Answer:
CH3(CH2)8CH3 > CH3CH2CH2CH3 > CH3CH3 (Option f)
Explanation:
Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones. In a larger atom or molecule, the valence electrons are, on average, farther from the nuclei than in a smaller atom or molecule. They are less tightly held and can more easily form temporary dipoles.
CH3CH3 has a molar mass of 30.07 g/mol
CH3(CH2)8CH3 has a molar mass of 142.28 g/mol
CH3CH2CH2CH3 has a molar mass of 58.12 g/mol
CH3(CH2)8CH3 > CH3CH2CH2CH3 > CH3CH3 (Option f)
The compounds should be ordered by increasing strength of their London dispersion forces, which relate to their molecular size. The correct order is CH3CH3 < CH3CH2CH2CH3 < CH3(CH2)8CH3, reflecting the increasing number of electrons and molecule size from ethane, to butane, to decane. Option e) is correct.
When placing the compounds CH3CH3, CH3(CH2)8CH3, and CH3CH2CH2CH3 in order of increasing strength of intermolecular forces (IMFs), it is important to consider the types of intermolecular forces present and the size of the molecules. All three compounds are nonpolar and primarily exhibit London dispersion forces. Because dispersion forces increase with the number of electrons and, thus, with the size of the molecule, the compound with the longest carbon chain will have the strongest intermolecular forces.
Thus, the correct ordering from weakest to strongest intermolecular forces is:
CH3CH3 (Ethane) - smallest molecule with the fewest electrons and hydrocarbonsCH3CH2CH2CH3 (Butane) - larger molecule with more electrons and hydrocarbons than ethaneCH3(CH2)8CH3 (Decane) - largest molecule with the most electrons and hydrocarbons, thus the strongest dispersion forcesTherefore, the correct answer is e. CH3CH3 < CH3CH2CH2CH3 < CH3(CH2)8CH3.
Concerning the 10.0 mL of 0.50 M NaCl to 100 mL of solution:
Does dilution change the concentration?
Yes
No
Answer:
Yes
Explanation:
Before dilution:
Volume of NaCl = 10 mL
Concentration of NaCl = 0.5 M
Number of moles = Molarity*Volume = 0.5*10 = 5 millimoles.
Note that number of moles of NaCl does not change on dilution as we are only adding water.
After dilution:
Volume of NaCl = 100 mL
Number of moles = 5 millimoles (no change)
New Concentration = Number of moles per volume in litres = [tex]\frac{5\times10^{-3}}{100\times10^{-3}}[/tex]= 0.05 Molar
Hence the concentration became one-tenth of the initial concentration after dilution.
Final answer:
Yes, dilution does change the concentration of a solution. By diluting a 0.50 M NaCl solution to a total volume of 100 mL, the concentration is reduced to 0.050 M, because the same amount of NaCl is spread out in a larger volume of water.
Explanation:
Concerning dilution, yes, it does change the concentration of the solution. When you add more solvent to a solution, the concentration of the solute decreases. For example, if you have a 0.50 M NaCl solution and add enough water to make the total volume 100 mL, the concentration of NaCl changes because the same amount of solute (NaCl) is now dispersed in a greater volume of solvent (water).
Step-by-step Calculation:
Determine the initial amount of NaCl in moles by multiplying the initial volume by the concentration: Moles of NaCl = 0.50 M × 0.010 L = 0.005 moles.
Keep in mind that the amount of NaCl does not change during dilution.
Calculate the final concentration by dividing the moles of NaCl by the final volume of the solution after dilution: Final concentration = 0.005 moles / 0.100 L = 0.050 M NaCl.
Follow the Carbon in Citric Acid Cycle: 1. How many cycles does it take to convert Acetyl CoA to CO2
Each round of Citric Acid Cycle produce 2 molecules of carbon dioxide. So one cycle of TCA cycle is enough to convert Acetyl CoA to carbon dioxide.
Explanation:Citric Acid Cycle or Kreb's Cycle or the TCA cycle is the 1st dedicated step towards the aerobic respiration. The end product of glycolysis is Pyruvate which is a three carbon compound. It's acted upon by Pyruvate Decarboxylase to produce a 2 carbon compound Acetyl CoA and a molecule of carbon dioxide. This Acetyl CoA now reacts with oxaloacetate to produce Citric Acid which is the 1st step of Citric Acid Cycle. This now produce several intermediates and a lot of reduced electron carriers along with 2 molecules of carbon dioxide and ends up being oxaloacetate again. So one cycle of Citric Acid Cycle is necessary to convert Acetyl CoA to CO2.
For the reaction X2 + Y + Z -> XY + XZ it is found that doubling the concentration of X2 doubles the reaction rate. triplingthe concentration on triples the rate, and doubling the concentration of 2 has no effect, (a) What is the rate law for thisreaction? (b) Why is it that the change in the concentration of 2 has no effect on the rate? (c) Suggest a mechanism for thereaction that is consistent with the rate law.
Answer:
(a) R=k[X2][Y] (b) reaction is zero order wrt Z (c) X2 + Y --- XY + X (slow step)
X + Z --- XZ (fast step)
Explanation:
(a) Suppose the reaction rate R with respect to a component X2 with concentration [X2] is generally expressed as follows:
R = k [X2]^n (1)
Where k is the rate constant and n is the order of reaction with respect to X2.
When the reaction rate is found to double by doubling the concentration of X2, the following equation could be developed:
2R = k *(2[X2])^n (2)
Dividing equation (2) by (1) yields
2R/R= k *(2[X2])^n / k *[X2]^n
2=2^n , n = 1
Thus, the reaction is first order with respect to X2.
In the same manner,
Tripling the concentration of Y triples the rate,
R = k [Y]^m (3)
Where k is the rate constant and m is the order of reaction with respect to Y.
When the reaction rate is found to triple by tripling the concentration of Y, the following equation could be developed:
2R = k *(3[Y])^m (4)
Dividing equation (4) by (3) yields
3R/R= k *(3[Y])^m / k *[Y]^m
3=3^m , m = 1
Thus, the reaction is first order with respect to Y.
Therefore,the rate law for this reaction is R = k [X2] [Y]
(b)
Provided that doubling the concentration of Z has no effect, we perform similar analysis as above with p representing order of reaction wrt Z:
R = k [Z]^p (5)
R = k (2*[Z]^p (6)
R/R= k *(2[Z])^p / k *[Z]^p
1=2^p , p = 0
Thus, the reaction is zero order wrt Z. This explains why the change in concentration has no effect on the rate.
(c) A suggested mechanism for the reaction that is consistent with the rate law will therefore be
X2 + Y ---- XY + X (slow step)
X + Z ---- XZ (fast step)
The rate law is determined by the slow step, and this is very consistent with the experimentally observed data.
Besides, addition of the two step yields the overall reaction, and both steps are reasonable.
Final answer:
The rate law is Rate = k[X2][Y] indicating first order dependencies on X2 and Y. Z has no effect on the rate, suggesting it's not involved in the rate-determining step, which implies a reaction mechanism with an initial slow step not involving Z.
Explanation:
Rate Law and Reaction Mechanism
Based on the given information, we can deduce the rate law for the reaction. As the rate doubles when the concentration of X2 is doubled, and it triples when the concentration of Y is tripled, this suggests first order dependencies with respect to X2 and Y. There is no effect on the rate when the concentration of Z is doubled, which means Z is zero order in the rate law. Therefore, the rate law can be expressed as:
Rate = k[X2][Y]
The change in concentration of Z has no effect on the rate because it is not involved in the rate-determining step of the mechanism. The reaction mechanism likely involves a slow initial step that does not include Z, followed by a rapid step that includes it. Thus, the overall rate of the reaction is controlled by the first, slower step in the mechanism.
Select the set of quantum numbers that represents each electron in a ground‑state Be atom. n = 2 , ???? = 0 , m ???? = 0 , m s = − 1 2 n = 1 , ???? = 0 , m ???? = 0 , m s = + 1 2 n = 2 , ???? = 0 , m ???? = 0 , m s = + 1 2 n = 2 , ???? = 1 , m ???? = 0 , m s = − 1 2 n = 1 , ???? = 0 , m ???? = − 1 , m s = + 1 2 n = 1 , ???? = 0 , m ???? = 0 , m s = − 1 2 n = 2 , ???? = 0 , m ???? = 0 , m s = − 1 n = 2 , ???? = 1 , m ???? = 1 , m s = + 1 2
Answer:
n: 1, ℓ: 0, ml: 0, ms:+1/2
n: 1, ℓ: 0, ml: 0, ms:-1/2
n: 2, ℓ: 0, ml: 0, ms:+1/2
n: 2, ℓ: 0, ml: 0, ms:-1/2
Explanation:
Beryllium has 4 electrons and its electron configuration is 1s² 2s².
The principal quantum number (n) describes the level of energy. Then, the first two electrons have n = 1, and the second 2 electrons have n = 2.
The azimuthal number (ℓ) describes the subshell of energy. All the 4 electrons are in s subshells, which correspond to ℓ = 0.
The magnetic quantum number (ml) describes the orbital of the subshell. The s subshell has only 1 s orbital, so the only possible value for ml is 0.
The spin quantum number (ms) describes the spin of the electron and can take 2 values: +1/2 or -1/2.
Considering these rules, the quantum numbers for these 4 electrons are:
n: 1, ℓ: 0, ml: 0, ms:+1/2
n: 1, ℓ: 0, ml: 0, ms:-1/2
n: 2, ℓ: 0, ml: 0, ms:+1/2
n: 2, ℓ: 0, ml: 0, ms:-1/2
Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an electrolytic cell. How many hours will it take to plate 15.0 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A ? Assume the copper in the electrolytic solution is present as Cu 2 + .
Final answer:
To electroplate 15.0 kg of copper onto the cathode using 34.5 A, first calculate the moles of copper needed then convert this to the required charge using Faraday's constant. Finally, divide the total charge by the current to find the time in seconds and convert to hours, resulting in approximately 367 hours.
Explanation:
To calculate the time required to electroplate 15.0 kg of copper using a current of 34.5 A, we need to use Faraday's laws of electrolysis. First, we need to determine the number of moles of copper to be plated. The molar mass of copper is approximately 63.55 g/mol, so:
15,000 g / 63.55 g/mol = 236.025 mol Cu
Since each copper ion (Cu2+) requires two electrons to be reduced to copper metal, the number of moles of electrons needed is twice the number of moles of copper:
2 × 236.025 mol = 472.05 mol e-
Each mole of electrons corresponds to a charge of 96,485 coulombs (Faraday's constant), so:
472.05 mol e- × 96,485 C/mol = 45,562,240.25 C
Now, we can calculate the time required to deliver this charge at a rate of 34.5 A (since 1 A = 1 C/s), using:
Time (s) = Total Charge (C) / Current (A)
Time (s) = 45,562,240.25 C / 34.5 A = 1,320,643 s
Converting seconds to hours:
1,320,643 s / (60 s/min) / (60 min/h) ≈ 367 h
Therefore, it will take approximately 367 hours to electroplate 15.0 kg of copper onto the cathode with a constant current of 34.5 A.
alculate the percent by mass (percent composition) of hydrogen in methane (CH4). Round your answer to the nearest whole number.
Answer:
25
Explanation:
In one mole of methane [tex](CH_{4})[/tex] there are 4 moles of hydrogen and one mole of carbon atom.
Mass of 1 mole of hydrogen atom = 1 g
Mass of 4 moles of hydrogen atom = 4 g
Mass of 1 mole of carbon atom = 12 g
Mass of 1 mole of methane = 12+4 = 16 g
Mass percent of hydrogen in methane = [tex]\frac{mass\ of\ 4\ moles\ of\ hydrogen\ atom}{mass\ of\ 1\ mole\ of\ methane}[/tex]
[tex]=\frac{4}{16}\times100=25[/tex]
Answer:25%
Explanation:
The total molecular mass of methane (CH4) = 12+4 =16
Hydrogen has a total mass of 4 out of the 16. Now to calculate the percentage of hydrogen there, we have (4/16) x 100 = 25
A solid sample of Rb2SO3 weighing 6.24 g reacts with 1.38 L gaseous HBr, measured at 75°C and 0.953 atm pressure. The solid RbBr, extracted from the reaction mixture and purified, has a mass of 7.32 g.
(a) What is the limiting reactant?
(b) What is the theoretical yield of RbBr, assuming com- plete reaction?
(c) What is the actual percentage yield of product?
Answer:
(a) HBr;
(b) 7.61 g;
(c) 96.2 %
Explanation:
Firstly, write the balanced chemical equation:
[tex]Rb_2SO_3 (aq) + 2 HBr (aq)\rightarrow 2 RbBr (aq) + SO_2 (g) + H_2O (l)[/tex]
(a) Find moles of each reactant dividing the mass by the molar mass of rubidium sulfite, then applying the ideal gas law for HBr:
[tex]n_{Rb_2SO_3}=\frac{6.24 g}{251.00 g/mol} = 0.02486 mol[/tex]
[tex]pV_{HBr}=n_{HBr}RT[/tex]
[tex]\therefore n_{HBr} = \frac{pV_{HBr}}{RT} = \frac{0.953 atm\cdot 1.38 L}{0.08206 \frac{L atm}{mol K}\cdot 348.15 K} = 0.04603 mol[/tex]
Find the limiting reactant by dividing each moles by the stoichiometric coefficients and comparing the two numbers:
[tex]eq._{Rb_2SO_3} = \frac{0.02486 mol}{1} = 0.02486 mol[/tex]
[tex]eq._{HBr} = \frac{0.04603 mol}{2} = 0.02302 mol[/tex]
That said, the equivalent of HBr is lower, so it's the limiting reactant.
(b) According to the balanced equation, the moles of HBr are equal to the moles of RbBr, so moles of RbBr theoretically are equal to:
[tex]n_{RbBr} = 0.04603 mol[/tex]
Using the molar mass of RbBr, convert this into mass:
[tex]m_{RbBr} = 0.04603 mol\cdot 165.372 g/mol = 7.61 g[/tex]
(c) To find the percent yield, divide the actual mass produced by the theoretical mass calculated in (b) and multiply by 100 %:
[tex]\%_{yield} =\frac{7.32 g}{7.61 g}\cdot 100\% = 96.2 \%[/tex]
When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of iron(III) nitrate are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for iron(III) nitrate in . Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.
Answer:
i = 3,5
Explanation:
There are missing the following values:
132 g Alanine
1150g of X
4,4°C the first freezing point dercreasing
132g of Iron(III) nitrate
5,6°C the second freezing point decreasing.
The freezing point depression is a colligative property that describes the decrease of the freezing point of a solvent on the addition of a non-volatile solute.
The formula is:
ΔT = i kf mb
Where ΔT is freezing point decreasing, i is Van't Hoff factor, kf, is cryoscopic constant and mb is molality of solution.
For alanine Van't Hoff factor is 1 (Ratio between particles in dissolution and before dissolution), molality is:
132g×(1mol/89,09g) = 1,48mol / 1,150kg = 1,29 mol/kg
Replacing:
4,4K = 1 kf 1,29mol/kg
kf = 3,41 K·kg/mol
Now, for Iron(III) nitrate molality is:
132g×(1mol/241,86g) = 0,546mol / 1,150kg = 0,475 mol/kg
Replacing:
5,6K = i×3,41 K·kg/mol×0,475 mol/kg
i = 3,5
I hope it helps!
The human body obtains 1062 kJ from a candy bar.If this energy were used to vaporize water at 100 ∘C, how much water in liters could be vaporized? (Assume that the density of water is 1.0 g/mL.) The heat of vaporization of water at 100 ∘C is 40.7 kJ/mole.
Answer:
The Answer Of this question is 0.4052 L ...
Dimethyl sulfoxide is an important polar aprotic solvent that can dissolve both polar and nonpolar compounds and is miscible in a wide range of organic solvents as well as water. Because it penetrates the skin very readily, it is sometimes used as a vehicle for topical application of pharmaceuticals.
Draw the structure of dimethyl sulfoxide. Include any nonbonding electrons on sulfur, and minimize formal charges by allowing sulfur to expand its octet.
Answer:
The structure is shown below.
Explanation:
To draw a structure first we need to know its molecular formula, which is C2H6SO for dimethyl sulfoxide. The central atom is sulfur, which is bonded to an oxygen and with two methyl groups (CH3).
Sulfur has 6 electrons in its valence shell, as so oxygen. To complete the octet of oxygen, 2 electrons will be shared by sulfur with it. So, it remains 4 electrons at the central atom. Carbon has 4 electrons in its valence shell, so it needs more 4 to be stable, and is already sharing 3 electrons with the hydrogens, thus, sulfur will share one electron with each one of them.
So, it will remain 2 nonbonding electrons in the central atom. According to the VSPER theory, to minimize formal charges, the structure would be a trigonal pyramid, but, the double bonding with oxygen has a large volume, then the geometry will be trigonal, as shown below.
Two changes are described below. a. Ice melts to form water. b. Sugar cubes dissolve in hot coffee. Which statement is true about the two changes?a. Both are chemical changes because there is a change in the composition of ice and sugar.b. Both are physical changes because there is a change in the physical states of ice and sugar.c. A is a physical change due to a change of state, but B is a chemical change because new molecules are formed.d. A is a chemical change due to a change of state, but B is a physical change because new molecules are formed.
Answer: Option (b) is the correct answer.
Explanation:
A change that does not lead to any difference in chemical composition of a substance is known as a physical change.
For example, shape, size, mass, volume, density, boiling point, etc of a substance are all physical properties.
As ice melts to form water shows that only the state of matter is changing. Hence, it is a physical change. Similarly, sugar cubes dissolve in hot coffee is also a physical change as no new compound has formed.
On the other hand, changes that lead to bring change in chemical composition of a substance is known as a chemical change.
For example, exploding dynamite, rotting cheese etc are all chemical changes.
Thus, we can conclude that both are physical changes because there is a change in the physical states of ice and sugar.
After a polypeptide chain has been synthesized, certain amino acids in the peptide may become modified. For each modified amino acid, identify the standard amino acid from which it is derived. Enter the unabbreviated name of the standard amino acid. The amino acid structure shows the fully ionized form. The side chain is C H 2, C H 2, C H (O H), C H 2, N H 3 (plus). standard amino acid: Identify the modification that yields the first amino acid depicted.
Answer:
The name of the amino acid is lysine.
The number five carbon in lysine is the carbon that is hydroxylated. The modification you ask is when adding hydroxyl group (C-OH bonds). These links are made by an enzyme called hydroxylase, vitamin C acting as a cofactor. This reaction is one of the most fundamental post-translational modifications.
Explanation:
Hydroxyproline is derived from the standard amino acid proline through the addition of a hydroxyl group, and it plays a role in the structure of collagen.
Explanation:The modified amino acid depicted in the question is hydroxyproline, which is derived from the standard amino acid proline.
During the modification process, proline is hydroxylated by adding a hydroxyl group (-OH) to the side chain. This results in the formation of hydroxyproline. Hydroxyproline plays an important role in the structure and stability of collagen, a protein found in connective tissues.
In summary, hydroxyproline is derived from proline through the addition of a hydroxyl group, and it is involved in the structure of collagen.
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Research and draw (on paper) the isomers of another coordination compound that has octahedral geometry.
Answer:
Sulfur hexafluoride SF6
Explanation:
In chemistry, the shape of the compounds in which six ligands (atoms, molecules or ions) are arranged around a central atom or ion, defining the vertices of an octahedron, is called octahedral molecular geometry or Oh. It is a very common structure, and it is very studied for its importance in the coordination chemistry of transition metals. From it, other important molecular geometries are derived by continuous deformation, such as the elongated octahedron, the flat octahedron, the square-based pyramid and the flat square. Indirectly, it is also related to tetrahedral molecular geometry.
The concept of octahedral coordination geometry was developed by Alfred Werner to explain the stoichiometry and isomeries in the coordination compounds. An example of a strictly octahedral compound is SF6 sulfur hexafluoride, but chemists use the term in a lax form, so that it is applied to compounds that are not mathematically octahedra, such as cobalt hexaamine (III).
Isomers are molecules that have the same molecular formula but different structure. It is classified as structural isomers and stereoisomers. Structural isomers differ in the way of joining their atoms and are classified into chain, position and function isomers.
The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H 2 O H2O that can be produced by combining 52.3 g 52.3 g of each reactant? 4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Answer:
35.3124 g is the maximum mass of [tex]H_2O[/tex] that can be produced.
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
For [tex]NH_3[/tex] :-
Mass of [tex]NH_3[/tex] = 52.3 g
Molar mass of [tex]NH_3[/tex] = 17.031 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{52.3\ g}{17.031\ g/mol}[/tex]
[tex]Moles\ of\ NH_3= 3.0709\ mol[/tex]
For [tex]O_2[/tex] :-
Given mass of [tex]O_2[/tex]= 52.3 g
Molar mass of [tex]O_2[/tex] = 31.9898 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{52.3\ g}{31.9898\ g/mol}[/tex]
[tex]Moles\ of\ O_2=1.6349\ mol[/tex]
According to the given reaction:
[tex]4NH_3+5O_2\rightarrow 4NO_4+6H_2O[/tex]
4 moles of [tex]NH_3[/tex] reacts with 5 moles of [tex]O_2[/tex]
1 mole of [tex]NH_3[/tex] reacts with 5/4 moles of [tex]O_2[/tex]
Also,
3.0709 moles of [tex]NH_3[/tex] reacts with [tex]\frac{5}{4}\times 3.0709[/tex] moles of [tex]O_2[/tex]
Moles of [tex]O_2[/tex] = 3.8386 moles
Available moles of [tex]O_2[/tex] = 1.6349 moles
Limiting reagent is the one which is present in small amount. Thus, [tex]O_2[/tex] is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
5 moles of [tex]O_2[/tex] on reaction forms 6 moles of [tex]H_2O[/tex]
1 mole of [tex]O_2[/tex] on reaction forms 6/5 moles of [tex]H_2O[/tex]
Thus,
1.6349 mole of [tex]O_2[/tex] on reaction forms [tex]\frac{6}{5}\times 1.6349[/tex] moles of [tex]H_2O[/tex]
Moles of [tex]H_2O[/tex] = 1.9618 moles
Molar mass of [tex]H_2O[/tex] = 18 g/mol
Mass of sodium sulfate = Moles × Molar mass = 1.9618 × 18 g = 35.3124 g
35.3124 g is the maximum mass of [tex]H_2O[/tex] that can be produced.
Answer:
35.3 g
Explanation:
From the balanced equation given we can say:
4 moles of NH3 reacts with 5 moles of O2 to give 6 moles of H2O.
4*17 g of NH3 reacts with 5*32 g of O2 to give 6*18 g of H2O.
68 g of NH3 reacts with 160 g of O2 to give 108 g of H2O.
Here the limiting reagent is O2 and excess reagent is NH3.
52.3 g of O2 will react with [tex]\frac{68}{160}\times52.3=22.23\ g\ of\ NH_{3}[/tex] to give :
[tex]\frac{108}{160}*52.3=35.3\ g\ of\ H_{2}O[/tex]
Hence the maximum mass of H2O that can be produced by 52.3 g of reactants is 35.3 g.
Express the equilibrium constant for the following reaction.P(g) + 3/2 Cl2(g) ↔ PCl3(g)K = [PCl3]^2/[P]^2[Cl2]^3K = [PCl3]/[P][Cl2]^3/2K = [P][Cl2]^3/2/[PCl3]K = [P]^1/2[Cl2]^1/3/[PCl3]^1/2K = [PCl3]^2/[P]2[Cl2]3
The correct expression for the equilibrium constant (K) is:
K = [PCl₃]² / [P] × [Cl₂](3/2)
The correct expression for the equilibrium constant (K) of the given reaction is:
K = [PCl₃]² / [P] × [Cl₂](3/2)
In this expression:
[PCl₃] represents the molar concentration of PCl₃ (gaseous product).
[P] represents the molar concentration of P (reactant P in its gaseous state).
[Cl₂] represents the molar concentration of Cl₂ (reactant Cl₂ in its gaseous state).
The exponents in the expression correspond to the stoichiometric coefficients in the balanced chemical equation:
P(g) + 3/2 Cl₂(g) ↔ PCl₃(g)
The stoichiometric coefficients of P and Cl₂ are both 1, and the coefficient of PCl₃ is 1. To make the equation balanced, we need to multiply Cl₂ by 3/2 (1.5).
So, the correct expression for the equilibrium constant (K) is:
K = [PCl₃]² / [P] × [Cl₂](3/2)
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A thermos bottle (Dewar vessel) has an evacuated space between its inner and outer walls to diminish the rate of transfer of thermal energy to or from the bottle’s contents. For good insulation, the mean free path of the residual gas (air; average molecular mass 5 29) should be at least 10 times the distance between the inner and outer walls, which is about 1.0 cm. What should be the maximum residual gas pressure in the evacuated space if T 5 300 K? Take an average diameter of d 5 3.1 × 10210 m for the molecules in the air.
There are quite a bunch of typo errors in the question; here is the correct question below:
A thermos bottle (Dewar vessel) has an evacuated space between its inner and outer walls to diminish the rate of transfer of thermal energy to or from the bottle’s contents. For good insulation, the mean free path of the residual gas (air; average molecular mass = 29) should be at least 10 times the distance between the inner and outer walls, which is about 1.0 cm. What should be the maximum residual gas pressure in the evacuated space if T = 300 K? Take an average diameter of d = 3.1 × 10⁻¹⁰ m for the molecules in the air.
Answer:
9.57 × 10⁻⁷ atm
Explanation:
The mean free path ( λ ) can be illustrated by the equation:
λ = [tex]\frac{1}{\sqrt{2} \pi d{^2}N/V }[/tex] ---------- (1)
N/V = [tex]\frac{N_AP}{RT}[/tex] ------------- (2)
From the above relation, we can deduce that;
P= [tex]\frac{RT}{\sqrt{2}\pi d{^2}N{_A}λ }[/tex] -------------(3)
let I= λ
From the above equations;
d= diameter of the atom
[tex]{N_A}[/tex] = avogadro's constant
P= pressure
R= universal rate constant which is given by 0.08206 L atm mol⁻¹ k⁻¹
T= temperature
From the question, we are given that the mean free path of the residual air molecules ( d = 3.1 × 10⁻¹⁰ m) is equal to 10cm = 0.1m
Therefore, we can determine the pressure using equation (3)
i.e
P= [tex]\frac{RT}{\sqrt{2}\pi d{^2}N{_A}λ }[/tex]
= [tex]\frac{(8.314J{K^-^1)(300K)}}{\sqrt{2}(3.14)(3.1*10^{-10}m)^2(6.022*10^{23}mol^{-1})(0.1m) }[/tex]
=97.06 × 10⁻³ Pa × [tex]\frac{1atm}{1.01325*10^5Pa}[/tex]
=9.57 × 10⁻⁷ atm
Therefore, the maximum residual gas pressure in the calculated space is; 9.57 × 10⁻⁷ atm
After the stannous chloride solution has been prepared, a few pieces of tin metal, Sn, are dropped into the bottle, to prevent oxidation by air.
Determine the standard cell potential for the following reaction: Sn(s)+Sn4+(aq)→2Sn2+(aq)
Express your answer with the appropriate units.
Answer:
E° = 0.29 V
Explanation:
Let's consider the following redox reaction.
Sn(s) + Sn⁴⁺(aq) → 2 Sn²⁺(aq)
We can identify both half-reactions:
Reduction (cathode): Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Oxidation (anode): Sn(s) → Sn²⁺(aq) + 2 e⁻ Ered = -0.14 V
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.15 V - (-0.14 V) = 0.29 V
Dinitrogen tetroxide partially decomposes according to the following equilibrium: N2O4 (g) → 2 NO2 (g)
A 1.000-L flask is charged with 3.00 × 10-2 mol of N2O4. At equilibrium, 2.36 × 10-2 mol of N2O4 remains.
Keq for this reaction is __________.
Answer:
Keq for this reaction is 6.94x10⁻³
Explanation:
The equilibrium equation is this one:
N₂O₄ (g) ⇄ 2NO₂ (g)
Initially we have 0.03 moles from the dinitrogen tetroxide and nothing from the dioxide.
In the reaction, some amount of compound (x) has reacted.
As ratio is 1:2, we have double x in products.
Finally in equilibrium we have:
N₂O₄ (g) ⇄ 2NO₂ (g)
0.03 - x 2x
And we know [N₂O₄] in equilibrium so:
0.03 - x = 0.0236
x = 0.03 - 0.0236 → 6.4x10⁻³
As this is the amount that has reacted, in equilibrium I have produced:
6.4x10⁻³ .2 = 0.0128 moles of NO₂
This is the expression for K,
[NO₂] ² / [N₂O₄]
0.0128² / 0.0236 = 6.94x10⁻³
Answer:
6.94x10-3
Explanation:
Of the atoms listed below, which one will have at least one electron in its d orbital?
A. Mg
B. K
C. S
D. Cr
E. none of the above
Answer:
D. Cr
Explanation:
In order to determine which atom has at least one electron in its d orbital, we have to write their theoretical electron configurations.
₁₂Mg 1s² 2s² 2p⁶ 3s²₁₉K 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹₁₆S 1s² 2s² 2p⁶ 3s² 3p⁴₂₄Cr 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴Cr has 4 electrons in d orbitals. Cr belongs to the d-block in the periodic table.
Which of the following are characteristics of reverse-phase chromatography?I The stationary phase is polar.II The mobile phase is more polar than the stationary phase.III Less polar mobile phase has a lower eluent strength.IV More polar mobile phase has a higher eluent strength.V The stationary phase is nonpolar.
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The correct options would be:
The mobile phase is more polar than the stationary phase
The stationary phase is nonpolar
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The characteristics of reversed phase chromatography are The mobile phase is more polar than the stationary phase and The stationary phase is non polar.
Hence, option II and V are correct.
What is RPC?RPC is Reversed Phase Chromatography. In reversed phase chromatography the stationary phase is non polar [hydrophobic] and the mobile phase is very polar [hydrophilic].
Now lets check all the option one by one:
Option (I): The stationary phase is non polar in reverse phase chromatography.
So it is incorrect option.
Option (II): The mobile phase is more polar than the stationary phase because in polar mobile phase has high affinity towards the polar solute.
So, it is correct option.
Option (III): Less polar mobile phase has higher eluent strength not lower eluent strength.
So, it is incorrect option.
Option (IV): More polar mobile phase has less eluent strength.
So, it is incorrect option.
Option (V): The stationary phase is non polar [hydrophobic] is reverse phase chromatography.
So, it is correct option.
Thus, from above conclusion we can say that the The mobile phase is more polar than the stationary phase and The stationary phase is non polar are the characteristics of reverse phase chromatography.
Hence option II and V are correct.
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What element is being oxidized in the following redox reaction?
MnO4 (aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g)
Answer:
C is the element thats has been oxidized.
Explanation:
MnO₄⁻ (aq) + H₂C₂O₄ (aq) → Mn²⁺ (aq) + CO₂(g)
This is a reaction where the manganese from the permanganate, it's reduced to Mn²⁺.
In the oxalic acid, this are the oxidation states:
H: +1
C: +3
O: -2
In the product side, in CO₂ the oxidation states are:
C: +4
O: -2
Carbon from the oxalate has increased the oxidation state, so it has been oxidized.
At 10 K Cp,m(Hg(s)) = 4.64 J K−1 mol−1. Between 10 K and the melting point of Hg(s), 234.3 K, heat capacity measurements indicate that the entropy increases by 57.74 J K−1 mol−1. The standard enthalpy of fusion of Hg(s) is 2322 J mol−1 at 234.3 K. Between the melting point and 298.0 K, heat capacity measurements indicate that the entropy increases by 6.85 J K−1 mol−1. Determine the Third-Law standard molar entropy of Hg(l) at 298 K.
Answer:
S°m,298K = 85.184 J/Kmol
Explanation:
∴ T = 10 K ⇒ Cp,m(Hg(s)) = 4.64 J/Kmol
∴ 10 K to 234.3 K ⇒ ΔS = 57.74 J/Kmol
∴ T = 234.3 K ⇒ ΔHf = 2322 J/mol
∴ 234.3 K to 298.0 K ⇒ ΔS = 6.85 J/Kmol
⇒ S°m,298K = S°m,0K + ∫CpdT/T(10K) + ΔS(10-234.3) + ΔHf/T(234.3K) + ΔS(234.3-298)
⇒ S°m,298K = 0 + 10.684 J/Kmol + 57.74 J/Kmol + 9.9104 J/Kmol + 6.85 J/kmol
⇒ S°m,298K = 85.184 J/Kmol
To find the Third-Law standard molar entropy of Hg(l) at 298 K, we sum up the entropy changes between 10K and 234.3 K, the entropy change due to fusion, and then the entropy change between the melting point and 298K. The sum gives us a final value of 74.51 J K−1 mol−1.
Explanation:The Third-Law standard molar entropy of Hg(l) at 298 K can be calculated by summing up the entropy changes that occur from 10K to 298K.
First calculate the entropy up to the melting point from 10 K which can be determined using the equation ΔS = ∫(Cp,mdT)/T.
This integral can be approximated as a rectangle from 10K to 234.3 K, hence ΔS₁ = (234.3-10)(4.64 J K−1 mol−1)/10 = 57.74 J K−1 mol−1.
Next, calculate the entropy change associated with the fusion process using the equation, ΔS = ΔH/T, giving ΔS₂ = 2322 J mol−1 / 234.3 K = 9.92 J K−1 mol−1.
Finally, add the entropy increase from the melting point to 298.0 K, which is given as 6.85 J K−1 mol−1.
Summing these values gives the Third-Law standard molar entropy of Hg(l) at 298 K: 57.74 J K−1 mol−1 + 9.92 J K−1 mol−1 + 6.85 J K−1 mol−1 = 74.51 J K−1 mol−1.
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How many grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to cool the coffee and the cup from 95.0°C to 45.0°C ? Assume the coffee has the same thermal properties as water and that the average heat of vaporization is 2340 kJ/kg (560 kcal/g). Neglect heat losses through processes other than evaporation, as well as the change in mass of the coffee as it cools. Do the latter two assumptions cause your answer to be higher or lower than the true answe
Answer:
31.3 g
The answer is higher than the true answer.
Explanation:
By neglecting the heat lost by other processes, the energy conservation states that:
Qcooling + Qevaporate = 0
The cooling process happens without phase change, so the heat can be calculated by:
Qcooling = m*c*ΔT
Where m is the mass, c is the heat capacity (cwater = 4184 J/kg.K), and ΔT is the temperature variation (final - initial).
The evaporate process happen without changing of temperature (pure substance), and the heat can be calculated by:
Qevaporate = m*L
Where m is the mass evaporated and L is the heat of evaporation (2340000 J/kg).
0.350*4184*(45 - 95) + m*2340000 = 0
2340000m = 73220
m = 0.0313 kg
m = 31.3 g
Because of the assumptions made, the real mass is not that was calculated. There'll be changing mass when the coffee is cooling, and there'll be heat loses by other processes because the system is not isolated. Also, the substance is not pure. So, there'll be more factors at the energy equation, thus, the answer is higher than the true answer.
To cool 350 g of coffee in a 100-g glass cup from 95.0°C to 45.0°C, 33.2 grams of coffee must evaporate.
To solve this problem, we first need to calculate the total heat that needs to be removed from the coffee and the cup.
Steps to Calculate:
Calculate the heat loss needed to cool the coffee:Thus, 33.2 grams of coffee must evaporate to cool the coffee and the cup from 95.0°C to 45.0°C. Neglecting other heat losses means this answer is slightly larger than the true answer.
Suppose caffeine has a partition coefficient of 2.5 between dichloromethane and water (2.5/1 = dichloromethane/water). You are given a solution of 10 grams of caffeine dissolved in 100 mL water. You extract your caffeine solution 1 x 100 mL dichloromethane, and evaporate your dichloromethane layer. How much caffeine would you expect to isolate?A) 8.0 grams B) 7.5 grams C) 7.1 grams D) 6.5 grams E) 2.9 grams
Answer: Option (C) is the correct answer.
Explanation:
It is given that partition coefficient between dichloromethane and water is 2.5. Let us assume that "x" grams of caffeine is present in 100 ml.
Hence, find the value of x as follows.
2.5 = [tex]\frac{\frac{x}{100}}{\frac{(10 - x)}{100}}[/tex]
x = 25 - 2.5x
x = 7.14
or, x = 7.1
Therefore, we can conclude that caffeine extracted is 7.1 grams.
Compare and contrast the three different types of friction. Explain one way they are all alike, and at least one way they are different.
Answer:
The correct answer is static, sliding and rolling friction.
Explanation:
There are three different types of friction. Static, sliding and rolling friction occur between solid surfaces. Static friction is strongest, followed by sliding friction and then rolling friction, which is the weakest of the three. The equality of these three types of friction is that they produce heat and make movement difficult. The difference is in their magnitude and in the conditions that produce each type of friction. Static is produced when a body at rest begins to move, sliding is produced when this body is already in motion and rolling is produced when a body rolls on a surface, deforming one or both of them.
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What mass of water can be obtained from 4.0 g of H2 and 16 g of O2?2 H2 + O2 ---> 2 H2O18 g36 g54 g9 g
Answer:
18 g is the mass produced by 4 g of H₂ and 16 g of O₂
Explanation:
The reaction is:
2H₂ + O₂ → 2H₂O
So, let's find out the limiting reactant as we have both data from the reactants.
Mass / Molar mass = moles
4 g/ 2g/m = 2 moles H₂
16g / 32 g/m = 0.5 moles O₂
2 moles of hydrogen react with 1 mol of oxygen, but I have 0.5, so the O₂ is the limiting.
1 mol of O₂ produces 2 mol of water.
0.5 mol of O₂ produce (0.5 .2)/1 = 1 mol of water.
1 mol of water weighs 18 grams.
Answer:
18 grams of [tex]H_2O[/tex]
Explanation:
The balanced equation of the reaction is:
[tex]H_2+\frac{1}{2}O_2 -->H_2O[/tex]
From the balanced equation we can say 1 mole of H2 reacts with 0.5 moles of O2 to give one mole of H2O.
Number of moles of H2 = [tex]\frac{Given\ mass}{Molar\ mass}=\frac{4}{2}=2\ moles[/tex]
Number of moles of O2 = [tex]\frac{Given\ mass}{Molar\ mass}=\frac{16}{32}=0.5\ moles[/tex]
We have 2 moles H2 and 0.5 moles of O2.
Not all H2 reacts because the amount of O2 is limited.
Since only 0.5 moles of O2 is available only 1 mole of H2 reacts according to the balanced equation.
Hence 1 mole of H2O is formed which is 18 grams.
Predict the shape, state the hybridizationof the central atom,and give the ideal bond angle(s) and any expected deviations ineach of the following.
a) BiF52-
Shape:
Hybridization:
Ideal bond angle(s):
Deviation from ideal angle(s):
b) BrO3-
Shape:
Hybridization:
Ideal bond angle(s):
Deviation from ideal angle(s):
c) IF4+
Shape:
Hybridization:
Ideal bond angle(s):
Deviation from ideal angle(s):
Answer:
BiF5
Shape- trigonal bipyramid
Hybridisation-sp3d
Ideal bond angle-120 and 90
BrO3-
Shape- tetrahedral
Hybridisation-sp3
Ideal bond angle 109o28'
IF4+
Shape- trigonal bipyramid
Hybridisation-sp3d
Ideal bond angle-120o and 90o
Explanation:
The shape adopted by a molecule according to VSEPR is such as minimizes repulsion between electron pairs. The shape of a molecule depends on electron pairs present on the outermost shell of the central atom. The bond angles are such that electrons are positioned as far apart in space as possible, given the number of electron pairs present.
BiF52- has a square pyramidal shape, sp3d2 hybridization, with ideal bond angles of 90° and 120°. BrO3- has a trigonal pyramidal shape, sp3 hybridization, with an ideal bond angle of 109.5°. IF4+ has a square planar shape, sp3d2 hybridization, with ideal bond angles of 90° and 180°.
Explanation:The shape, hybridization of the central atom, ideal bond angle(s), and any expected deviations for the molecules are as follows:
BiF52-: The shape is square pyramidal. The hybridization of the central atom, Bi, is sp3d2. The ideal bond angles are 90° and 120°. Due to the presence of lone pairs, there can be deviations from these angles.BrO3-: This molecule has a trigonal pyramidal shape. The hybridization of the central Br atom is sp3. The ideal bond angle is 109.5°. Since there is one lone pair of electrons, deviations from the ideal angle occur.IF4+: The shape of this molecule is square planar. The central atom I is in sp3d2 hybridization. The ideal bond angles are 90° and 180°. As it is a symmetrical molecule with no lone pairs, there aren't deviations from this angle.Learn more about Molecule Structures here:https://brainly.com/question/31544427
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vWould radiation with a wavelength 0.91 nm or a frequency of 5.9×1011 s−1 be detected by an X-ray detector? Would radiation with a wavelength 0.91 or a frequency of 5.9×1011 be detected by an X-ray detector? Only the radiation with a wavelength 0.91 nm can be observed by an X-ray detector. Only the radiation with a frequency of 5.9×1011 s−1 can be observed by an X-ray detector. Both types of radiation can be observed by an X-ray detector. Neither type of radiation can be observed by an X-ray detector.
Answer:
Only the radiation with a wavelength 0.91 nm can be observed by an X-ray detector.
Explanation:
To answer this question we need to consult the ranges in which x rays are in the electromagnetic spectrum:
The X radiation in the electromagnetic spectrum fall in the region of:
frequency: 3 x 10¹⁶ Hz to 3 x 10¹⁹ Hz (1Hz = 1s⁻¹)
wavelengt: 1 pm to 10 nm
Comparing the values in our question,
0.91 nm will be detected
5.9 x 10¹¹ Hz will not be detected.
The radiation with a wavelength 0.91 or a frequency of [tex]5.9*10^{11}[/tex] will have:
A. Only the radiation with a wavelength 0.91 nm can be observed by an X-ray detector.
Electromagnetic spectrum:X-rays are both types of high energy (high frequency) electromagnetic radiation. They are packets of energy that have no charge or mass (weight).
The X radiation in the electromagnetic spectrum fall in the region of:
Frequency : 3 x 10¹⁶ Hz to 3 x 10¹⁹ Hz (1Hz = 1s⁻¹)
Wavelength : 1 pm to 10 nm
On comparing to the values given in the question:
0.91 nm will be detected
5.9 x 10¹¹ Hz will not be detected.
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A chemist must prepare 400 mL of 1.00M of aqueous potassium iodide working solution. He'll do this by pouring out 1.82 mol/L some aqueous potassium iodide stock solution into a graduated cylinder and diluting it with distilled water. Calculate the volume in of the potassium iodide stock solution that the chemist should pour out. Round your answer to significant digits.
Answer:
220mL
Explanation:
The dilution formular was applied to obtain the volume of stock solution required to prepare the desired concentration of solution in the desired volume. Details are found in the image attached.
The following (unbalanced) reaction is one of the steps to producing acid rain [as H2SO4(aq)] from sulfur-containing coal. In a study of this reaction, a flask contains an equilibrium concentration of the three substances SO2(g), O2(g) and SO3(g) held at a constant temperature.
SO2 (g) + O2 (g) <-----> SO3 (g)
The concentrations at equilibrium are found to be: [SO2] = 3.61 x 10-3M [O2] = 6.11 x 10-4 M [SO3] = 1.01 x 10-2 M
(a) Write the Equilibrium Constant expression, Kc for this specific reaction.
(b) Calculate the value of Kc for this temperature.
(c) Would you classify this as a product-favored reaction?
Answer:
a) kc= [SO3 ]/([SO2 ][O2 ])
b) kc= 2.27*10⁶ M⁻¹
v) the reaction is product-favored
Explanation:
for the reaction, the equilibrium constant is
SO2 (g) + O2 (g) <-----> SO3 (g)
he equilibrum constant is
kc= [SO3 ]/([SO2 ]*[O2 ])
replacing values
kc= [SO3 ]/([SO2 ]*[O2 ]) = 1.01*10⁻² M/(3.61*10⁻³M*6.11 x 10⁻⁴ M) = 2.27*10⁶ M⁻¹
since kc>>1 the reaction is product-favored