The quantity of protein in the solution with an absorbance of 0.234 is approximately 12.472 micrograms (μg).
We have,
The given calibration curve equation is:
A = 0.0182x + 0.007
Where:
A is the corrected absorbance
x is the quantity of protein in the solution in micrograms (μg)
You're given that the absorbance of the blank solution (without protein) is 0.055.
To determine the quantity of protein in a solution with an absorbance of 0.234, we need to rearrange the equation to solve for x:
A = 0.0182x + 0.007
Substitute the absorbance value (A) and solve for x:
0.234 = 0.0182x + 0.007
0.234 - 0.007 = 0.0182x
0.227 = 0.0182x
x = 0.227 / 0.0182
x ≈ 12.472
Thus,
The quantity of protein in the solution with an absorbance of 0.234 is approximately 12.472 micrograms (μg).
Learn more about equations here:
https://brainly.com/question/17194269
#SPJ12
The quantity of protein in the solution is determined by first correcting the absorbance of the solution by subtracting the absorbance of the blank solution. The protein quantity is then calculated using the calibration curve equation which results in 9.45μg.
Explanation:To determine the quantity of protein in the solution, we first need to correct the measured absorbance by subtracting the absorbance of the blank solution. In this case, the absorbance of the solution is 0.234 and the absorbance of the blank solution is 0.055. Therefore, the corrected absorbance (A) is 0.234 - 0.055 = 0.179.
We can then use the equation for the calibration curve, A = 0.0182x + 0.007, to find the quantity of protein (x) in the solution. Substituting A = 0.179 into the equation gives 0.179 = 0.0182x + 0.007. Solving this equation for x gives x = (0.179 - 0.007) / 0.0182 = 9.45 μg. Therefore, the solution contains 9.45 μg of protein.
Learn more about Calibration Curve here:https://brainly.com/question/32826496
#SPJ3
A sample of steam with a mass of 0.535 g at a temperature of 100 ∘C condenses into an insulated container holding 4.40 g of water at 5.0 ∘C. (For water, ΔH∘vap=40.7 kJ/mol and Cwater=4.18 J/(g⋅∘C).)
Completed question:
A sample of steam with a mass of 0.535 g and at a temperature of 100 ∘C condenses into an insulated container holding 4.40 g of water at 5.0 ∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C)
Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?
Answer:
73.9°C
Explanation:
The steam is already at the water boiling point (100°C at 1 atm), so it will first lose heat to be condensed, by the equation:
Q1 = -n*ΔH∘vap*1000
Where n is the number of moles, ΔH∘vap is the enthalpy of evaporation, and the minus sign indicates that the heat is being lost. The equation is multiplied by 1000 because ΔH∘vap is in kJ, and the result must be in J.
Then, when it is in a liquid state, it will change heat with the cold water presented. The hot one will lose heat (Q2) and the cold one will gain heat (Q3) as the equation:
Q2 = mh*c*ΔTh
Q3 = mc*c*ΔTc
Where mh is the mass of the hot water, ΔTh is the variation of the temperature of the hot water (final - initial), mc is the mass of the cold water, and ΔTc is the variation of the temperature of the cold water.
At equilibrium they both will have the same final temperature (T), and, because the system doesn't change heat with the surroundings, the sum of all those heats must be 0.
n = mass/molar mass
Molar mass of water: 18g/mol
n = 0.535/18 = 0.0297 mol
Q1 + Q2 + Q3 = 0
-0.0297*40.7*1000 + 0.535*4.18*(T - 100) + 4.40*4.18*(T - 5.0) = 0
-1208.79 + 2.2363T - 223.63 + 18.392T - 91.96 = 0
20.6283T = 1524.38
T = 73.9°C
The final temperature of the mixture is mathematically given as
T = 73.9°C
What is the final temperature of the mixture?Question Parameter(s):
Generally, the equation for the heat is mathematically given as
Q1 = -n*dH∘vap*1000
Q2 = mh*c*dTh
Q3 = mc*c*dTc
Therefore
Q1 + Q2 + Q3 = 0
-0.0297*40.7*1000 + 0.535*4.18*(T - 100) + 4.40*4.18*(T - 5.0) = 0
20.6283T = 1524.38
T = 73.9°C
In conclusion, The temperature is
T = 73.9°C
Read more about Energy
https://brainly.com/question/13439286
Complete question:
A sample of steam with a mass of 0.535 g and at a temperature of 100 ∘C condenses into an insulated container holding 4.40 g of water at 5.0 ∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C) Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?
__CuSO4*5H2O(s)--->_____+_____
Answer:
CuSO4.5H2O —> CuSO4 + 5H2O
Explanation:
This reaction shows how hydrate copper sulphate losses its water of crystallization to become anhydrous copper sulphate
CuSO4.5H2O —> CuSO4 + 5H2O
Which represent potential energy?
A. A candy bar
B. A stretched rubber band
C. Wind
D. A roller coaster at the top of a hill
E. Rowing a boat
F. A ball rolling down a hill
Answer: Option A. A candy bar
Option B. stretched rubber band
Option D. A roller coaster at the top of a hill
Explanation:
Answer:
The correct answers are options A. "A candy bar", B. "A stretched rubber band" and D. "A roller coaster at the top of a hill".
Explanation:
Potential energy is defined as the force that any body for its location at a certain position. Three examples of objects having potential energy are a candy bar, a stretched rubber band and a roller coaster at the top of a hill. These three objects are not in motion and its energy is only "potential", rising from the arrangement of its particles, the gravitational force, electric charges, among other factors.
Name the mineral group for each of the following minerals: Group of answer choices Kyanite (Al2SiO5) Ilmenite (FeTiO3) Rhodochrosite (MnCO3) Celestite (SrSO4) Chalcocite (Cu2S)
Answer:
Kyanite (Al2SiO5) - silicate
Ilmenite (FeTiO3) - Oxides
Rhodochrosite (MnCO3) - carbonate
Celestite (SrSO4) - sulphate
Chalcocite (Cu2S) - sulphide
Explanation:
Minerals are classified according to their chemical composition. For example those that hve the CO32- ion are called carbonates and those with the SO42- ion are called sulphates while the ones with S2- ion are called sulphides
In a student experiment, the empirical formula of a copper halide was found by adding aluminum metal to an aqueous solution of the halide, displacing copper metal. The copper metal was filtered, washed with distilled water, dried, and weighed; three separate determinations were performed. The copper halide solution contained 42.62 g of copper chloride per liter. The student recorded the following experimental data.Trial Volume of copper chloride solution Mass of filter paper Mass of filter paper with copper (ml) (g) (g)A 49.6 0.908 1.694B 48.3 0.922 1.693C 42.2 0.919 1.588Write the empirical formula of copper chloride based on the experimental data.
Answer: The empirical formula for the given compound is [tex]CuCl_3[/tex]
Explanation:
We are given:
Mass of copper chloride in 1 L or 1000 mL of solution = 42.62 grams
Taking Trial A:
Volume of solution = 49.6 mL
Applying unitary method:
In 1000 mL of solution, the mass of copper chloride present is 42.62 grams
So, in 49.6 mL of solution, the mass of copper chloride will be = [tex]\frac{42.62}{1000}\times 49.6=2.114g[/tex]
We are given:
Mass of filter paper = 0.908 g
Mass of filter paper + copper = 1.694 g
Mass of copper = [1.694 - 0.908] g = 0.786 g
Mass of chlorine in the sample = [2.114 - 0.786]g = 1.328 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Copper =[tex]\frac{\text{Given mass of Copper}}{\text{Molar mass of Copper}}=\frac{0.786g}{63.5g/mole}=0.0124moles[/tex]
Moles of Chlorine = [tex]\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{1.328g}{35.5g/mole}=0.0374moles[/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 0.0124 moles.
For Copper = [tex]\frac{0.0124}{0.0124}=1[/tex]
For Chlorine = [tex]\frac{0.0374}{0.0124}=3.02\approx 3[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of Cu : Cl = 1 : 3
Hence, the empirical formula for the given compound is [tex]CuCl_3[/tex]
Based on the experimental data, the empirical formula of the copper chloride is CuCl₂.
To determine the empirical formula of the copper chloride, we'll use the provided experimental data for three trials.
First, calculate the mass of copper produced in each trial:
Trial A: Mass of copper = Mass of filter paper with copper - Mass of filter paper = 1.694 g - 0.908 g = 0.786 gTrial B: Mass of copper = 1.693 g - 0.922 g = 0.771 gTrial C: Mass of copper = 1.588 g - 0.919 g = 0.669 gNext, calculate the moles of copper produced using its molar mass (Cu = 63.55 g/mol):
[tex]\text{Trial A:} & \quad \text{Moles of copper} = \frac{0.786 \, \text{g}}{63.55 \, \text{g/mol}} \approx 0.01237 \, \text{mol} \\[/tex][tex]\text{Trial B:} & \quad \text{Moles of copper} = \frac{0.771 \, \text{g}}{63.55 \, \text{g/mol}} \approx 0.01213 \, \text{mol} \\[/tex][tex]\text{Trial C:} & \quad \text{Moles of copper} = \frac{0.669 \, \text{g}}{63.55 \, \text{g/mol}} \approx 0.01053 \, \text{mol}[/tex]Now, find the moles of copper chloride (assuming CuCl₂):
1. [tex]\text{Trial A:} & \quad \text{Volume of solution} = 49.6 \, \text{mL}, \\[/tex]
[tex]& \quad \text{Moles of } \text{CuCl}_2 = \frac{42.62 \, \text{g/L} \times 0.0496 \, \text{L}}{134.45 \, \text{g/mol}} \approx 0.01572 \, \text{mol} \\[/tex]2. [tex]\text{Trial B:} & \quad \text{Volume of solution} = 48.3 \, \text{mL}, \\[/tex]
[tex]& \quad \text{Moles of } \text{CuCl}_2 = \frac{42.62 \, \text{g/L} \times 0.0483 \, \text{L}}{134.45 \, \text{g/mol}} \approx 0.01535 \, \text{mol} \\[/tex]3. [tex]\text{Trial C:} & \quad \text{Volume of solution} = 42.2 \, \text{mL}, \\[/tex]
[tex]& \quad \text{Moles of } \text{CuCl}_2 = \frac{42.62 \, \text{g/L} \times 0.0422 \, \text{L}}{134.45 \, \text{g/mol}} \approx 0.01337 \, \text{mol}[/tex]Lastly, the moles of chlorine (as Cl2) can be determined as the difference between the moles of Cu and total moles of CuCl₂:
Trial A: Moles of Cl = 0.01572 mol - 0.01237 mol = 0.00335 molTrial B: Moles of Cl = 0.01535 mol - 0.01213 mol = 0.00322 molTrial C: Moles of Cl = 0.01337 mol - 0.01053 mol = 0.00284 molAveraging these values, the ratio of moles of Cu to Cl is approximately 1:2, suggesting an empirical formula of CuCl₂.
Complete Question: -
In a student experiment, the empirical formula of a copper halide was found by adding aluminum metal to an aqueous solution of the halide, displacing copper metal. The copper metal was filtered, washed with distilled water, dried, and weighed. Three separate determinations were performed. The copper halide solution contained 42.62 g of copper chloride per liter. The student recorded the following experimental data:
Trial A:
Volume of copper chloride solution (mL): 49.6Mass of filter paper (g): 0.908Mass of filter paper with copper (g): 1.694Trial B:
Volume of copper chloride solution (mL): 48.3Mass of filter paper (g): 0.922Mass of filter paper with copper (g): 1.693Trial C:
Volume of copper chloride solution (mL): 42.2Mass of filter paper (g): 0.919Mass of filter paper with copper (g): 1.588Write the empirical formula of copper chloride based on the experimental data.
A flashlight battery is hooked to a toy motor, and then the battery and the connections are sprayed with a water-proof coating. The battery is immersed in a beaker holding 100 mL of water. When the toy motor drives a device that raises a weight of 1.00 kg a distance of 10.0 m, the temperature of the water falls by 0.024 C. Assuming that the heat capacity of the battery is negligible compared to that of the water, Calculate the change in the energy of the battery contents as a result of the chemical reaction that took place in the battery.
Explanation:
Formula to calculate work done by motor is as follows.
Work done by motor = [tex]mass \times g \times height[/tex]
where, g = gravitational constant = 10 [tex]m/s^{2}[/tex]
Therefore, work done by motor is as follows.
Work done by motor = [tex]1.00 kg \times 10 m/s^{2} \times 10.0 m[/tex]
= 100.0 J
Now, heat lost by water will be calculated as follows.
q = [tex]mC \times \Delta T[/tex]
= [tex]g \times 4.184 J/g^{o}C \times 0.024^{o}C[/tex]
= 10.0 J
Hence, heat gained by motor = heat lost by water
As, heat gained by motor = 10.0 J
So, change in energy = heat gained - work done
Therefore, change in energy will be calculated as follows.
Change in energy = heat gained - work done
= (10.0 J) - (100.0 J)
= -90.0 J
Thus, we can conclude that change in the energy of the battery contents is -90.0 J.
Write and balance a complete chemical reaction equation for the combustion of butane. Butane has carbon atoms. Butane has hydrogen atoms. Butane is considered . Which compound is needed for the complete combustion of butane
Answer:
The balanced reaction is this:
2 C₄H₁₀ (g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O (g)
Explanation:
Combustion is a chemical reaction whose reagents are oxygen, usually in excess and a hydrocarbon to generate carbon dioxide and water in the form of steam, as products.
Butane is considered as a reactant and it is a sort of alkane, in this case with 4 C (prefix but).
O₂ is needed for the complete combustion of butane.
Some diamonds appear yellow because they contain nitrogen compounds that absorb purple light of frequency 7.23 x 10¹⁴ Hz. Calculate the wavelength (in nm and Å) of the absorbed light.
Answer: The wavelength of the absorbed light is 415 nm or [tex]4150\AA[/tex]
Explanation:
To calculate the wavelength of light, we use the equation:
[tex]\lambda=\frac{c}{\nu}[/tex]
where,
[tex]\lambda[/tex] = wavelength of the light
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\nu[/tex] = frequency of light = [tex]7.23\times 10^{14}Hz=7.23\times 10^{14}s^{-1}[/tex]
Putting the values in above equation, we get:
[tex]\lambda=\frac{3\times 10^8m/s}{7.23\times 10^{14}s^{-1}}=4.15\times 10^{-7}m[/tex]
Converting this into nanometers, we use the conversion factor:
[tex]1m=10^9nm[/tex]
So, [tex]4.15\times 10^{-7}m\times (\frac{10^9nm}{1m})=415nm[/tex]
Converting this into angstroms, we use the conversion factor:
[tex]1m=10^{10}\AA[/tex]
So, [tex]4.15\times 10^{-7}m\times (\frac{10^{10}\AA}{1m})=4150\AA[/tex]
Hence, the wavelength of the absorbed light is 415 nm or [tex]4150\AA[/tex]
The absorbed light that gives some diamonds a yellow color has a wavelength of 415 nm, or 4150 Å, calculated using the speed of light and the given frequency.
Explanation:The color of light, such as that absorbed by a diamond, is determined by its wavelength. Given the frequency, we can calculate the wavelength using the formula c = λv, where 'c' is the speed of light (3.00 x 10^8 m/s), 'λ' is the wavelength, and 'v' is the frequency.
In this case, the frequency 'v' is given as 7.23 x 10^14 Hz, so we solve for 'λ' to get: λ = c/v = (3.00 x 10^8 m/s) / (7.23 x 10^14 Hz) = 4.15 x 10^-7 m. Converting this to nm (nanometers) and Å (angstroms) gives us 415 nm and 4150 Å, respectively.
Learn more about Wavelength of Absorbed Light here:https://brainly.com/question/33569910
#SPJ3
Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:
(a) Rb (b) N (c) Br
Answer :
(a) The charge and full ground-state electron configuration of the monatomic ion is, (+1) and [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
(b) The charge and full ground-state electron configuration of the monatomic ion is, (-3) and [tex]1s^22s^22p^6[/tex]
(c) The charge and full ground-state electron configuration of the monatomic ion is, (-1) and [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
Explanation :
For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.
When an unequal number of electrons and protons then it leads to the formation of ionic species.
Ion : An ion is formed when an atom looses or gains electron.
When an atom looses electrons, it will form a positive ion known as cation.
When an atom gains electrons, it will form a negative ion known as anion.
(a) The given element is, Rb (Rubidium)
As we know that the rubidium element belongs to group 1 and the atomic number is, 37
The ground-state electron configuration of Rb is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1[/tex]
This element will easily loose 1 electron and form [tex]Rb^+[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Rb ion is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
(b) The given element is, N (Nitrogen)
As we know that the nitrogen element belongs to group 15 and the atomic number is, 7
The ground-state electron configuration of N is:
[tex]1s^22s^22p^3[/tex]
This element will easily gain 3 electrons and form [tex]N^{3-}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of N ion is:
[tex]1s^22s^22p^6[/tex]
(c) The given element is, Br (Bromine)
As we know that the bromine element belongs to group 17 and the atomic number is, 35
The ground-state electron configuration of Rb is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^5[/tex]
This element will easily gain 1 electron and form [tex]Br^-[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Br ion is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
The most likely monatomic ions formed by Rb, N, and Br are Rb+, N3-, and Br-, respectively.
Explanation:(a) The most likely monatomic ion formed by Rb is Rb+. The ground-state electron configuration of Rb is [Kr]5s1, and when it loses one electron to form the ion, its configuration becomes [Kr].
(b) The most likely monatomic ion formed by N is N3-. The ground-state electron configuration of N is 1s22s22p3, and when it gains three electrons to form the ion, its configuration becomes 1s22s22p6.
(c) The most likely monatomic ion formed by Br is Br-. The ground-state electron configuration of Br is [Ar]4s23d104p5, and when it gains one electron to form the ion, its configuration becomes [Ar]4s23d104p6.
Learn more about Ground-state electron configuration here:https://brainly.com/question/33895511
#SPJ3
Write the condensed electron configuration of each transition metal ion, and predict whether it is paramagnetic:
(a) V³⁺ (Z = 23)
(b) Ni²⁺ (Z = 28)
(c) La³⁺ (Z = 57)
Final answer:
The condensed electron configurations for V³⁺, Ni²⁺, and La³⁺ are [Ar]3d², [Ar]3d⁸, and [Xe] respectively. V³⁺ and Ni²⁺ are paramagnetic with unpaired d electrons, while La³⁺ is not paramagnetic.
Explanation:
To determine the condensed electron configuration of transition metal ions and their magnetic properties, we consider the electron configuration of the neutral atom and then remove electrons corresponding to the ion's charge, starting with the outermost shell.
V³⁺ (Vanadium 3+): Vanadium has an atomic number of 23, so the electron configuration of neutral V is [Ar]3d³4s². Removing 3 electrons for the V³⁺ ion, typically from the 4s and then 3d orbitals, gives a condensed electron configuration of [Ar]3d². This ion has two unpaired electrons in the 3d orbital and is paramagnetic.Ni²⁺ (Nickel 2+): Nickel has an atomic number of 28, and the electron configuration of neutral Ni is [Ar]3d⁸ 4s². For Ni²⁺, we remove two electrons to get [Ar]3d⁸. This ion also has unpaired electrons in the 3d orbital, making it paramagnetic.La³⁺ (Lanthanum 3+): Lanthanum has an atomic number of 57, with the electron configuration of [Xe]5d¹ 6s² for the neutral atom. Removing 3 electrons from the 5d and 6s orbitals for La³⁺ gives us [Xe]. There are no unpaired electrons in the 5d or 6s orbitals, so it is not paramagnetic.Draw the structure of silicon tetrahydride according to Lewis theory. What would be its associated molecular geometry?
Answer : The molecular geometry of the molecule is, tetrahedral.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, [tex]SiH_4[/tex]
As we know that silicon has '4' valence electrons and hydrogen has '1' valence electron.
Therefore, the total number of valence electrons in [tex]SiH_4[/tex] = 4 + 4(1) = 8
According to Lewis-dot structure, there are 8 number of bonding electrons and 0 number of non-bonding electrons.
Now we have to determine the hybridization of the given molecules.
Formula used :
[tex]\text{Number of electron pair}=\frac{1}{2}[V+N-C+A][/tex]
where,
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
The given molecule is, [tex]SiH_4[/tex]
[tex]\text{Number of electrons}=\frac{1}{2}\times [4+4]=4[/tex]
The number of electron pair are 4 that means the hybridization will be [tex]sp^3[/tex] and the electronic geometry and the molecular geometry of the molecule will be tetrahedral.
Hence, the molecular geometry of the molecule is, tetrahedral.
Silicon tetrahydride has a tetrahedral molecular geometry with the silicon atom being sp³ hybridized and bond angles approximately 109.5°.
Explanation:The structure of silicon tetrahydride, also known as silane, can be represented according to Lewis theory by a central silicon atom surrounded by four hydrogen atoms, each bonded to the silicon with a single bond. The Lewis structure reflects that all the valence electrons of silicon are used to form the bonds with hydrogen atoms, leaving no lone pairs on the silicon. The associated molecular geometry of silicon tetrahydride is tetrahedral, as silicon is sp³ hybridized, with the four regions of electron density (the four Si-H bonds) arranged in a tetrahedral fashion. This geometry results in bond angles that are approximately 109.5°.
How fast must a 142-g baseball travel in order to have a de Broglie wavelength that is equal to that of an x-ray photon with λ = 100. pm?
Answer:
4.7 x 10⁻²³
Explanation:
The strategy here is to use de Broglie relation to answer this question:
mv = h/λ
where m is the mass, h is Planck´s constant, and λ is the wavelength:
v = h/ mλ
h = 6.626 x 10⁻³⁴ J·s
m = 142 g = 0.142 kg ( we are working in the metric system )
λ = 100 pm = 100 pm x ( 1 x 10⁻¹² m/pm ) = 1 x 10⁻¹⁰ m
⇒ v = 6.626 x 10⁻³⁴ J·s / ( 0.142 kg x 1 x 10⁻¹⁰ m ) = 4.7 x 10⁻²³ m/s
This calculation shows why we do not talk about quantum effects at the macro level, notice the extreme low velocity that the baseball will have to have a wavelength equal to that of an x-ray photon.
A baseball with a mass of 142 grams would need to travel at a velocity of about 4.67 x 10^7 m/s in order to have a de Broglie wavelength equal to that of an x-ray photon with a wavelength of 100 pm.
Explanation:To calculate the required velocity for the baseball to have a de Broglie wavelength equal to an x-ray photon, we use the de Broglie equation given as lambda = h / (m * v), where lambda is the de Broglie wavelength, h is Planck's constant (6.626 x 10^-34 kg m²/s), m is the mass of the particle, and v is the velocity. Given the wavelength lambda = 100 pm = 100 x 10^-12 m, and the mass of the baseball 142 g = 0.142 kg, we can rearrange the equation for velocity to get v = h / (m * lambda). Using the values, we get a velocity of approximately 4.67 x 10^7 m/s.
The seemingly large velocity can be attributed to the fact that de Broglie wavelengths are typically observable at small scales (in the quantum realm). For a macroscopic object like baseball, the velocities required for corresponding de Broglie wavelengths become unfeasibly high.
Learn more about de Broglie wavelength here:https://brainly.com/question/32576779
#SPJ3
What is the speed of an electron that has a de Broglie wavelength of 100. nm?
Answer: The speed of the electron is [tex]7.24\times 10^3m/s[/tex]
Explanation:
To calculate the speed of electron for given wavelength, we use the equation given by De-Broglie's, which is:
[tex]\lambda=\frac{h}{mv}[/tex]
where,
[tex]\lambda[/tex] = De-Broglie's wavelength = [tex]100nm=100\times 10^{-9}m[/tex]
h = Planck's constant = [tex]6.6\times 10^{-34}Js[/tex]
m = mass of the electron = [tex]9.11\times 10^{-31}kg[/tex]
v = speed of the electron = ?
Putting values in above equation, we get:
[tex]100\times 10^{-9}m=\frac{6.6\times 10^{-34}Js}{9.11\times 10^{-31}kg\times v}\\\\v=\frac{6.6\times 10^{-34}Js}{9.11\times 10^{-31}kg\times 100\times 10^{-9}m}=7.24\times 10^3m/s[/tex]
Hence, the speed of the electron is [tex]7.24\times 10^3m/s[/tex]
The speed of an electron with a de Broglie wavelength of 100 nm is approximately 7.28 × 105 m/s, calculated by rearranging the de Broglie wavelength formula and using known values for Planck's constant and the electron's mass.
Explanation:The task here is to calculate the speed of an electron given its de Broglie wavelength. The de Broglie wavelength λ is related to the momentum p of a particle via the equation λ = h / p, where h is Planck's constant (6.62607015 × 10−34 m² kg / s). Momentum can be expressed as p = mv, where m is the mass of the electron (9.11 × 10−31 kg) and v is the velocity we need to find.
Reordering the de Broglie equation to solve for velocity, v = h / (λm), and substituting the given values, we calculate the electron's speed for a de Broglie wavelength of 100 nm (1 × 10−9 meters).
Calculation:v = λm / h = 6.62607015 × 10−34 / (9.11 × 10−31 × 1 × 10−9)
= 6.62607015 × 10−5 / 9.11
= 7.28 × 105 m/s.
The speed of the electron is approximately 7.28 × 105 meters per second (m/s).
If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the answer to one decimal place. pKa = 4.87
Answer:
pH = 4.543
Explanation:
CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+pKa = - Log Ka∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
∴ pKa = 4.87
⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
added 300 mL 0f 0.02 M NaOH:
⇒ C CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)
⇒ C CH3CH2COOH = 0.048 M
⇒ C NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M
mass balance:
⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)
charge balance:
⇒ [H3O+] + [Na+] = [CH3CH2COO-]
∴ [Na+] = 0.02 M
⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)
(2) in (1):
⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]
replacing in Ka:
⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])
⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]
⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]
⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0
⇒ [H3O+ ] = 2.867 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = 4.543
Aragonite, with a density of 2.9 g/cm3, has exactly the same chemical composition as calcite, which has a density of 2.7 g/cm3. Other things being equal, which of these two minerals formed under higher pressure?
Answer:
Aragonite is most likely be formed under high pressure.
Explanation:
Relating the densities of Aragonite and Calcite which both has the same chemical composition, it shows that the density of Aragonite is more than Calcite which says to some large extent that there is every possibility that Aragonite is most likely to be formed under high pressure.
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr (at the moment of mixing). The system then re-equilibrates. The appropriate chemical equation is PCl3(g)+Cl2(g) <-->PCl5(g) Calculate the new partial pressures after equilibrium is reestablished in torr
Answer:
The new partial pressures after equilibrium is reestablished:
[tex]PCl_3,p_1'=6.798 Torr[/tex]
[tex]Cl_2,p_2'=26.398 Torr[/tex]
[tex]PCl_5,p_3'=223.402 Torr[/tex]
Explanation:
[tex]PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g) [/tex]
At equilibrium before adding chlorine gas:
Partial pressure of the [tex]PCl_3=p_1=13.2 Torr[/tex]
Partial pressure of the [tex]Cl_2=p_2=13.2 Torr[/tex]
Partial pressure of the [tex]PCl_5=p_3=217.0 Torr[/tex]
The expression of an equilibrium constant is given by :
[tex]K_p=\frac{p_1}{p_1\times p_2}[/tex]
[tex]=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245[/tex]
At equilibrium after adding chlorine gas:
Partial pressure of the [tex]PCl_3=p_1'=13.2 Torr[/tex]
Partial pressure of the [tex]Cl_2=p_2'=?[/tex]
Partial pressure of the [tex]PCl_5=p_3'=217.0 Torr[/tex]
Total pressure of the system = P = 263.0 Torr
[tex]P=p_1'+p_2'+p_3'[/tex]
[tex]263.0Torr=13.2 Torr+p_2'+217.0 Torr[/tex]
[tex]p_2'=32.8 Torr[/tex]
[tex]PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g) [/tex]
At initail
(13.2) Torr (32.8) Torr (13.2) Torr
At equilbriumm
(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr
[tex]K_p=\frac{p_3'}{p_1'\times p_2'}[/tex]
[tex]1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}[/tex]
Solving for x;
x = 6.402 Torr
The new partial pressures after equilibrium is reestablished:
[tex]p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr[/tex]
[tex]p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr[/tex]
[tex]p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr[/tex]
Be sure to answer all parts. Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everything from fats and waxes to airplane glue and nail polish. At high temperatures, it decomposes in a first-order process to methane and ethene (CH2═C═O). At 600°C, the rate constant is 8.7 × 10−3 s−1. (a) What is the half-life of the reaction
Answer:
Half life = 79.67 sec
Explanation:
Given that:
k = [tex]8.7\times 10^{-3}\ s^{-1}[/tex]
The expression for half life is shown below as:-
[tex]t_{1/2}=\frac{\ln2}{k}[/tex]
Where, k is rate constant
So,
[tex]t_{1/2}=\frac{\ln2}{8.7\times 10^{-3}\ s^{-1}}[/tex]
[tex]t_{1/2}=114.94252\ln \left(2\right)\ sec=79.67\ sec[/tex]
Half life = 79.67 sec
A balloon filled with 0.500 L of air at sea level is submerged in the water to a depth that produces a pressure of 3.25 atm. What is the volume of the balloon at this depth? a. 0.154 L b. 6.50 L c. 0.615 L d. 1.63 L d. None of the above
Answer:
Option a . 0.154L
Explanation:
P₁ . V₁ = P₂ . V₂
when we have constant temperature and constant moles for a certain gas.
At sea level, pressure is 1 atm so:
0.5 L . 1atm = V₂ . 3.25 atm
(0.5L . 1atm) / 3.25 atm = 0.154 L
"0.154 L" is the volume of the balloon.
Given:
Pressure,
[tex]P_1 = 1 \ atm[/tex][tex]P_2 = 3.25 \ atm[/tex]Volume,
[tex]V_1 = 0.5 \ L[/tex][tex]V_2 = ?[/tex]As we know,
→ [tex]P_1. V_1 = P_2 .V_2[/tex]
or,
→ [tex]V_2 = \frac{P_1. V_1}{P_1}[/tex]
By substituting the values, we get
[tex]= \frac{0.5\times 1}{3.25}[/tex]
[tex]= \frac{0.5}{3.25}[/tex]
[tex]= 0.154 \ L[/tex]
Thus the above answer i.e., "option a" is correct.
Learn more:
https://brainly.com/question/16365001
An FM station broadcasts music at 93.5 MHz (megahertz, or 10⁶ Hz). Find the wavelength (in m, nm, and Å) of these waves.
Answer:
The wavelength in m = 3.208 m
The wavelength in nm =[tex]3.208\times 10^9 nm[/tex]
The wavelength in Å = [tex]3.208\times 10^{10} \AA[/tex]
Explanation:
To calculate the wavelength of light, we use the equation:
[tex]\lambda=\frac{c}{\nu}[/tex]
where,
[tex]\lambda[/tex] = wavelength of the radiation
c = speed of light = [tex]3.0\times 10^8m/s[/tex]
[tex]\nu[/tex] = frequency of wave
We have :
Frequency of the wave = [tex]\nu =93.5MHz=93.5\times 10^6 Hz[/tex]
[tex]1 Hz = 1s^{-1}[/tex]
Wavelength of the wave = [tex]\lambda [/tex]
[tex]\lambda =\frac{c}{\nu}[/tex]
[tex]\lambda=\frac{3\times 10^8 m/s}{93.5\times 10^6 s^{-1}}=3.208 m[/tex]
[tex]1 m = 10^9 nm[/tex]
[tex]3.208 m= 3.208\times 10^9 nm[/tex]
[tex]1 m = 10^{10} \AA[/tex]
[tex]3.2086 m= 3.208\times 10^{10} \AA[/tex]
The wavelength in m = 3.208 m
The wavelength in nm =[tex]3.208\times 10^9 nm[/tex]
The wavelength in Å = [tex]3.208\times 10^{10} \AA[/tex]
The wavelength in m = 3.208 m
The wavelength in nm = [tex]3.208*10^9 nm[/tex]
The wavelength in Å = [tex]3.208*10^{10}[/tex] Å
What is Wavelength?It is equal to the speed (v) of a wave train in a medium divided by its frequency (f). It is given by:
[tex]\lambda=\frac{c}{f}[/tex]
where,
[tex]\lambda[/tex] = wavelength of the radiation
c = speed of light = [tex]3.0*10^8m/s[/tex]
f = frequency of wave = 93.5 MHz = [tex]93.5*10^6Hz[/tex]
Calculation of wavelength:
[tex]\lambda=\frac{c}{f}\\\\\lambda=\frac{3.0*10^8m/s}{93.5*10^6s^{-1}} \\\\\lambda=3.208m=3.208*10^9nm=3.208*10^{10}[/tex]Å.
Thus, The value for wavelength in different units are:
The wavelength in m = 3.208 m
The wavelength in nm = [tex]3.208*10^9 nm[/tex]
The wavelength in Å = [tex]3.208*10^{10}[/tex] Å
Find more information about Wavelength here:
brainly.com/question/10728818
The following unbalanced chemical equations are provided in the PhET simulation: Make Ammonia : N2 + H2 → NH3 Separate Water : H2O → H2 + O2 Combust Methane : CH4 + O2 → CO2 4- H2ODetermine for which elements the specified atoms are balanced or unbalanced in each of these chemical equations when every substance in the reaction is given a coefficient of one (1).
Answer:
For 1: Nitrogen and hydrogen atoms are unbalanced.
For 2: Hydrogen atom is balanced and oxygen atom is unbalanced.
For 3: Carbon atom is balanced and oxygen and hydrogen atom is unbalanced.
Explanation:
For 1:The given chemical equation follows:
[tex]N_2+H_2\rightarrow NH_3[/tex]
On the reactant side:
Number of nitrogen atoms = 2
Number of hydrogen atoms = 2
On the product side:
Number of nitrogen atoms = 1
Number of hydrogen atoms = 3
As, the number of nitrogen and hydrogen atoms are not same on both the sides of the reaction. So, these elements are unbalanced.
For 2:The given chemical equation follows:
[tex]H_2O\rightarrow H_2+O_2[/tex]
On the reactant side:
Number of oxygen atoms = 1
Number of hydrogen atoms = 2
On the product side:
Number of oxygen atoms = 1
Number of hydrogen atoms = 3
As, the number of oxygen atoms are not same on both the sides of the reaction. So, this element is unbalanced.
Number of hydrogen atoms are same on both the sides of the reaction. So, this element is balanced.
For 3:The given chemical equation follows:
[tex]CH_4+O_2\rightarrow CO_2+H_2O[/tex]
On the reactant side:
Number of carbon atoms = 1
Number of oxygen atoms = 2
Number of hydrogen atoms = 4
On the product side:
Number of carbon atoms = 1
Number of oxygen atoms = 3
Number of hydrogen atoms = 2
As, the number of oxygen and hydrogen atoms are not same on both the sides of the reaction. So, these element are unbalanced.
Number of carbon atoms are same on both the sides of the reaction. So, this element is balanced.
Nitrogen and hydrogen atoms are unbalanced in first reaction. Hydrogen atom is balanced and oxygen atom is unbalanced in second reaction. Carbon atom is balanced and oxygen and hydrogen atom is unbalanced for third reaction.
A balanced reaction is a chemical equation in which the number of atoms of each element on both sides of the equation is the same.
[tex]\rm N_2 + H_2 \rightarrow NH_3[/tex]
On the reactant side:
Number of nitrogen atoms = 2
Number of hydrogen atoms = 2
On the product side:
Number of nitrogen atoms = 1
Number of hydrogen atoms = 3
Because the number of nitrogen and hydrogen atoms on both sides of the reaction is not the same. As a result, these elements are imbalanced.
[tex]\rm H_2O \rightarrow H_2 + O_2[/tex]
On the reactant side:
Number of oxygen atoms = 1
Number of hydrogen atoms = 2
On the product side:
Number of oxygen atoms = 1
Number of hydrogen atoms = 3
Because the number of oxygen atoms on both sides of the reaction is not the same. As a result, this element is imbalanced. The number of hydrogen atoms on both sides of the reaction is the same. As a result, this element is balanced.
[tex]\rm CH_4 + O_2 \rightarrow CO_2 + H_2O[/tex]
On the reactant side:
Number of carbon atoms = 1
Number of oxygen atoms = 2
Number of hydrogen atoms = 4
On the product side:
Number of carbon atoms = 1
Number of oxygen atoms = 3
Number of hydrogen atoms = 2
Because the number of oxygen and hydrogen atoms on both sides of the reaction is not the same. As a result, these elements are imbalanced. The number of carbon atoms on both sides of the reaction is the same. As a result, this element is balanced.
To know more about balanced and unbalanced reaction, here:
https://brainly.com/question/14258973
#SPJ3
Rank the following photons in terms of decreasing energy:
(a) IR (v = 6.5 x 10¹³ s⁻¹)
(b) microwave (v = 9.8 x 10¹¹ s⁻¹)
(c) UV (v = 8.0 x 10¹⁵ s⁻¹)
Answer:
The order of the energy of the photons of given wave will be
= Ultraviolet waves > infrared waves > microwaves
Explanation:
[tex]E=h\nu =\frac{h\times c}{\lambda}[/tex]
where,
E = energy of photon
[tex]\nu [/tex] = frequency of the radiation
h = Planck's constant = [tex]6.63\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] = wavelength of the radiation
We have :
(a) Frequency of infrared waves = [tex]\nu _1=6.5\times 10^{13} s^{-1}[/tex]
(b) Frequency of microwaves= [tex]\nu _2=9.8\times 10^{11} s^{-1}[/tex]
(c) Frequency of ultraviolet waves = [tex]\nu _3=8.0\times 10^{15} s^{-1}[/tex]
So, the decreasing order of the frequencies of the waves will be :
[tex]\nu _3> \nu _1> \nu _2[/tex]
As we can see from the formula that energy is directly proportional to the frequency of the wave.
[tex]E\propto \nu [/tex]
So, the order of the energy of the photons of given wave will be same as their order of frequencies:
[tex]E_3>E_1>E_2[/tex]
= Ultraviolet waves > infrared waves > microwaves
How many grams of NO2 are theoretically produced if we start with 1.20 moles of S and 9.90 moles of HNO3? Reaction: S + 6HNO3 → H2SO4 + 6NO2 + 2H2O
Answer:
331.2 grams of [tex]NO_2[/tex] are theoretically produced .
Explanation:
[tex]S + 6HNO_3\rightarrow H_2SO_4 + 6NO_2 + 2H_2O[/tex]
Moles of sulfur = 1.20 mol
Moles nitric acid = 9.90 moles
According to reaction ,1 mole of S reacts with 6 moles of sulfuric acid.
Then 1.20 moles of S will react with :
[tex]\frac{6}{1}\times 1.20 mol=7.2 moles[/tex] of nitric acid
This means that S is in limiting amount and nitric acid is in excessive amount.
So, amount of [tex]NO_2[/tex] gas will depend upon amount of S.
According to reaction, 1 m,ole of S gives 6 moles of [tex]NO_2[/tex] gas .
Then 1.20 moles of S will give:
[tex]\frac{6}{1}\times 1.20 mol=7.2 moles[/tex] of [tex]NO_2[/tex]
Mass of 7.2 moles of [tex]NO_2[/tex]
7.2 g × 46 g/mol = 331.2 g
331.2 grams of [tex]NO_2[/tex] are theoretically produced .
The question involves a stoichiometry problem on the production of Nitrogen dioxide (NO2) from Sulfur (S) and Nitric acid (HNO3). The theoretical yield of NO2 is constrained by the amount of Sulfur, the limiting reactant in this case. With given quantities, according to stoichiometric calculations, 331.27 grams of NO2 are theoretically produced.
Explanation:The subject of this question is stoichiometry, a concept in chemistry which is the basis of predicting how much product can be made in a chemical reaction based on the quantity of reactants.
The reaction in the question is: S + 6HNO3 → H2SO4 + 6NO2 + 2H2O, according to which 1 mole of S will react with 6 moles of HNO3 to produce 6 moles of NO2. If we start with 1.20 moles of S, the reaction theoretically produces 6*1.20 = 7.20 moles of NO2.
However, when we consider the moles of HNO3, 9.90 moles of it can theoretically produce 9.90 moles of NO2 as per the balanced equation. However, this is more than what can be produced by the available S. Hence, S becomes the limiting reactant here. Therefore, the theoretical yield of NO2 is constrained by the amount of S and is 7.20 moles. When converting from moles to grams, using the molar mass of NO2 which is approximately 46.01 g/mol, we find that 7.20 moles of NO2 equals 331.27 grams.
Thus, the amount of NO2 that can be theoretically produced from these amounts of reactants is 331.27 grams.
Learn more about Stoichiometry here:https://brainly.com/question/30218216
#SPJ3
Five lines in the H atom spectrum have wavelengths (in Å): (a) 1212.7; (b) 4340.5; (c) 4861.3; (d) 6562.8; (e) 10,938. Three lines result from transitions to nfinal = 2 (visible series). The other two result from transitions in different series, one with nfinal = 1 and the other with nfinal = 3. Identify ninitial for each line.
Answer:
(a) n₂ = 2
(b) n₂ = 5
(c) n₂ = 4
(d) n₂ = 3
(e) n₂ = 6
Explanation:
The Rydberg equation give us the wavelength of the transition between energy levels according to the formula:
1/λ = Rh x ( 1/n₁² - 1/n₂² )
where n₁ is the final state and n₂ is the initial state.
The strategy here, since we are given the wavelength, is to solve for λ, and then by substituting for n₂ combinations find which ones match our question.
λ = 1 / [ Rh x ( 1/n₁² - 1/n₂² ) ]
Lets express Rh in 1/Angstrom
1.097 x 10 ⁷ / [m x ( 1 m/ 10¹⁰ A) ] = 0.011 / Å
⇒ λ = 1 / [0.011 A x ( 1/n₁² - 1/n₂² )] = 911.6 Å / ( 1/n₁² - 1/n₂² )
For n₁ = 2 n₂ = 3, 4, 5,.......
λ ( n₂ = 3 ) = 911.6 A / ( 1/2² - 1/3² ) = 6563.5 Å
λ ( n₂ = 4 ) = 911.6 A / ( 1/2² - 1/4² ) = 4861.9 Å
λ ( n₂ = 5 ) = 911.6 A / ( 1/2² - 1/5² ) = 4341.0 Å
So we have matched three of the transitions
Now for n₁ = 1 n₂ = 2, 3, 4....
λ ( n₂ = 2 ) = 911.6 A / ( 1/1² - 1/2² ) = 1215.5 Å
For n₁ = 3 n₂ = 4, 5, 6....
λ ( n₂ = 4 ) = 911.6 A / ( 1/3² - 1/4² ) = 18752.9 Å
λ ( n₂ = 5 ) = 911.6 A / ( 1/3² - 1/5² ) = 12819.4 Å
λ ( n₂ = 6 ) = 911.6 A / ( 1/3² - 1/6² ) = 10939.2 Å
a. For [tex]\lambda=1212.7\;\rm \r{A}[/tex], initial transition level is 2.
b. For [tex]\lambda=4340.5\;\rm \r{A}[/tex], initial transition level is 5.
c. For [tex]\lambda=4861.3\;\rm \r{A}[/tex], initial level of transition is 4.
d. For [tex]\lambda=6562.8\;\rm \r{A}[/tex], initial level of transition is 3.
e. For [tex]\lambda=10,938\;\rm \r{A}[/tex], the initial level of transition has been 6.
The wavelength from the transition has been given by the Rydberg equation as:
[tex]\dfrac{1}{\lambda}=\text Rh\dfrac{1}{N_1}-\dfrac{1}{N_2}[/tex]
Where, wavelength of the radiation, [tex]\lambda[/tex]
The initial transition level, [tex]N_1[/tex]
The final transition level, [tex]N_2[/tex]
The constant [tex]Rh=1.097\;\times\;10^7[/tex], or in Armstrong it can be given as, [tex]0.011/\r{\rm A}[/tex]
The wavelength can be given as:
[tex]\lambda=\dfrac{1}{0.011\;\r{\rm A}\;\times\;(\frac{1}{N_1^2} -\frac{1}{N_2^2}) }[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{N_1^2}-\frac{1}{N_2^2} }[/tex]
a. The initial level ([tex]N_2[/tex]) of transition has been given for [tex]\lambda=1212.7\;\rm \r{A}[/tex], and [tex]N_2=2[/tex] has been given, with substituting [tex]N_1=1[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{1^2}-\frac{1}{2^2} }\\\lambda=1215.5\;\r{A}[/tex]
Thus, for [tex]\lambda=1212.7\;\rm \r{A}[/tex], initial transition level is 2.
b. For, [tex]\lambda=4340.5\;\rm \r{A}[/tex], and [tex]N_2=2[/tex], [tex]N_1=5[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{2^2}-\frac{1}{5^2} }\\\lambda=4341.0\;\r{A}[/tex]
Thus, for [tex]\lambda=4340.5\;\rm \r{A}[/tex], initial transition level is 5.
c. For, [tex]\lambda=4861.3\;\rm \r{A}[/tex], the final transition has been, [tex]N_2=2[/tex], the initial level has been substituted as [tex]N_1=4[/tex]:
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{2^2}-\frac{1}{4^2} }\\\lambda=4861.3\;\r{A}[/tex]
Thus, for [tex]\lambda=4861.3\;\rm \r{A}[/tex], initial level of transition is 4.
d. For [tex]\lambda=6562.8\;\rm \r{A}[/tex], the final level has been 2, the initial level has been substituted as, [tex]N_1=3[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{2^2}-\frac{1}{3^2} }\\\lambda=6562.8\;\r{A}[/tex]
Thus, for [tex]\lambda=6562.8\;\rm \r{A}[/tex], initial level of transition is 3.
e. For [tex]\lambda=10,938\;\rm \r{A}[/tex], the initial level has been substituted as, [tex]N_1=6[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{3^2}-\frac{1}{6^2} }\\\lambda=10,938\;\r{A}[/tex]
Thus, for [tex]\lambda=10,938\;\rm \r{A}[/tex], the initial level of transition has been 6.
For more information about transition levels, refer to the link:
https://brainly.com/question/4511550
The equilibrium concentrations of the reactants and products are [ HA ] = 0.260 M [HA]=0.260 M , [ H + ] = 2.00 × 10 − 4 M [H+]=2.00×10−4 M , and [ A − ] = 2.00 × 10 − 4 M [A−]=2.00×10−4 M . Calculate the value of p K a pKa for the acid HA HA .
Answer:
pKa of the acid HA with given equilibrium concentrations is 6.8
Explanation:
The dissolution reaction is:
HA ⇔ H⁺ + A⁻
So at equilibrium, Ka is calculated as below
Ka = [H⁺] x [A⁻] / [HA] = 2.00 x 10⁻⁴ x 2.00 x 10⁻⁴ / 0.260
= 15.38 x 10⁻⁸
Hence, by definition,
pKa = -log(Ka) = - log(15.38 x 10⁻⁸) = 6.813
Calculate the number of pounds of CO2 released into the atmosphere when a 12.0-gallon tank of gasoline is burned in an automobile engine. Assume that gasoline is primarily octane, C8H18 and that the density of gasoline is 0.692 g.mL-1 (this assumption ignores additives). Also assume complete combustion.
Useful conversion factors:
1 gallon = 3.785L
1 Kg = 2.204 lb
______________ lb
Answer:
213.89 lb of CO2.
Explanation:
Equation for the reaction:
C8H18 + 25/2O2(g) --> 8CO2(g) + 9H2O(g)
Given:
Volume of gasoline = 12 gallon
Converting gallon to ml,
12 gallon * 3.785 l/1 gallon * 1000 ml/1 l
= 45420 ml
Density of the gasoline = 0.692 g/ml
Mass = density * volume
= 45420 * 0.692
= 31430 g
Molar mass of octane = (8*12) + (18*1)
= 114 g/mol.
Number of moles = mass/molar mass
= 31430/114
= 275.702 mol.
From the above equation, 1 mole of octane was completed burnt to give off 8 moles of CO2.
By stoichiometry,
Number of moles of CO2 = 275.702 * 8
= 2205.614 mol of CO2.
Molar mass of CO2 = 12 + (2*16)
= 44 g/mol
Mass of CO2 = number of moles * molar mass
= 2206.614 * 44
= 97047.02 g
Converting g to pound,
= 97047.02 g *1 kg/1000 g * 2.204 lb/1kg
= 213.89 lb of CO2.
What is the maximum pressure at which solid CO2 (dry ice) can be converted into CO2 gas without melting?
Answer:
Explanation:
The triple point of carbon dioxide is 5.11 atmosphere at -56.6 degree celsius . At pressure greater than 5.11 , solid carbon dioxide liquefies , as it is warmed. At pressure lesser than 5.11 atmosphere , it will go into gaseous state without liquefying . Excessive pressure helps liquification process.
So maximum pressure required is 5.11 atmosphere. Beyond this pressure , solid CO2 will liquify.
Final answer:
Dry ice (solid CO₂) sublimes without melting at pressures below its triple point of 5.11 atm, transitioning directly from a solid to a gas at a temperature of -78.5°C, notably at atmospheric pressure of 1 atm.
Explanation:
The maximum pressure at which solid CO₂ (dry ice) can be converted into CO₂ gas without melting is at pressures below the substance's triple point. The triple point of CO₂ is at -56.6°C and 5.11 atm, which is the condition where the solid, liquid, and gaseous phases of CO₂ coexist in equilibrium. At pressures below 5.11 atm, solid CO₂ does not melt but instead undergoes sublimation, transitioning directly from a solid to a gas.
Specifically, at standard atmospheric pressure of 1 atm, dry ice sublimes at a temperature of -78.5°C. Thus, solid CO₂ will sublime at any pressure below its triple point without melting. Hence, the answer to the student's question is that solid CO₂ can be converted into gas without melting at pressures below 5.11 atm, and this will especially be observed at the standard atmospheric pressure where sublimation occurs directly.
Three of the reactions that occur when the paraffin of a candle (typical formula C21H44) burns are as follows:
(1) Complete combustion forms CO2 and water vapor.
(2) Incomplete combustion forms CO and water vapor.
(3) Some wax is oxidized to elemental C (soot) and water vapor.
(a) Find ΔH∘rxn of each reaction (ΔH∘f of C21H44=−476kJ/mol; use graphite for elemental carbon).
(b) Find q (in kJ) when a 254-g candle bums completely.
(c) Find q (in kJ) when 8.00% by mass of the candle burns incompletely and another 5.00% undergoes soot formation.
(a) ΔH∘rxn: (1) -1471.1 kJ/mol, (2) 3491.9 kJ/mol, (3) -480 kJ/mol. (b) Complete burn: -1259.6 kJ. (c) Incomplete burn: 387.8 kJ, Soot: -53.3 kJ.
How to find q (in kJ) when 8.00% by mass of the candle burns incompletely and another 5.00% undergoes soot formationLet's work through the given questions step by step.
(a) Calculating ΔH∘rxn for each response:
The responses you gave are combustion responses of paraffin wax ([tex]C_2_1H_4_4[/tex]). To discover the ΔH∘rxn for each response, we'll utilize the given ΔH∘f for [tex]C_2_1H_4_4[/tex] and the standard enthalpies of arrangement for the items.
Response (1): Total combustion
[tex]C_2-1H_4_4 + 32 O_2 = 21CO_2 + 22H_2O[/tex]
ΔH∘rxn =( (ΣΔH∘f(products)) - (ΣΔH∘f(reactants)))
ΔH∘rxn = ((21) × (ΔH∘f[tex](CO_2[/tex])) + ((22) × (ΔH∘f([tex]H_2O[/tex])) -(ΔH∘f([tex]C_2_1H_4_4[/tex]))
ΔH∘f(CO2) =( -393.5 kJ/mol and (ΔH∘f(H2O)) = -285.8 kJ/mol:
ΔH∘rxn(1) = ((21) × (-393.5 kJ/mol) + (22) × (-285.8 kJ/mol)) - -476 kJ/mol
ΔH∘rxn(1) = (-8233.5 kJ/mol + 6288.4 kJ/mol) - (-476 kJ/mol)
ΔH∘rxn(1) = (-1471.1 kJ/mol)
Response (2): Complete combustion
[tex]C_2_1H_4_4 + 22O_2 → 21CO + 22H_2O[/tex]
Using the same approach:
(ΔH∘rxn(2)) = ((21) × ΔH∘f(CO)) + (22) × ΔH∘f([tex]H_2O[/tex])] - (ΔH∘f([tex]C_2_1H_4_4[/tex]))
Given ΔH∘f(CO) = -110.5 kJ/mol:
ΔH∘rxn(2) = [21 × (-110.5 kJ/mol) + 22 × (-285.8 kJ/mol)] - (-476 kJ/mol)
ΔH∘rxn(2) = -2320.5 kJ/mol + 6288.4 kJ/mol - (-476 kJ/mol)
ΔH∘rxn(2) = 3491.9 kJ/mol
Response (3): Formation of soot
[tex]C_2_1H_4_4 + 11O_2[/tex] → 21 C (graphite) +[tex]22H_2O[/tex]
ΔH∘rxn(3) = [21 × ΔH∘f(C) + 22 × ΔH∘f([tex]H_2O[/tex])] - ΔH∘f([tex]C_2_1H_4_4[/tex])
Given ΔH∘f(C) = kJ/mol (since graphite is the standard reference state):
ΔH∘rxn(3) = [21 × kJ/mol + 22 × (-285.8 kJ/mol)] - (-476 kJ/mol)
ΔH∘rxn(3) = -6292.6 kJ/mol + 6288.4 kJ/mol - (-476 kJ/mol)
ΔH∘rxn(3) = -480 kJ/mol
(b) Calculating q for complete combustion:
To calculate q (warm discharged) when the candle burns totally, ready to utilize the condition:
q = n × ΔH∘rxn
Given the mass of the candle is 254 g and the molar mass of [tex]C_2_1H_4_4[/tex] is around 296.66 g/mol:
n = 254 g / 296.66 g/mol = 0.856 mol
q = 0.856 mol × (-1471.1 kJ/mol) = -1259.6 kJ
So, when the candle burns totally, it discharges around 1259.6 kJ of warm.
(c) Calculating q for fragmented combustion and sediment arrangement:
For inadequate combustion, the response is the same as response (2), and for sediment arrangement, the reaction is the same as response (3). We have to calculate the warm discharged for each of these cases independently.
To begin with, calculate the mass of the candle included in each response:
Deficient combustion: 8.00% by mass
Sediment arrangement: 5.00% by mass
Add up to mass burned = 8.00% + 5.00% = 13.00% = 0.13 (decimal frame)
Presently, for fragmented combustion:
n_incomplete = 0.13 ×254 g / 296.66 g/mol = 0.111 mol
q_incomplete = 0.111 mol × 3491.9 kJ/mol = 387.8 kJ
For sediment arrangement:
n_soot = 0.13 × 254 g / 296.66 g/mol = 0.111 mol
q_soot = 0.111 mol × (-480 kJ/mol) = -53.3 kJ
So, when 8.00% of the candle burns not completely, it discharges roughly 387.8 kJ of heat, and when 5.00% experiences sediment arrangement, it assimilates roughly 53.3 kJ of warm.
If it's not too much trouble note that the negative sign for q_soot demonstrates that warm is ingested, which is anticipated for an endothermic prepare-like sediment arrangement.
Learn more about mass here:
https://brainly.com/question/86444
#SPJ1
The given problems ask to calculate the enthalpy change of the full and partial combustion of a paraffin candle to CO2, CO and elemental carbon (soot), and the heat released in these combustion reactions. The calculated ΔHrxn of each reaction is used with the given or calculated mass to get the heat released.
Explanation:For the question asked, first we look at the combustion reactions of a candle wax paraffin formula C21H44.
The complete combustion reaction of paraffin is represented by the equation: C21H44 + 32O2 ---> 21CO2 + 22H2O The incomplete combustion reaction of paraffin is represented by the equation: C21H44 + 32O2 ---> 2CO + 21H2O The soot formation reaction (elemental carbon oxidation) of paraffin is represented by the equation: 2C21H44 + 31O2 ---> 42C(s) + 22H2O
By using the standard enthalpy of formation (ΔHf) values, we can calculate ΔHrxn for each reaction. By mass of candle burnt, we can find q by using the formula q = mass x specific heat x ΔT; q can also be calculated from the calculated ΔHrxn multiplied by the moles involved.
For the final part (c), if 8.00% by mass of the candle burns incompletely and another 5.00% undergoes soot formation, q for each partial scenario can be calculated from the relevant ΔHrxn multiplied by the moles involved (mass burnt / molar mass), and adding these two values together.
Learn more about Combustion of Candle here:https://brainly.com/question/31483144
#SPJ11
Summarize the rules for the allowable values of the four quantum numbers of an electron in an atom.
Final answer:
The quantum numbers n, l, m_l, and m_s describe the energy level, shape of the orbital, orbital orientation, and spin of an electron, respectively. No two electrons can have the same set of four quantum numbers due to the Pauli exclusion principle, allowing only two with opposite spins in the same orbital.
Explanation:
Quantum Numbers and Electron Configuration
Electrons in atoms have quantized energies, and their states are described by four quantum numbers. These quantum numbers include:
The principal quantum number (n), which specifies the energy level or shell of an electron within an atom and can have positive integer values (1, 2, 3, ...).The angular momentum quantum number (l), which identifies the shape of the orbital and can have values from 0 to n-1 for each value of n.The magnetic quantum number (ml), which indicates the orientation of the orbital in space and can take on integer values between -l and +l, including zero.The spin quantum number (ms), which describes the intrinsic spin of the electron within an orbital and can only be ±½.According to the Pauli exclusion principle, no two electrons in the same atom can have identical sets of these quantum numbers. This means that within a single orbital defined by n, l, and ml, only two electrons can exist and they must have opposite spins (ms).
Final answer:
The four quantum numbers with their allowed values are the principal quantum number (n) with positive integers, the azimuthal quantum number (l) with values from 0 to n-1, the magnetic quantum number ([tex]m_l[/tex]) with values between -l and +l, and the spin quantum number ([tex]m_s[/tex]) with values of (+1/2) or (-1/2).
Explanation:
Allowed Values for Quantum Numbers
Each electron in an atom is described by four quantum numbers: the principal quantum number (n), the azimuthal quantum number (l), the magnetic quantum number ([tex]m_l[/tex]), and the spin quantum number ([tex]m_s[/tex]). The allowed values for these quantum numbers are:
n: This can be any positive integer (1, 2, 3, ...). It denotes the energy level and size of the orbital.l: This can take on any integer value ranging from 0 to n-1, where n is the principal quantum number. It represents the shape of the orbital.[tex]m_l[/tex]: For a given azimuthal quantum number l, this can be any integer between -l and +l, including zero. This number determines the orientation of the orbital in space.[tex]m_s[/tex]: This has only two possible values, (+1/2) or (-1/2), indicating the two opposite directions of electron spin.These quantum numbers are based on the solutions to the Schrödinger Equation for atoms and are fundamental to understanding electron configurations in atoms.
The mass of a single molybdenum atom is 1.59×10-22 grams. How many molybdenum atoms would there be in 40.6 milligrams of molybdenum?
Answer:
2.55×10²⁰ atoms of Mo are contained in 40.6 mg
Explanation:
1 atom of Mo has a mass of 1.59×10⁻²² g
So let's think a rule of three for this question.
Firstly, we convert the mass in mg to g
40.6 mg / 1000 = 0.0406 g
In the mass of 1.59×10⁻²² g, there is 1 atom of Mo
In the mass of 0.0406 g there are (0.0406 / 1.59×10⁻²²) = 2.55×10²⁰ atoms
The number of molybdenum atoms in 40.6 milligrams of molybdenum is approximately 2.55 x 10^22.
Explanation:To calculate the number of molybdenum atoms in 40.6 milligrams of molybdenum, you need to first convert the milligrams to grams, as the given mass of a single molybdenum atom is in grams. So, 40.6 milligrams equals to 0.0406 grams. Next, divide the total mass (0.0406 grams) by the mass of a single atom (1.59×10-22 grams) to find the number of atoms.
This gives us:
0.0406 grams / 1.59×10-22 grams = 2.55 x 1022 atoms.
So, there are approximately 2.55 x 1022 atoms of molybdenum in 40.6 milligrams of molybdenum.
Learn more about atomic mass calculation here:
https://brainly.com/question/32288931
#SPJ3
Which of the following statements is incorrect concerning the thermochemical equation below? 2SO3(g) → 2SO2(g) + O2(g); ΔH° = 198 kJ a. The enthalpy of the reactants exceeds that of the products. b. For the reaction 2SO2(g) + O2(g) → 2SO3(g), ΔH° = –198 kJ. c. The reaction is endothermic. d. The external pressure is 1 atm. e. For every mole of SO3(g) consumed, 99 kJ of heat at constant pressure is consumed as well.
Answer:
The incorrect statement is a.
Explanation:
[tex]2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)[/tex] ΔH° = 198 kJ
Endothermic reactions are defined as the reactions in which energy is absorbed by the chemical reaction.The enthalpy of the reaction [tex](\Delta H)[/tex] positive.
This can be also understood as the reactions in which the energy of products is more than the energy of the reactants.
In the given thermochemical equation,enthalpy of the products exceeds that of the reactants
The incorrect statement regarding the thermochemical equation 2SO₃(g) → 2SO₂(g) + O₂(g); ΔH° = 198 kJ is that the enthalpy of the reactants exceeds that of the products, which is falls under an endothermic reaction. Therefore the option a is incorrect for this reaction.
The question revolves around the correctness of statements related to a given thermochemical equation: 2SO₃(g) → 2SO₂(g) + O₂(g); ΔH° = 198 kJ. The incorrect statement is:
The enthalpy of the reactants exceeds that of the products.
Since the ΔH° is positive (198 kJ), it indicates that the reaction is endothermic. This means energy is absorbed, thus the enthalpy of the products exceeds that of the reactants, making option (a) incorrect. Conversely, for the reverse reaction (2SO₂(g) + O₂(g) → 2SO₃(g)), ΔH° would indeed be – 198 kJ, illustrating that the reaction is exothermic in the reverse direction, as more energy is released than absorbed.
Options b and c are correct as they accurately describe the thermodynamics of the reaction and its reverse. The statement about external pressure (d) is not directly related to thermochemical equations but generally assumes standard conditions unless otherwise specified. Lastly, option e correctly divides the total heat absorbed by the number of moles of SO₃(g) consumed to arrive at the amount of heat consumed per mole. Therefore, the incorrect statement concerns the comparison of enthalpy between reactants and products.