Answer:
concentration of CrO4²⁻ ions in the final solution = 6.53 × 10⁻⁵ mol /L
Explanation:
First we calculate the number of moles of AgNO₃:
number of moles = molar concentration × volume
number of moles = 0.5 × 39.20 = 19.6 mmoles = 0,0196 moles AgNO₃
Then we calculate the number of moles of K₂CrO₄:
number of moles = mass / molar weight
number of moles = 5.832 / 194 = 0.03 moles K₂CrO₄
The chemical reaction will look like this:
2 AgNO₃ + K₂CrO₄ → Ag₂CrO₄ + 2 KNO₃
Now we devise the following reasoning:
if 2 moles of AgNO₃ are reacting with 1 mole of K₂CrO₄
then 0,0196 moles of AgNO₃ are reacting with X moles of K₂CrO₄
X = (0.0196 × 1) / 2 = 0.0098 moles of K₂CrO₄
now the the we calculate the amount of unreacted K₂CrO₄:
unreacted K₂CrO₄ = 0.03 - 0.0098 = 0.0202 moles
now the molar concentration of CrO4²⁻ ions:
molar concentration = number of moles / solution volume (L)
molar concentration = 0.0202 / (39.20 + 270) = 6.53 × 10⁻⁵ mol /L
The specific gravity of gasoline is approximately 0.70.
(a) Estimate the mass (kg) of 50.0 liters of gasoline.
(b) The mass flow rate of gasoline exiting a refinery tank is 1150 kg/min. Estimate the volumetric flow rate in liters/s.
(c) Estimate the average mass flow rate lbm/min delivered by a gasoline pump when the gas pump has a pump rate of 10 gallons per 2 minutes.
Answer:
a) mass=35 kg
b) volumetric flow rate= 27.37 [tex]\frac{liters}{s}[/tex]
c) average mass flow rate=29.21 [tex]\frac{lbm}{min}[/tex]
Explanation:
Specific gravity is defined as the relation between the density of one substance and the density of another reference substance (it is usual that water is used in this case). In this case the specific gravity is the relationship between the density of gasoline and the density of water. So:
[tex]0.7=\frac{densitygasoline}{densitywater}[/tex]
Now you can know the density of gasoline with a simple mathematical operation and knowing that densitywater≅ 1000 [tex]\frac{kg}{m^{3} }[/tex]:
densitygasoline=0.7*densitywater
densitygasoline=0.7*1 [tex]\frac{kg}{liters}[/tex]
densitygasoline=0.7 [tex]\frac{kg}{liters}[/tex]
a)
Now that the density is known, you can calculate the mass in 50 liters of gasoline.
By definition of density, you know that in 1 liter there are 700 kg of gasoline. So using The Rule of Three( tool that allows you to solve problems based on proportions), you can calculate the mass in 50 liters:
[tex]\frac{700 kg}{1 liter} =\frac{mass}{50 liters}[/tex]
[tex]mass=\frac{0.7 kg* 50 liters}{1 liter}[/tex]
mass=35 kg
b)
First, it is convenient to change units: from kg / min to kg / seconds. For that you know that 1 min= 60 seconds. So:
[tex]\frac{1150 kg}{min} =\frac{1150 kg}{60 s}[/tex]
[tex]\frac{1150 kg}{min} =\frac{19.16 kg}{s}[/tex]
By definition, density is the relationship between the mass and volume of a substance: This is: [tex]density=\frac{mass}{volume}[/tex]
Applied to this case, it would be [tex]density=\frac{mass flow rate}{volumetric flow rate}[/tex]
The mass flow rate and the density of the gasoline are known, so you can calculate the volumetric flow rate:
[tex]volumetric flow rate= \frac{mass flow rate}{density gasoline}[/tex]
volumetric flow rate= 27.37 [tex]\frac{liters}{s}[/tex]
c)
You know that the pump rate is 10 gallons per 2 minute. This is:
[tex]\frac{10 gallons}{2 minutes} = 5 \frac{gallons}{minute}[/tex]
To calculate the average mass flow, you must relate this data to density and do the necessary unit conversions. Conversions are done similarly to what was previously applied in the previous exercises.
You have to know that:
1 gallon = 3.785 liters1 kg= 2.205 kgSo:
[tex]\frac{5 gallons*3.785 liters}{1gallons*min} =\frac{18.925 liters}{min}[/tex]
You know that the density is [tex]density=\frac{mass flow rate}{volumetric flow rate}[/tex]
The volumetric flow rate (pump rate) and the density of the gasoline are known, so you can calculate the average mass flow rate:
average mass flow rate=density*volumetric flow rate
[tex]average mass flow rate=\frac{0.7 kg}{liters} *\frac{18.925 liters}{min}[/tex]
average mass flow rate=13.2475 [tex]\frac{kg}{min}[/tex]
Converting the units:
[tex]average mass flow rate=\frac{13.2475 kg*2.205 lbm}{1 kg*min}[/tex]
average mass flow rate=29.21 [tex]\frac{lbm}{min}[/tex]
On a website devoted to answering engineering questions, viewers were invited to determine how much power a 100-MW power plant generates annually. The answer declared to be best was submitted by a civil engineering student, who stated, "It produces 100 MW/hr so over the year that's 100*24*365.25 & do the math." a. Carry out the calculation, showing all the units. Answer b. What is wrong with the statement of the question? c. Why was the student wrong in saying that the plant produces 100 MW/hr?
Answer:
(a) The plant generates 3,153,600,000 MJ a year.
(b) The problem with the question is that "power" is not "generated". What is generated is Energy, and Power is the rate of generation of Energy.
(c) The student is wrong because the plants generates 100 MJ of energy a second. "MW/hr" is not a unit of energy or power, so it has no sense.
Maybe he get confused with MW-h, which is a unit of energy, used to measure electrical consumption.
Explanation:
(a) The power of the plant (100 MW) is the rate at which electrical energy is produced. It has units of [energy]/[time].
In this case, the plant produces 100 MJ/s. The energy produced can also be expressed in other units, like MJh.
To calculate the energy generated in one year, we have
[tex]Energy = Power * Time = \\\\Energy = 100 MW*1year\\\\Energy = (100 \frac{MJ}{s})*(1 year*\frac{365 days}{1year}* \frac{24hours}{1day}*\frac{3600s}{1hour})\\\\Energy=(100 \frac{MJ}{s})*(31,536,000s)=3,153,600,000MJ[/tex]
The plant generates 3,153,600,000 MJ a year.
(b) The problem with the question is that "power" is not "generated". What is generated is Energy, and Power is the rate of generation of Energy.
(c) The student is wrong because the plants generates 100 MJ of energy a second. "MW/hr" is not a unit of energy or power, so it has no sense.
Maybe he get confused with MW-h, which is a unit of energy, used to measure electrical consumption. That unit represents the amount of energy consumed or generated by a 1 MW unit in one hour. It is equivalent to 3600 MJ (1 MW-h = 3600 MJ).
In this unit, the 100 MW plant generates 876,000 MWh.
What is the temperature (°C) of 1.75 g of O2 gas occupying 3.10 L at 1.00 atm? Enter your answer in the box provided.
Answer:
691.29 K or 418.14 °C
Explanation:
Hello, at first the moles of oxygen gas are required:
[tex]n_{O_2}=1.75 g * \frac{1mol O_2}{32 g O_2} =0.0547 mol[/tex]
Now, based on the ideal gas equation, we solve for the temperature:
[tex]PV=nRT\\T=\frac{PV}{nR}\\T=\frac{1atm * 3.10 L}{0.0547 mol*0.082 \frac{atm*L}{mol*K} }\\T=691.29 K[/tex]
Best regards.
A turbine operating adiabatically is fed with steam at 400 °C and 8.0 MPa at the rate of 1000 kg/h. Process steam saturated at 0.5 MPa is withdrawn from an intermediate location in the turbine at a rate of 300 kg/h, and the remaining steam leaves the turbine saturated at 0.1 MPa. The kinetic energies and differences in potential energies of all streams are negligible. What is the power output of the turbine?
Explanation:
We select the enthalpy of steam at state 1 at [tex]800^{o}C[/tex] and 8.0 MPa from the steam tables as follows.
[tex]h_{1}[/tex] = 3138 kJ/kg
Also, we select the enthalpy of steam at state 2 at 0.5 MPa from the steam tables as follows.
[tex]h_{2}[/tex] = 2748.6 kJ/kg
At state 3 also, from the steam tables at state 3 at 0.1 MPa.
[tex]h_{3}[/tex] = 2675.1 kJ/kg
Hence, calculate the mass flow rate at state 3 as follows.
[tex]m_{3} = m_{1} - m_{2}[/tex]
= 1000 kg/h - 300 kg/h
= 700 kg/h
Now, we will calculate the power output of the turbine as follows.
[tex]W_{r} = m_{1}(h_{1} - h_{2}) + m_{3}(h_{2} - h_{3})[/tex]
= 1000 kg/h (3138 kJ/kg - 2748.6 kJ/kg) + 700 kg/h (2784.6 kJ/kg - 2675.1 kJ/kg)
= 440850 kJ/h
It is known that 1 kJ/h = 0.000278 kW.
Therefore, [tex]440850 kJ/h \times \frac{0.000278 kW}{1 kJ/hr}[/tex]
= 122.56 kW
Thus, we can conclude that the power output of the turbine is 122.56 kW.
Express each of the following values in the given base unit. 1st attempt Part 1 (2.5 points) m = 2.54 cm Part 2 (2.5 points) S = 316 MS
Answer:
1) 2.54 cm = 0.0254 m
2) 316 Ms = [tex]3.16\times 10^{8} s[/tex]
Explanation:
1) 2.54 cm
In 1 centimeter is there are 0.01 meters.
1 cm = 0.01 m
[tex]2.54 cm = 2.54\times 0.01 m = 0.0254 m[/tex]
2) 316 Ms
In 1 mega second is there are 1 million seconds.
[tex]1 Ms = 10^{6} s[/tex]
[tex]316 Ms= 316\times 10^{6} s= 3.16\times 10^{8} s[/tex]
For light with a frequency of 5.371 x 10^14 s^-1, what is the energy (in J)?
Answer:
The energy is 3.559 * 10⁻¹⁹ J
Explanation:
For this problem we need to use Planck's equation, this equation describes the relationship between energy radiated by light (a photon, to be precise) and its frequency. The equation is:
E = h * v
Where E is energy (in Joules), h is Planck's constant (6.626 * 10⁻³⁴ J·s), and v is the frequency (in s⁻¹).
Now we solve the equation, using the data given in the problem:
E = 6.626 * 10⁻³⁴ J·s * 5.371 * 10¹⁴ s⁻¹ = 3.559 * 10⁻⁴⁷ J
Thus the energy is 3.559 * 10⁻¹⁹ J
If 37.1 mL AgNO3 solution reacts with excess potassium chloride solution to yield 1.56 g of AgCl precipitate, what is the molarity of silver ion in the original solution? AgCl = 143.3 g/mol Enter your answer in decimal format with three decimal places and no units.
Answer:
Molarity of silver ion: 0.296 M
Explanation:
Reaction:
AgNO₃ + KCl → AgCl↓ + K⁺ + NO₃⁻
From the reaction, we know that the moles of AgCl produced will be the same as the moles of initial silver.
First, let´s calculate the number of moles of AgCl produced:
1.56 g AgCl was produced, that is, (1.56 g AgCl * 1 mol AgCl/143.3 g AgCl) 0.011 moles AgCl.
The moles of silver ion present in the original solution was 0.011 mol. Since this number of moles was present in a 37.1 ml solution, then, in 1000 ml:
moles of silver ion per liter = 1000 ml * 0.011 mol / 37.1 ml = 0.296 mol
Molarity of silver ion = 0.296 M
The molarity of silver ion in the original AgNO3 solution is 0.293 M.
Explanation:To determine the molarity of silver ion in the original AgNO3 solution when 37.1 mL reacts to yield 1.56 g of AgCl, we first need to calculate the moles of AgCl produced using the molar mass of AgCl.
The molar mass of AgCl is 143.3 g/mol. So, the moles of AgCl formed are calculated as follows:
(1.56 g AgCl) / (143.3 g/mol) = 0.010878 mol AgCl
Since AgNO3 reacts with KCl in a 1:1 mole ratio to produce AgCl, the moles of AgNO3 that reacted is also 0.010878 mol. To find the molarity of AgNO3, we divide the moles by the volume of solution in liters:
(0.010878 mol AgNO3) / (0.0371 L) = 0.293 M
Which of the following non-hydrogen atom transitions does the photon have at its long wavelength?
Move from Level n=6 to Level n=1
Move from Level n=4 to Level n=1
Move from Level n=3 to Level n=2
Move from Level n=5 to Level n=4
Please explain thae way, how I can solve such questions and how to calculate (If needed) in future?
Answer:
The move from Level n=3 to Level n=2 has the long wavelength.
Explanation:
First, due to the selection rules, only transitions between adjacent levels are allowed, thus, only a transition between Level n=3 to Level n=2 or Level n=5 to Level n=4 are allowed. The two first options are wrong.
Second, analyzing the transition between Level n=3 to Level n=2 and the transition between Level n=5 to Level n=4 it is necessary to think in terms of the equation of the difference of energy for these type of transitions:
Δ[tex]E = \frac{h^{2}}{8.m.L}(n_{LUMO}^{2} -n_{HOMO}^{2} )[/tex] (1)
The difference in energy (ΔE) is directly proportional to the quadratic difference between the 'n' levels of transition. Therefore, If the transition occurs between smaller 'n' levels the difference of energy will be smaller too.
Also, the energy (ΔE) is inversely proportional to the wavelength (λ) so a smaller energy means a larger wavelength.
ΔE = c / λ (2)
Hence, the move from Level n=3 to Level n=2 has a long wavelength.
In order to calculate this wavelength is necessary to replace the data on equation (1) and (2).
Which compound has the bigger lattice energy? K Br CaBr2 o O Na, NaF Rb20 0 Rb, S 0
Answer:
The answer is NaF.
Explanation:
Lattice energy:
It can be define as "the amount of energy released when ions combine to from the ionic solids or energy required to break the compound into its ions"
Example:
Consider the example of ionic solid NaCl. The sodium chloride is formed when ions like Na+ and Cl - are combine. During the formation of NaCl crystals the energy is released which is called lattice energy of sodium chloride.
Dependence of lattice energy:
Lattice energy depend upon following factors:
1. Charges of the ions
2. Sizes of the ions
As the charge of ion increase the lattice energy also increase. There is a direst relationship between them.
The size of the ions and lattice energy have inverse relation. As the size decreases, lattice energy increases. Therefor, the compound NaF consist of smallest cation (Na+) and the smallest anion (F-) have largest lattice energy.
F= q1 × q2/ r²
As the size of ions decreases the distance between the oppositely charged ions in a ionic compound decreases. They will more tightly packed in a crystal and their center will more closer to each others hence increase the lattice energy.
The Prandtl number, Pr, is a dimensionless group important in heat transfer. It is defined as Pr = Cp*mu/k where Cp is the heat capacity of a fluid, mu is the fluid viscosity, and k is the fluid thermal conductivity. For a given fluid, Cp = 0.58 J/(g* deg C), k = 0.28 W/(m * deg C), and mu = 1934 Ibm / (ft * h). Determine the value of the Prandtl number for this fluid.
Answer:
The value of the Prandlt number for this fluid is: 1656.04
Explanation:
As it is stated in the problem : Pr = Cp*mu/k
where:
cp: heat capacity of the fluid
mu: viscosity of the fluid
k: thermal conductivity of the fluid
Now for a given fluid we have
cp= 0.58 J/(g* deg C)
mu=1934 Ibm / (ft * h)
k = 0.28 W/(m * deg C)
If we put these values in the ecuation of the Prandlt number we have:
Pr = (0.58 J/(g* deg C)) × (1934 Ibm / (ft * h) / 0.28 W/(m * deg C)) =
As we can see we have to convert the units so we can operate all the values in the same units of measurement and then cancel them so as to obtain a dimensionless result.
Converting the value of: mu = 1934 Ibm / (ft * h)
1 ft= 0.3048 m
1 h= 3600 s
1 lbm= 453.59 g
mu= 1934 Ibm / (ft * h) × (453,59 g/ lbm) × (1h/3600 s) × (1 ft/0.3048 m) = 799.47 g/ (m *s).
Pr = (0.58 J/(g* deg C)) × (799.47 g/ (m *s) / 0.28 W/(m * deg C)) = 1656.04
What are the major causes of environmental problems?
Answer: This world is rapidly moving towards to some serious enviornmental problems, the cause of this problems are mentioned below;
(1) The increase in average temperature of Earth.
(2) Deforestation
(3) Over Population
(4) Mining
Because of these reasons mentioned above there are many enviornmental issue occurring, and many problems we will face in future for example, water crisis, lack of resources.
A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After all the metal had reacted, the remaining acid required 0.03340 L of 0.5000 M NaOH solution for complete neutralization. Calculate the molar mass of the metal and identify the element.
Answer: The metal having molar mass equal to 26.95 g/mol is Aluminium
Explanation:
To calculate the number of moles for given molarity, we use the equation:[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)
Molarity of NaOH solution = 0.5000 M
Volume of solution = 0.03340 L
Putting values in equation 1, we get:
[tex]0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol[/tex]
The chemical equation for the reaction of NaOH and sulfuric acid follows:[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of NaOH reacts with 1 mole of sulfuric acid
So, 0.01670 moles of NaOH will react with = [tex]\frac{1}{2}\times 0.01670=0.00835mol[/tex] of sulfuric acid
Excess moles of sulfuric acid = 0.00835 moles
Calculating the moles of sulfuric acid by using equation 1, we get:Molarity of sulfuric acid solution = 0.5000 M
Volume of solution = 127.9 mL = 0.1279 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol[/tex]
Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles
The chemical equation for the reaction of metal (forming [tex]M^{3+}[/tex] ion) and sulfuric acid follows:[tex]2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2[/tex]
By Stoichiometry of the reaction:
3 moles of sulfuric acid reacts with 2 moles of metal
So, 0.0556 moles of sulfuric acid will react with = [tex]\frac{2}{3}\times 0.0556=0.0371mol[/tex] of metal
To calculate the molar mass of metal for given number of moles, we use the equation:[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Mass of metal = 1.00 g
Moles of metal = 0.0371 moles
Putting values in above equation, we get:
[tex]0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol[/tex]
Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium
explain why sodium (metal) is soft and can be bend, whereas NaCl is hard and brittle.
Answer:
Sodium chloride has a crystalline face-centered cubic structure whereas metallic sodium body-centered cubic structure.
Explanation:
Hello, atomic arrangements provide the molecules different features and behaviors, since the sodium metal has a body-centered cubic structure (https://en.wikipedia.org/wiki/Cubic_crystal_system#/media/File:Cubic-body-centered.svg) the lack of inner atoms, allows the material to be soft and bendable. On the other hand the compacted sodium chloride's face-centered cubic structure (https://en.wikipedia.org/wiki/Cubic_crystal_system#/media/File:Cubic-face-centered.svg), provides it a crystalline structure which is hard and brittle since the atoms are closer.
Best regards!
A sample of nitric acid has a mass of 8.2g. It is dissolved in 1L of water. A 25mL aliquot of this acid is titrated with NaOH. The concentration of the NaOH is 0.18M. What titre volume was added to the aliquot to achieve neutralisation?
Answer:
18.075 mL of NaOH solution was added to achieve neutralization
Explanation:
First, let's formulate the chemical reaction between nitric acid and sodium hydroxide:
NaOH + HNO3 → NaNO3 + H2O
From this balanced equation we know that 1 mole of NaOH reacts with 1 mole of HNO3 to achieve neutralization. Let's calculate how many moles we have in the 25 mL aliquot to be titrated:
63.01 g of HNO3 ----- 1 mole
8.2 g of HNO3 ----- x = (8.2 g × 1 mole)/63.01 g = 0.13014 moles of HNO3
So far we added 8.2 grams of nitric acid (0.13014 moles) in 1 L of water.
1000 mL solution ---- 0.13014 moles of HNO3
25 mL (aliquot) ---- x = (25 mL× 0.13014 moles)/1000 mL = 0.0032535 moles
So, we now know that in the 25 mL aliquot to be titrated we have 0.0032535 moles of HNO3. As we stated before, 1 mole of NaOH will react with 1 mole of HNO3, hence 0.0032535 moles of HNO3 have to react with 0.0032535 moles of NaOH to achieve neutralization. Let's calculate then, in which volume of the given NaOH solution we have 0.0032535 moles:
0.18 moles of NaOH ----- 1000 mL Solution
0.0032535 moles---- x=(0.0032535moles×1000 mL)/0.18 moles = 18.075mL
As we can see, we need 18.075 mL of a 0.18 M NaOH solution to titrate a 25 mL aliquot of the prepared HNO3 solution.
In the soap testing experiment, what is the purpose of adding some MgCl2 to the soap solution at the end?
Explanation:
The water is hard due to the presence of ions of Mg²⁺. Now in hard water soaps are ineffective . Since in hard water Mg²⁺ ion forms precipitate , which concludes less number of soap molecules are present in the solution and less amount of froth which disables the cleansing property of the soap .
Now the experiment , the very last step is the hard water test where some amount of MgCl₂ is added ,
Now this magnesium salt act as a source of Mg²⁺ ion and now the soap action can be determined , whether it is able to form froth or not .
MgCl2 is added to the soap solution during a soap testing experiment to simulate the conditions of hard water. This allows for assessment of the soap's effectiveness as a cleansing agent in hard water, as soaps often form insoluble precipitates when reacting with minerals like magnesium.
In a soap testing experiment, MgCl2 (magnesium chloride) is added to the soap solution as a way to simulate hard water conditions because hard water contains high levels of magnesium and other minerals. One property of soap is its ability to react with ions present in water. In hard water, soaps react with calcium or magnesium ions to form insoluble precipitates, which makes the soap less effective as a cleaning agent. This is notable because soaps form insoluble calcium and magnesium compounds in hard water, while detergents form water-soluble products.
The behavior of soap in this soap-magnesium chloride solution, therefore, would give an indication of its prospective behaviour in hard water conditions.
Learn more about Soap Testing here:
https://brainly.com/question/13263724
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Give the ΔH value for the combustion of butane as shown in the reaction 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)+5315 kJ.
Express your answer using four significant figures. If the value is positive, do not include the + sign in your answer.
The ΔH value for the combustion of butane is given as 5315 kJ for the reaction as written, which is for 2 moles of butane. To find the enthalpy of combustion per mole, divide this number by 2, yielding -2658 kJ/mol.
Explanation:The student asked to give the ΔH value for the combustion of butane as shown in the reaction 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) + 5315 kJ. The ΔH value for the reaction is given as +5315 kJ, indicating the amount of energy released during the combustion of butane.
Now, the enthalpy of combustion is typically expressed on a per mole basis, referring to the amount of heat released when one mole of a substance is burned. Since the reaction above shows the combustion for 2 moles of butane, we can calculate the enthalpy of combustion per mole by dividing the total energy by 2. So, the enthalpy of combustion for one mole of butane would be 5315 kJ / 2 = 2657.5 kJ/mol, which is typically reported in kilojoules per mole (kJ/mol).
Expressed with four significant figures, the enthalpy of combustion per mole of butane would be -2658 kJ/mol (the negative sign indicates that energy is released).
You're a newly appointed Engineer in APE Chemical Sdn Bhd. and your first task is to design a 0.35m vessel to be used to store saturated vapor ethanol at 480°C and 6000 kPa. For these conditions, analyze (6) The molar volume of the saturated vapor ethanol. . (15 marks) (6) The mass of the saturated vapor ethanol in the vessel
Answer:
molar volume of vapor ethanol is [tex]V = 1.043 \times 10^{-3} m^3/mol[/tex]
mass of ethanol is 15430.22 g
Explanation:
By using ideal gas equation
PV = nRT
Where P is pressure , R is gas constant
so, volume is
[tex]V = \frac{RT}{P} = \frac{8.314\times (480+273)}{6000\times 10^3}[/tex]
[tex]V = 1.043 \times 10^{-3} m^3/mol[/tex]
molar volume of vapor ethanol is
[tex]V = 1.043 \times 10^{-3} m^3/mol[/tex]
b)
volume of vessel is given [tex]0.35 m^3[/tex]
therefore total moles of ethanol in given vessel will be
[tex]n =\frac{V_{vessel}}{V}[/tex]
[tex]n =\frac{0.35}{1.043\times10^{-3}}[/tex]
n = 335.44 mol
we know that
mole is given as [/tex]n = \frac{mass}{moleculae weight}[/tex]
weight of ethanol is 46 g/mol
n\times 46 = mass
[tex]335.44\times 46 = 15430.22 g[/tex]
mass of ethanol is 15430.22 g
How many atoms of Phosphorus (symbol P) are in 45.67g of P? (Your answer should just be a number. Do not include atoms as the unit. For scientific notation use the "^" symbol before the exponent. For example 2.54 x 10^5)
Answer:
8.882 × 10^23
Explanation:
The molar mass of phosphorus is 30.97 g/mol. The moles of P corresponding to 45.67 g are:
45.67 g × (1 mol/30.97 g) = 1.475 mol
In 1 mole of P, there are 6.022 × 10²³ atoms of P (Avogadro's number). The number of atoms of P in 1.475 moles are:
1.475 mol × (6.022 × 10²³ atoms/1 mol) = 8.882 × 10²³ atoms
Suppose the gas above the soda in a bottle of soft drink
ispure CO2 at a pressure of 2atm. Calculate [CO2] at 25
degreeC.
Henry's law at 25C of CO2 is 0.033363 (mol/L*atm).
Answer: The concentration of carbon dioxide in bottle of soft drink is [tex]6.6\times 10^{-2}mol/L[/tex]
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{CO_2}=K_H\times p_{liquid}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]0.033363mol/L.atm[/tex]
[tex]p_{CO_2}[/tex] = partial pressure of gas in a bottle of soft drink = 2 atm
Putting values in above equation, we get:
[tex]C_{CO_2}=0.033363mol/L.atm\times 2atm\\\\C_{CO_2}=0.066726mol/L=6.6\times 10^{-2}[/tex]
Hence, the concentration of carbon dioxide in bottle of soft drink is [tex]6.6\times 10^{-2}mol/L[/tex]
a 2.60 g sample of titanium metal chemically combines
withchlorine gas to form 10.31g of a titanium chloride.
a. What is the empirical formula of the titaniumchloride
b. What is the percent by mass of titanium and chlorine in
thesample?
Answer:
For a: The empirical formula for the given compound is [tex]TiCl_4[/tex]
For b: The percent by mass of titanium and chlorine in the sample is 25.55 % and 74.78 % respectively.
Explanation:
For a:We are given:
Mass of Titanium = 2.60 g
Mass of sample = 10.31 g
Mass of Chlorine = 10.31 - 2.60 = 7.71 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of titanium =[tex]\frac{\text{Given mass of Titanium}}{\text{Molar mass of Titanium}}=\frac{2.60g}{47.867g/mole}=0.054moles[/tex]
Moles of Chlorine = [tex]\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{7.71g}{35.5g/mole}=0.217moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.054 moles.
For Titanium = [tex]\frac{0.054}{0.054}=1[/tex]
For Chlorine = [tex]\frac{0.217}{0.054}=4.01\approx 4[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of Ti : Cl = 1 : 4
Hence, the empirical formula for the given compound is [tex]TiCl_4[/tex]
For b:To calculate the percentage by mass of substance in sample, we use the equation:
[tex]\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100[/tex] .......(1)
For Titanium:Mass of sample = 10.31 g
Mass of titanium = 2.60 g
Putting values in above equation, we get:
[tex]\%\text{ composition of titanium}=\frac{2.60g}{10.31g}\times 100=25.22\%[/tex]
For Chlorine:Mass of sample = 10.31 g
Mass of chlorine = 7.71 g
Putting values in above equation, we get:
[tex]\%\text{ composition of chlorine}=\frac{7.71g}{10.31g}\times 100=74.78\%[/tex]
Hence, the percent by mass of titanium and chlorine in the sample is 25.55 % and 74.78 % respectively.
How many degrees of freedom does each of the following systems have? (Answer as a number, i.e., 1, 2, 3, etc.)
1. Liquid water in equilibrium with its vapor?
2. Liquid water in equilibrium with a mixture of water vapor and nitrogen?
3. A liquid solution of alcohol in water in equilibrium with its vapor?
Answer:
Explanation:
Hello, since the Gibbs' phase rule states the following equation:
[tex]F=C-P+2[/tex]
Whereas C is the number of components and P the present phases, you answers are:
1. F=1-2+2=1.
2. F=2-2+2=2.
3. F=2-2+2=2.
Best regards.
Determine the temperature, volume, and quality of 1kg of H2O under the following conditions
A) U=2900kj/kg, P= 1.7MPa
B) U= 350 kj/kg P= 0.3 Mpa
Which of the following statements is not true regarding molecular orbital theory? O Bonding molecular orbitals have lower energy than the atomic orbitals from which they are formed. O Molecular orbitals are formed by the combination of atomic orbitals. Two atomic orbitals combine to form one molecular orbital. O Molecular orbitals are regions of a molecule where the electrons are most likely to be found.
Answer:
Option c, Two atomic orbitals combine to form one molecular orbital
Explanation:
Molecular orbitals are formed by linear combination of atomic orbitals.
Some of the important facts of molecular orbital theories are as follows:
No. of the molecular orbitals formed are equal to the no. of atomic orbitals participated.Half of the molecular orbitals are bonding molecular orbitals and half of the molecular orbitals are anti bonding molecular orbitals.Anti bonding molecular orbitals have energy higher than participating atomic orbitals. Bonding molecular orbitals have energy lower than participating atomic orbitals.Molecular orbitals are that region in the molecule where electrons are most likely to found.So, among given, option c which is 'atomic orbitals combine to form one molecular orbital' is incorrect.
All the statements provided are true with regards to Molecular Orbital Theory, except for 'Two atomic orbitals combine to form one molecular orbital.' In fact, two atomic orbitals combine to form two molecular orbitals, one being a lower-energy bonding molecular orbital and the other one being a higher-energy antibonding orbital.
Explanation:In the context of Molecular Orbital Theory, three out of the four statements made in the question are correct. Namely, molecular orbitals are indeed formed by the combination of atomic orbitals, and they represent regions of a molecule where electrons are most likely to be found. Additionally, bonding molecular orbitals do have lower energy than the atomic orbitals from which they are formed.
However, the statement suggesting that 'Two atomic orbitals combine to form one molecular orbital' is not accurate. In reality, when two atomic orbitals combine, they form two molecular orbitals: one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital.
This nuanced concept is a cornerstone of MO theory and helps explain a range of phenomena, from the paramagnetism of the oxygen molecule to violations of the octet rule. Understanding this allows us to further appreciate the differences between bonding and antibonding orbitals and their roles in the structure and stability of molecules.
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Draw the Lewis Structure for CH4.
Explanation:
Methane (CH₄)
Valence electrons of carbon = 4
Valence electrons of hydrogen = 1
The total number of the valence electrons = 4 + 4(1) = 8
The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete. So,
The Lewis structure is:
H
:
H : C : H
:
H
To draw the Lewis Structure for CH4, follow the steps to determine the total number of valence electrons, place the atoms and lone pairs, and draw the final structure. Carbon in CH4 forms 4 single bonds with Hydrogen, resulting in a tetrahedral arrangement of Hydrogen atoms around the Carbon atom.
Explanation:The Lewis Structure for CH4, which represents the arrangement of electrons in the molecule, can be drawn using the following steps:
Determine the total number of valence electronsDetermine the central atom, which is Carbon (C) in this case, and surround it with the Hydrogen (H) atomsPlace the remaining electrons as lone pairs on the central atom and the outer atomsCheck if the central atom has an octet of electrons. If not, move a lone pair from an outer atom to form a double bond with the central atomDraw the final Lewis structureFor CH4, Carbon has 4 valence electrons and Hydrogen has 1 valence electron each. The total number of valence electrons is 4 + (4 x 1) = 8. Following the steps, Carbon is surrounded by 4 Hydrogen atoms, and each Hydrogen atom shares its electron with Carbon. Carbon ends up with a complete octet by sharing electrons with the Hydrogen atoms, forming 4 single bonds. The final Lewis structure for CH4 is a tetrahedral arrangement of the Hydrogen atoms around the Carbon atom, represented by a molecular formula, structural formula, or other models.
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An antibacterial ointment with 8% muciprocin as active ingredient contains _ mupirocin per 25 g of the ointment? g
Answer:
2 grams
Explanation:
The percentage in weight/weight is equivalent to the solute mass divided by the solution mass and multiplied by 100. For example, if you have 25g/100g then the percentage will be 25%. In this case you just need to multiply 8*25 and divided it by 100, so the answer is 2 grams of mupirocin per 25 g of the ointment.
How many mL of concentrated hydrochloric acid should be added to 48.3 g of sodium nitrite to prepare 2.50 L of a buffer solution with a pH of 2.60?
Answer:
47 mL
Explanation:
The equilibrium of nitrous acid is:
HNO₂ ⇄ NO₂⁻ + H⁺ Ka = 4,5x10⁻⁴
If desire pH is 2,60 the [H⁺] concentration in equilibrium is:
[H⁺] = [tex]10^{-2,6}[/tex] =2,51x10⁻³ M
Initial molarity of sodium nitrite is:
43,8g × [tex]\frac{1mol}{68,9953 g}[/tex]÷ 2,5 = 0,254 M
Thus, in equilibrium the concentration of chemicals is:
[NO₂⁻] = 0,254 - x
[HNO₂] = x
[H⁺] = 2,51x10⁻³ = Y-x Where Y is initial concentration.
Equilibrium formula is:
4,5x10⁻⁴ = [tex]\frac{[2,51x10^{-3}[0,254 - x] ]}{[x]}[/tex]
Solving, x = 0,215
Thus, initial [H⁺] concentration is:
0,215 + 2,51x10⁻³ = 0,2175 M
If total volume is 2,50 L:
2,50L ×[tex]\frac{0,2175 mol}{L}[/tex] = 0,5438 mol of HCl
As molarity of concentrated hydrochloric acid is 11,65 mol per liter:
0,5438 mol HCl ×[tex]\frac{1L}{11,65}[/tex] = 0,047 L ≡ 47 mL
I hope it helps!
How many electrons are there in the valence shell of the oxygen atom of water? O a. 2 O 6.4 OC.6 O d. 8
Answer:
d. 8
Explanation:
Valence electrons of oxygen = 6
Valence electrons of hydrogen = 1
The total number of the valence electrons = 6 + 2(1) = 8
The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete.
In the water molecule,
Oxygen has 2 bond pairs and two lone pairs which means that the total electrons in the valence shell is 8.
Define "Hydrogen Bond"
Answer:
Hydrogen bond is a special type of dipole-dipole interaction between the hydrogen atom in a polar bond, such as N-H, O-H, or F-H, and electronegative O,N or F atom. The interaction is written:
A-B . . . B or A-H . . . .A
A and B represent O,N or F; A-H is one molecule or part of a molecule and B is a part of another molecule, and the dotted represents the hydrogen bond. So, hydrogen bond is a type of intermolecular forces (attractive forces between molecules, NOT between atoms of the same molecule), and only a few elements can participate on hydrogen bond formation.
Note: dipole-dipole interactions are forces between polar molecules, that is, between molecules that possess dipole moments.
Final answer:
A hydrogen bond is a weak intermolecular attraction where a partially positive hydrogen atom, bonded to an electronegative atom, is attracted to a lone pair of electrons on a nearby electronegative atom.
Explanation:
A hydrogen bond is a type of intermolecular or intramolecular attractive force. It occurs when a hydrogen atom, covalently bonded to a highly electronegative atom such as nitrogen, oxygen, or fluorine, develops a partial positive charge. This partially positive hydrogen atom is then attracted to the lone pair of electrons on a neighboring electronegative atom, often in a different molecule. This attraction is relatively weak compared to covalent bonds but is stronger than other dipole-dipole interactions. In water, hydrogen bonds are a crucial part of the molecule's properties, including its high boiling point and surface tension.
A mixture of Fe2O3 and FeO was found to contain 72.00% Fe
bymass. What is the mass of Fe2O3 in 0.500 g of this mixture...how
doi work this..i dont even know what to look for as the given
asneeded?
Answer:
The mass of Fe₂O₃ in 0.500 g of mixture is 0.367 g.
Explanation:
First off, we know that 72% of the mass of the mixture is iron. The information also tells us that the remaining 28% of the mass is oxygen.
Now we calculate the total mass of iron and the total mass of oxygen in the mixture:
0.500 g * 0.72 = 0.360 g of Fe0.500 g * 0.28 = 0.140 g of OWith the mass of each element we can calculate the number of moles of each atom, using the atomic weight:
0.360 g Fe * 1 mol / 55.845 g = 0.00645 moles of Fe
0.140 g O * 1 mol / 16 g = 0.00875 moles of O
The number of moles of Fe in the mixture is equal to the number of moles of FeO plus two times the number of moles of Fe₂O₃:
0.00645 = [tex]2*n_{Fe2O3} +n_{FeO}[/tex] eq A
The number of moles of O in the mixture is equal to the number of moles of FeO plus three times the number of moles of Fe₂O₃:
0.00875 = [tex]3*n_{Fe2O3} +n_{FeO}[/tex] eq B
So now we have a system of two equations and two unknowns, we solve for [tex]n_{Fe2O3}[/tex]:
From eq A:
[tex]n_{FeO3}=0.00645-2*n_{Fe2O3}[/tex]
Replacing in eq B:
[tex]0.00875=3*n_{Fe2O3} + (0.00645-2*n_{Fe2O3})\\0.00230=n_{Fe2O3}[/tex]
Now we just need to convert moles of Fe₂O₃ into grams, using the molecular weight:
0.00230 moles * 159.66 g/mol = 0.367 g Fe₂O₃
A mixture of methanol and methyl acetate contains 7.0 weight percent methanol. Determine the number of gmols of methanol in 300.0 kilograms of the mixture.
Answer:
656,25 moles of CH3OH
Explanation:
If the mixture has 7% of weight in methanol (CH3OH), it means that for every 100 kg of the mixture there are 7 kg of methanol.
To solve the problem we just use this relation and convert the kilograms of methanol to gmoles of methanol using the molecuar weight of the methanol (32g/mol):
[tex]300 kgmixture*\frac{7kgCH3OH}{100kg}*\frac{1000g}{1kg}*\frac{1molCH3OH}{32g}=656,25molesCH3OH[/tex]