The apparent "atomic mass" of the contaminated iodine is approximately 1111.9285 amu.
To determine the apparent "atomic mass" of contaminant iodine, we need to take into account the contribution of both the naturally occurring iodine ([tex]\rm ^1^2^7I[/tex]) and the man-made radioisotope ([tex]\rm ^1^2^9I[/tex]).
The formula for calculating the atomic mass is:
Atomic mass = (mass of isotope 1 * abundance of isotope 1) + (mass of isotope 2 * abundance of isotope 2) + ...
Lets calculating the abundances of [tex]\rm ^1^2^7I[/tex] and [tex]\rm ^1^2^9I[/tex]:
Abundance of [tex]\rm ^1^2^7I[/tex] = 1 - Abundance of [tex]\rm ^1^2^9I[/tex]
Given:
Mass of [tex]\rm ^1^2^7I[/tex] = 126.9045 amu
Mass of [tex]\rm ^1^2^9I[/tex] = 128.9050 amu
We have two isotopes: [tex]\rm ^1^2^7I[/tex] and [tex]\rm ^1^2^9I[/tex]. The masses of these isotopes and their abundances need to be considered:
Atomic mass = (mass of [tex]\rm ^1^2^7I[/tex] * abundance of [tex]\rm ^1^2^7I[/tex]) + (mass of [tex]\rm ^1^2^9I[/tex]* abundance of [tex]\rm ^1^2^9I[/tex])
We need to calculate the abundances of [tex]\rm ^1^2^7I[/tex] and [tex]\rm ^1^2^9I[/tex] before we can calculate the atomic mass.
Mass of contaminated iodine sample = 12.3849 g + 1.00070 g = 13.3856 g
Now we can calculate the abundances:
Abundance of [tex]\rm ^1^2^7I[/tex] = (mass of [tex]I-127[/tex] in sample) / (total mass of sample)
Abundance of [tex]\rm ^1^2^7I[/tex] = (12.3849 g * 126.9045 amu) / (13.3856 g) = 117.0997 amu / 13.3856 g ≈ 8.7411
Abundance of [tex]\rm ^1^2^9I[/tex] = (mass of I-129 in sample) / (total mass of sample)
Abundance of [tex]\rm ^1^2^9I[/tex]= (1.00070 g * 128.9050 amu) / (13.3856 g) = 128.9050 amu / 13.3856 g ≈ 9.6175
We can calculate the atomic mass:
Atomic mass = (126.9045 amu * 8.7411) + (128.9050 amu * 9.6175) ≈ 1111.9285
Therefore, the apparent "atomic mass" of the contaminated iodine is approximately 1111.9285 amu.
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The apparent atomic mass of the contaminated iodine is 126.9045 amu.
Explanation:
To calculate the apparent atomic mass of the contaminated iodine, we can use the formula:
Apparent atomic mass = (mass of naturally occurring iodine * atomic mass of naturally occurring iodine + mass of synthetic radioisotope * atomic mass of synthetic radioisotope) / total mass of contaminated iodine
Given:
Mass of naturally occurring iodine = 12.3849 gAtomic mass of naturally occurring iodine = 126.9045 amuMass of synthetic radioisotope = 1.00070 gAtomic mass of synthetic radioisotope = 128.9050 amuSubstituting the given values into the formula:
Apparent atomic mass = (12.3849 g * 126.9045 amu + 1.00070 g * 128.9050 amu) / (12.3849 g + 1.00070 g)
Simplifying the expression:
Apparent atomic mass = (1570.6078055 + 128.905070) / 13.3856
Apparent atomic mass = 1699.5128755 / 13.3856
Apparent atomic mass = 126.9045 amu
Therefore, the apparent atomic mass of the contaminated iodine is 126.9045 amu.
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Atomic hydrogen produces well-known series of spectral lines in several regions of the electromagnetic spectrum. Each series fits the Rydberg equation with its own particular n₁ value. Calculate the value of n₁ (by trial and error if necessary) that would produce a series of lines in which:
(a) The highest energy line has a wavelength of 3282 nm.
(b) The lowest energy line has a wavelength of 7460 nm.
Answer: a) The value of n₁ that would produce a series of lines in which the highest energy line has a wavelength of 3282 nm is 6.
b) The value of n₁ that would produce a series of lines in which the lowest energy line has a wavelength of 7460 nm is 5.
Explanation:
The formula that relates wavelength of emissions to Rydberg's constant and the n₁ values is
(1/λ) = R ((1/(n₁^2)) - (1/(n2^2))
Where λ = wavelength, R = (10.972 × 10^6)/m, n2 = ∞ (since they're emitted out of the atom already)
a) n₁ = ?
λ = 3282 nm = (3.282 × 10^-6)m
(1/(3.282 × 10^-6)) = (10.972 × 10^6) ((1/(n₁^2) (since 1/∞ = 0)
n₁^2 = (3.282 × 10^-6) × (10.972 × 10^6) = 36
n₁ = 6.
The value of n₁ that would produce a series of lines in which the highest energy line has a wavelength of 3282 nm is 6.
b) n₁ = ?
λ = 7460 nm = (7.46 × 10^-6)m
(1/(7.46 × 10^-6)) = (10.972 × 10^6) ((1/(n₁^2)) - (1/(n2^2)) for lowest energy line, n2 = n₁ + 1
(n₁^2)((n₁+1)^2))/(2n₁+1) = (7.46 × 10^-6) × (10.972 × 10^6) = 81.85
(n₁^2)((n₁+1)^2))/(2n₁+1) = 81.85
Solving the quadratic eqn,
n₁ = 5.
The value of n₁ that would produce a series of lines in which the lowest energy line has a wavelength of 7460 nm is 5.
QED!
The net change in the multistep biochemical process of photosynthesis is that CO₂ and H₂O form glucose (C₆H₁₂O₆) and O₂. Chlorophyll absorbs light in the 600 to 700 nm region. (a) Write a balanced thermochemical equation for formation of 1.00 mol of glucose. (b) What is the minimum number of photons with λ = 680. nm needed to prepare 1.00 mol of glucose?
Answer:(a) 6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)
(b) 5.55*10^37photons
Explanation:
(a) Here we have to write a balanced thermochemical equation for formation of 1.00 mol of glucose.
In this question it has been given that Chlorophyll absorbs light in the 600 to 700 nm region,
1st we will write a chemical equation for biochemical process of photosynthesis is that CO2 and H2O form glucose (C6H12O6) and O2.
6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)
The heat change off the reaction can be calculated as,
={(1 mol)(6 mol) }- {(6 mol) [H2O]}
=[1 1273.3 kJ + 6(0)] - [6 (-39.5 kJ) + 6 (-285.840 kJ)]
= 2802.74 or 2802.7 kJ
Thus the balanced equation can be written as,
6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g) = 2802.7 kJ for 1.00 mol of glucose.
A 0.10 M solution of Na2HPO4 could be made a buffer solution with all of the following EXCEPT ________. View Available Hint(s) A 0.10 solution of could be made a buffer solution with all of the following EXCEPT ________. K3PO4 Na3PO4 H3PO3 NaH2PO4
Final answer:
The answer to the question is Na3PO4, as it provides the same anion as Na2HPO4 without a conjugate acid or base, making it unable to form a buffer solution with Na2HPO4.
Explanation:
A buffer solution is formed from a weak acid and its conjugate base, or a weak base and its conjugate acid. Therefore, a 0.10 M solution of Na2HPO4 (which is sodium hydrogen phosphate) could be made into a buffer solution with another compound that either provides its conjugate acid or its conjugate base. Na2HPO4 can act as both a weak acid (donating H+) and a weak base (accepting H+).
A buffer solution with Na2HPO4 could be made using the following combinations:
H3PO4 and Na2HPO4 (H3PO4 is the conjugate acid of Na2HPO4)NaH2PO4 (NaH2PO4 can provide the conjugate acid of Na2HPO4)K3PO4 (K3PO4 can provide the conjugate base of Na2HPO4)The one that cannot be used to form a buffer with Na2HPO4 is Na3PO4, because Na3PO4 is the fully deprotonated form and provides the same anion as Na2HPO4 without an accompanying conjugate acid or base. Therefore, the correct answer is Na3PO4.
Half of the first 18 elements have an odd number of electrons, and half have an even number. Show why atoms of these elements aren’t half paramagnetic and half diamagnetic.
Answer:
Diamagnetism in atom occurs whenever two electrons in an orbital paired equalises with a total spin of 0.
Paramagnetism in atom occurs whenever at least one orbital of an atom has a net spin of electron. That is a paramagnetic electron is just an unpaired electron in the atom.
Here is a twist even if an atom have ten diamagnetic electrons, the presence of at least one paramagnetic electron, makes it to be considered as a paramagnetic atom.
Simply put paramagnetic elements are one that have unpaired electrons, whereas diamagnetic elements do have paired electron.
The atomic orbital and radius increases by gaining electron linearly so even electron numbered atoms are diamagnetic while the odd electron numbered atoms are paramagnetic.
Running through the first 18 elements one can observe that there is an alternative odd number of electrons and an even number proofing that that half of the first 18 elements shows paramagnetism and diamagnetism respectively.
Explanation:
Rank the following compounds in order of increasing acidity. A: h5ch17p19b1 B: h5ch17p19a1 C: h5ch17p19c1
Answer: Least Acidic: A
Moderate Acidic: C
Most Acidic: B
Explanation:
Given compounds' acidity cannot be determined as the molecular formulas seem incorrect. Normally, acidity is determined by factors such as the presence of hydrogen atoms and their ability to be donated influenced by bond polarity and molecular structure.
Explanation:The acidity of the given compounds cannot be determined as the given molecular formulas (h5ch17p19b1, h5ch17p19a1, h5ch17p19c1) appear to be incorrect or non-standard. Typically, the acidity of a compound is influenced by factors like the presence of hydrogen atoms, how easily these can be donated (as determined by bond polarity and structure of the molecule), and the stability of the conjugate base after a hydrogen atom has been donated.
In common terminology, acidity refers to the ability of a substance to donate a proton (H+) in a chemical reaction. The traditional scale for measuring acidity is the pH scale, where lower pH values indicate higher acidity.
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Phthalic acid is a diprotic acid having the formula HO2CC6H4CO2H that can be converted to a salt by reaction with base. Which of the following is expected to be most soluble in water? A) HO2CC6H4CO2H B) HO2CC6H4CO2Na C) HO2CC6H4CO2K D) NaO2CC6H4CO2Kand why?
Answer:
D) NaO2CC6H4CO2K
Explanation:
Water is a polar solvent and tends to solvate polar molecules. This allows solute molecules to interact with the solvent and that is why the solubility of a molecule in water increases with the increase in its polarity. So, the salt of phthalic acid is more soluble in water than phthalic acid itself. Although the monosodium and monopotassium salts are also more soluble than phthalic acid, the dialkali phthalate salt (NaO2CC6H4CO2K) is the most soluble due to the highest polarity.
Calculate the boiling point (in degrees C) of a solution made by dissolving 7 g of naphthalene {C10H8} in 14.4 g of benzene. The Kbp of the solvent is 2.53 K/m and the normal boiling point is 80.1 degrees C. Enter your answer using 2 decimal places.
Answer:
The boiling point = 89.69 °C
Explanation:
Step 1: Data given
Mass of naphthalane = 7.0 grams
Mass of benzene = 14.4 grams
The Kbp of the solvent = 2.53 K/m
The normal boiling point is 80.1°C
Naphthalene, C10H8 , is a non-electrolyte, which means that the van't Hoff factor for this solution will be 1
Step 2: Calculate moles naphthalene
Moles naphthalene = mass / molar mass
Moles naphthalene = 7.0 grams / 128.17 g/mol
Moles naphthalene = 0.0546 moles
Step 3: Calculate molality
Molality = moles naphthalene / mass water
Molality = 0.0546 moles / 0.0144 kg
Molality = 3.79 molal
Step 4:
ΔT = i*Kb*m
ΔT = 1*2.53 K/m * 3.79 molal
ΔT = 9.59 °C
The boiling point = 80.1 °C + 9.59 °C = 89.69 °C
The boiling point of a solution made by dissolving 7 g of naphthalene in 14.4 g of benzene, with a Kbp of 2.53 K/m, is 89.68 degrees C.
Explanation:To calculate the boiling point of a solution made by dissolving naphthalene in benzene, we can use the boiling point elevation formula: \(\Delta T = i \cdot K_{bp} \cdot m\), where \(\Delta T\) is the boiling point elevation, \(i\) is the van't Hoff factor (which is 1 for non-electrolytes like naphthalene), \(K_{bp}\) is the ebullioscopic constant of the solvent, and \(m\) is the molality of the solution.
The molality (\(m\)) is calculated using the formula: \(m = \frac{moles\ of\ solute}{kilograms\ of\ solvent}\). Naphthalene's molar mass is 128.17 g/mol. Thus, the moles of naphthalene are \(\frac{7\ g}{128.17\ g/mol} = 0.0546\ moles\). The mass of benzene is 14.4 g, which is 0.0144 kg. So, the molality is \(\frac{0.0546\ moles}{0.0144\ kg} = 3.79\ m\).
Now, we can find the boiling point elevation: \(\Delta T = 1 \cdot 2.53\ K/m \cdot 3.79\ m = 9.58\ K\). Convert K to \(\degree C\) by using the normal boiling point of benzene (80.1 \(\degree C\)) plus the boiling point elevation: \(80.1 \(\degree C\) + 9.58 \(\degree C\) = 89.68 \(\degree C\)\).
The boiling point of this solution is 89.68 degrees C.
You have 716.7 ml of 3.86 M HCl. Using a volumetric pipet, you take 119.56 ml of that solution and dilute it to 969.88 ml in a volumetric flask. Now you take 100.00 ml of that solution and dilute it to 145.62 ml in a volumetric flask. What is the concentration of hydrochloric acid in the final solution? Enter to 4 decimal places.
Answer:
0.327 M is the concentration of hydrochloric acid in the final solution.
Explanation:
Case 1:
716.7 ml of 3.86 M HCl. Using a volumetric pipet, you take 119.56 ml of that solution and dilute it to 969.88 ml in a volumetric flask.
Before dilution :
[tex]M_1=3.86 M, V_1=119.56 ml[/tex]
After dilution :
[tex]M_2=?, V_2=969.88 ml[/tex]
[tex]M_1V_1=M_2V_2[/tex] ( dilution )
[tex]M_2=\frac{M_1V_1}{V_2}=\frac{3.86 M\times 119.56 mL}{969.88 mL}[/tex]
[tex]M_2=0.4758 M[/tex]
Case 2:
Now you take 100.00 ml of case 1 solution and dilute it to 145.62 ml in a volumetric flask
Before diluting it further :
[tex]M_2=0.4758 M, V_1=100.00 ml[/tex]
After dilution :
[tex]M_3=?, V_3=145.62 ml[/tex]
[tex]M_3V_3=M_2V_2[/tex] ( dilution )
[tex]M_3=\frac{M_2V_2}{V_3}=\frac{0.4758 M\times 100.00 mL}{145.62 mL}[/tex]
[tex]M_3=0.327 M[/tex]
0.327 M is the concentration of hydrochloric acid in the final solution.
Suppose you are presented with a clear solution of sodium thiosulfate, Na2S2O3 . How could you determine whether the solution is unsaturated, saturated, or supersaturated?
Explanation:
A solution is said to saturated when it cannot dissolve any extra solute in it. The extra solute put remains undissolved.
A solution is said to unsaturated, when the concentration of solute is less as compared to solubility of the solution it is said to be unsaturated.
A solution is said to be super saturated when it contains more of the solute than the solvent can dissolve under normal conditions is called super saturated.
We can determine whether the solution is unsaturated, saturated, or supersaturated by knowing the amount of solute in the solution.
What is unsaturated, saturated, or supersaturated?A solution is said to be saturated when it cannot dissolve any extra solute in it, a solution is said to be unsaturated solution, when the concentration of solute is less as compared to solubility of the solution and the solution is able to dissolve more solute.
Whereas, a solution is said to be super saturated when it contains more of the solute than the solvent can dissolve under normal conditions.
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For each electronic configuration given, choose the electronic configuration of the element that would match its chemical properties. (1) 1s22s22p63s2 a. 1s22s22p63s23p3b. 1s22s22p63s23p64s23d104p6c. 1s22s2d. 1s22s22p3(2) 1s22s22p63s23p64s23d104p6 a.1s22s22p63s23p3 1s22s22p63s2 1s22s2 1s22s22p6(c) 1s22s22p3 1s22s22p63s23p64s23d104p6 1s22s22p6 1s22s2 1s22s22p63s23p3
Answer:1s22s22p63s2- Magnesium
1s22s22p63s23p3- phosphorus
1s22s22p63s23p64s23d104p6- krypton
1s22s2- Beryllium
1s22s22p6- neon
1s22s22p3- nitrogen
Explanation:
The identity of an element is deducible from its electronic configuration by looking out for the outermost shell configuration and counting the total number of electrons present. for example; phosphorus 15 electrons and an outermost shell configuration of ns2 np3. Any electronic configuration written above which reflects these characteristics must belong to phosphorus.
The question is about identifying elements based on given electron configurations and then matching these configurations to known ones. The elements matching the given configurations have the same number of electrons in their outermost energy level, indicating the possibility of similar chemical properties.
Explanation:The question is regarding the identification of elements based on their electron configurations and comparing them to known configurations. The electron configuration describes the distribution of electrons in an atom's electron shells and subshells.
For instance, the given first electron configuration (1) 1s22s22p63s2 corresponds to the element Neon. The element that would match this configuration is one whose electron configuration is 1s22s22p6, which is also for Neon in a neutral state.
For the second case (1s22s22p63s23p64s23d104p6), it corresponds to the element Krypton. The element that would match this configuration is also Krypton in a neutral state. These elements are said to have similar chemical properties because they have the same number of electrons in their outermost energy level, making their bonding patterns similar.
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Consider the generic acid, HA, and how it interacts with water:HA (aq) + H2O (l) ⇌ H3O+ (aq) + A-Consider the following statements and answer true or false. The stronger A- is as a weak base, the stronger HA will be as a weak acid.a. True b. False Having a greater number of electron withdrawing groups is typically the best way to stabilize a conjugate base.a. True b. False If HA is a strong acid, A- will be a relatively strong weak conjugate base.a. True b. False An acid with a Ka value of 1.2 x 10-4 is a weaker acid than an acid with a Ka value of 1.5 x 10-8.a. True b. False
Answer:
B. False
A. True
B. False in
Explanation:
1.
From Bronsted-Lowrys definition of Acids and bases, a strong acid is a substance that gives up a proton (to form a weak conjugate base), while a strong base is one that willingly accepts a proton.
Therefore in the reaction,
HA(aq) + H2O (l) ⇌ H3O+ (aq) + A-
The stronger HA is, the weaker the A- and vice versa; the weaker HA is, the stronger A-. Example, HCl is a strong acid and its conjugate base, Cl- is a weak base.
B. False
2.
Electron drawing groups are molecules and/or atoms that enable the release of a proton from a specie. They cause inductive as well as mesomeric effects. Examples, -NO2, -COOH, -OH etc.
Fluoride ion is the most stable in this series because it's the most electronegative while carbon is the least stable because it's the least electronegative. Because of this, we were able to say that H-F was the most acidic, because it had the most stable conjugate base.
A. True.
3.
pKa1 = -log[Ka1]
= -log[1.2 x 10-4]
= 3.92
pKa2 = -log[Ka2]
= -log[1.5 x 10-8]
= 7.82
Using Bronsted-Lowry definition, the smaller the pKa value the more ease the acid loses its proton, that is, the smaller the pKa value, the stronger the acid.
Therefore, pKa2 > pKa1, so Ka value of 1.2 x 10-4 is a stronger acid than Ka value of 1.5 x 10-8
B. False.
1. B. False
2. A. True
3. B. False
1.
Bronsted Lowry Concept:
According to Bronsted Lowry, an acid is a proton (H⁺) donor, and a base is a proton acceptor.
The given reaction,
[tex]HA(aq) + H_2O (l)[/tex] ⇌ [tex]H_3O^+ (aq) + A^-[/tex]
The stronger HA is, the weaker the A- and vice versa; the weaker HA is, the stronger A-.
Example, HCl is a strong acid and its conjugate base, Cl- is a weak base.
Thus the given statement is False.2.
Electron drawing groups are molecules and/or atoms that enable the release of a proton from a species. Examples, -NO₂, -COOH, -OH etc.
Fluoride ion is the most stable in this series because it's the most electronegative while carbon is the least stable because it's the least electronegative. Because of this, we were able to say that H-F was the most acidic, because it had the most stable conjugate base.
Thus the given statement is True.3.
[tex]p_{Ka_1} = -log[Ka_1]\\\\p_{Ka_1}= -log[1.2 * 10^{-4}]\\\\p_{Ka_1}= 3.92[/tex]
[tex]p_{Ka_2} = -log[Ka_2]\\\\p_{Ka_2}= -log[1.5 * 10^{-8}]\\\\p_{Ka_2}= 7.82[/tex]
Using Bronsted-Lowry definition, the smaller the pKa value the more ease the acid loses its proton, that is, the smaller the pKa value, the stronger the acid.
Therefore, [tex]p_{Ka_2} > p_{Ka_1}[/tex], so Ka value of[tex]1.2 * 10^{-4}[/tex] is a stronger acid than Ka value of [tex]1.5 * 10^{-8}[/tex]
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A drop of liquid tends to have a spherical shape due to the property of 1. viscosity. 2. capillary action. 3. surface tension. 4. vapor pressure. 5. close packing.
Answer:
surface tension.
Explanation:
Draw the partial (valence-level) orbital diagram, and write the symbol, group number, and period number of the element:
(a) [Kr] 5s²4d¹⁰
(b) [Ar] 4s²3d⁸
The given configurations correspond to the elements Cadmium (Cd) and Nickel (Ni) respectively. Cd is in group 12, period 5 and Ni is in group 10, period 4.
Explanation:The requested electron configurations correspond to specific elements in the periodic table.
(a) The configuration [Kr] 5s²4d¹⁰ corresponds to the element Cadmium (Cd). Its symbol is Cd, it is in group 12, and period 5. The partial valence-level orbital diagram is as follows:
5s: ↑↓ 4d: ↑↓|↑↓|↑↓|↑↓|↑↓
(b) The configuration [Ar] 4s²3d⁸ corresponds to the element Nickel (Ni). Its symbol is Ni, it is in group 10, and period 4. The partial valence-level orbital diagram is as follows:
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The configurations [Kr] 5s²4d¹⁰ and [Ar] 4s²3d⁸ represent Cadmium and Nickel respectively. Cadmium is in the 12th group, 5th period, and Nickel is in the 10th group, 4th period. Both are transition metals with distinct chemical reactions.
Explanation:The element with the electron configuration [Kr] 5s²4d¹⁰ is Cadmium (Cd). It belongs to the 12th group and 5th period. Its valence electron configuration diagram shows that there are 2 electrons in the 5s subshell and 10 electrons in the 4d subshell.
On the other hand, the element with the electron configuration [Ar] 4s²3d⁸ is Nickel (Ni). This element belongs to the 10th group and 4th period. Its valence electron configuration diagram shows it has 2 electrons occupying the 4s subshell and 8 electrons in the 3d subshell.
These identified elements, Cadmium and Nickel, represent their unique chemical and physical properties. For instance, they are both transition metals and behave similarly in many chemical reactions.
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A molecule from a new organism contains adenine, cytosine, guanine, and thymine. What molecules is being looked at?
Since, the options are not given the question is incomplete. The complete question is as follows:
A molecule from a new organism contains adenine, cytosine, guanine, and thymine. What is this unknown molecule?
DNA
lipid
carbohydrate
RNA
protein
Answer: DNA
Explanation:
The DNA short for deoxyribonucleic acid is a genetic material that can be found in majority of the living beings on earth. It is composed of two strands that are coiled around each other. Each strand consists of nucleotides. Each nucleotide is composed of four nitrogen exhibiting nucleobases like guanine, cytosine, adenine and thymine along with the deoxyribose sugar and phosphate group. In DNA there are two groups of nitrogenous bases these includes the pyrimidines and purines. The pyrimidines are cytosine and thymine and the purines are guanine and adenine.
According to the given situation, a molecule from a new organism consists of adenine, cytosine thymine and guanine these all are nitrogenous bases which can be found in DNA.
A liquid mixture of 0.400 mole fraction ethanol and 0.600 methanol was placed in an evacuated (i.e., no air) bottle and after many days is now in equilibrium with its vapor. Assuming Raoult's Law applies (actually, both activity coefficients are within 0.02 of unity), what is the mole fraction of each compound in the vapor at 25C? at 40C?
Answer:
mole fraction methanol = 0.76
mole fraction ethanol = 0.24
Explanation:
Raoult´s law gives us the partial vapor pressure of a component in solution as the product of the mole fraction of the component and the value of its pure pressure:
PA = X(A) x Pº(A)
where PA is the partial vapor pressure of component A, X(A) is the mole fraction of A, and Pº(A) its pure vapor pressure.
From reference literature the pure pressures of methanol, and ethanol are at 25 ºC :
PºCH₃OH = 16.96 kPa
PºC₂H₅OH = 7.87 kPa
Given that we already have the mole fractions, we can calculate the partial vapor pressures as follows:
PCH₃OH = 0.600 x 16.96 kPa = 10.18 kPa
PC₂H₅OH = 0.400 x 7.87 kPa = 3.15 kPa
Now the total pressure in the gas phase is:
Ptotal = PCH₃OH + PC₂H₅OH = 10.18 kPa + 3.15 kPa = 13.33 kPa
and the mole fractions in the vapor will be given by:
X CH₃OH = PCH₃OH / Ptotal = 10.18 kPa/ 13.33 kPa = 0.76
X C₂H₅OH = 1 - 0.76 = 0.24
Which element in each of the following sets would you expect to have the lowest IE₃?
(a) Na, Mg, Al (b) K, Ca, Sc (c) Li, Al, B
Answer:
(a) AL
(b) Sc
(c)Al
Explanation:
Ionization Energy is the energy required to remove electrons from the outer most orbitals of atom.
The higher the electron is on energy level the farther its from nucleus and more loosely bonded thus need lesser energy.
By looking at electron configuration we can figure out which electron will need more energy.
(a)Na, Mg, Al1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹
Na₁₁ ⇒ 1s², 2s², 2p⁶, 3s¹
Mg₁₂ ⇒ 1s², 2s², 2p⁶, 3s²
Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹
Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.
(b) K, Ca, ScK₁₉⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²
Ca₂₀⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²
Sc₂₁⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹
Sc will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.
(c) Li, Al, BLi₃ ⇒ 1s², 2s¹
Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹
B₅ ⇒ 1s², 2s², 2p¹
Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.
Answer:
A. Al
B. Sc
C. Al
Explanation:
The third ionisation energy is the energy required to an extra electron from a +2 ion or the energy required to remove the third electron from an element.
Lithium - 1s2 2s1
Sodium - 1s2 2s2 2p6 3s1
Magnesium - 1s2 2s2 2p6 3s2
Aluminium - 1s2 2s2 2p6 3s2 3p1
Potassium - 1s2 2s2 2p6 3s2 3p6 4s1
Calcium- 1s2 2s2 2p6 3s2 3p6 4s2
Boron - 1s2 2s2 2p1
Scandium - 1s2 2s2 2p6 3s2 3p6 3d1 4s2
Removing 2 electrons,
Li2+- 1s1
Na2+ - 1s2 2s2 2p5
Mg2+ - 1s2 2s2 2p6
Al2+ - 1s2 2s2 2p6 3s1
K2+ - 1s2 2s2 2p6 3s2 3p5
Ca2+ - 1s2 2s2 2p6 3s2 3p4
Boron - 1s2 2s1
Scandium - 1s2 2s2 2p6 3s2 3p6 4s1
So comparing,
A. Na, Mg, Al
The third electron is lost from a p- orbital and the energy level of p- is less than s- orbital but 3s is way less than the 2p so the lowest third ionisation energy is Al
B. K, Ca, Sc
The third electrons are lost from the 3p orbital in K and Ca but in 4s in Sc and if you remember, 4s has a lesser energy level than 3p orbital. So, Sc has the lowest third ionisation energy.
C. Li, Al, B
Al has the lowest third ionisation energy because Li loses its from 1s which is closest to the nucleus and B from 2s which is also close to the nucleus.
There are three sets of sketches below, showing the same pure molecular compound (water, molecular formula H_2 O) at three different temperatures. The sketches are drawn as if a sample of water were under a microscope so powerful that individual atoms could be seen. Only one sketch in each set is correct. Use the slider to choose the correct sketch in each set. You may need the following information: melting point of H_2 O: 0.0 degree C boiling point of H_2 O: 100.0 degree C
Answer:
Only sketch B has the water molecules in the right form/state that the temperature presented predicts!
Explanation:
N.B - With the initial assumption that all the processes or water states exist at normal conditions of atmospheric pressure and temperature!
In the image attached to this solution, sketch A is at -23°C, sketch B is at 237°C and sketch C is at 60°C.
But for water, it's boiling point is 100°C; meaning that the this is the temperature where water molecules change form from fairly free to move around, almost incompressible liquid state to the gaseous state in which the water molecules (now called steam) are totally free to move around.
Its melting point is 0°C; that is, this is the temperature where the water molecules change form from the orderly solid form (called ice) where motion is totally restricted to only vibrations into the more free liquid state.
This explanation indicates that water molecules at temperatures below 0°C exist in the orderly solid form.
Water molecules at temperatures between 0°C and 100°C exist as the fairly free liquid and at temperatures higher than 100°C, the water molecules exist in the free to move about gaseous state.
In the sketches attached to this solution, sketch A evidently shows the water molecules in the fairly free to move about form (that is, liquid form), but matches this state with a temperature of -23°C which corresponds more to the solid, orderly state of water molecules shown in sketch C. Hence, that is a mismatch.
Sketch B shows water molecules in the very freeing state of gaseous form and rightly matches that form with a temperature way above the boiling point of water, 237°C. Thereby indicating a correct match between temperature and the sketch.
Sketch C however shows water molecules in their very organized solid form but mismatches this form to 60°C which corresponds more to the liquid state sketch in sketch A.
Only sketch B has the water molecules in the right form/state that the temperature presented predicts!
Hope this helps!!!
How many moles are contained in 70. milliliters of a 0.167 M solution of p-toluidine hydrochloride? Enter only the number to two significant figures.
Answer: The amount of p-toluidine hydrochloride contained is 2.4 moles.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Molarity of p-toluidine hydrochloride solution = 0.167 M
Volume of solution = 70. mL
Putting values in equation 1, we get:
[tex]0.167=\frac{\text{Moles of p-toluidine hydrochloride}\times 1000}{70}\\\\\text{Moles of p-toluidine hydrochloride}=\frac{(0.167\times 70}{1000}=2.38mol=2.4mol[/tex]
Hence, the amount of p-toluidine hydrochloride contained is 2.4 moles.
Calculate the mass of sucrose necessary to make a 5% by mass sucrose solution if the solution contains 50.0 ml of distilled water.
Answer: 2.5g
Explanation:
5 % sucrose solution means that 5 % of the weight of the solution is sucrose.
If 1 liter of water weighs 1000 grams. To prepare 5% sucrose solution 5/100 x 1000 = 50 grams. Since 1 liter equals 1000ml, thus a 5 % solution has 50 grams of solute dissolved in one liter.
To prepare 5% sucrose solution in 50mls
=5/100 x 50
= 0.05 x 50
= 2.5g
Therefore to prepare 5% sucrose solution in 50mls we dissolve 2.5g of sucrose in 50ml of water
Final answer:
To prepare a 5% by mass sucrose solution with 50.0 ml of distilled water, approximately 2.63 g of sucrose is required, which is calculated using the percent by mass formula and assumes that water's density is 1 g/ml.
Explanation:
The student's question asks to calculate the mass of sucrose necessary to prepare a 5% by mass sucrose solution using 50.0 ml of distilled water. The concept involved here is the percent by mass calculation which is used in preparing solutions in chemistry. To calculate the mass of the sucrose needed, one must use the formula:
Percent by mass = (mass of solute / mass of solution) × 100%
Given that we want a 5% sucrose solution, we can rearrange the formula to solve for the mass of sucrose:
Mass of sucrose = (Percent by mass × mass of solution) / 100%
First we need to convert the volume of water to mass, assuming the density of water is approximately 1 g/ml:
Mass of water (solvent) = 50.0 ml × 1 g/ml = 50.0 g
The total mass of the solution will be the mass of the water plus the mass of the sucrose, which we can call 'x':
Mass of solution = mass of water + x
Now, plugging in the known values and solving for 'x' gives us:
x = (5% × (50.0 g + x)) / 100%
Solving this equation for 'x' yields:
0.05 × 50.0 g + 0.05x = x
2.5 g + 0.05x = x
2.5 g = x - 0.05x
2.5 g = 0.95x
x = 2.5 g / 0.95
x = 2.6316 g
Therefore, the mass of sucrose necessary to make a 5% sucrose solution with 50.0 ml of distilled water is approximately 2.63 g.
(4 points) The following lead compound for a pharmaceutical drug contains a rotatable bond. Using the principles of rigidification, draw two analogs that would enable testing of two different conformations
Answer:
Explanation:
The solution has been attached
Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for any weak species:
1. LiOH
2. HF
3. HCl
4. NH3
Ka expression: ___________
Answer:
1. LiOH : strong base
2. HF : weak acid
3. HCl : strong acid
4. NH3 : weak base
Ka expression: Ka = [A- ] * [H+] / [HA]
Explanation:
The Acidity Constant:
The acid dissociation constant, Ka, (or acidity constant, or acid ionization constant) is a measure of the strength of a weak acid (which is not completely dissociated):
HA ↔ A- + H +
HA is a generic acid that dissociates into A- (the conjugate base of the acid), and the hydrogen or proton ion, H +.
The dissociation constant Ka is written as the ratio of equilibrium concentrations (in mol / L):
Ka = [A- ] * [H+] / [HA]
When we write in square brackets, we refer to the concentration of that element.
Final answer:
LiOH is a strong base, HF is a weak acid with a Kₐ expression of Kₐ = [H⁺][F⁻]/[HF], HCl is a strong acid, and NH₃ is a weak base with a Kₐ expression for its conjugate acid NH₄⁺ as Kₐ = [NH₄⁺][OH⁻]/[NH₃].
Explanation:
Compounds can be identified as strong acids, weak acids, strong bases, or weak bases depending on their ability to dissociate in solution. The classification is based on the strength of the acids and bases, which is a measure of their tendency to donate or accept protons.
LiOH (Lithium hydroxide) is a strong base. Strong bases like LiOH completely dissociate into ions in an aqueous solution.HF (Hydrofluoric acid) is a weak acid. Weak acids do not fully dissociate in solution. The Kₐ expression for HF is as follows: Kₐ = [H⁺][F⁻]/[HF].HCl (Hydrochloric acid) is a strong acid. Strong acids fully dissociate into their constituent ions in aqueous solution.NH₃ (Ammonia) is a weak base. The Kₐ expression for the hydrolysis of the ammonium ion (NH₄⁺) is relevant here and would be: Kₐ = [NH₄⁺][OH⁻]/[NH₃].A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration of a solution equals the mass of what's dissolved divided by the total volume of the solution.
Here's how the student prepared the solution:
She put some solid sodium hydroxide into the graduated cylinder and weighed it. With the sodium hydroxide added, the cylinder weighed. She added water to the graduated cylinder and dissolved the sodium hydroxide completely. Then she read the total volume of the solution from the markings on the graduated cylinder. The total volume of the solution was .
What concentration should the student write down in her lab notebook?
Answer:
The concentration the student should write down in her lab is 2.2 mol/L
Explanation:
Atomic mass of the elements are:
Na: 22.989 u
S: 32.065 u
O: 15.999 u
Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.
Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.
For mole(s) of Na2S2O3 = (mass taken)/(molar mass)
= (17.240 g)/(158.105 g/mol) = 0.1090 mole.
Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)
= 0.05029 L.
To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:
= (moles of sodium thiosulfate)/(volume of solution in L)
= (0.1090 mole)/(0.05029 L)
= 2.1674 mol/L
A solution containing potassium bromide is mixed with one containing lead acetate to form a solution that is 0.013 M in KBr and 0.0035 M in Pb(C2H3O2)2 . Part A Will a precipitate form in the mixed solution
Answer: The precipitate will not be formed in the above solution.
Explanation:
The chemical equation for the reaction of potassium bromide and lead acetate follows:
[tex]2KBr(aq.)+Pb(CH_3COO)_2(aq.)\rightarrow PbBr_2(s)+2CH_3COOK(aq.)[/tex]
We are given:
Concentration of KBr = 0.013 M
Concentration of lead acetate = 0.0035 M
1 mole of KBr produces 1 mole of potassium ions and 1 mole of bromide ions
So, concentration of bromide ions = 0.013 M
1 mole of lead (II) acetate produces 1 mole of lead (II) ions and 2 moles of acetate ions
So, concentration of lead (II) ions = 0.0035 M
The salt produced is lead (II) bromide
The equation follows:
[tex]PbBr_2(s)\rightleftharpoons Pb^{2+}(aq.)+2Br^-(aq.)[/tex]
The expression of [tex]Q_{sp}[/tex] for above equation follows:
[tex]Q_{sp}=[Pb^{2+}]\times [Br^-]^2[/tex]
Putting values of the concentrations in above expression, we get:
[tex]Q_{sp}=(0.0035)\times (0.013)^2\\\\Q_{sp}=5.92\times 10^{-7}[/tex]
We know that:
[tex]K_{sp}[/tex] for lead (II) bromide = [tex]4.67\times 10^{-6}[/tex]
There are 3 conditions:
When [tex]K_{sp}>Q_{sp}[/tex]; the reaction is product favored. (No precipitation)When [tex]K_{sp}<Q_{sp}[/tex]; the reaction is reactant favored. (Precipitation occurs)When [tex]K_{sp}=Q_{sp}[/tex]; the reaction is in equilibrium. (sparingly soluble)As, the [tex]Q_{sp}[/tex] is less than [tex]K_{sp}[/tex]. The above reaction is product favored. This means that no salt or precipitate will be formed.
Hence, the precipitate will not be formed in the above solution.
5. Consider the process where nA mol of gas A initially at 1 bar pressure mix with nB mol of gas B also at 1 bar to form 1 mol of a uniform mixture of A and B at a final total pressure of 1 bar, and all at constant temperature T. Assume that all gases behave ideally. a. Show that the entropy change, mixSm, for this process is given by mixSm
Answer:
ΔSmix,m = 5.7628 J/K.mol
Explanation:
mix: A + B
∴ nA = x mol A
∴ nB = y mol B
⇒ n mix = x + y = 1 mol
∴ P total = 1 bar
∴ T: constant
entropy of gases when mixing:
ΔSmix = - nA*R*LnXA - nB*R*LnXB∴ XA = x/1 = x
∴ XB = y/1 = y
⇒ ΔSmix = - x*R*Lnx - y*R*Lny
assuming: x = y = 0.5 mol
⇒ ΔSmix = - (0.5)(R)(- 0.693) - (0.5)(R)(- 0.693)
⇒ ΔSmix = (0.3465)(R) + (0.3465)(R)
⇒ ΔSmix = (0.6931)(R)
∴ R = 8.314 J/K.mol
⇒ ΔSmix,m = (0.6931)(8.314 J/K.mol)
⇒ ΔSmix,m = 5.7628 J/K.mol
If the molecular weight of a semiconductor is 27.9 grams/mole and the diamond lattice constant is 0.503 nm, what is the density of the semiconductor in grams/cc ? Two significant digits, fixed point notation.
Explanation:
The given data is as follows.
Mass = 27.9 g/mol
As we know that according to Avogadro's number there are [tex]6.023 \times 10^{26}[/tex] atom present in 1 mole. Therefore, weight of 1 atom will be as follows.
1 atoms weight = [tex]\frac{38}{6.023 \times 10^{26}}[/tex]
In a diamond cubic cell, the number of atoms are 8. So, n = 8 for diamond cubic cell.
Therefore, total weight of atoms in a unit cell will be as follows.
= [tex]\frac{8 \times 27.9 g/mol}{6.023 \times 10^{26}}[/tex]
= [tex]37.06 \times 10^{-26}[/tex]
Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.
= [tex]a^{3}[/tex]
= [tex](0.503 \times 10^{-9})^{3}[/tex]
= [tex]0.127 \times 10^{-27} m^{3}[/tex]
Formula to calculate density of diamond cell is as follows.
Density = [tex]\frac{mass}{volume}[/tex]
= [tex]\frac{37.06 \times 10^{-26}}{0.127 \times 10^{-27} m^{3}}[/tex]
= 2918.1 [tex]g/m^{3}[/tex]
or, = 0.0029 g/cc (as 1 [tex]m^{3} = 10^{6} cm^{3}[/tex])
Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.
The vapor pressure of ethanol is 54.68 mm Hg at 25°C. How many grams of estrogen (estradiol), C18H24O2, a nonvolatile, nonelectrolyte (MW = 272.4 g/mol), must be added to 239.0 grams of ethanol to reduce the vapor pressure to 54.11 mm Hg ?
Answer: The mass of estrogen that must be added is 2.83 grams
Explanation:
The equation used to calculate relative lowering of vapor pressure follows:
[tex]\frac{p^o-p_s}{p^o}=i\times \chi_{solute}[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex] = relative lowering in vapor pressure
i = Van't Hoff factor = 1 (for non electrolytes)
[tex]\chi_{solute}[/tex] = mole fraction of solute = ?
[tex]p^o[/tex] = vapor pressure of pure ethanol = 54.68 mmHg
[tex]p_s[/tex] = vapor pressure of solution = 54.11 mmHg
Putting values in above equation, we get:
[tex]\frac{54.68-54.11}{54.68}=1\times \chi_{\text{estrogen}}\\\\\chi_{\text{estrogen}}=0.0104[/tex]
This means that 0.0104 moles of estrogen are present in the solution
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of estrogen = 0.0104 moles
Molar mass of estrogen = 272.4 g/mol
Putting values in above equation, we get:
[tex]0.0104mol=\frac{\text{Mass of estrogen}}{272.4g/mol}\\\\\text{Mass of estrogen}=(0.0104mol\times 272.4g/mol)=2.83g[/tex]
Hence, the mass of estrogen that must be added is 2.83 grams
In a typical fireworks device, the heat of the reaction between a strong oxidizing agent, such as KClO₄, and an organic compound excites certain salts, which emit specific colors. Strontium salts have an intense emission at 641 nm, and barium salts have one at 493 nm. (a) What colors do these emissions produce? (b) What is the energy (in kJ) of these emissions for 5.00 g each of the chloride salts of Sr and Ba? (Assume that all the heat released is converted to light emitted.)
Answer:
a) The wavelength 641nm of strontium emits a red color in visible spectrum of strontium saltsThe wavelength 493nm of Barium emits a green color in visible spectrum of barium salts.Explanation:
The detailed and step by step calculation is as shown in the attachment.
To enhance glycogen storage after exercise, an athlete weighing 175 lb should consume how many grams of carbohydrate every hour for 4 hours postexercise?
Answer:
Explanation:hiii
80 to 95 grams of carbohydrate should be consumed every hour for 4 hours post-exercise.
What is a carbohydrate?Carbohydrates are biomolecules that are made up of carbon, hydrogen, and oxygen atoms.
Examples of carbohydrates are starch, sugar, fiber.
Carbohydrate is the main component of our food which gives us energy.
If an athlete weighs 175 lb and has to exercise every four hours.
He will need a regular amount of carbohydrates to get energy.
Thus, the amount of carbohydrate required is 80 to 95 grams.
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200.0 mL of 0.200 M HCl is titrated with 0.050 M NaOH. What is the pH after the addition of 100. mL of the NaOH solution
Answer : The pH of the solution is, 0.932
Explanation :
First we have to calculate the moles of HCl and NaOH.
[tex]\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.200mole/L\times 0.200L=0.040mole[/tex]
[tex]\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.050mole/L\times 0.100L=0.0050mole[/tex]
The balanced chemical reaction will be,
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
From the balanced reaction we conclude that,
As, 1 mole of NaOH neutralizes by 1 mole of HCl
So, 0.0050 mole of NaOH neutralizes by 0.0050 mole of HCl
Thus, the number of neutralized moles = 0.0050 mole
Remaining moles of HCl = 0.040 - 0.0050 = 0.035 moles
Total volume of solution = 200.0 mL + 100.0 mL = 300.0 mL = 0.300 L
Now we have to calculate the concentration of HCl(acid).
[tex]Concentration=\frac{Moles}{Volume}=\frac{0.035mol}{0.300L}=0.117M[/tex]
As we know that, 1 mole of HCl dissociates to give 1 mole of hydrogen ion and 1 mole of chloride ion.
So, concentration of [tex]H^+[/tex] = 0.117 M
Now we have to calculate the pH of solution.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (0.117)[/tex]
[tex]pH=0.932[/tex]
Thus, the pH of the solution is, 0.932
"The pH after the addition of 100.0 mL of 0.050 M NaOH to 200.0 mL of 0.200 M HCl is approximately 2.68.
To find the pH, we first need to determine the moles of HCl and NaOH involved in the reaction:
Moles of HCl = volume (L) — concentration (mol/L)
Moles of HCl = 0.200 L — 0.200 mol/L = 0.0400 mol
Moles of NaOH = volume (L) — concentration (mol/L)
Moles of NaOH = 0.100 L — 0.050 mol/L = 0.0050 mol
The balanced equation for the reaction between HCl and NaOH is:
[tex]\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \][/tex]
From the equation, we see that the reaction between HCl and NaOH is 1:1. Therefore, the moles of NaOH added will neutralize an equal number of moles of HCl.
Moles of HCl remaining after neutralization = moles of HCl initially - moles of NaOH added
Moles of HCl remaining = 0.0400 mol - 0.0050 mol = 0.0350 mol
The new concentration of HCl after the addition of NaOH is:
[tex]\[ \text{Concentration of HCl} = \frac{\text{moles of HCl remaining}}{\text{total volume}} \] \[ \text{Concentration of HCl} = \frac{0.0350 \text{ mol}}{0.200 \text{ L} + 0.100 \text{ L}} = \frac{0.0350 \text{ mol}}{0.300 \text{ L}} \] \[ \text{Concentration of HCl} = 0.1167 \text{ M} \][/tex]
Since HCl is a strong acid, it dissociates completely in water. Therefore, the concentration of HCl is equal to the concentration of H+ ions in the solution.
[tex]\[ \text{pH} = -\log[\text{H}^+] \] \[ \text{pH} = -\log(0.1167) \] \[ \text{pH} \approx 2.68 \][/tex]
Thus, the pH of the solution after the addition of 100.0 mL of 0.050 M NaOH is approximately 2.68."
1.35 Draw structures for all constitutional isomers with the following molecular formulas: (a) C6H14 (b) C2H5Cl (c) C2H4Cl2 (d ) C2H3Cl3
I hope this will help ;)
The structures for all constitutional isomers with the following molecular formulas a) CH₃-CH₃-CH₃-CH₃-C₂H₂ b) CH₃-CH₂-Cl c) CH₃-CH₂-Cl d) CH₂-CH-Cl .
What are isomers?The isomers are those compounds which contain same molecular formula but different molecular structures but the physical and chemical properties will be same like melting and boiling points.
The isomers for the formula will be,
(a) C6H14 = CH₃-CH₃-CH₃-CH₃-C₂H₂
(b) C2H5Cl = CH₃-CH₂-Cl
(c) C2H4Cl2 = CH₃-CH₂-Cl
(d ) C2H3Cl3 = CH₂-CH-Cl .
The compounds are same in molecular formula but the representation of structure is different.
Therefore, a) CH₃-CH₃-CH₃-CH₃-C₂H₂ b) CH₃-CH₂-Cl c) CH₃-CH₂-Cl d) CH₂-CH-Cl . structures for all constitutional isomers with the following molecular formulas.
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