The initial reaction rate for the elementary reaction 2A + B → 4C was measured as a function of temperature when the concentration of A was 2 M and that of B was 1.5 M. (a) What is the activation energy? (b) What is the frequency factor? (c) What is the rate constant as a function of temperature using Equation (S3-5) and T0 = 27°C as the base case?
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
a) The activation energy is 124.776[tex]\frac{kJ}{mole}[/tex]
b) The frequency factor is 1.77 ×[tex]10^{18}[/tex]
c) The rate constant is 0.00033 [tex](\frac{dm^{3} }{mole} )^{2}[/tex][tex]\frac{1}{s}[/tex]
Explanation:
From the question the elementary reaction for A and B is given as
2A + B → 4C
The rate equation the elementary reaction is
-[tex]r_{A}[/tex] = [tex]k[/tex][tex][A]^{2}[/tex][tex][B][/tex]
= [tex]k[2]^{2}[1.5][/tex]
= 6k
[tex]k = \frac{-r_{A} }{6}[/tex]
When temperature changes, the rate constant change an this causes the rate of reaction to change as shown on the second uploaded image.
The relationship between temperature and rate constant can be deduced from these equation
[tex]k = Aexp(-\frac{E_{a} }{RT} )[/tex]
taking ln of both sides we have
[tex]lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}[/tex]
Considering the graph for the rate constant [tex]ln k[/tex] and [tex](\frac{1}{T} )[/tex] the slope from the equation is [tex]-(\frac{E_{a} }{R})[/tex] and the intercept is [tex]ln A[/tex]
From the given table we can generate another table using the equation above as shown on the third uploaded image
The graph of [tex]ln k[/tex] vs [tex](\frac{1}{T} )[/tex] is shown on the fourth uploaded image
From the graph we can see that the slope is [tex]-(\frac{E_{a} }{R} ) = - 15008[/tex]
Now we can obtain the activation energy [tex]E_{a}[/tex] by making it the subject in the equation also generally R which is the gas constant is [tex]8.145 \frac{J}{kmole}[/tex]
[tex]E_{a} = 15008 × 8,3145\frac{J}{molK}[/tex]
[tex]= 124\frac{KJ}{mole}[/tex]
Hence the activation energy is [tex]= 124\frac{KJ}{mole}[/tex]
b) From the graph its intercept is [tex]ln A = 42.019[/tex]
[tex]A = exp(42.019)[/tex]
[tex]=1.77 × 10^{18}[/tex]
Hence the frquency factor A is [tex]=1.77 × 10^{18}[/tex]
c) From the equation of rate constant
[tex]lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}[/tex]
We have
[tex]ln k = 42.019 - 15008 * (\frac{1}{300} )[/tex]
[tex]k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}[/tex]
Hence the rate constant is [tex]k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}[/tex]
Related Questions
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.
Answers
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.
As a science major, you are assigned to the "Atomic Dorm" at college. The first floor has one suite with one bedroom that has a bunk bed for two students. The second floor has two suites, one like the one on the first floor, and the other with three bedrooms, each with a bunk bed for two students. The third floor has three suites, two like the ones on the second floor, and a third suite that has five bedrooms, each with a bunk bed for two students. Entering students choose room and bunk on a first-come, first-serve basis by criteria in the following order of importance:
(1) They want to be on the lowest available floor.
(2) They want to be in the smallest available dorm room.
(3) They want to be in a lower bunk if available.
(a) Which bunk does the 1st student choose?
(b) How many students are in top bunks when the 17th student chooses?
(c) Which bunk does the 21st student choose?
(d) How many students are in bottom bunks when the 25th student chooses?
Answers
Explanation:
A.
The first student will be on the lower bunk on the first floor because 1. They want on the lowest available floor and 2. They want to be in a lower bunk if available.
B.
7 students are in the TOP bunks because 1. They want on the lowest available floor and 2. They want to be in a lower bunk if available. Therefore, all the rooms up till the third floor (Remember, third floor has 3 suites), so the first floor is filled - 1 person on the top bunk, 2 floor is filled- 4 persons and the third floor; the first suite is filled - 1 person and the second suite is a little partially filled- 1 person.
C.
Following the criteria 1, 2 and 3, the 21st student occupies the third suite on the third floor because all the floors (1 and 2) are occupied so the third suite on the third floor is still vacant.
D.
From the criteria there are therefore 10 persons at the TOP bunk. All the rooms up till the third floor are filled, so the first floor is filled - 1 person on the top bunk, second floor is filled (2 suites) - 4 persons and the third floor; the first suite and second suite is filled - 4 persons; the thirs suite has 6 persons present so 1 person is at the top bunk.
An FM station broadcasts music at 93.5 MHz (megahertz, or 10⁶ Hz). Find the wavelength (in m, nm, and Å) of these waves.
Answers
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
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Make a plot of arsenate species vs. pH for a 25 mM arsenate solution. Vary pH from 0-14. Neglect activity corrections. A groundwater contaminated with arsenate has a pH range of 7.0 to 8.5. What is/are the dominant arsenate species in this range?
Answers
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.
Answers
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.
The density of platinum is 21.45 g/cm3. What is the volume of a platinum sample with a mass of 11.2 g?
Answers
Final answer:
To find the volume of a platinum sample with a mass of 11.2 g, divide the mass by the density of platinum, which is 21.45 g/cm3, resulting in a volume of approximately 0.5221 [tex]cm^3[/tex].
Explanation:
To determine the volume of a platinum sample with a mass of 11.2 g, given the density of platinum is 21.45 g/cm3, we can use the formula for density which is:
Density = Mass / Volume.
By rearranging the formula to solve for volume, we get:
Volume = Mass / Density.
Substituting the given values:
Volume = 11.2 g / 21.45 g/cm3 = 0.5221 [tex]cm^3[/tex].
Therefore, the volume of the platinum sample is approximately 0.5221 [tex]cm^3[/tex].
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?
Answers
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.
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For main-group elements, are outer electron configurations similar or different within a group? Within a period? Explain.
Answers
Answer:
Same within the groups...
Different within the periods...
Explanation:
The outer configuration is same with a group because of their arrangement in the periodic table. If we see group 1a of alkalies, all of them have 1 electron in their outer shell and this order goes on increasing up-to noble gases that have 8 electrons in their outer shell each.
This arrangement is called periodicity of the periodic table and elements are arranged according to the periodic law.
Along a period, these properties of outer configuration goes on increasing than within the group.
A 7.73 mass % aqueous solution of ethylene glycol (HOCH2CH2OH) has a density of 1.19 g/mL. Calculate the molarity of the solution. Give your answer to 2 decimal places.
Answers
Answer:
Molarity for the solution is 1.48 M
Explanation:
Molarity involves the moles of solute that are contained in 1L of solution. It is a sort of concentration. The most usual.
7.73 by mass involves the mass of solute that are contained in 100 g of solution, so this one of ethylene glycol contains 7.73 g of it.
Let's determine the moles of solute
7.73 g / molar mass of ethylene glycol = moles
7.73 g / 62 g/mol = 0.124 mol
If the mass of solution is 100 g, we can determine the volume with density.
Density = Mass / Volume
Volume = Mass / Density
Volume = 100 g / 1.19 g/mL = 84.03 mL
In conclusion, 0.124 moles are contained in 84.03 mL
Molarity is mol/L, so let's convert the volume in L
84.03 mL . 1L / 1000 mL = 0.08403 L
0.124 mol / 0.08403 L = 1.48 M
tyrosine kinase inhibitor binds and inhibits BTK. As a result of the experiment, you are able to elute BTK from the column, but in a mixture of other tyrosine kinases. Why are tyrosine kinases other than BTK present in the eluate?
Answers
Answer: It is because tyrosine kinases and BTK have similar solubilities
Explanation:
In column chromatography, components of a mixture are seperated based on their relative solubilities in two non-mixing phases.
In essence, tyrosine kinases and BTK are present in the eluate due to their similar solubility rates that arise from the similar chemical structure both possess (otherwise it would be impossible for the inhibitor meant for Tyrosine kinase to bind and also inhibits BTK)
Thus, the similar solubilities of both groups is the reason they could elute out of the column without being adsorped.
Calculate the molalities of the following aqueous solutions:________. (a) 1.22 M sugar (C12H22O11) solution (density of solution = 1.12 g/ml), (b) 0.87 M NaOH solution (density = 1.04 g/ml), (c) 5.24 M NaHCO3 solution (density = 1.19 g/ml).
Answers
a) The molality of the 1.22 M sugar solution is approximately [tex]\(1.089 \, \text{mol/kg}\).[/tex]
b) The molality of the 0.87 M NaOH solution is approximately [tex]\(0.8365 \, \text{mol/kg}\).[/tex]
c) The molality of the 5.24 M NaHCO3 solution is approximately [tex]\(4.4034 \, \text{mol/kg}\).[/tex]
(a) 1.22 M sugar (C12H22O11) solution:
Given:
- Molarity (\(M\)) of sugar solution = 1.22 M
- Density of solution = 1.12 g/ml
We'll assume that the density given is for the solution and not just for water.
1. Calculate the mass of 1 L (1000 ml) of solution:
[tex]\[ \text{mass of solution} = \text{density} \times \text{volume} \][/tex]
[tex]\[ \text{mass of solution} = 1.12 \, \text{g/ml} \times 1000 \, \text{ml} \][/tex]
[tex]\[ \text{mass of solution} = 1120 \, \text{g} \][/tex]
2. Convert the mass of the solution to kilograms:
[tex]\[ \text{mass of solution in kg} = \frac{{1120 \, \text{g}}}{{1000}} = 1.12 \, \text{kg} \][/tex]
3. Calculate the moles of sugar (C12H22O11):
[tex]\[ \text{moles of sugar} = \text{Molarity} \times \text{volume in L} \][/tex]
[tex]\[ \text{moles of sugar} = 1.22 \, \text{mol/L} \times 1 \, \text{L} = 1.22 \, \text{mol} \][/tex]
4. Calculate the molality ([tex]\(m\)[/tex]):
[tex]\[ m = \frac{{\text{moles of sugar}}}{{\text{mass of solvent in kg}}} \][/tex]
Let's perform the calculation.
For the 1.22 M sugar solution:
1. Calculate the mass of 1 L (1000 ml) of solution:
[tex]\[ \text{mass of solution} = \text{density} \times \text{volume} \][/tex]
[tex]\[ \text{mass of solution} = 1.12 \, \text{g/ml} \times 1000 \, \text{ml} \][/tex]
[tex]\[ \text{mass of solution} = 1120 \, \text{g} \][/tex]
2. Convert the mass of the solution to kilograms:
[tex]\[ \text{mass of solution in kg} = \frac{{1120 \, \text{g}}}{{1000}} = 1.12 \, \text{kg} \][/tex]
3. Calculate the moles of sugar (C12H22O11):
[tex]\[ \text{moles of sugar} = \text{Molarity} \times \text{volume in L} \][/tex]
[tex]\[ \text{moles of sugar} = 1.22 \, \text{mol/L} \times 1 \, \text{L} = 1.22 \, \text{mol} \][/tex]
4. Calculate the molality ([tex]\(m\)[/tex]):
[tex]\[ m = \frac{{\text{moles of sugar}}}{{\text{mass of solvent in kg}}} \][/tex]
[tex]\[ m = \frac{{1.22 \, \text{mol}}}{{1.12 \, \text{kg}}} \][/tex]
[tex]\[ m \approx 1.089 \, \text{mol/kg} \][/tex]
So, the molality of the 1.22 M sugar solution is approximately [tex]\(1.089 \, \text{mol/kg}\).[/tex]
(b) 0.87 M NaOH solution:
Given:
- Molarity ([tex]\(M\)[/tex]) of NaOH solution = 0.87 M
- Density of solution = 1.04 g/ml
We'll follow the same steps as above to calculate the molality.
Let's calculate for the NaOH solution.
For the 0.87 M NaOH solution:
1. Calculate the mass of 1 L (1000 ml) of solution:
[tex]\[ \text{mass of solution} = \text{density} \times \text{volume} \][/tex]
[tex]\[ \text{mass of solution} = 1.04 \, \text{g/ml} \times 1000 \, \text{ml} \][/tex]
[tex]\[ \text{mass of solution} = 1040 \, \text{g} \][/tex]
2. Convert the mass of the solution to kilograms:
[tex]\[ \text{mass of solution in kg} = \frac{{1040 \, \text{g}}}{{1000}} = 1.04 \, \text{kg} \][/tex]
3. Calculate the moles of NaOH:
[tex]\[ \text{moles of NaOH} = \text{Molarity} \times \text{volume in L} \][/tex]
[tex]\[ \text{moles of NaOH} = 0.87 \, \text{mol/L} \times 1 \, \text{L} = 0.87 \, \text{mol} \][/tex]
4. Calculate the molality (\(m\)):
[tex]\[ m = \frac{{\text{moles of NaOH}}}{{\text{mass of solvent in kg}}} \][/tex]
[tex]\[ m = \frac{{0.87 \, \text{mol}}}{{1.04 \, \text{kg}}} \][/tex]
[tex]\[ m = \frac{{0.87 \, \text{mol}}}{{1.04 \, \text{kg}}} \][/tex]
[tex]\[ m \approx 0.8365 \, \text{mol/kg} \][/tex]
So, the molality of the 0.87 M NaOH solution is approximately [tex]\(0.8365 \, \text{mol/kg}\).[/tex]
(c) 5.24 M NaHCO3 solution:
Given:
- Molarity (\(M\)) of NaHCO3 solution = 5.24 M
- Density of solution = 1.19 g/ml
Let's follow the same steps as above to calculate the molality for the NaHCO3 solution.
For the 5.24 M NaHCO3 solution:
1. Calculate the mass of 1 L (1000 ml) of solution:
[tex]\[ \text{mass of solution} = \text{density} \times \text{volume} \][/tex]
[tex]\[ \text{mass of solution} = 1.19 \, \text{g/ml} \times 1000 \, \text{ml} \][/tex]
[tex]\[ \text{mass of solution} = 1190 \, \text{g} \][/tex]
2. Convert the mass of the solution to kilograms:
[tex]\[ \text{mass of solution in kg} = \frac{{1190 \, \text{g}}}{{1000}} = 1.19 \, \text{kg} \][/tex]
3. Calculate the moles of NaHCO3:
[tex]\[ \text{moles of NaHCO3} = \text{Molarity} \times \text{volume in L} \][/tex]
[tex]\[ \text{moles of NaHCO3} = 5.24 \, \text{mol/L} \times 1 \, \text{L} = 5.24 \, \text{mol} \][/tex]
4. Calculate the molality (\(m\)):
[tex]\[ m = \frac{{\text{moles of NaHCO3}}}{{\text{mass of solvent in kg}}} \][/tex]
[tex]\[ m = \frac{{5.24 \, \text{mol}}}{{1.19 \, \text{kg}}} \][/tex]
[tex]\[ m = \frac{{5.24 \, \text{mol}}}{{1.19 \, \text{kg}}} \][/tex]
[tex]\[ m \approx 4.4034 \, \text{mol/kg} \][/tex]
So, the molality of the 5.24 M NaHCO3 solution is approximately [tex]\(4.4034 \, \text{mol/kg}\).[/tex]
A 1.85 mass % aqueous solution urea (CO(NH2)2)has a density of 1.05 g/mL. Calculate the molality of the solution. Give your answer to 2 decimal places.
Answers
Answer:
Molality for this solution is 0.31 m
Explanation:
Molality → Mol of solute / kg of solvent
Mass of solute = 1.85 g (CO(NH₂)₂
% by mass, means mass of solute in 100 g of solution
Mass of solution = Mass of solute + Mass of solvent
As 1.85 g is the mass of solute, the mass of solvent would be 98.15 g to complete the 100g of solution.
Let's convert the mass of solvent (g to kg)
1 g = 1×10⁻³ kg
98.2 g . 1×10⁻³ kg / 1g = 0.0982 kg
Let's convert the mass of solute to moles (mass / molar mass)
1.85 g / 60 g/mol = 0.0308 mol
Molality = 0.0308 mol / 0.0982 kg → 0.31 m
To find the molality of the solution, we calculate the number of moles of urea (0.031 mol) and the mass of water (0.10315 kg). The molality is then computed as moles of urea divided by the mass of water, which comes out to be approximately 0.30 m.
To calculate the molality of the solution, we need to first know the number of moles of the solute (urea) and the mass of the solvent (water) in kilograms.
The mass % of the solution is given as 1.85%, meaning that we have 1.85g of urea in 100g of the solution. Considering that the molar mass of urea (CO(NH2)2) is about 60.06 g/mol, the number of moles of urea can be calculated as follows: Moles of urea = mass / molar mass = 1.85g / 60.06 g/mol ≈ 0.031 mol.
Since the density of the solution is given as 1.05 g/mL, the mass of the solution for 100 mL (or 100g since density = mass/volume) would be 100 mL * 1.05 g/mL = 105g. In this 105g of the solution, 1.85g is urea, so the mass of water (solvent) would be 105g - 1.85g = 103.15g or 0.10315 kg (since 1g = 0.001 kg).
Molality is the number of solute's moles divided by the mass of solvent (in kg). Thus, the molality of the solution would be: Molality = moles of urea / mass of water in kg = 0.031 mol / 0.10315 kg ≈ 0.30 mol/kg or 0.30 m.
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What is the speed of an electron that has a de Broglie wavelength of 100. nm?
Answers
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).
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).
Answers
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.
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Consider the following types of electromagnetic radiation:
(1) Microwave
(2) Ultraviolet
(3) Radio waves
(4) Infrared
(5) X-ray
(6) Visible
(a) Arrange them in order of increasing wavelength.
(b) Arrange them in order of increasing frequency.
(c) Arrange them in order of increasing energy.
Answers
Electromagnetic radiation can be arranged by wavelength, frequency, or energy. In increasing order, by wavelength, it is: X-ray, Ultraviolet, Visible, Infrared, Microwave, Radio waves. In increasing order, by frequency and energy, it is: Radio waves, Microwave, Infrared, Visible, Ultraviolet, X-ray.
Explanation:The types of electromagnetic radiation listed can be arranged by wavelength, frequency, and energy. The relationship among these three properties is that as wavelength increases, frequency and energy decrease, and vice versa.
In order of increasing wavelength, the arrangement is: X-ray, Ultraviolet, Visible, Infrared, Microwave, Radio waves.In order of increasing frequency, the arrangement is the inverse: Radio waves, Microwave, Infrared, Visible, Ultraviolet, X-ray. In order of increasing energy, the arrangement is the same as frequency, because energy and frequency are directly proportional: Radio waves, Microwave, Infrared, Visible, Ultraviolet, X-ray.Learn more about Electromagnetic Radiation here:https://brainly.com/question/2334706
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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
Answers
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
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
Answers
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
Consider the reaction where A reacts with B to give C and D according to the rate equation: rate = k[A][B] If the concentration of both reagents is doubled, what happens to the rate of the reaction?
Answers
Answer: the rate of the reaction will increase.
Explanation: when the concentration of a reaction is increased, this will increase the effective collisions of reactants which in turn increases the rate of the reaction
Doubling the concentration of both reactants A and B in the rate equation rate = k[A][B], which is first order with respect to each reactant, will quadruple the reaction rate. This is a characteristic of second-order reactions.
Explanation:The student is asking about the effect on the reaction rate when concentrations of reactants in a given rate equation are altered. In the scenario where A reacts with B to give C and D, described by the rate equation rate = k[A][B], if the concentration of both A and B is doubled, the reaction rate will change. This is because the given reaction is first order with respect to both A and B, meaning it is a second-order reaction overall.
If you double the concentration of A to 2[A], the rate equation becomes k(2[A])[B], thereby doubling the reaction rate. Similarly, if B's concentration is also doubled to 2[B], the rate equation now reads k(2[A])(2[B]), which means you multiply the initial rate by four. Thus, doubling the concentrations of both A and B will quadruple the reaction rate due to the multiplicative effect in the rate law.
Remember, rate laws are determined experimentally, and in experiments comparing different trials, doubling the concentration often leads to a reaction rate that is proportional to that change. This reflects a direct relationship between reactant concentrations and reaction rates, as observed in first-order reactions for each reactant.
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Calculate the volume in milliliters of a sodium carbonate solution that contains of sodium carbonate . Be sure your answer has the correct number of significant digits.
Answers
Answer:
25.0 g of sodium carbonate are present in 220 ml of the solution.
Explanation:
Hi there!
I have found the complete question on the web:
Calculate the volume in milliliters of a 1.07 M sodium carbonate solution that contains 25.0g of sodium carbonate Na2CO3. Be sure your answer has the correct number of significant digits.
First, let's find how many moles of sodium carbonate have a mass of 25.0 g.
The molar mass of Na₂CO₃ is 106 g/mol.
So, if 106 g of sodium carbonate is equal to 1 mol, then 25.0 g will be:
25.0 g · (1 mol / 106 g) = 0.236 mol Na₂CO₃
The solution contains 1.07 mol sodium carbonate in 1 liter.
So, if 1.07 mol sodium carbonate is present in 1 l solution, then, 0.236 mol will be present in:
0.236 mol · (1000 ml / 1.07 mo) = 221 ml (220 ml without any intermediate rounding).
25.0 g of sodium carbonate are present in 220 ml of the solution.
Final answer:
To calculate the volume in milliliters of a sodium carbonate solution, you need to know the concentration of the solution.
Explanation:
To calculate the volume in milliliters of a sodium carbonate solution, we need to know the concentration of the solution. Without that information, we are unable to provide an accurate calculation. Please provide the concentration of the sodium carbonate solution in the question.
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)
Answers
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
Two classrooms, labeled A and B, are of equal volume and are connected by an open door. Classroom A is warmer than classroom B (maybe it has a sunny window). Derive a formula that relates the masses of air in each room MA and MB to the temperatures TA and TB. Which room contains a greater mass of air
Answers
Answer:
Room B contains the greater mass of air
Explanation:
According to the given data, the volume of both rooms is the same, so let's consider both volumes as V. An open door connects both the room, which infer that pressure for both the rooms is also the same, let it be P.
Considering the above conditions, the general gas equation for the air in both rooms can be given as:
[tex]PV=n_{A}RT_{A}\\ \\ PV=n_{B}RT_{B}[/tex]
Here, n represents the moles of air, R is the gas constant, and T is the temperature. Taking the ratio of both the above equations we get
[tex]\frac{PV}{PV}= \frac{n_{A}RT_{A}}{n_{B}RT_{B}}\\[/tex]
moles of the gas are mass per molecular mass, as the molecular mass is the same in both the cases so n can be replaced by m, as follows
[tex]\frac{PV}{PV}= \frac{m_{A}RT_{A}}{m_{B}RT_{B}}\\[/tex]
By simplifying we get,
[tex]\frac{m_{A}}{m_{B}}=\frac{T_{B}}{T_{A}}[/tex]
According to the given conditions, TA is greater than TB. So from the derived relation, it can be seen that the mass of air in room B is greater than the mass of air in room A.
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.
Answers
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.
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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.
Answers
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.
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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
Answers
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.
A graduated cylinder contains 15.0 mLmL of water. What is the new water level after 34.0 gg of silver metal with a density of 10.5 g/mLg/mL is submerged in the water?
Answers
Answer: The new water level when silver metal is submerged is 18.24 mL
Explanation:
We are given:
Volume of graduated cylinder = 15.0 mL
To calculate volume of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of silver metal = 10.5 g/mL
Mass of silver metal = 34.0 g
Putting values in above equation, we get:
[tex]10.5g/mL=\frac{34.0g}{\text{Volume of silver metal}}\\\\\text{Volume of silver metal}=\frac{34.0g}{10.5g/mL}=3.24mL[/tex]
New water level = Volume of silver metal + Volume of graduated cylinder
New water level = (3.24 + 15.0) mL = 18.24 mL
Hence, the new water level when silver metal is submerged is 18.24 mL
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?
Answers
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.
There are four types of Chemical Warfare Agents (CWA). Which CWA prevents blood from providing oxygen to tissues and organs?
Answers
Answer:
Blood agent
Explanation:
chemical warfare agents are of four major types:
nerve, blister, choking and blood agents
Nerve agents - it attacks the central nervous system and can enter the body through inhalation
Blister agents - it attacks skins and it is rapidly absorbed by the skin cells
chocking agents - as name denoted, it is related to the respiratory system.
Blood agent- it attacks the circulatory system of the body. it prevents the circulation of blood into the tissue and organs
Rank the following bonds from strongest to weakest and provide the bond energy: the bond between hydrogen and oxygen in a water molecule; the bond between sodium and chloride in the NaCl molecule; the bond between atoms in a metal; the van der Waals bond between adjacent hydrogen atoms.
Answers
The ranking of the bonds from strongest to weakest is: bond between atoms in a metal, bond between sodium and chloride in NaCl, bond between hydrogen and oxygen in water, and the van der Waals bond between adjacent hydrogen atoms. The bond energies for each bond are different, with the bond between atoms in a metal having the highest bond energy, followed by the bond between sodium and chloride in NaCl, the bond between hydrogen and oxygen in water, and the van der Waals bond between adjacent hydrogen atoms having the lowest bond energy.
Explanation:The ranking of the bonds from strongest to weakest is as follows:
The bond between atoms in a metal (strongest) The bond between sodium and chloride in the NaCl molecule The bond between hydrogen and oxygen in a water molecule The van der Waals bond between adjacent hydrogen atoms (weakest)
The bond energy, also known as bond dissociation energy, is the energy required to break a bond. Here are the bond energies for each bond:
The bond between atoms in a metal: high bond energy (specific values vary) The bond between sodium and chloride in the NaCl molecule: high bond energy (~788 kJ/mol) The bond between hydrogen and oxygen in a water molecule: moderate bond energy (~464 kJ/mol) The van der Waals bond between adjacent hydrogen atoms: low bond energy (negligible) Learn more about bond strength here:
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The strongest bond is the bond between atoms in a metal, followed by the bond between sodium and chloride in NaCl. The bond between hydrogen and oxygen in water is weaker than the previous two, and the van der Waals bond between adjacent hydrogen atoms is the weakest. The bond energy is high for the bonds in a metal and NaCl, moderate for the bond in water, and low for the van der Waals bond between hydrogen atoms.
Explanation:The bonds can be ranked from strongest to weakest as follows:
Bond between atoms in a metal: This bond is the strongest among the given options.Bond between sodium and chloride in the NaCl molecule: This bond is stronger than the bond between hydrogen and oxygen in a water molecule.Bond between hydrogen and oxygen in a water molecule: This bond is weaker than the bond between sodium and chloride.Van der Waals bond between adjacent hydrogen atoms: This bond is the weakest among the given options.The bond energy for each bond is:
Bond between atoms in a metal: High bond energyBond between sodium and chloride in the NaCl molecule: High bond energyBond between hydrogen and oxygen in a water molecule: Moderate bond energyVan der Waals bond between adjacent hydrogen atoms: Low bond energyLearn more about bond strength here:https://brainly.com/question/30226871
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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)
Answers
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.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
Answers
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.