Explanation:
Refraction is the phenomenon of change in speed of light when the light when it passes from the medium of one optical density to a medium of another optical density. The light rays also show a bending when passing through the interface of such medium.Diffraction is the process of spreading of the light or any system of wave after passing through a narrow gap of the order of wavelength of the wave.Dispersion is the phenomenon of acquiring the velocity corresponding to its wavelength by a wave.Interference of waves is the phenomenon of interaction of of two waves when they meet and superimpose on one another to give a wave of resultant properties of both the source waves.How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?
Answer:
The concentration of methyl isonitrile will become 15% of the initial value after 10.31 hrs.
Explanation:
As the data the rate constant is not given in this description, However from observing the complete question the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .
Now the ratio of two concentrations is given as
[tex]ln (\frac{C}{C_0})=-kt[/tex]
Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.
k is the rate constant which is given as [tex]5.11 \times 10^{-5} \, s^{-1}[/tex]
So time t is given as
[tex]ln (\frac{C}{C_0})=-kt\\ln(0.15)=-5.11 \times 10^{-5} \times t\\t=\frac{ln(0.15)}{-5.11 \times 10^{-5} }\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs[/tex]
So the concentration will become 15% of the initial value after 10.31 hrs.
Complete question:
The rearrangement of methyl isonitrile (CH3NC) to acetonitrile (CH3NC) is a first-order reaction and has a rate constant of 5.11x10-5s-1 at 472k . If the initial concentration of CH3NC is 3.00×10-2M. How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?
Answer:
The time taken for the concentration of methyl isonitrile to drop to 15.0 % of its initial value is 10.3 hours
Explanation:
The initial concentration of CH3NC is 3.00 X 10⁻²M =0.03M
The rate constant K= 5.11 X 10⁻⁵s⁻¹
If the concentration of methyl isonitrile to drop to 15.0 %;
The new concentration of methyl isonitrile becomes 0.15 X 0.03 = 0.0045 M
The time taken to drop to 0.0045 M, can be calculated as follows:
[tex]t = -ln[\frac{(CH_3NC)}{(CH_3NC)_0}]/K[/tex]
[tex]t = (-ln[\frac{0.0045}{0.03}]/5.11 X 10^{-5})X(\frac{1 min}{60 s}) = 618.8 mins[/tex]
→ 618.8 mins X 1hr/60mins = 10.3 hours
Therefore, the time taken for the concentration of methyl isonitrile to drop to 15.0 % of its initial value is 10.3 hours
What is the electric potential, in volts, due to the proton in an orbit with radius 0.530 × 10–10 m?
Answer : The electric potential on proton is, 27.1 V
Explanation :
To calculate the electric potential we are using the formula.
[tex]V=\frac{1}{4\pi \epsilon_0}\frac{q}{r}[/tex]
where,
V = electric potential
q = charge on proton = [tex]1.6\times 10^{-19}C[/tex]
r = radius = [tex]0.530\times 10^{-10}m[/tex]
[tex]\frac{1}{4\pi \epsilon_0}=8.99\times 10^9[/tex]
Now put all the given values in the above formula, we get:
[tex]V=(8.99\times 10^9)\times \frac{1.6\times 10^{-19}}{0.530\times 10^{-10}}[/tex]
[tex]V=27.1V[/tex]
Thus, the electric potential on proton is, 27.1 V
cooling a sample of matter from 70°c to 10°c at constant pressure causes its volume to decrease from 873.6 to 712.6 cm3. classify the material as a nearly ideal gas, a nonideal gas, or condensed.
Explanation:
Expression for the coefficient of thermal expansion is as follows.
[tex]\alpha = \frac{1}{V}(\frac{\Delta V}{\Delta T})[/tex]
where, V = initial volume
[tex]\Delta V[/tex] = Final volume - initial volume
= (712.6 - 873.6) [tex]cm^{3}[/tex]
= -161 [tex]cm^{3}[/tex]
Now, we will calculate the change in temperature as follows.
[tex]\Delta T[/tex] = Final temperature - Initial temperature
= (10 + 273) K - (70 + 273) K
= 283 K - 343 K
= -60 K
Substituting these values into the equation as follows.
[tex]\alpha = \frac{1}{873.6} \times (\frac{161}{60}) K^{-1}[/tex]
= 0.00307 [tex]K^{-1}[/tex]
It is known that for non-ideal gases the value of alpha is 0.366% which is 0.00366 per Kelvin. As it is close to our result, hence the given sample of gas is a non-ideal gas.
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.
A 1.0223 g sample of an unknown nonelectrolyte dissolved in 10.2685 g of benzophenone produces a solution that freezes at 31.7°C. If the pure benzophenone melted at 47.5°C, what is the molecular weight of the unknown compound?
Answer: The molecular weight of unknown non-electrolyte is 61.75 g/mol
Explanation:
Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.
[tex]\Delta T_f=\text{Freezing point of benzophenone}-\text{Freezing point of solution}[/tex]
To calculate the depression in freezing point, we use the equation:
[tex]\Delta T_f=iK_fm[/tex]
or,
[tex]\text{Freezing point of benzophenone}-\text{Freezing point of solution}=iK_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]where,
i = Vant hoff factor = 1 (for non-electrolyte)
[tex]K_f[/tex] = molal freezing point depression constant = 9.80°C/m
[tex]m_{solute}[/tex] = Given mass of unknown non-electrolyte = 1.0223 g
[tex]M_{solute}[/tex] = Molar mass of unknown non-electrolyte = ?
[tex]W_{solvent}[/tex] = Mass of solvent (benzophenone) = 10.2685 g
Putting values in above equation, we get:
[tex]47.5-31.7=1\times 9.80\times \frac{1.0223\times 1000}{M_{solute}\times 10.2685}\\\\M_{solute}=\frac{1\times 9.80\times 1.0223\times 1000}{15.8\times 10.2685}=61.75g/mol[/tex]
Hence, the molecular weight of unknown non-electrolyte is 61.75 g/mol
The molecular weight of the unknown nonelectrolyte dissolved in benzophenone can be calculated using the formula for freezing point depression and the information provided in the question. After performing the necessary calculations, the molecular weight of the unknown compound is found to be 331 g/mol.
Explanation:To calculate the molecular weight of the unknown compound, we use the formula for freezing point depression:
ΔT = iKfm
Where ΔT is the change in temperature, i is the van't Hoff factor which is 1 for nonelectrolytes, Kf is the cryoscopic constant for benzophenone, which, in this case, is 5.12°C kg/mol, and m is the molality of the solution. The change in temperature is the difference between the freezing points of pure benzophenone and the solution: 47.5°C - 31.7°C = 15.8°C.
The molality can be calculated by rearranging the formula:
m = ΔT / (iKf) = 15.8 / (1 * 5.12) = 3.086 mol/kg
Since we are given grams and want to convert to kilograms:
1.0223g of unknown compound / molar mass (unknown) = 3.086 mol/kg
Rearrange to find the molar mass (molecular weight) of the compound:
molar mass (unknown) = 1.0223g / 3.086 kg = 0.331 kg/mol = 331 g/mol
Hence the molecular weight of the unknown nonelectrolyte dissolved in benzophenone is 331 g/mol.
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A 49.48-mL sample of an ammonia solution is analyzed by titration with HCl. It took 38.73 mL of 0.0952 M HCl to titrate the ammonia. What is the concentration of the original ammonia solution?
Answer:
M₂ = 0.0745 M
Explanation:
In case of titration , the following formula can be used -
M₁V₁ = M₂V₂
where ,
M₁ = concentration of acid ,
V₁ = volume of acid ,
M₂ = concentration of base,
V₂ = volume of base .
from , the question ,
M₁ = 0.0952 M
V₁ = 38.73 mL
M₂ = ?
V₂ = 49.48 mL
Using the above formula , the molarity of ammonia , can be calculated as ,
M₁V₁ = M₂V₂
0.0952 M * 38.73 mL = M₂* 49.48 mL
M₂ = 0.0745 M
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.
You find a mysterious bottle in the laboratory labeled 0.01 M Sodium Hydroxide. What is the pH of the solution?
Answer: pH of the solution is 12
Explanation:
NaOH —> Na+ + OH-
NaOH = 0.01M
Na+ = 0.01M
OH- = 0.01M
pOH = —log [OH-] = —log [0.01]
pOH = 2
pH + pOH = 14
pH = 14 — pOH = 14 — 2 = 12
pH = 12
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.
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Identify each element below, and give the symbols of the other elements in its group:
(a) [Ar] 4s²3d¹⁰4p⁴
(b) [Xe] 6s²4f¹⁴5d²
(c) [Ar] 4s²3d⁵
Answer:
In explanation
Explanation:
a. Selenium, atomic number 34
Argon is 18, adding the other electrons gives 34.
Other elements include: oxygen O , Sulphur S , Tellurium Te and Polonium Po
b. Hafnium, element 72
Xe is 54, adding the other electrons makes the total 72.
Other elements; Titanium Ti, Zirconium Zr, Rutherfordium Rf
c. Manganese, element 25
Other group members are Technetium Tc , Rhenium Re and Bohrium Bo
__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
Rank the ions in each set in order of increasing size:
(a) CI⁻, Br⁻, F⁻
(b) Na⁺, Mg²⁺, F⁻
(c) Cr²⁺, Cr³⁺
Answer:
a) F⁻ < Cl⁻ < Br⁻
b) Mg²⁺ < Na⁺ < F⁻
c) Cr³⁺ < Cr²⁺
Explanation:
The ions in (a) part of the question belong to a halogen group. F, Cl, and Br are present in periods 2, 3, and 4 respectively. As we move down the group the size of atoms increases hence their ions will be in the same order. (ion from top to bottom of group 7)
The ions in (b) part of the question are isoelectronic. The relative size of such species can be estimated by the charge on their nucleus. Lower the nucleus charge greater will be the size of the ion.
Nuclear charge of Mg²⁺ = no. of protons = 12
Nuclear charge of Na⁺ = no. of protons = 11
Nuclear charge of F⁻ = no. of protons = 9
The ions in (c) part are the two oxidized states of chromium. In such cases, higher the number of nuclear charge smaller will be the ion.
If the density of water is 1.00 g/mL and the density of mercury is 13.6 g/mL, how high a column of water in meters can be supported by standard atmospheric pressure?
Answer:
The answer to this is
The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters
Explanation:
To solve this we first list out the variables thus
Density of the water = 1.00 g/mL =1000 kg/m³
density of mercury = 13.6 g/mL = 13600 kg/m³
Standard atmospheric pressure = 760 mmHg or 101.325 kilopascals
Therefore from the equation for denstity we have
Density = mass/volume
Pressure = Force/Area and for a column of water, pressure = Density × gravity×height
Therefore where standard atmospheric pressure = 760 mmHg we have for Standard tmospheric pressure= 13600 kg/m³ × 9.81 m/s² × 0.76 m = 101396.16 Pa
This value of pressure should be supported by the column of water as follows
Pressure = 101396.16 Pa = kg/m³×9.81 m/s² ×h
∴ [tex]h = \frac{101396.16}{(1000)(9.81)}[/tex] = 10.336 meters
The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters
Final answer:
A column of water about 10.33 meters high can be supported by standard atmospheric pressure, based on the relationship between pressure, fluid density, gravitational acceleration, and height of the fluid column.
Explanation:
To calculate the height of a column of water supported by standard atmospheric pressure, we can use the pressure exerted by a column of liquid formula, which is p = hρg where p is the pressure, h is the height of the liquid column, ρ (rho) is the density of the liquid, and g is the acceleration due to gravity. The standard atmospheric pressure is 101,325 Pa, and the density of water is 1.00 g/cm³ or 1000 kg/m³.
Since 1 atm is equivalent to the pressure exerted by a 760 mm column of mercury, and mercury is 13.6 times denser than water, a water column would need to be 13.6 times taller. Thus:
101325 Pa = (h)(1000 kg/m³)(9.81 m/s²)
h = 101325 Pa / (1000 kg/m³ × 9.81 m/s²) = 10.33 m
Therefore, a column of water about 10.33 meters high can be supported by standard atmospheric pressure.
Determine the moles of H+ reacting with the metal based on the following experimental data. 10.00 mL of 1.00 M HCl solution is added to a sample of Mg metal. The reaction goes to completion and all of the Mg metal is gone. 21.80 mL of 0.30 M NaOH solution is required to reach the end point when titrating the remaining HCl.
Answer:
0.01 mol of H+.
Explanation:
Equation of reaction:
Mg(s) + 2HCl(aq) --> MgCl2(aq) + H2(g)
Molar concentration = number of moles/volume
= 1*0.01
= 0.01 mol.
Dissociation:
2HCl --> 2H+ + 2Cl-
By stoichiometry,
If 2 mole of HCl dissociates to guve 2 moles of H+
Number of moles of H+ = 0.01 mol.
Which of the following has the largest number of molecules? a 1 g of benzene, C6H6 b 1 g of formaldehyde, CH2O c 1 g of TNT, C7H5N3O6 d 1 g of naphthalene, C10H8 e 1 g of glucose, C6H12O6
Answer: 1g of formaldehyde
Explanation:
Molar Mass of benzene (C6H6) = (12x6) + (6x1) = 78g/mol
1mole(78g) of benzene contains 6.02x10^23 molecules.
1g of benzene with contain = (6.02x10^23) /78 = 7.72x10^21 molecules
Molar Mass of formaldehyde (CH2O) = 12 +2+16 = 30g/mol
1mole (30g) of formaldehyde contains 6.02x10^23 molecules.
1g of formaldehyde will contain = (6.02x10^23) /30 = 2.01x10^22 molecules
Molar Mass of TNT(C7H5N3O6) = (12x7)+(5x1)+(14x3)+(16x6) = 227g/mol
1mole(227g) of TNT contains 6.02x10^23 molecules.
1g of TNT will contain = (6.02x10^23)/227 = 2.65x10^21 molecules
Molar Mass of naphthalene (C10H8) = (12x10) +8 =128g/mol
1mole(128g) of naphthalene contains 6.02x10^23 molecules.
1g of naphthalene will contain = (6.02x10^23)/128 = 4.7x10^21 molecules
Molar Mass of glucose(C6H12O6) = (12x6) + 12 +(16x6) = 180g/mol
1mole(180g) of glucose contains 6.02x10^23 molecules.
1g of glucose will contain = (6.02x10^23)/180 = 3.34x10^21 molecules
Therefore, 1g of each of the compound contains the following:
Benzene = 7.72x10^21 molecules
Formaldehyde = 2.01x10^22 molecules
TNT = 2.65x10^21 molecules
Naphthalene = 4.7x10^21 molecules
Glucose = 3.34x10^21 molecules
From the above, we see clearly that formaldehyde has the largest number of molecules
Final answer:
The compound with the largest number of molecules is formaldehyde with 2.00 x 10^22 molecules.
Explanation:
The substance with the largest number of molecules can be determined by calculating the number of moles of each substance and then using Avogadro's number to convert moles to molecules. The formula to calculate the number of moles is moles = mass / molar mass. By using this formula:
For benzene: moles = 1g / 78.11g/mol = 0.0128 moles
For formaldehyde: moles = 1g / 30.03g/mol = 0.0333 moles
For TNT: moles = 1g / 227.13g/mol = 0.0044 moles
For naphthalene: moles = 1g / 128.2g/mol = 0.0078 moles
For glucose: moles = 1g / 180.16g/mol = 0.0055 moles
Then, we can use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules:
For benzene: molecules = 0.0128 mol x 6.022 x 10^23 molecules/mol = 7.72 x 10^21 molecules
For formaldehyde: molecules = 0.0333 mol x 6.022 x 10^23 molecules/mol = 2.00 x 10^22 molecules
For TNT: molecules = 0.0044 mol x 6.022 x 10^23 molecules/mol = 2.65 x 10^21 molecules
For naphthalene: molecules = 0.0078 mol x 6.022 x 10^23 molecules/mol = 4.68 x 10^21 molecules
For glucose: molecules = 0.0055 mol x 6.022 x 10^23 molecules/mol = 3.31 x 10^21 molecules
Therefore, the compound with the largest number of molecules is formaldehyde with 2.00 x 10^22 molecules.
Consider the following equilibria in aqueous solution (1) Ag++ Cl-ーAgCl(aq) (2) AgCl(aq)CAgC12 (3) AgCls)Ag*CI K = 20-103 K = 93 K= 1.8.10-10 (a) Find K for the reaction AgCI(s)AgCl(aq). The species AgCl(aq) is an ion pair consisting of Ag and Cl associated with each other in solution. (b) Find [AgCl(aq)] in equilibrium with excess AgCl(s).
Explanation:
(a) Chemical reaction equation is given as follows.
[tex]Ag^{+} + Cl^{-} \rightarrow AgCl(aq) \rightarrow K_{1}[/tex]
Also,
[tex]AgCl(s) \rightarrow Ag^{+} + Cl^{-} \rightarrow K_{3}[/tex]
Therefore, the net reaction equation is as follows.
[tex]AgCl(s) \rightarrow AgCl(aq)[/tex]
Now, we will calculate the value of K for this reaction as follows.
K = [tex]K_{1} \rightarrow K_{2}[/tex]
= [tex]2.0 \times 10^{3} \times 1.8 \times 10^{-10}[/tex]
= [tex]3.6 \times 10^{-7 }[/tex]
Hence, the value of K for the given reaction is [tex]3.6 \times 10^{-7 }[/tex].
(b) As the reaction is given as follows.
[tex]AgCl(s) \rightarrow AgCl(aq)[/tex]
Therefore, when excess of AgCl(s) is added then the amount of [AgCl(aq)] present in equilibrium is as follows.
K = [AgCl(aq)] = [tex]3.6 \times 10^{-7 }[/tex]
Thus, the value of [AgCl(aq)] in equilibrium with excess AgCl(s) is [tex]3.6 \times 10^{-7 }[/tex].
(a) The equilibrium constant K for the reaction[tex]\(\text{AgCl(s)} \rightleftharpoons \text{AgCl(aq)}\) is \( 3.6 \times 10^{-9} \).[/tex]
(b) The concentration of AgCl(aq) in equilibrium with excess AgCl(s) is [tex]\( 3.6 \times 10^{-9} \, \text{M} \).[/tex]
Part (a): Finding K for the reaction-
We are given the following equilibria and their constants:
1. [tex]\(\text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(aq)} \quad K_1 = 20\)[/tex]
2. [tex]\(\text{AgCl(aq)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad K_2 = 93\)[/tex]
3. [tex]\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad K_3 = 1.8 \times 10^{-10}\)[/tex]
We need to find K for the reaction:
[tex]\[ \text{AgCl(s)} \rightleftharpoons \text{AgCl(aq)} \][/tex]
We can see that we need to combine these equilibria in a way that gets us from AgCl(s) to AgCl(aq)
Let's write the reactions again:
1. [tex]\(\text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(aq)} \quad K_1 = 20\)[/tex]
2. [tex]\(\text{AgCl(aq)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad K_2 = 93\)[/tex]
3. [tex]\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad K_3 = 1.8 \times 10^{-10}\)[/tex]
The relationship between these equilibria can be described as follows:
- [tex]\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad (K_3)\)[/tex]
- [tex]\(\text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(aq)} \quad (K_1)\)[/tex]
If we multiply K₃ by K₁, we will get the desired equilibrium:
[tex]\[ \text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad (K_3) \][/tex]
[tex]\[ \text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(aq)} \quad (K_1) \][/tex]
Multiplying these two equilibria together, the intermediate [tex]\(\text{Ag}^+\)[/tex] and [tex]\(\text{Cl}^-\)[/tex] ions cancel out, giving us:
[tex]\[ \text{AgCl(s)} \rightleftharpoons \text{AgCl(aq)} \][/tex]
The equilibrium constant for this reaction is:
[tex]\[ K = K_3 \times K_1 \][/tex]
Substituting the given values:
[tex]\[ K = (1.8 \times 10^{-10}) \times 20 \] \\[/tex]
[tex]\[ K = 3.6 \times 10^{-9} \][/tex]
So, the equilibrium constant K for the reaction [tex]\(\te{AgCl(s)} \rightleftharpoons \text{AgCl(aq)}\)[/tex] is [tex]\( 3.6 \times 10^{-9} \).[/tex]
Part (b): Finding AgCl(aq) in equilibrium with excess AgCl(s)
When AgCl(s) is in equilibrium with AgCl(aq), the equilibrium expression for this reaction is:
[tex]\[ K = [\text{AgCl(aq)}] \][/tex]
From Part (a), we found that:
[tex]\[ K = 3.6 \times 10^{-9} \][/tex]
Therefore, the concentration of [tex]\(\text{AgCl(aq)}\)[/tex] in equilibrium with excess [tex]\(\text{AgCl(s)}\)[/tex] is:
[tex]\[ [\text{AgCl(aq)}] = 3.6 \times 10^{-9} \, \text{M} \][/tex]
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.
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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
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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?
Arrange each set in order of decreasing atomic size:
(a) Ge, Pb, Sn (b) Sn, Te, Sr (c) F, Ne, Na (d) Be, Mg, Na
Answer: The decreasing order of atomic size:
For a: Pb > Sn > Ge
For b: Sr > Sn > Te
For c: Na > F > Ne
For d: Mg > Na > Be
Explanation:
Atomic radius of an atom is defined as the total distance from the nucleus to the outermost shell of the atom.
As moving from top to bottom in a group, there is an addition of shell around the nucleus and the outermost shell gets far away from the nucleus and hence, the distance between the nucleus and outermost shell increases. Thus, increasing the atomic radii of the atom.
As moving from left to right in a period, more and more electrons get added up in the same shell and the attraction between the last electron and nucleus increases, which results in the shrinkage of size of an atom. Thus, decreasing the atomic radii of the atom on moving towards right of the periodic table.
For the given options:
Option a: Ge, Pb, SnGermanium (Ge) is present in Group 14 and period 4 of the periodic table.
Lead (Pb) is present in Group 14 and period 6 of the periodic table.
Tin (Sn) is present in Group 14 and period 5 of the periodic table.
So, the decreasing order of atomic size is:
Pb > Sn > Ge
Option b: Sn, Te, SrTin (Sn) is present in Group 14 and period 5 of the periodic table.
Tellurium (Te) is present in Group 16 and period 5 of the periodic table.
Strontium (Sr) is present in Group 2 and period 5 of the periodic table.
So, the decreasing order of atomic size is:
Sr > Sn > Te
Option c: F, Ne, NaFluorine (F) is present in Group 17 and period 2 of the periodic table.
Neon (Ne) is present in Group 18 and period 2 of the periodic table.
Sodium (Na) is present in Group 1 and period 3 of the periodic table.
So, the decreasing order of atomic size is:
Na > F > Ne
Option d: Be, Mg, NaBeryllium (Be) is present in Group 2 and period 2 of the periodic table.
Magnesium (Mg) is present in Group 2 and period 3 of the periodic table.
Sodium (Na) is present in Group 1 and period 3 of the periodic table.
So, the decreasing order of atomic size is:
Mg > Na > Be
The decreasing order of atomic size are the following:
a: Pb > Sn > Ge
b: Sr > Sn > Te
c: Na > F > Ne
d: Mg > Na > Be
Trend of atomic size in periodic tableAtomic radius of an atom is defined as the total distance from the nucleus to the outermost shell of the atom. As move from top to bottom in a group, addition of shell occurs and the atomic radii of the atom increases.
While on the other hand. as we move from left to right, atomic size of an atom decreases due to addition of nuclear charge in the nucleus.
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You are interested in thymol in mouthwash. The manufacturer lists the active ingredients as thymol (0.064%), menthol (0.042%), eucalyptol (0.092%) and methyl salicylate. The inactive ingredients include caramel, water and alcohol (21.6%) in addition to sodium benzoate, color, poloxamer 407 and benzoic acid.
(a) Does thymol represent a major, minor or trace concentration?
(b) of the ingredients listed with concentration, which ones can be considered in trace amounts?
(c) Which substance(s) is/are the sample?
(d) Which substance(s) is/are the analyte?
(e) Which substance(s) is/are the matrix?
(f) If you were to prepare a blank for your measurement, what substance(s) would it contain?
Answer:
a) Thymol represents a minor concentration.
b) Of the ingredients listed, the ones that qualify as being contained in trace amounts are the ones with concentrations less than 0.01%, 100ppm or 0.1mg/L. And in this question, none of the given concentrations is less than 0.01%. Maybe one of the other constituents whose concentrations weren't given is in trace amounts.
c) The mouthwash brought for analysis is the sample.
d) Thymol is the analyte.
e) The matrix is every component of the mouthwash sample apart from the analyte constituent, thymol.
Menthol, eucalyptol, methyl salicylate, caramel, water, alcohol in addition to sodium benzoate, color, poloxamer 407 and benzoic acid are all member substances of the matrix.
f) The blank can contain every constituent apart from the analyte constituent, thymol.
Explanation:
a) The composition range for major, minor and trace components are given thus.
1-100% Major
0.01-1% Minor
1 ppb to 100ppm Trace
<1ppb ultra trace
In Chemistry terms,
•Major constituents: Substances in concentrations over 1mg/L
•Minor constituents: Substances in concentrations between 1mg/L and 0.1 mg/L
•Trace constituents: Substances in concentrations under 0.1 mg/L
Hence, thymol with a concentration of 0.064% represents a minor constituent.
b) Of the ingredients listed, the ones that qualify as being contained in trace amounts are the ones with concentrations less than 0.01%, 100ppm or 0.1mg/L. And in this question, none of the given concentrations is less than 0.01%. Maybe one of the other constituents whose concentrations weren't given is in trace amounts.
c) In chemical analysis or Analytical chemistry, a sample is a portion of material selected from a larger quantity of material.
Therefore, for this question, the mouthwash is the sample.
d) In Analytical Chemistry, an analyte or analyte component is a substance or chemical constituent that is of interest in an analytical procedure.
For this question, the analyte is thymol; the constituent the analyser is interested in.
e) In chemical analysis, matrix refers to the components of a sample other than the analyte of interest.
Therefore, the matrix is every component of the mouthwash sample apart from the analyte constituent, thymol.
Menthol, eucalyptol, methyl salicylate, caramel, water, alcohol in addition to sodium benzoate, color, poloxamer 407 and benzoic acid are all member substances of the matrix.
f) A blank solution is a solution containing little to no analyte of interest, so, the blank can contain every constituent apart from thymol.
Hope this helps!
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]
Select the NMR spectrum that corresponds best to p-bromoaniline. (see Hint for the structure.) The selected tab will be highlighted in blue. Click the tab number to toggle among them. All spectra are taken in CDCl3 and the peak at 0.0 ppm is trimethylsilane, which is used as a standard to calibrate chemical shifts.
The best NMR spectrum that corresponds to p-bromoaniline can be determined by analyzing the chemical shift values and multiplicity of the peaks. Tab 2 is the correct spectrum that matches the expected chemical shifts for p-bromoaniline.
Explanation:The NMR spectrum that corresponds best to p-bromoaniline can be determined by analyzing the chemical shift values and multiplicity of the peaks. In the NMR spectrum of p-bromoaniline, we would expect to see a peak for the bromine atom at a lower chemical shift value and a peak for the amino group at a higher chemical shift value.
The chemical shift values for p-bromoaniline would be different from those of the other molecules given in the choices. By comparing the chemical shift values and multiplicity pattern, we can identify the correct spectrum that matches the expected chemical shifts for p-bromoaniline.
Based on the given information, the best NMR spectrum that corresponds to p-bromoaniline would be Tab 2. This tab shows the expected chemical shift values and multiplicity pattern for p-bromoaniline.
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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
What is the volume of a 0.5 mol/L NaOH solution required to completely react with 20 mL of a 2.0 mol/L HCl solution? What is the pH value of the produced solution? (Hint: Write and balance the chemical reaction between HCl and NaOH, and then use the stoichiometric relation (molar relation) between NaOH and HCl to calculate the minimum volume of NaOH solution.) Group of answer choices
Answer:
The volume of NaOH required is 80mL
pH of the resulting solution is 7 since reaction produces a normal salt
Explanation:
First, we generate a balanced equation for the reaction as shown below:
NaOH + HCl —> NaCl + H2O
From the equation,
nA = 1
nB = 1
From the question,
Ma = 2.0 mol/L
Va = 20 mL
Mb = 0.5 mol/L
Vb =?
MaVa / MbVb = nA /nB
2x20 / 0.5 x Vb = 1
0.5 x Vb = 2 x 20
Divide both side by 0.5
Vb = (2 x 20)/0.5
Vb = 80mL
The volume of NaOH required is 80mL
Since the reaction involves a strong acid and a strong base, a normal salt solution will be produced which is neutral to litmus paper. So since the salt solution is neutral to litmus paper then the pH of the resulting solution is 7
The pH of a solution prepared by mixing HCl and NaOH, we need to determine the concentration of the resulting solution. Since HCl and NaOH react in a 1:1 ratio to form water and salt (NaCl), we can use the concept of neutralization to find the resulting concentration.
Moles of HCl = volume (in L) × molarity = 0.020 L × 0.30 mol/L = 0.006 mol
Moles of NaOH = volume (in L) × molarity = 0.015 L × 0.60 mol/L = 0.009 mol
Since HCl and NaOH react in a 1:1 ratio, the limiting reagent is HCl. This means that all of the HCl will react with NaOH, and we will have an excess of NaOH. Therefore, the number of moles of HCl remaining will be zero.
Total volume = 20.0 mL + 15.0 mL = 35.0 mL = 0.035 L
Concentration of resulting solution = (Moles of HCl + Moles of NaOH) / Total volume
= (0.006 mol + 0.009 mol) / 0.035 L
= 0.429 mol/L
pH = -log[H+]
pH = -log(0.429)
≈ 0.366
Therefore, the pH of the solution prepared by mixing 20.0 mL of 0.30 M HCl with 15.0 mL of 0.60 M NaOH is approximately 0.366.
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How many directions of cleavage does fluorite exhibit (see animation in placemark)? a. eight b. four c. two d. three
Answer:
Option (B)
Explanation:
In the field of mineralogy, cleavage is usually defined as a surface along which a mineral has the tendency to break. These surfaces are the plane of weaknesses that are formed due to deformation as well as metamorphism.
Fluorite is a mineral that has a hardness of 4, according to the Mohs hardness scale. It is comprised of four cleavage directions. This means that it has four sets of perfect cleavage, and this mineral often breaks into small pieces that are octahedron in shape.
Thus, the correct answer is option (B).
Gold has a density of 0.01932 kg/cm3. What is the mass (in kg) of a 92.5 cm3 sample of gold?A) 1.79B) 0.560C) 92.5D) 0.000209E) 4790
Answer:
A. 1.79kg
Explanation:
The value of the mass of a particular substance could be obtained if there is adequate information about the volume of that particular substance and the density of that substance.
Mathematically expressed, the mass of a substance is the product of the density and the volume .
In this particular question, the density is 0.01932kg/cm3 while the volume is 92.5cm3
The mass is thus = 92.5 * 0.01932 = 1.7871kg which is approximately 1.79kg
The mass of a 92.5 cm3 sample of gold is approximately 1.79 kg.
Explanation:The mass of an object can be calculated by multiplying its density by its volume. In this case, the density of gold is given as 0.01932 kg/cm3, and the volume of the gold sample is 92.5 cm3. To find the mass, we can use the formula:
mass = density * volume
Substituting the values, we have:
mass = 0.01932 kg/cm3 * 92.5 cm3
Calculating this, we find that the mass of the gold sample is approximately 1.79 kg.
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A ground-state H atom absorbs a photon of wavelength 94.91 nm, and its electron attains a higher energy level. The atom then emits two photons: one of wavelength 1281 nm to reach an intermediate level, and a second to reach the ground state.
(a) What higher level did the electron reach?
(b) What intermediate level did the electron reach?
(c) What was the wavelength of the second photon emitted?
A) The higher level that the electron reached from ground state is; n = 5
B) The intermediate level that the electron reached from ground state is; n = 3
C) The wavelength of the second photon emitted is; λ = 103 nm
A) We are given;
Wavelength of photon absorbed by ground state H atoms; λ_g = 94.91 nm = 94.91 × 10⁻⁹ m
Formula to get the higher level is Rydberg's formula;
1/λ = R(1/n₁² - 1/n₂²)
where;
R is rydberg constant = 1.097 × 10⁷ m⁻¹
Thus;
1/(94.91 × 10⁻⁹) = 1.097 × 10⁷(1/1² - 1/n₂²)
0.9605 = 1 - 1/n₂²
1/n₂² = 1 - 0.9605
1/n₂² = 0.0395
n₂ = √(1/0.0395)
n₂ ≈ 5
B) We want to find the intermediate level where wavelength = 1281 nm = 1281 × 10⁻⁹ m
Thus;
1/(1281 × 10⁻⁹) = 1.097 × 10⁷(1/n₂² - 1/5²)
0.0712 = 1/n₂² - ¹/₂₅
1/n₂² = 0.0712 + ¹/₂₅
1/n₂² = 0.1112
n₂ = √(1/0.1112)
n₂ ≈ 3
C) Formula for energy of photon is;
E = hc/λ
where;
h is Planck's constant = 6.626 × 10⁻³⁴ m².kg/s
c is speed of light = 3 × 10⁸ m/s
Thus;
E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(1281 × 10⁻⁹)
E = 1.552 × 10⁻¹⁹ J
The energy at ground state is;
E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(94.91 × 10⁻⁹)
E = 20.944 × 10⁻¹⁹ J
Thus;
Energy of second photon = (20.944 × 10⁻¹⁹) - (1.552 × 10⁻¹⁹)
Energy of second photon = 19.352 × 10⁻¹⁹ J
Wavelength of second photon emitted is;
λ = hc/E
λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/19.352 × 10⁻¹⁹
λ = 103 nm
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The electron moves to the third energy level after the initial photon absorption, then drops to the fourth energy level releasing a photon of 1281 nm wavelength, and finally back to the ground state with a recorded emission of 97.08 nm wavelength.
Explanation:To solve this, we can use the Rydberg formula for hydrogen: 1/λ = R (1/n1^2 - 1/n2^2), where λ is the wavelength of the light, R is the Rydberg constant (1.096776 x 10^7 m^-1), n1 is the lower energy level, and n2 is the higher energy level.
(a) The photon absorbed takes the electron to a higher energy level, so we use the wavelength 94.91 nm, and n1 = 1 (ground state). This calculates to n2 = 2.32, meaning the electron moves to roughly the third energy level as energy levels are only in whole numbers.
(b) The first emitted photon has wavelength 1281 nm, which brings the electron to an intermediate level. By substituting the values, we calculate n1 = 3 and get n2 = 4. So, the intermediate energy level is 4.
(c) For the second photon emitted, we know the electron goes from n = 4 to the ground state n = 1. Rearranging the Rydberg formula, we find λ = 97.08 nm, which is the wavelength of the second photon emitted.
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When collecting your solid, why should the filter paper in the Buchner funnel be wetted with some of the cold recrystallization solvent before a recrystallization mixture is filtered?
Answer: It is important to wet the filter paper in the Buchner funnel first with cold re crystallization solvent before the re crystallization mixture being filtered to minimize gaps around the edges of the filter paper which can prevent mechanical impurities from passing through. This gives better filtration where most impurities can be filtered. Furthermore, it provides good vacuum.
Draw the structures of the 4 isomers of C8H18 that contain 2 methyl branches on separate carbons of the main chain.
Explanation:
1. 2,3-dimethylhexane
2. 2,4-dimethylhexane
3. 2,5-dimethylhexane
4. 3,4-dimethylhexane
Below are the structures of the isomers.
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.