Answer:
Approximately [tex]1.29 \times 10^3[/tex] grams.
Explanation:
Let [tex]x[/tex] represent the number of grams of aluminum iodide required to yield that 73.75 grams of aluminum.
In most cases, the charge on each aluminum ion would be +3 while the charge on each iodide ion would be -1. For the charges to balance, there needs to be three iodide ions for every aluminum ion. Hence, the empirical formula for aluminum iodide would be [tex]\rm AlI_3[/tex].
How many moles of formula units in that [tex]x[/tex] grams of [tex]\rm AlI_3[/tex]? Start by calculating its formula mass [tex]M(\mathrm{AlI_3})[/tex]. Look up the relative atomic mass of aluminum and iodine on a modern periodic table:
Al: 26.982.I: 126.904.[tex]M(\mathrm{AlI_3}) = 1\times 26.982 + 3\times 126.904 = 410.694\; \rm g \cdot mol^{-1}[/tex].
[tex]n(\mathrm{AlI_3}) = \displaystyle \frac{m}{M} = \frac{x}{410.694}\;\rm mol[/tex].
Since there's one aluminum ion in every formula unit,
[tex]n(\mathrm{Al}) = n(\mathrm{AlI_3}) = \displaystyle \frac{x}{410.694}\; \rm mol[/tex].
How many grams of aluminum would that be?
[tex]m(\mathrm{Al}) = n \cdot M = \displaystyle \frac{x}{410.694}\; \times 26.982 = \frac{26.982}{410.694}\, x\; \rm g[/tex].
However, since according to the question, the percentage yield (of aluminum) is only [tex]86.8\%[/tex]. Hence, the actual yield of aluminum would be:
[tex]\begin{aligned}&\text{Actual Yield} \\ &= \text{Percentage Yield} \times \text{Theoretical Yield} \\ &= 86.8\% \times \frac{26.982}{410.694}\, x \\ &= 0.868 \times \frac{26.982}{410.694}\, x \\ &\approx 0.0570263\, x\; \rm g\end{aligned}[/tex].
Given that the actual yield is 73.75 grams,
[tex]0.0570263\, x = 73.75[/tex].
[tex]\displaystyle x = \frac{73.75}{0.0570263} \approx 1.29 \times 10^3\; \rm g[/tex].
To yield 73.75 grams of aluminum with a percent yield of 86.8%, approximately 1113.33 grams of aluminum iodide would be required.
Explanation:To determine the amount of aluminum iodide required to yield 73.75 grams of aluminum with a percent yield of 86.8%, we need to use stoichiometry and the concept of percent yield.
First, we need to determine the theoretical yield of aluminum iodide using the balanced chemical equation: 2 Al + 3 I2 → 2 AlI3
From the equation, we can see that 2 moles of Al react with 3 moles of I2 to produce 2 moles of AlI3. Therefore, the molar ratio between Al and AlI3 is 2:2.
To calculate the theoretical yield of AlI3, we need to convert the mass of Al to moles using the molar mass of Al. Then, we use the mole ratio to find the moles of AlI3, and finally, convert it back to grams using the molar mass of AlI3.
Using the molar mass of Al (26.98 g/mol), we find that the moles of Al is 73.75 g / 26.98 g/mol = 2.73 moles.
Since the molar ratio between Al and AlI3 is 2:2, the moles of AlI3 produced is also 2.73 moles.
Finally, we can calculate the mass of AlI3 using the molar mass of AlI3 (407.7 g/mol): 2.73 moles * 407.7 g/mol = 1113.33 g.
Therefore, approximately 1113.33 grams of aluminum iodide would be required to yield an actual amount of 73.75 grams of aluminum with a percent yield of 86.8%.
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name each ionic compound. In each of these compounds, the metal forms only one type of ion. a)CeCl b)SrBr2 c) K2O d)LiF
Explanation:
A. CeCl
Cerium chloride.
Metal: Ce+
B. SrBr2
Strontium chloride.
Metal: Sr2+
C. K2O
Potassium oxide.
Metal: K+
D. LiF
Lithuim fluoride.
Metal: Li+
Chlorine (Cl) creates an anion with a -1 charge, but the metal cerium (Ce) only forms one type of ion with a +3 charge. As a result, the substance is known as cerium(III) chloride.
a) CeCl: Cerium(III) chloride
Whereas bromine (Br) generates an anion with a -1 charge, the metal strontium (Sr) only forms one sort of ion with a +2 charge in this combination. As a result, the substance is known as strontium bromide.
b) SrBr2: Strontium bromide
In this molecule, oxygen (O) generates an anion with a -2 charge whereas the metal potassium (K) only produces one sort of ion with a +1 charge. The substance is referred to as potassium oxide as a result.
c) K2O: Potassium oxide
Lithium is the only element present in this combination (Li). As lithium only ever produces an ion with a positive charge, it is known simply as lithium.
d) Li: Lithium
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Summarize the trend in metallic character as a function of position in the periodic table. Is it the same as the trend in atomic size? Ionization energy?
Answer:
The trend in metallic character as a function of position in the periodic table is that the metallic character increases as you go down a group. Since the ionization energy decreases going down a group (or increases going up a group), the increased ability for metals lower in a group to lose electrons makes them more reactive.
This is not the same for the atomic size, as you go down a column of the periodic table, the atomic radii increase. This is because the valence electron shell is getting a larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus.
Similarly, it is also different for the ionization energy trend, as you go down the periodic table, it becomes easier to remove an electron from an atom (i.e., IE decreases) because the valence electron is farther away from the nucleus.
Final answer:
The metallic character in the periodic table decreases across a period and increases down a group. It trends similarly to atomic size but oppositely to ionization energy.
Explanation:
Metallic character: The metallic trend follows the trend of the atomic radius. It increases within a group of the periodic table from the top to the bottom and decreases within a period from left to right. Metallic character relates to the ease of losing an electron in a chemical reaction and is opposite to the trend of ionization energy.
Atomic size shows a trend that parallels the metallic character. Atomic size increases down a group because of the increase in electron shells, which makes the valence electrons less tightly held. Conversely, atomic size decreases from left to right within a period due to an increasing effective nuclear charge, which draws electrons closer to the nucleus, reducing the size of the atom.
In summary, the metallic character and atomic size increase from right to left in a period and from top to bottom in a group, while ionization energy generally shows the opposite trend. Hence, the trend in metallic character is similar to the trend in atomic size but opposite to the trend in ionization energy.
For some reaction carried out at constant atmospheric pressure and at a constant temperature of 25◦C, it is found that ∆H = −38.468 kJ/mol and ∆S = +51.4 J mol · K . What is the value of ∆G for this reaction under these conditions?
Answer: ΔG =23.169kJ/mol
Explanation:
Solution
To Calculate Gibbs free energy ΔG for the reaction above we use the equation ΔG=ΔH−TΔS.
Where
ΔH= 38.468 kJ/mol = 38468 J/mol
∆S = +51.4 J mol−1 K−1).
T = 25◦C =298k
ΔG= 38468J/mol−298k(51.4 J mol−1 K−1).
ΔG = 38468 J/mol - 15317.2J/mol
ΔG = 23168.8J/mol
ΔG =23.169kJ/mol
The sulfur atom of sulfur dioxide is considered to be sp2-hybridized. The expected bond angle is 120°, but is actually slightly smaller (119°). Write down the correct statement that explains the smaller bond angle.
Answer: The bonds are intermediate between double and single bonds
Explanation:
A closer look at the diagram below shows that the bonds in sulphur IV oxide are intermediate between double and single bonds. Hence they do not have the exact bond angle of single bonds. This is why the bond angle is not exactly 120°. There are two resonance structures in the diagram that clearly show this point.
What is the percent yield of a reaction in which 51.5 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce metallic tungsten and 5.76 mL of water (d = 1.00 g/mL)?
Answer:
The percent yield of a reaction is 48.05%.
Explanation:
[tex]WO_3+3H_2\rightarrow W+3H_2O[/tex]
Volume of water obtained from the reaction , V= 5.76 mL
Mass of water = m = Experimental yield of water
Density of water = d = 1.00 g/mL
[tex]M=d\times V = 1.00 g/mL\times 5.76 mL=5.76 g[/tex]
Theoretical yield of water : T
Moles of tungsten(VI) oxide = [tex]\frac{51.5 g}{232 g/mol}=0.2220 mol[/tex]
According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:
[tex]\frac{3}{1}\times 0.2220 mol=0.6660 mol[/tex]
Mass of 0.6660 moles of water:
0.666 mol × 18 g/mol = 11.988 g
Theoretical yield of water : T = 11.988 g
To calculate the percentage yield of reaction , we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
[tex]=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%[/tex]
The percent yield of a reaction is 48.05%.
Answer:The percent yield of a reaction is 48.05%.
Explanation:
Draw the structure of ozone according to VSEPR theory. What would be its associated molecular geometry?
Ozone, according to the VSEPR theory, has a bent or 'V' shaped geometry due to the repulsion of electron pairs. This is because it has one lone pair and two bonding domains.
O
/ \
O O
Explanation:The structure of ozone, or O₃, can be drawn according to the VSEPR theory. The central atom is one oxygen atom while the other two oxygen atoms are attached to the central one. Then, there is one lone pair on the central atom, creating a 'bent' or 'V' shape in its geometry.
The VSEPR (Valence Shell Electron Pair Repulsion) theory suggests that electron pairs will repel each other as much as possible, resulting in specific molecular geometries. Ozone is a molecular geometry example of a molecule with 3 total sites of electrons, 2 bonding domains, and one non-bonding domain. This leads to a 'bent' or 'V' shape because the non-bonding pair pushes the two bonding domains closer together.
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he decomposition of acetaldehyde, CH3CHO, was determined to be a second order reaction with a rate constant of 0.0771 M-1 s-1. If the initial concentration of acetaldehyde is 0.358 M , what will the concentration be after selected reaction times
Answer:
The concentration is [-1 + sqrt(1+0.11t)]/0.1542 M
Explanation:
Let the concentration of CH3CHO after selected reaction times be y
Rate = Ky^2 = change in concentration of CH3CHO/time
K = 0.0771 M^-1 s^-1
Change in concentration of CH3CHO = 0.358 - y
0.0771y^2 = 0.358-y/t
0.0771ty^2 = 0.358 - y
0.0771ty^2 + y - 0.358 = 0
The value of y must be positive and is obtained in terms of t using the quadratic formula
y = [-1 + sqrt(1^2 -4(0.0771t)(-0.358)]/2(0.0771) = [-1 + sqrt(1+0.11t)]/0.1542 M
Final answer:
The question involves calculating the instantaneous rate of a second order decomposition reaction of acetaldehyde using the given rate constant and concentration values.
Explanation:
The question deals with a second order reaction describing the decomposition of acetaldehyde (CH3CHO) into methane (CH4) and carbon monoxide (CO). A second order reaction rate is dependent on the square of the concentration of one reactant or the product of two reactants concentrations. The rate constant provided (0.0771 M-1 s-1 or 4.71 × 10-8 L mol-1 s-1) is used alongside the concentration of acetaldehyde to determine the instantaneous rate of reaction or, in some scenarios, to deduce the remaining concentration of acetaldehyde at a given time.
Let the concentration of CH3CHO after selected reaction times be y
Rate = Ky^2 = change in concentration of CH3CHO/time
K = 0.0771 M^-1 s^-1
Change in concentration of CH3CHO = 0.358 - y
0.0771y^2 = 0.358-y/t
0.0771ty^2 = 0.358 - y
0.0771ty^2 + y - 0.358 = 0
The value of y must be positive and is obtained in terms of t using the quadratic formula
y = [-1 + sqrt(1^2 -4(0.0771t)(-0.358)]/2(0.0771) = [-1 + sqrt(1+0.11t)]/0.1542 M
If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of solution required to reach the equivalence point is 5.125 g, how many moles of hydrogen peroxide molecules are present?
Answer:
0.004522 moles of hydrogen peroxide molecules are present.
Explanation:
Mass by mass percentage of hydrogen peroxide solution = w/w% = 3%
Mass of the solution , m= 5.125 g
Mass of the hydrogen peroxide = x
[tex]w/w\% = \frac{x}{m}\times 100[/tex]
[tex]3\%=\frac{x}{5.125 g}\times 100[/tex]
[tex]x=\frac{3\times 5.125 g}{100}=0.15375 g[/tex]
Mass of hydregn pervade in the solution = 0.15375 g
Moles of hydregn pervade in the solution :
[tex]=\fraC{ 0.15375 g}{34 g/mol}=0.004522 mol[/tex]
0.004522 moles of hydrogen peroxide molecules are present.
To determine the number of moles of hydrogen peroxide molecules present in the solution, multiply the mass of the solution by the mass percent of hydrogen peroxide. Then, convert the mass of hydrogen peroxide to moles using its molar mass. The number of moles of hydrogen peroxide molecules is approximately 0.00452.
Explanation:To determine the number of moles of hydrogen peroxide molecules present, we first need to calculate the mass of hydrogen peroxide in the solution. We can do this by multiplying the mass of the solution (5.125 g) by the mass percent of hydrogen peroxide (3.00% or 0.03).
Mass of hydrogen peroxide = 5.125 g × 0.03 = 0.15375 g
Next, we need to convert the mass of hydrogen peroxide to moles using its molar mass. The molar mass of hydrogen peroxide (H2O2) is 34.0146 g/mol.
Moles of hydrogen peroxide = 0.15375 g ÷ 34.0146 g/mol ≈ 0.00452 mol
Therefore, there are approximately 0.00452 moles of hydrogen peroxide molecules present.
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The H atom and the Be3³⁺ ion each have one electron. Does the Bohr model predict their spectra accurately? Would you expect their line spectra to be identical? Explain.
Explanation:
a) Bohr model is perfect for atoms that have single electron and fortunately both Be3+ ion and H atom have one electron so, Bohr model can easily and accurately applied to predict the spectrum of Be3+ and H atom.
b) The energy of an atom in Bohr model is given by
[tex]E= \frac{-13.6z^2}{n^2}[/tex]
the values of z for H atom and Be3+ ion are 1 and 4 respectively. Hence, energy of atoms would be different for both atoms. Hence, line spectra to be identical is not possible.
The density of mercury is 13.6 g/cm3 . What volume (in quarts) is occupied by 100. g of Hg? (1 L = 1.06 qt)
Answer:
0.00077 qt
Explanation:
Density -
Density of a substance is given by the mass of the substance divided by the volume of the substance .
Hence , d = m / V
V = volume
m = mass ,
d = density ,
From the question ,
The mass mercury = 100 g
Density of mercury = 13.6 g/cm³ .
Hence , by using the above formula ,and putting the corresponding values , the volume of mercury is calculated as -
d = m / V
13.6 g/cm³ = 100 g / V
V = 7.35 cm³
1 cm³ = 0.001 L
V = 7.35 * 0.001 L = 0.0073 L
Since ,
1 L = 1.06 qt
V = 0.0073* 1.06 qt = 0.0077 qt
The vapor pressure of cobalt is 400 mm Hg at 3.03x10^3 K.
Assuming that its molar heat of vaporization is constant at 450 kJ/mol, the vapor pressure of liquid Co is _____ mm Hg at a temperature of 3.07x10^3 K.
We can calculate the vapor pressure of liquid cobalt at a given temperature by using the Clausius-Clapeyron equation and the given vapor pressure at another temperature. This involves substituting known values into the equation and solving for the desired vapor pressure.
The question is asking for the calculated vapor pressure of liquid cobalt at a certain temperature based on its known vapor pressure at another temperature. This involves using the Clausius-Clapeyron equation, which describes the relationship between the vapor pressure of a substance and its temperature. Let's denote the initial conditions (i.e., 400 mm Hg at 3.03x10^3 K) as P1 and T1, and the conditions we want to find (i.e., vapor pressure at 3.07x10^3K) as P2 and T2.
First, convert the molar heat of vaporization from kJ/mol to J/mol by multiplying by 1000, which gives 450000 J/mol. Next, the Clausius-Clapeyron equation can be rearranged to solve for P2:
P2 = P1 * exp [ -ΔHvap (1/T2 - 1/T1) / R ]
where ΔHvap is the molar heat of vaporization, R is the ideal gas constant (8.314 J/mol·K). Substituting all known values into this equation will give the vapor pressure of liquid Co at the desired temperature.
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The vapor pressure of liquid cobalt at 3.07x10^3 K is approximately 3748.64 mm Hg.
The vapor pressure of liquid cobalt at a temperature of 3.07x10^3 K can be determined using the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature. The equation is given by:
[tex]\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \][/tex]
First, we need to convert [tex]\( \Delta H_{\text{vap}} \)[/tex] from kJ/mol to J/mol to match the units of [tex]\( R \)[/tex]:
[tex]\[ \Delta H_{\text{vap}} = 450 \text{ kJ/mol} \times 1000 \text{ J/kJ} = 450,000 \text{ J/mol} \][/tex]
Now we can plug the values into the Clausius-Clapeyron equation:
[tex]\[ \ln\left(\frac{P_2}{400 \text{ mm Hg}}\right) = -\frac{450,000 \text{ J/mol}}{8.314 \text{ J/(mol·K)}}\left(\frac{1}{3.07x10^3 \text{ K}} - \frac{1}{3.03x10^3 \text{ K}}\right) \][/tex]
Solving for [tex]\( P_2 \):[/tex]
[tex]\[ \ln\left(\frac{P_2}{400}\right) = -\frac{450,000}{8.314}\left(\frac{1}{3.07x10^3} - \frac{1}{3.03x10^3}\right) \] \[ \ln\left(\frac{P_2}{400}\right) = -\frac{450,000}{8.314}\left(\frac{3.03x10^3 - 3.07x10^3}{(3.07x10^3)(3.03x10^3)}\right) \] \[ \ln\left(\frac{P_2}{400}\right) = -\frac{450,000}{8.314}\left(\frac{-40}{3.07x10^3x3.03x10^3}\right) \] \[ \ln\left(\frac{P_2}{400}\right) = -\frac{450,000}{8.314}\left(\frac{-40}{9.3051x10^6}\right) \][/tex]
[tex]\[ P_2 \approx 3748.64 \text{ mm Hg} \][/tex]
Therefore, the vapor pressure of liquid cobalt at 3.07x10^3 K is approximately 3748.64 mm Hg.
In the mid-17th century, Isaac Newton proposed that light existed as a stream of particles, and the wave-particle debate continued for over 250 years until Planck and Einstein presented their revolutionary ideas. Give two pieces of evidence for the wave model and two for the particle model.
Answer:
Light as a wave
1. Young's Double Slit Experiment
2. Davisson-Germer Experiment
Light as a particle
1. Einsteins Photoelectric Effect Phenomenon
2. Diffraction Phenomenon of Particles
Evidence for light as a wave includes diffraction and interference patterns, while evidence for the particle model includes the photoelectric effect and emission spectra.
The mid-17th century debate on whether light is a wave or a particle extended well into the 20th century until revolutionary concepts by Planck and Einstein. There is evidence for both the wave model and the particle model of light. Two pieces of evidence for the wave nature of light are diffraction and interference patterns, as seen in Thomas Young's double-slit experiment.
Diffraction occurs when light encounters an obstacle, spreading out as a result, and interference patterns occur when waves overlap and combine in constructive or destructive ways. Conversely, evidence for the particle nature includes phenomena like the photoelectric effect, as explained by Einstein, where light knocks electrons from a material, and the behavior of emission spectra, where individual energy quanta or "photons" are emitted from atoms.
A buffer solution is made by mixing a weak acid with its conjugate base. If the ratio of conjugate base to acid is 4, and the pH of the buffer is 7.2, what is the pKa of the weak acid? Round the answer to one decimal place.
Answer:
6.6 is the [tex]pK_a[/tex] of the weak acid.
Explanation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant =?
The ratio of conjugate base to acid is = [tex]\frac{[salt]}{acid}=4[/tex]
pH = 7.2
Putting values in above equation, we get:
[tex]7.2=pK_a+\log(4)[/tex]
[tex]pK_a=7.2-\log(4)=6.598\approx 6.6[/tex]
6.6 is the [tex]pK_a[/tex] of the weak acid.
Identify each element below, and give the symbols of the other elements in its group:
(a) [He] 2s²2p¹
(b) [Ne] 3s²3p⁴
(c) [Xe] 6s²5d¹
Answer:
Answer in explanation
Explanation:
a. Boron , element 5
Helium has 2 electrons, add to the other 3 to give 5.
Other group members are : Aluminum Al, Gallium Ga, Indium In , Thallium Tl and Nihonium Nh
b. Sulphur, element 16
Neon is 10 , add other 6 electrons to make 16
Other group members are: Oxygen O, selenium Se , Tellurium Te and Polonium Po
c. Lanthanum, element 57
Xenon is 54, add the other 3 electrons to give 57.
Other elements in group : Scandium Sc , Yttrium Y , Actinium Ac, Lutetium Lu and/or Lawrencium Lr
Calculate the concentration of H3O+ of a solution if the concentration of OH- at 25°C is 3.8 × 10-5 M and determine if the solution is acidic or basic.
Answer:
[H₃O⁺] = 2.63×10⁻¹⁰ M
As pH = 9.57, the solution is basic
Explanation:
We must know this knowledge:
[OH⁻] . [H₃O⁺] = 1×10⁻¹⁴
3.8×10⁻⁵ . [H₃O⁺] = 1×10⁻¹⁴
[H₃O⁺] = 1×10⁻¹⁴ / 3.8×10⁻⁵ → 2.63×10⁻¹⁰ M
Let's determine the pH to state if the solution is acidic or basic
pH < 7 → acidic ; pH > 7 → basi
pH = - log [H₃O⁺]
pH = - log 2.63×10⁻¹⁰ → 9.57
A sodium hydroxide solution that contains 24.8 grams of NaOH per L of solution has a density of 1.15 g/mL. Calculate the molality of the NaOH in this solution.
Final answer:
The molality of the sodium hydroxide (NaOH) solution is 0.54 mol/kg.
Explanation:
The molality of a solution is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is sodium hydroxide (NaOH) and the solvent is water.
To calculate the molality, we need to first convert the given mass of NaOH to moles using its molar mass, which is 40.0 g/mol. Then, we need to convert the mass of the solution to kilograms using the density of the solution, which is 1.15 g/mL.
Using the given information:
Mass of NaOH = 24.8 g/L
Density of solution = 1.15 g/mL
Molar mass of NaOH = 40.0 g/mol
The molality can be calculated as follows:
Convert mass of NaOH to moles: 24.8 g/L x (1 mol NaOH / 40.0 g NaOH) = 0.62 mol/LConvert density of solution to mass of solution: 1.15 g/mL x 1000 mL/L = 1150 g/LConvert mass of solution to kilograms: 1150 g/L ÷ 1000 = 1.15 kg/LCalculate molality: 0.62 mol/L ÷ 1.15 kg/L = 0.54 mol/kgAccording to the Bohr model of the atom, when an electron goes from a higher-energy orbit to a lower-energy orbit, it ________ electromagnetic energy with an energy that is equal to the ________ between the two orbits.
Answer:
emits (radiates) , energy difference
Explanation:
According to the Bohr theory, when an electron jumps from higher orbital to the lower orbital, it radiates energy which is equal to the energy difference between the orbitals.
Mathematically, it can be shown as:-
The expression for Bohr energy is shown below as:-
[tex]E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules[/tex]
For transitions:
[tex]Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]
[tex]\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]
Also, [tex]\Delta E=\frac {h\times c}{\lambda}[/tex]
Where,
h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
According to the Bohr model of the atom, when an electron transitions from a higher-energy orbit to a lower-energy orbit, it emits electromagnetic energy equal to the energy difference between the two orbits.
Explanation:The Bohr model of the atom states that when an electron transitions from a higher-energy orbit to a lower-energy orbit, it emits electromagnetic energy with an energy equal to the difference between the two orbits. This energy is released in the form of a photon.
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Why does the malachite green dye elute first? What physical properties does it have that affect it’s interaction with alumina and how are those different from crystal violet?
Answer:
MG is less polar
Explanation:
The structures of crystal violet (CV) and malachite green (MG) are shown in Figures 1 and 2, respectively.
The obvious difference is that CV has an extra dimethylamino group (polar).
Alumina is a polar adsorbent, so it retains the more polar substances more strongly and they are eluted last.
MG is less polar than CV, so it is retained less strongly and is eluted first.
A solution contains 0.25 M Ni(NO3)2 and 0.25 M Cu(NO3)2. Can the metal ions be separated by slowly adding Na2CO3? Assume that for successful separation 99% of the metal ion must be precipitated before the other metal ion begins to precipitate, and assume no volume change on addition of Na2CO3.
Explanation:
Ksp of NiCO3 = 1.4 x 10^-7
Ksp of CuCO3 = 2.5 x 10^-10
Ionic equations:
NiCO3 --> Ni2+ + CO3^2-
CuCO3 --> Cu2+ + CO3^2-
[Cu2+][CO3^2-]/[Ni2+][CO3^2-]
= (2.5* 10^-10)/(1.4* 10^-7)
= 0.00179.
[Cu2+]/[Ni2+]
= 0.00179
= 0.00179*[Ni2+]
If all of Cu2+ is precipitated before Na2CO3 is added.
= 0.00179 * (0.25)
The amount of Cu2+ not precipitated = 0.000448 M
The percent of Cu2+ precipitated before the NiCO3 precipitates = concentration of Cu2+ unprecipitated/initial concentration of Cu2+ * 100
= 0.000448/0.25 * 100
= 0.18%
Therefore, percentage precipitated = 100 - 0.18
= 99.8%
The two metal ions can be separated by slowly adding Na2CO3. Thus that is the unpptd Cu2+.
The metal ions can be separated by slowly adding Na2CO3 based on the relative solubilities of their carbonates. Nickel carbonate (NiCO3) is less soluble than copper carbonate (CuCO3), allowing the selective precipitation of nickel ions before copper ions.
Explanation:To determine if the metal ions can be separated by slowly adding Na2CO3, we need to consider the solubility of the metal carbonates. Nickel carbonate (NiCO3) and copper carbonate (CuCO3) both have low solubilities, but it is crucial to examine their relative solubilities. If one carbonate is significantly less soluble than the other, it can be selectively precipitated first.
In this case, NiCO3 is less soluble than CuCO3. Therefore, by slowly adding Na2CO3 to the solution, we can precipitate the majority of the Ni2+ ions as NiCO3 before CuCO3 begins to precipitate. This satisfies the condition that 99% of the metal ion must be precipitated before the other metal ion begins to precipitate.
Therefore, it is possible to separate the nickel and copper ions in the solution by slowly adding Na2CO3.
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Suppose that a certain drug company manufactured a compound that had nearly the same structure as a substrate for a certain enzyme but that could not be acted upon chemically by the enzyme. What type of interaction would the compound have with the enzyme
Answer: Reversible competitive inhibition
Explanation:
In the case of reversible competitive inhibition, an inhibitor molecule competes with the substrate for binding to the active site of the enzyme. The inhibitor blocks the active site of the enzyme. Thus the enzyme substrate complex do not form. The structure of the inhibitor is similar to the substrate thus also have the binding affinity with the enzyme. The process is reversible because the inhibitor will leave the enzyme it exerts no permanent effect on the enzyme.
The given situation is the example of reversible competitive inhibition as substrate remain unchanged and the enzyme was not able to act on the substrate chemically may be due to inhibition of the function of the enzyme.
You need to prepare a solution with a specific concentration of Na+Na+ ions; however, someone used the end of the stock solution of NaClNaCl, and there isn’t any NaClNaCl to be found in the lab. You do, however, have some Na2SO4Na2SO4. Can you substitute the same number of grams of Na2SO4Na2SO4 for the NaClNaCl in a solution? Why or why not?
Explanation:
Ionic equation
NaCl(aq) --> Na+(aq) + Cl-(aq)
Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)
In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.
Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)
= 142 g/mol
Molecular weight of NaCl = 23 + 35.5
= 58.5 g/mol
Masses
% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100
= 46/142 * 100
= 32.4%
% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100
= 23/58.5 * 100
= 39.3%
Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.
You cannot substitute Na2SO4 directly for NaCl based on mass since they have different molar masses. The same mass of Na2SO4 will provide more Na+ ions than NaCl, leading to a change in the Na+ ion concentration.
Explanation:No, you cannot substitute the same number of grams of Na2SO4 for the NaCl in a solution. This is because NaCl and Na2SO4 have different molar masses and therefore different numbers of moles per gram. The concentration of a solution is determined by the number of moles of solute per unit volume of solvent, not the mass. Hence, using the same mass of a different compound would alter the concentration of Na+ ions in the solution.
For instance, if one mole of NaCl gives us one mole of Na+, one mole of Na2SO4 will provide two moles of Na+. In other words, the same mass of Na2SO4 contains more Na+ ions than the same mass of NaCl. So using the same mass of Na2SO4 in place of NaCl will result in a solution with a higher Na+ ion concentration.
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Data has been collected to show that at a given wavelength in a 1 cm pathlength cell, Beer's Law for the absorbance of Co2 is linear. If a 0.135 M solution of Co2 has an absorbance of 0.350, what is the concentration of a solution with an absorbance of 0.420?
Answer : The concentration of a solution with an absorbance of 0.420 is, 0.162 M
Explanation :
Using Beer-Lambert's law :
[tex]A=\epsilon \times C\times l[/tex]
As per question, at constant path-length there is a direct relation between absorbance and concentration.
[tex]\frac{A_1}{A_2}=\frac{C_1}{C_2}[/tex]
where,
A = absorbance of solution
C = concentration of solution
l = path length
[tex]A_1[/tex] = initial absorbance = 0.350
[tex]A_2[/tex] = final absorbance = 0.420
[tex]C_1[/tex] = initial concentration = 0.135 M
[tex]C_2[/tex] = final concentration = ?
Now put all the given value in the above relation, we get:
[tex]\frac{0.350}{0.420}=\frac{0.135}{C_2}[/tex]
[tex]C_2=0.162M[/tex]
Thus, the concentration of a solution with an absorbance of 0.420 is, 0.162 M
Silver Mining is opening a new mineral extraction facility in the local town and will employ several thousand people. They have decided to install scrubbers on the smokestacks of their facility in order to protect the environment, even though they are not required by the law to install them. This is an example of:Business ethical behaviorLegal Behavior
Answer:
Ethical Behavior
Explanation:
It's the ethical behavior how companies work or do business that have the positive impact on the community. They not only think about making money but also about the welfare of the society. They are concerned about the products they made and it's impact on the environment. Ethical behaviour is based on the human perception of right and wrong. That kind of behaviour whichis not required by the law, but is done for the betterment of the society is ethical behaviour.
Silver Mining's voluntary decision to install environmentally friendly devices, despite no legal requirement, exemplifies business ethical behavior.
Explanation:In this instance, Silver Mining's decision to install scrubbers on the smokestacks of their new facility, even though not legally required, is a clear example of business ethical behavior. This action demonstrates the company prioritizing environmental protection over potential costs. While the act also aligns with legal behavior, it is not driven by legal necessity. It's a voluntary measure taken for the greater good, thus making it an ethical business decision.
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Determine the equilibrium pH and speciation (concentration of each species) of the following two solutions. Neglect activity corrections. Species added Total concentration (solution a) HCl 10-3 M (solution b) NaCl 10-3 g.
Answer:
HCl solution - H30+ and Cl ions. pH 3
NaCl - Na+ and Cl-. pH 7
Explanation:
a) HCl solution - the hydrogen ion combines with water molecule to form the hydronium molecule which is responsible for acidity. The chloride ion is also found in solution.
pH = -log [H+] = -log(10^-3) = 3
b) NaCl 10-3 g. The solid dissocates in water forming the Na+ and Cl- ions. None of these ions affect pH
If a buffer solution is 0.250 M 0.250 M in a weak base ( K b = 8.0 × 10 − 5 ) Kb=8.0×10−5) and 0.540 M 0.540 M in its conjugate acid, what is the pH ?
Answer:
9.57
Explanation:
Given that:
[tex]pK_{b}=-\log\ K_{b}=-\log(8.0\times 10^{-5})=4.1[/tex]
Considering the Henderson- Hasselbalch equation for the calculation of the pOH of the basic buffer solution as:
[tex]pOH=pK_b+log\frac{[conjugate\ acid]}{[base]}[/tex]
So,
[tex]pOH=4.1+\log\frac{0.540}{0.250}=4.43[/tex]
pH + pOH = 14
So, pH = 14 - 4.43 = 9.57
What is the wavelength (in nm) of the least energetic spectral line in the infrared series of the H atom?
Answer:
The least energetic spectral line in the infrared series of the H atom is 656.1 nm
Explanation:
Photon wavelength is inversely proportional to energy. To obtain the least energetic spectral line of the hydrogen atom (H), we determine the longest wavelength possible.
[tex]\frac{1}{\lambda} = R_H[\frac{1}{n_f^2} -\frac{1}{n^2}][/tex]
Where;
nf = 2
n = 3
RH is Rydberg constant = 1.09737 × 10⁷m⁻¹
λ is the wavelength of the least energetic spectral line
Substituting the above values into the equation, we will have
[tex]\frac{1}{\lambda} = 1.09737 X 10^7[\frac{1}{2^2} -\frac{1}{3^2}][/tex]
[tex]\frac{1}{\lambda} = 1.09737 X 10^7[\frac{1}{4} -\frac{1}{9}][/tex]
[tex]\frac{1}{\lambda} = 1.09737 X 10^7[0.25 -0.1111][/tex]
[tex]\frac{1}{\lambda} = 1.09737 X 10^7[0.1389][/tex]
[tex]\frac{1}{\lambda} = 1524246.93[/tex]
[tex]\lambda} = \frac{1}{1524246.93}[/tex]
[tex]\lambda} = 6.561 X10^{-7} m[/tex]
λ = 656.1 X10⁻⁹ m
In (nm): λ = 656.1 nm
Therefore, the least energetic spectral line in the infrared series of the H atom is 656.1 nm
The wavelength of the least energetic spectral line in the infrared series of the hydrogen atom is approximately 18,400 nanometers (nm).
To find the wavelength of the least energetic spectral line in the infrared series of the hydrogen atom, we can use the Rydberg formula for the hydrogen atom:
1 / λ = R_H * (1/n₁² - 1/n₂²)
Where:
λ is the wavelength of the spectral line.
R_H is the Rydberg constant for hydrogen, approximately 1.097 x 10^7 m⁻¹.
n₁ is the principal quantum number of the initial energy level.
n₂ is the principal quantum number of the final energy level.
For the least energetic line in the infrared series, we need to consider the transition where the electron moves from a higher energy level (n₂) to a lower energy level (n₁). In this case, n₂ > n₁.
Since we are interested in the infrared series, we'll consider transitions ending in the n₁ = 3 energy level. We want the least energetic line, so we'll choose the smallest value for n₂.
Let's take n₂ = 4 and calculate:
1 / λ = 1.097 x 10^7 m⁻¹ * (1/3² - 1/4²)
1 / λ = 1.097 x 10^7 m⁻¹ * (1/9 - 1/16)
1 / λ = 1.097 x 10^7 m⁻¹ * (7/144)
Now, solve for λ:
λ = 144 / (1.097 x 10^7 m⁻¹ * 7)
λ ≈ 0.0184 meters or 18.4 millimeters
To express the wavelength in nanometers (nm), we can convert millimeters to nanometers:
λ ≈ 18.4 mm * 1,000,000 nm/mm = 18,400 nm
So, the wavelength of the least energetic spectral line in the infrared series of the hydrogen atom is approximately 18,400 nm.
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Given that a chlorine-oxygen bond in ClO2(g) has an enthalpy of 243 kJ/molkJ/mol , an oxygen-oxygen bond has an enthalpy of 498 kJ/molkJ/mol , and the standard enthalpy of formation of ClO2(g) is? ΔH∘f=102.5kJ/molΔHf∘=102.5kJ/mol , use Hess's law to calculate the value for the enthalpy of formation per mole of ClO(g).
The answer & explanation for this question is given in the attachment below.
What determines the types of chemical reactions that an atom participates in? A. the number of electrons in the outermost electron shell B. the number of electrons in the innermost electron shell C. its atomic mass the number of protons D. it contains its atomic number
Answer:
The answer would be A. the number of electrons in the outermost electron shell.
Explanation:
These are called valence electrons which are transferred, shared, and rearranged by creating covalent bonds producing new substances.
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The type of chemical reaction an atom chooses is determined by the number of the outermost electrons in the outermost shell of an atom.
The reactivity of an atom is determined by the number of electrons in the outermost shell of the particular atom.The number of electrons is used to determine the type of bond formed by that atom in a chemical reaction.The outermost shell of the atom is called the valence shell and the number of outermost electrons is thus called the valence electrons.About the importance of electrons the below points should be noted;
The electron is the major constituent of an atom which determines the reactivity of an atom.The outermost electron is more used for any reaction to occur than the innermost electronsTherefore the answer is the number of electrons in the outermost electron shell.
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2. Assume that a sample of 10.00 g of a solid unknown is dissolved in 25.0 g of water. Assuming that pure water freezes at 0.0 oC and the solution freezes at -5.58 oC, what is the molal concentration of the solution
Answer:
m = 3 moles/kg
Explanation:
This is a problem of freezing point depression, and the formula or expression to use is the following:
ΔT = i*Kf¨*m (1)
Where:
ΔT: Change of temperature of the solution
i: Van't Hoff factor
m: molality of solution
Kf: molal freezing point depression of water (Kf = 1.86 °C kg/mol)
Now, the value of i is the number of moles of particles obtained when 1 mol of a solute dissolves. In this case, we do not know what kind of solution is, so, we can assume this is a non electrolyte solute, and the value of i = 1.
Let's calculate the value m, which is the molality solving for (1):
m = ΔT/Kf (2)
Finally, let's calculate ΔT:
ΔT = T2 - T1
ΔT = 0 - (-5.58)
ΔT = 5.58 °C
Now, let's replace in (2):
m = 5.58/1.86
m = 3 moles/kg
This is the molality of solution.
The other data of mass, can be used to calculate the molecular mass of this unknown solid, but it's not asked in the question.
The d-metals can be mixed together to form a wide range of alloys because:
1. the range of d metal radii is not very great.
2. the d-electrons interact strongly with each other.
3. the d-metals have low melting points.
4. the d-metals have a wide range of metal radii.
5. the nucleus is well shielded by the d electrons.
Answer:
the range of d metal radii is not very great.
Explanation:
The difference in metallic radii are not great hence the metallic ions are almost similar in size across the series. As a result of this, they can easily take up positions in the lattice of other transition metals leading to the formation of transition metal alloys. This explains the wide range of transition metal alloys used for various purposes in industry.