation when the computer is 5 yrs old as-
[tex]x\times (1-0.15)^{5}=324.50[/tex]
or, [tex]x\times (0.85)^{5}=324.50[/tex]
or, [tex]x=\frac{324.50}{(0.85)^{5}}[/tex]
or, [tex]x=731.34[/tex]
So, the price of the new computer was [tex]\$ 731.34[/tex]
A balloon has a volume of 3.50 liters at 20.0°C. What will be the volume of the balloon at 100.0°C? Assume the pressure doesn't change and the amount of gas in the balloon doesn't change O a. 0.224 Ob.2.75L O c. 17.5L O d. 65.45 Oe. None of the above
Answer: The volume of balloon at 100°C is 4.46 L
Explanation:
To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
where,
[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.
[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.
We are given:
[tex]V_1=3.50L\\T_1=20^oC=(20+273)K=293K\\V_2=?L\\T_2=100^oC=(100+273)K=373K[/tex]
Putting values in above equation, we get:
[tex]\frac{3.5L}{293K}=\frac{V_2}{373K}\\\\V_2=4.46L[/tex]
Hence, the volume of balloon at 100°C is 4.46 L
At constant temperature 10.0L of N, at 0.983 atm is compressed to 2.88L.What is the final pressure of N,? 0.283 atm b) 1.51 atm c) 3.41 atm d 29.3 atm a)
Answer:
c) 3.41 atm
Explanation:
We can calculate the final pressure using Boyles Law, P₁V₁ = P₂V₂.
P₁V₁ = P₂V₂
(0.983 atm)(10.0 L) = P₂(2.88 L)
9.83 ÷ 2.88 = P₂
P₂ = 3.41 atm
The final pressure can be determined by using Boyles law and will be 3.41 atm.
What is pressure?The force delivered perpendicularly to a surface of the structure per unit area throughout whom that force would be dispersed is known as pressure.
What is Boyles law?
According to Boyle's law, the relationship between a gas's pressure as well as volume seems to be inverse.
Given data:
[tex]P_{1} = 0.983 atm\\V_{1} = 10 L\\V_{2} = 2.88 L[/tex]
Calculation of pressure.
The formula of Boyles law:
[tex]P_{1} V_{1} = P_{2} V_{2}[/tex]
Put the value of given data:
[tex](0.983) (10 ) = P_{2} (2.88)\\P_{2} = 3.41 atm[/tex]
Therefore, the final pressure will be 3.41 atm.
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Write 274,541,005,000 in Engineering Notation with 3 significant figures.
Answer : The correct answer is, [tex]0.274\times 10^{12}[/tex]
Explanation :
Engineering notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form times 10 raise to the power. It is similar to the scientific notation but in engineering notation, the powers of ten are always multiples of 3.
The engineering notation written in the form:
[tex]a\times 10^b[/tex]
where,
a = the number which is greater than 0 and less than 999
b = an integer multiple of 3
For example : [tex]45.89\times 10^3[/tex] or [tex]56.45\times 10^{-6}[/tex]
If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.
As we are given the 274,541,005,000 in standard notation.
Now converting this into engineering notation, we get:
[tex]\Rightarrow 274,541,005,000=0.274\times 10^{12}[/tex]
As, the decimal point is shifting to left side, thus the power of 10 is positive.
Hence, the correct answer is, [tex]0.274\times 10^{12}[/tex]
Final answer:
To write 274,541,005,000 in Engineering Notation with 3 significant figures, we need to move the decimal point so that we have a number between 1 and 10 multiplied by a power of 10. The number becomes 2.75 × 10¹¹.
Explanation:
Engineering Notation is a way of expressing numbers in scientific notation but with the power of 10 always being a multiple of 3. To write 274,541,005,000 in Engineering Notation with 3 significant figures, we need to move the decimal point so that we have a number between 1 and 10 multiplied by a power of 10. The number becomes 2.75 × 10¹¹.
Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1°C. At what temperature does benzene boil when the external pressure is 445 torr?
Explanation:
The given data is as follows.
[tex]T_{1}[/tex] = [tex]80.1^{o}C[/tex] = (80 + 273) = 353 K
Atmospheric pressure = 445 torr
Heat of vaporization = 30.72 kJ/mol = [tex]30.72 kJ/mol \times \frac{1000 J/}{1 kJ}[/tex]
= 30720 J/mol
Now, according to Clausius-Clapereryon equation,
[tex]ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times \frac{1}{T_{2}} - \frac{1}{T_{1}}[/tex]
Putting the given values into the above equation as follows.
[tex]ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times \frac{1}{T_{2}} - \frac{1}{T_{1}}[/tex]
[tex]ln (\frac{445}{760}) = \frac{-30720 J/mol}}{8.314 J/mol K} \times \frac{1}{T_{2}} - \frac{1}{353 K}[/tex]
-0.5352 = [tex]-3694.9723 \times \frac{1}{T_{2}} - \frac{1}{353 K}[/tex]
[tex]\frac{1}{T_{2}}[/tex] = 0.002977
[tex]T_{2}[/tex] = 335.909 K
or, = [tex](335.909 - 273)^{o}C[/tex]
= [tex]62.909^{o}C[/tex]
Thus, we can conclude that benzene boils at [tex]62.909^{o}C[/tex] when the external pressure is 445 torr.
We have that for the Question "At what temperature does benzene boil when the external pressure is 445 torr" it can be said to be at
[tex]T_2=62.9^o[/tex]From the question we are told
Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1°C. At what temperature does benzene boil when the external pressure is 445 torr?
temperature of benzeneGenerally the equation for the Pressure ratio is mathematically given as
[tex]ln(p2/p1)=-(heat of vapour/R)*(1/t2-1/t1}\\\\Therefore\\\\ln(445/760)=-(30720/8.3k)*(\frac{1}{1/t2-1/353.1}\\\\T2=335.919k\\\\[/tex]
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Calculate how many g of ammonia will be produced from 9.35 g of nitrogen gas and excess hydrogen using the following equation: 2 NH(g) 3 H2(g) N(g) 4 answer
Answer: The mass of ammonia produced in the reaction is 11.36 g
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of nitrogen gas = 9.35 g
Molar mass of nitrogen gas = 28 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of nitrogen gas}=\frac{9.35g}{28g/mol}=0.334mol[/tex]
The chemical reaction for the formation of ammonia from hydrogen and nitrogen follows:
[tex]3H_2+N_2\rightarrow 2NH_3[/tex]
As, hydrogen gas is present in excess. So, it is considered as an excess reagent.
Nitrogen gas is considered as a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
1 mole of nitrogen gas is producing 2 moles of ammonia gas
So, 0.334 moles of nitrogen gas will produce = [tex]\frac{2}{1}\times 0.334=0.668mol[/tex] of ammonia gas.
Now, calculating the mass of ammonia gas by using equation 1, we get:
Moles of ammonia gas = 0.668 mol
Molar mass of ammonia gas = 17 g/mol
Putting values in equation 1, we get:
[tex]0.668mol=\frac{\text{Mass of ammonia gas}}{17g/mol}\\\\\text{Mass of ammonia gas}=11.36g[/tex]
Hence, the mass of ammonia produced in the reaction is 11.36 g
Final answer:
To calculate the mass of ammonia produced from 9.35 g of nitrogen gas, convert the mass of nitrogen to moles, use the stoichiometric relationship from the balanced reaction to find moles of ammonia, and then convert back to grams. The result is 11.37 g of NH3.
Explanation:
To calculate how many grams of ammonia (NH3) will be produced from 9.35 g of nitrogen gas (N2) using the chemical equation N₂(g) + 3H₂(g) → 2NH3(g), we need to perform stoichiometric calculations. First, we determine the molar mass of nitrogen gas (N2) which is 28.02 g/mol. Using this, we find that 9.35 g of N2 is equivalent to 9.35 g / 28.02 g/mol = 0.3338 mol of N2.
From the balanced chemical equation, we know that 1 mole of nitrogen gas produces 2 moles of ammonia. Therefore, 0.3338 mole N2 × (2 moles NH3 / 1 mole N2) = 0.6676 mole NH3. Finally, using the molar mass of ammonia, which is 17.03 g/mol, we can find the mass of ammonia produced: 0.6676 mol × 17.03 g/mol = 11.37 g of NH3.
How does a buffer resist change in pH upon addition of a strong acid? The strong acid reacts with the strong base in the buffer to form a salt, which produces few H+ ions in solution and therefore only a little change in pH. The strong acid reacts with the weak base in the buffer to form a weak acid, which produces few H+ ions in solution and therefore only a little change in pH. The strong acid reacts with the weak acid in the buffer to form a weak base, which produces few H+ ions in solution and therefore only a little change in pH.
Answer:
The strong acid reacts with the weak base in the buffer to form a weak acid, which produces few H+ ions in solution and therefore only a little change in pH.
Explanation:
When a strong acid is added to the buffer, the acid dissociates and furnish hydrogen ions which combine with the conjugate of the weak acid, forming weak acid. The weak acid dissociates to only some extent and can furnish only some protons and there is no significant change in the pH.
Hence, option B is correct.
In a buffer, a strong acid reacts with the weak base to form a weak acid, yielding fewer H+ ions, so only a slight change in pH is noted. Buffer solutions, including weak conjugate acid-base pairs, resist pH changes. However, the buffering ability may be disrupted if large amounts of acid or base are added, a concept known as buffer capacity.
Explanation:A buffer essentially resists changes in pH when an acid or base is added to the solution. This is due to the presence of a weak conjugate acid-base pair. In the buffer, a strong acid will react with the weak base to form a weak acid, thus producing fewer H+ ions. This, in essence, leads to only a slight change in pH.
A typical example of a buffer might include a solution of acetic acid and sodium acetate. This buffer consists of a weak acid and its salt. Similarly, a solution of ammonia and ammonium chloride could serve as a buffer consisting of a weak base and its salt.
It's crucial to note, however, that buffers do not have infinite capacity to resist pH changes. This is known as the buffer capacity. If a large amount of acid or base is added to the buffer solution, potentially lowering the concentration of the conjugate acid or base pair, the ability of the buffer to regulate pH may be disrupted.
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