All listed are common goals about energy in chemistry, including producing, conserving, storing, and using energy. However, these aren't the inherent goals of chemistry, which is more focused on the study of substances and their transformations.
Explanation:In the context of energy within the field of chemistry, all of the options you've listed are typical goals: finding ways to produce energy, conserve energy, store energy, and use energy. Chemistry plays a crucial role in these areas, such as in the development of new energy sources or improving energy efficiency. However, these aren't necessarily inherent goals of chemistry. The inherent goal of chemistry is to study the properties, composition, and structure of substances, and the transformations they undergo.
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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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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.
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 ...
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!
Compute the radius r of an impurity atom that will just fit into an FCC octahedral site in terms of the atomic radius R of the host atom, and compute the radius r of an impurity atom that will just fit into an BCC tetrahedral site in terms of the atomic radius R.
Answer:
FCC: r = 0.414R
BCC: r = 0.291R
Explanation:
For an FCC unit cell, the interstitial site is located at the middle of the edge. An atom that can occupy the interstitial site will have a diameter of 2*r. And we know that:
2*r = a - 2*R equation (1.0)
a = [tex]2*\sqrt{2}*R[/tex]
Therefore, substituting the expression for 'a' in equation (1.0)
2*r = [tex]2*\sqrt{2}*R[/tex] - 2*R
r = R*([tex]2\sqrt{2} - 2[/tex])/2 = 0.414R
For a BCC unit cell, there is a right-angle triangle formed by 3 arrows. Using the triangle, we have:
[tex]\frac{a^{2} }{2} +\frac{a^{2} }{4} = (R+r)^{2}[/tex] equation (2.0)
a = [tex]\frac{4R}{\sqrt{3} }[/tex]
replacing the expression for a in equation (2.0), we have:
[tex]\frac{4R^{2} }{2\sqrt{3} } + \frac{4R^{2} }{4\sqrt{3} } = R^{2} + 2Rr + r^{2}[/tex]
Further simplification and rearrangement, the expression above is simplified to:
[tex]r^{2} + 2Rr - 0.667R^{2} = 0[/tex]
Solving the above quadratic equation, we have:
[tex]r = \frac{-2R - 2.582R}{2}or\frac{-2R + 2.582R}{2}[/tex]
r = - 2.291R or 0.291 R
Since the value of r can only be positive, the correct answer is r = 0.291R
To find the radius of an impurity atom in a FCC octahedral site, use the length of the face diagonal and the atomic radius of the host atom. For a BCC tetrahedral site, consider the relationship between the cation and anion radii.
Explanation:In an FCC structure, the radius of an impurity atom that will just fit into an octahedral site can be calculated using the length of the face diagonal and the atomic radius of the host atom. The length of the diagonal is equal to four times the host atom radius, so we can use this information to find the radius of the impurity atom.
For a BCC structure, the radius of an impurity atom that will just fit into a tetrahedral site can be calculated by considering the relationship between the cation and anion radii. The cation radius is typically a certain percentage of the anion radius, and this information can be used to determine the radius of the impurity atom.
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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
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.
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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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.
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.
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.
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
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
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.
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.
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.
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.
Place the following in order of decreasing X-A-X bond angle, where A represents the central atom and X represents the outer atoms in each molecule.CS2 CF4 SCl2CS2 > SCl2 > CF4CF4 > SCl2 > CS2CF4 > CS2 > SCl2CS2 > CF4 > SCl2SCl2 > CF4 > CS2Step by Step
Answer:
[tex]CS_{2}[/tex] > [tex]CF_{4}[/tex] > [tex]SCl_{2}[/tex]
Explanation:
The X-A-X bond angle means the angle between the surrounding 'X' atoms and the central 'A' atom. The compound [tex]CS_{2}[/tex] has two bond pairs and it is linear in shape. Its bond angle is 180 degrees. The compound [tex]CF_{4}[/tex] has four bond pairs and it is tetrahedral in shape. Its bond angle is 109.5 degrees. The compound [tex]SCl_{2}[/tex] has a bond angle of approximately 109.5 degrees. Therefore the decreasing order of bond angle is:
[tex]CS_{2}[/tex] > [tex]CF_{4}[/tex] > [tex]SCl_{2}[/tex]
The correct order of decreasing X-A-X bond angle is CS2 > CF4 >SCl2.
What is bond angle?The term bond angle refers to the dihedral angle that exists between two atoms that are bonded to the same central atom. Usually, the central atom is the least electronegative atom of the three.
Looking at the compounds involved, we will see that the correct order of decreasing X-A-X bond angle, where A represents the central atom and X represents the outer atoms in each molecule is CS2 > CF4 >SCl2.
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Exactly 1.0 lb Hydrone, an alloy of sodium with lead, yields (at 0.0°C and 1.00 atm) 2.6 ft3 of hydrogen when it is treated with water. All the sodium reacts according to the following reaction: 2 Na 1 2 H O() 8n 2 NaOH(aq) 1 H (g) in alloy 2 2 and the lead does not react with water. Compute the per- centage by mass of sodium in the alloy.
Answer:
The percentage by mass of sodium in the alloy is 33.29%.
Explanation:
Volume of hydrogen gas = [tex]V = 2.6 ft^3=73.6237 L[/tex]
[tex]1 ft^3=28.3168 L[/tex]
Pressure of hydrogen gas = P = 1 atm
Temperature of the gas = T = 0.0°C =273.15 K
Moles of hydrogen gas = n
[tex]PV=nRT[/tex] (Ideal gas)
[tex]n=\frac{PV}{RT}=\frac{1atm \times 73.6237 L}{0.0821 atm L/mol K\times 273.15 K}[/tex]
n = 3.2830 mole
Moles of hydrogen gas = 3.280 mole
[tex]2 Na(s) +2H_2O(l)\rightarrow 2NaOH(aq)+ H_2 (g)[/tex]
According to reaction 1 mole of hydrogen is obtained from 2 moles of sodium.
Then 3.280 moles of hydrogen gas will be obtained from :
[tex]\frac{2}{1}\times 3.280 mol=6.566 mol[/tex]
Mass of 6.566 moles of sodium =
6.566 mol × 23 g/mol = 151.02 g
Mass of hydrone = 1.0 lb = 453.592 g
The percentage by mass of sodium in the alloy:
[tex]\frac{151.02 g}{ 453.592 g}\times 100=33.29\%[/tex]
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]
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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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.
We discussed the different types of intermolecular forces in this lesson. Which type would you expect to find in CO2?
Answer:
Dispersion forces.
Explanation:
CO2 contains dispersion forces, and covalent bonds. It is a linear molecule, and the bond angle of O-C-O is 180 degree. O is more electronegative than C, the C-O contains polar bond with the having negative end pointing towards the O.
CO contains two C-O bonds. They cancel each other out because of the dipoles point in opposite directions. Although, CO2 contains polar bonds, it is known as a nonpolar molecule. So, the only intramolecular forces which CO2 having are London dispersion forces.
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.
Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN3) into solid sodium and gaseous dinitrogen. Suppose 22.0 L of dinitrogen gas are produced by this reaction, at a temperature of 11.0 C and pressure of exactly 1 atm. Calculate the mass of sodium azide that must have reacted.
Answer:
39g
Explanation:
Details of the solution is shown below. From the information provided regarding the N2 produced, we could calculate the amount of N2 produced and use that to find the mass of sodium azide reacted.
The balanced chemical equation for the decomposition of solid sodium azide (NaN3) is 2 NaN3(s) = 2 Na(s) + 3 N2(g). Using the ideal gas law, we can calculate the number of moles of dinitrogen gas produced, which is 1.07 mol. From the balanced equation, we find that 1 mole of NaN3 decomposes to produce 3 moles of N2. Therefore, the mass of sodium azide that reacted is 23.40 g.
The balanced chemical equation for the decomposition of solid sodium azide (NaN3) into solid sodium and gaseous dinitrogen is:
2 NaN3(s) → 2 Na(s) + 3 N2(g)
Given that 22.0 L of dinitrogen gas are produced by this reaction at a temperature of 11.0°C and a pressure of exactly 1 atm, we can use the ideal gas law to calculate the number of moles of dinitrogen gas produced:
n = PV / RT = (1 atm)(22.0 L) / (0.0821 atm·L/mol·K)(11.0°C + 273.15 K) = 1.07 mol
From the balanced chemical equation, we can see that 1 mole of NaN3 decomposes to produce 3 moles of N2. Therefore, the number of moles of NaN3 that reacted is 1.07 / 3 = 0.36 mol.
To calculate the mass of sodium azide that must have reacted, we can use the molar mass of NaN3 which is 65.01 g/mol:
mass = moles × molar mass = 0.36 mol × 65.01 g/mol = 23.40 g
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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
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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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.
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