A current of 620 A for 90 seconds produces 5.209 g of aluminum.
The mass of pure aluminum produced in the Hall-Heroult process can be calculated by using Faraday's laws of electrolysis and the molar mass of aluminum. First, we determine the moles of electrons transferred using the given current and time. Then, using the stoichiometry of the reaction where 4 moles of Al are produced per 12 moles of electrons:
Calculate the charge (Q) passed: Q = current (I) x time (t) = 620 A x 90.0 s = 55800 C
Determine the moles of electrons: Moles of e- = Q / Faraday's constant (F) Assuming the Faraday constant is approximately 96500 C/mol, this gives Moles of e- = 55800 C / 96500 C/mol = 0.5784 mol
Calculate the moles of aluminum: Since 3 moles of electrons yield 2 moles of Al, 0.5784 mol of e- will yield (4/12) x 0.5784 mol of Al = 0.1931 mol of Al.
Compute the mass of aluminum produced: Mass of Al = moles of Al x molar mass of Al = 0.1931 mol x 26.98 g/mol = 5.209 g
The mass of pure aluminum produced is therefore 5.209 g, assuming 100% efficiency and no other side reactions.
To find the mass of pure aluminum produced, we calculate the charge passed, find the moles of electrons transferred, determine the moles of aluminum from the moles of electrons, and use the molar mass of aluminum to find that approximately 5.20 grams of aluminum is produced.
To calculate the mass of pure aluminum produced in the Hall-Heroult process when 620 A of current is passed through a Hall-Heroult cell for 90.0 seconds, we use Faraday's laws of electrolysis. The reduction of aluminum oxide to aluminum involves the transfer of three moles of electrons per mole of aluminum according to the equation:
Al₂O₃ + 3e⁻ → 2Al + 1.5O₂
Firstly, we calculate the amount of charge passed using the formula Q = It, where Q is the charge in coulombs, I is the current in amperes, and t is time in seconds. Then, using Faraday's constant (96485 C/mol), we determine the moles of electrons transferred. Since we need 3 moles of electrons to produce 1 mole of aluminum, we find the moles of aluminum formed by dividing the moles of electrons by 3. Finally, we calculate the mass using the molar mass of aluminum (26.98 g/mol).
Q = 620 A * 90.0 s = 55800 C
Moles of electrons = Q/F = 55800 C / 96485 C/mol ≈ 0.5785 mol
Moles of aluminum = Moles of electrons / 3 ≈ 0.1928 mol
Mass of aluminum = Moles of aluminum * Molar mass of Al
Mass of aluminum ≈ 0.1928 mol * 26.98 g/mol ≈ 5.20 g
The mass of pure aluminum produced is approximately 5.20 grams.
In the reaction Pb + Cu2+ → Pb2+ + Cu, the Cu2+
A. gains protons
B. loses protons
C. gains electrons
D. loses electrons
Answer:
The answer is C.
Explanation:
Cu 2+ is reduced to Cu by gaining 2 electrons, so reduction occurs .
Proton cannot be gain or lose .
How do substances differ after a chemical change?
The physical properties of the substance stay the same.
The physical property of the substance will change.
The physical property of the substance will become better.
Answer:
B. The physical property of the substance will change.
Explanation:
About Answer A:
The physical properties of the substance can not stay the same after a chemical change. If the physical properties stay the same, this is just a physical change.
..
About Answer B:
After a chemical change, the physical properties of the substance can become better or worse depend on about what chemical reaction we are talking about.
What is the pressure of 1.78 g of nitrogen gas confined to a volume of 0.118 L at 25 C?
The pressure of the nitrogen gas confined to a volume of 0.118 L is 13.16 atm.
The pressure of nitrogen gas can be calculated by using the ideal gas equation.
PV =nRT
P = pressure
V = volume = 0.118 L
n = moles of nitrogen gas
moles = [tex]\rm\dfrac{weight}{molecular\;weight}[/tex]
moles of nitrogen gas = [tex]\rm \dfrac{1.78}{28}[/tex]
Moles of nitrogen gas = 0.0635 moles
R = constant = 0.0821
T = temperature = 25[tex]\rm ^\circ C[/tex] = 298 K
Substituting the values:
P [tex]\times[/tex] 0.118 = 0.0635 [tex]\times[/tex] 0.0821 [tex]\times[/tex] 298
P [tex]\times[/tex] 0.118 = 1.5535
P = 13.16 atm.
The pressure of the nitrogen gas confined to a volume of 0.118 L is 13.16 atm.
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The pressure of the nitrogen gas confined to a volume of 0.118 L at 25°C is approximately 13.16 atmospheres.
To solve this problem, we will use the ideal gas law, which is given by the equation:
[tex]\[ PV = nRT \][/tex]
where:
- P is the pressure of the gas,
- V is the volume of the gas,
- n is the number of moles of the gas,
- R is the ideal gas constant, and
- T is the temperature in Kelvin.
First, we need to calculate the number of moles of nitrogen gas [tex](\( N_2 \))[/tex]using the molar mass of nitrogen, which is approximately 28.02 g/mol.
[tex]\[ T = 25^\circ \text{C} + 273.15 = 298.15 \text{ K} \][/tex]
Next, we convert the temperature from Celsius to Kelvin by adding 273.15 to the given temperature:
[tex]\[ T = 25^\circ \text{C} + 273.15 = 298.15 \text{ K} \][/tex]
Now, we can rearrange the ideal gas law to solve for pressure P:
[tex]\[ P = \frac{nRT}{V} \][/tex]
The ideal gas constant R is approximately 0.0821 L·atm/(mo·K). Plugging in the values we have:
[tex]\[ P = \frac{(0.0635 \text{ mol})(0.0821 \text{ L·atm/(mol·K)})(298.15 \text{ K})}{0.118 \text{ L}} \] \[ P \approx \frac{(0.0635)(0.0821)(298.15)}{0.118} \] \[ P \approx \frac{1.553}{0.118} \] \[ P \approx 13.16 \text{ atm} \][/tex]
how many grams of ammonium carbonate are needed to decompose in order to produce 6.52 g of carbon dioxide?
Answer:
= 14.24g of [tex](NH4)_{2}CO_{3}[/tex] is required.
Explanation:
Reaction equation:
[tex](NH4)_{2}CO_{3}[/tex] → [tex]2NH_{3} + CO_{2} + H_{2} O[/tex]
Mole ratio of ammonium carbonate to carbon dioxide is 1:1
1 mole of CO2 - 44g
?? mole of CO2 - 6.52g
= 6.52/44 = 0.148 moles was produced from this experiment.
Therefore, if 1 mole of [tex](NH4)_{2}CO_{3}[/tex] - 96.09 g
0.148 mol of [tex](NH4)_{2}CO_{3}[/tex] -- ?? g
=0.148 × 96.09
= 14.24g of [tex](NH4)_{2}CO_{3}[/tex] is required.
The age of the Hawaiian islands gets older as you move from East to west. If the half life of k-40 is 1.3 billion years old, approximate the age of Maui if a sample rock on the island has 99.96% of its parent k-40 remaining.
A. 440,000 years
B. 750,000 years
C. 5.2 million years
D. 1.3 millions years
Answer:
B
Explanation:
From the calculation, we have the age of the sample to be 750,000 years. Option B
What is the half life?The half-life of a substance, usually associated with radioactive decay, is the time it takes for half of a quantity of that substance to decay. This concept is also used in various other contexts to describe the rate of decay or reduction of a substance's quantity over time.
If you have a radioactive substance, the half-life is the time it takes for half of the radioactive atoms to decay into other elements. After one half-life, half of the original substance will remain, and the other half will have transformed into a different element.
We know that;
[tex]N/No = (1/2)^t/t1/2\\N = 0.9996No\\ 0.9996No/No = (1/2)^t/1.3 * 10^3\\ 0.9996 = (1/2)^t/1.3 * 10^3\\ln 0.9996 = t/1.3 * 10^3 ln 0.5\\t/1.3 * 10^3 = ln 0.9996 / ln 0.5\\t = ln 0.9996 / ln 0.5 * 1.3 * 10^9[/tex]
t = 750,000 years
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Describe how the metal probably increases the reaction rate, identify whether this an example of homogeneous or heterogeneous catalysts, and explain how you know.
Explanation:
The metal probably helps to speed up the reaction rate a/c to collision theory, reactant molecules must collide with a reasonable direction by either weakening bonds in reactant molecules to make them extra reactive or by attaching reactant molecules in the exact direction to react.
This is known as an example of heterogeneous catalysis because the catalyst is solid and the reactants are liquid or gases mixture. In this catalysis, the catalyst is present in a different phase compare to the reactants.
Answer:
The metal probably increases reaction rate by either holding reactant molecules in the correct orientation to react or by weakening or breaking bonds in reactant molecules to make them more reactive.This is an example of heterogeneous catalysis. It is heterogeneous catalysis because the catalyst is a solid and the reactants are gases. In heterogeneous catalysis, the catalyst is in a different phase than the reactants.
Explanation:
What is the pH of a 0.45 M solution of hydrochloride acid (H
Answer:
0.35
Explanation:
pH = -log[H+]
[H+] = [HCl} = 0.45 M because HCl is a strong acid, and dissociate completely.
pH = - log[0.45] = 0.35
An enzyme with molecular weight of 310 kDa undergoes a change in shape when the substrate binds. This change can be characterized by ultracentrifugation. In the absence of any ligand, the s20,w = 11.7S, density of solvent is 1g/cm3and viscosity 1.005 cP; the specific volume of the protein is 0.732 cm3/g. a) Find the radius of the protein, assuming it is sphericalb) Upon binding of the substrate the sedimentation coefficient increases by 3.5%; what is the radius of the bound enzyme?
Answer:
(a) r = 6.26 * 10⁻⁷cm
(b) r₂ = 6.05 * 10⁻⁷cm
Explanation:
Using the sedimentation coefficient formula;
s = M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle
s = M ( 1 - Vρ) / N*6πnr
making r sbjct of formula, r = M (1 - Vρ) / N*6πnrs
Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s
r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)
r = 6.26 * 10⁻⁷cm
b. Using the formula r₂/r₁ = s₁/s₂
s₂ = 0.035 + 1s₁ = 1.035s₁
making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁
r₂ = 6.3 * 10⁻⁷cm / 1.035
r₂ = 6.05 * 10⁻⁷cm
Be sure to answer all parts. Which of the following ions possess a dipole moment? (a) ClF2+ (b) ClF2− has a dipole moment has no dipole moment cannot be determined has a dipole moment has no dipole moment cannot be determined (c) IF4+ has a dipole moment has no dipole moment cannot be determined (d) IF4− has a dipole moment has no dipole moment cannot be determined
Answer:
(A). Has Dipole moment, (B). No dipole moment, (C). Has dipole moment, (D). It has no Dipole moment.
Explanation:
In order to determine if a specie has dipole moment or not there is the need for us to draw the Lewis structure, please check attached file for the Lewis structures of each species.
(A). ClF2+: it HAS dipole moment because of Asymmetry. Note that the Fluorine atoms are more Electronegative that the chlorine atom.
(B). ClF2− : Normally, it should have a dipole moment because Fluorine atoms are more Electronegative that the chlorine atom on each side BUT due to its geometry (according to VSEPR theory) which makes it to have a linear geometry, the dipole moment will cancel out ,hence, NO DIPOLE MOMENT.
(C). IF4+ : it HAS dipole moment. Although, the dipole moment of two out of the four Fluorine will cancel out the other two will not cancel out making it to have Dipole moment.
(D). IF4-: it has no dipole moment because the four bonds are in opposite directions
ClF2+ and IF4+ has dipole moment whereas ClF2− and IF4− has no dipole moment.
ClF2+ has dipole moment because of Asymmetry i.e. the Fluorine atoms has higher electronegativity value that the chlorine atom.
ClF2− has no dipole moment due to its geometry i.e. it has a linear geometry, the dipole moment will cancel out and no dipole moment occur.
IF4+ has dipole moment because the dipole moment of two out of the four Fluorine will cancel out but the other two will not cancel out making it to have Dipole moment.
IF4- has no dipole moment because the four bonds are in opposite directions.
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Provide a specific example the relationship between processing, structure, and properties of an engineering material. Be specific with your example by including the material type, the processing involved, the structural changes, and specific properties that are being influenced. Be clear in how one thing impacts the other.Feel free to use equations, graphs, or sketches to support your example.One page maximum.
Answer:
An example of engineering material, are plastics, they are derived from organic, natural materials, such as cellulose, coal, natural gas, salt and, of course, oil. Oil is a complex mixture of thousands of compounds and must be processed before being used.
Explanation:
Plastic production begins with distillation at a refinery, where crude oil is separated into groups of lighter components, called fractions. Each fraction is a mixture of hydrocarbon chains (chemical compounds formed by carbon and hydrogen) that differ in terms of the size and structure of their molecules. One of those fractions, naphtha, is the essential compound for the production of plastic.
Two main processes are used to make plastic: polymerization and polycondensation, and both require specific catalysts. In a polymerization reactor, monomers like ethylene and propylene join to form long polymer chains. Each polymer has its own properties, structure and dimensions depending on the type of basic monomer that has been used.
Consider the hypothetical serine protease, which shows the specificity pockets. The S1 pocket has a glutamic acid in the bottom, the S 2 pocket is small and hydrophobic, and the S1 ' pocket is deep and hydrophobic. Suggest an amino acid sequence that this protease would cleave.
Answer:
See explaination
Explanation:
Hydrophobic literally means “the fear of water”. Hydrophobic molecules and surfaces repel water. Hydrophobic liquids, such as oil, will separate from water. Hydrophobic molecules are usually nonpolar, meaning the atoms that make the molecule do not produce a static electric field.
Please see attachment for the step by step solution of the given problem.
This protease would likely cleave peptide bonds adjacent to sequences with basic amino acids (P1), followed by hydrophobic residues (P1') such as Arginine-Serine-Leucine (RSL) or Lysine-Valine-Leucine (KVL).
The specificity of a serine protease is determined by the arrangement of amino acid residues in its active site, particularly in the specificity pockets. In the hypothetical serine protease described, with a glutamic acid in the S1 pocket, a small hydrophobic S2 pocket, and a deep hydrophobic S1' pocket, it would likely cleave peptide bonds adjacent to specific amino acid sequences that complement its active site.Amino acid sequences that could be cleaved by this protease might include those where the P1 position (immediately before the scissile bond) is a basic amino acid, such as arginine or lysine, as they can interact favorably with the acidic S1 pocket. Additionally, the P2 position should contain a hydrophobic amino acid, like leucine or phenylalanine, to fit into the small hydrophobic S2 pocket. The P1' position should also have a hydrophobic amino acid, like valine or isoleucine, to interact with the deep hydrophobic S1' pocket.A potential cleavage site could be represented as P2-Hydrophobic | P1-Basic | P1'-Hydrophobic, where the protease cleaves the peptide bond between P1 and P1'.For more such questions on amino acids
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A 5.00L tank at −2.4°C is filled with 16.7g of sulfur hexafluoride gas and 8.88g of dinitrogen monoxide gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction of each gas. Be sure each of your answer entries has the correct number of significant digits. gas mole fraction sulfur hexafluoride dinitrogen monoxide
Answer:
X(SF₆) = 0.361
X(N₂O) = 0.639
Explanation:
Step 1: Calculate the moles of each gas
We use the following expression.
n = m / M
where,
n: molesm: massM: molar massSF₆: 16.7 g / 146.06 g/mol = 0.114 mol
N₂O: 8.88 g / 44.01 g/mol = 0.202 mol
The total number of moles is 0.114 mol + 0.202 mol = 0.316 mol
Step 2: Calculate the mole fraction of each gas
We use the following expression.
X = moles of gas / total number of moles
X(SF₆) = 0.114 mol / 0.316 mol = 0.361
X(N₂O) = 0.202 mol / 0.316 mol = 0.639
Infrared radiation falls in the wavelength region of 1.00×10-6 to 1.00×10-3 meters.
What is the energy of infrared radiation that has a wavelength of 4.04×10-4 m?
Answer:4.920 X10^-22J
Explanation:
Energy of infrared radiation = hc/λ
where h=plank's constant=6.626X 10^-34js
c= speed of light = 3.0 x 10^8ms⁻¹
wavelength λ= 4.04x 10^-4m
E = 6.626X 10^-34js X 3.0 X 10^8 ms⁻¹/ 4.04 X 10^-4m
=4.920 X10^-22J
The _____ is one of the way paces where North America and Pacific Plates meet.
Answer:
The ring of fire
Explanation:
The ring of fire is where they meet or at the fault lines
the period of a ripple on a lake is half a second. what is the frequency of the wave
Answer:
2 Hertz
Explanation:
The frequency would be 2 Hertz.
The frequency of a wave is defined as the rate at which the particles of a medium vibrates when the wave is passed through it while the period of a wave is the time it takes the particles to make a complete cycle of vibration.
The frequency of a wave is inversely related to its period and is defined by the following equation:
f = 1/t, where f is the frequency (in hertz) and t is the period (in seconds).
Hence, if the period of a ripple is 1/2 or 0.5 seconds, the frequency becomes;
f = 1/0.5 = 2 Hertz
Final answer:
The frequency of a wave with a period of half a second is 2 Hz, meaning it oscillates twice every second.
Explanation:
If the period of a ripple on a lake is half a second, the frequency of the wave is calculated by taking the inverse of the period. Frequency (f) is defined as the number of cycles per unit time. To calculate the frequency when the period (T) is 0.5 seconds, use the formula f = 1/T. This means that the frequency of the wave is 2 Hz (hertz), as f = 1/0.5 s = 2 Hz. Therefore, the wave oscillates twice every second.
3. Most refrigerants utilize the energy involved in the liquid/gas phase change of a molecule, where an ideal refrigerant is noncorrosive, nonflammable, and has a boiling point around the desired temperature. Many fluorinated carbons meet all of these requirements, but the extremely high stability of some (such as CFCs) caused them to be phased out by the Montreal Protocol due to their ozone depleting potential. 3.a. R-12 (CCl2F2) was commonly used in household refrigerators and vehicle air conditioners prior to 1994. What is the correct Lewis structure for R-12
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
The compound R-12 (CCl2F2) has 32 valance electrons as shown in the Lewis structure attached to this answer.
The Lewis structure of a compound is a structure that shows the valence electrons in a molecule as dots or single lines to represent a shared pair of electrons in a covalent bond.
The compound R-12 (CCl2F2) has 32 valence electrons. The arrangement of these valence electrons have been shown in the Lewis structure attached to this answer.
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What carboxylic acid would be formed if the malonic ester synthesis (alkylation of diethyl malonate followed by hydrolysis with acidic workup and decarboxylation) were carried out with one equivalent of malonic ester, one equivalent of 1,5-dibromopentane, and two equivalents of hydroxide as the base
1. **Alkylation**: Diethyl malonate reacts with 1,5-dibromopentane to form the alkylated product.
2. **Hydrolysis**: The alkylated product undergoes acidic hydrolysis to yield a carboxylic acid.
3. **Decarboxylation**: Carboxylic acid formed undergoes decarboxylation to produce the final compound, 1,3-propanediol.
let's break down the reaction step by step:
1. **Alkylation with 1,5-dibromopentane**:
The malonic ester synthesis involves the alkylation of diethyl malonate. In this case, one equivalent of 1,5-dibromopentane is used.
The reaction proceeds via an S<sub>N2</sub> mechanism, where the nucleophilic oxygen attacks the electrophilic carbon of the alkyl halide. The bromine atom is displaced, forming a new carbon-carbon bond.
The reaction can be represented as:
[tex]\[ \text{CH}_2(\text{CO}_2\text{Et})_2 + \text{Br(CH}_2\text{)_4Br} \rightarrow \text{CH}_2(\text{CO}_2\text{Et})_3\text{CH}_2\text{Br} + \text{Br}^-\][/tex]
2. **Hydrolysis with acidic workup**:
The product from the alkylation step undergoes hydrolysis under acidic conditions. This step cleaves the ester groups, yielding carboxylic acids.
[tex]\[ \text{CH}_2(\text{CO}_2\text{Et})_3\text{CH}_2\text{Br} + 2\text{H}_3\text{O}^+ \rightarrow \text{CH}_2(\text{CO}_2\text{H})_3\text{CH}_2\text{OH} + 2\text{EtOH}\][/tex]
3. **Decarboxylation**:
Decarboxylation involves the removal of a carboxyl group from a molecule as carbon dioxide. In this case, since there are three carboxyl groups in the product, three equivalents of carbon dioxide will be released.
[tex]\[ \text{CH}_2(\text{CO}_2\text{H})_3\text{CH}_2\text{OH} \rightarrow \text{CH}_2(\text{CO}_2\text{H})_2\text{CH}_2\text{OH} + \text{CO}_2\][/tex]
The resulting compound is 1,3-propanediol.
So, the carboxylic acid formed in the reaction is glyceric acid (1,3-propanediol).
Justin added concentrated HCl solution in excess to 45.8 grams of magnesium metal and collected the hydrogen gas that was produced by the reaction.
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
If Justin collected 2.01 grams of H2, what was the percent yield of H2?
A 4.39%
B 46.9%
C 52.9%
D 93.8%
Answer:
52.9 %.
Explanation:
Mg + 2HCl ---> MgCl2 + H2
Using the relative atomic masses of magnesium and hydrogen:
Theoretically 24.3 g of Mg will produce 2.016 g of H2
so 45.8 g will produce (2.016 * 45.8) / 24.3 = 3.80 g H2.
So the % yield = 2.01 * 100 / 3.80
= 52.9%.
If Justin collected 2.01 grams of H[tex]_2[/tex], 52.9% was the percent yield of H[tex]_2[/tex]. Therefore, the correct option is option C.
What is percent yield?The % ratio of the theoretical yield to the actual yield is known as the percent yield. It is computed as the theoretical yield times by 100% divided by the experimental yield. The percent yield equals 100% if the theoretical and actual yields are equal. Because the real yield is frequently lower than the theoretical value, percent yield is typically lower than 100%.
If the percent yield is more than 100%, more sample than expected was retrieved from the reaction. This may have happened when other reactions took place and the product was also created.
Mg + 2HCl [tex]\rightarrow[/tex] MgCl[tex]_2[/tex] + H[tex]_2[/tex]
Theoretically 24.3 g of Mg will produce 2.016 g of H2
so, 45.8 g will produce (2.016 × 45.8) / 24.3 = 3.80 g H2.
% yield = 2.01 ×100 / 3.80
= 52.9%.
Therefore, the correct option is option C.
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Which are Arrhenius base examples? (SELECT ALL THAT APPLY)
Group of answer choices
NaOH
NH3
NH4
KOH
HF
LiOH
Answer:
Arrhenius bases: NaOH, KOH and LiOH.
Explanation:
Arrhenius bases are hydroxide ([tex]OH^{-}[/tex]) containing molecules which furnish [tex]OH^{-}[/tex] ion in aqueous solution.
So, clearly, NaOH, KOH and LiOH are arrhenius bases as they contain [tex]OH^{-}[/tex] ion as well as furnish [tex]OH^{-}[/tex] ion in aqueous solution.
Hydrolysis in water: [tex]NaOH(S)+H_{2}O(l)\rightarrow Na_{aq.}^{+}+OH_{aq.}^{-}[/tex]
[tex]KOH(S)+H_{2}O(l)\rightarrow K_{aq.}^{+}+OH_{aq.}^{-}[/tex]
[tex]LiOH(S)+H_{2}O(l)\rightarrow Li_{aq.}^{+}+OH_{aq.}^{-}[/tex]
The value of AH° for the reaction below is +128.1 kJ CH3OH (I)-CO (g) + 2H2 (g) How much heat is consumed when 87.1 g of hydrogen gas is formed? A) 2.76 x 103 kJ B) 5.52 x 103 kJ C) -5.52 x 103 kJ D) -2.76 x 103 kJ E) -128.1 kJ
The heat consumed in the formation of 87.1g of hydrogen gas, according to the given reaction, would be closest to 5.52 x 10^3 kJ.
Explanation:The given reaction, CH3OH(I) -> CO(g) + 2H2(g), reveals that every time the reaction occurs, 2 mol of hydrogen, H2, is produced and the reaction consumes 128.1 kJ of heat, indicated by the positive ΔH° value. This indicates that the reaction is endothermic - heat is absorbed during this chemical process. The question asks how much heat is consumed when 87.1 g of hydrogen gas is formed. To answer this, we first need to convert the mass of hydrogen gas to moles, using its molar mass (2 g/mol).
So, 87.1g of H2 is equivalent to 87.1/2 = 43.55 mol.
Since 2 mol of H2 consumes 128.1 kJ of heat, we can say that 1 mol of H2 would consume 128.1/2 = 64.05 kJ of heat.
Therefore, for 43.55 mol of H2, the heat consumed would be: 43.55 * 64.05 = ~2.79 x 10^3 kJ.
As a result, option B) 5.52 x 10^3 kJ would the closest to the calculated answer.
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The heat consumed when 87.1 g of hydrogen gas is formed during the given reaction is approximately 2.76 x 103 kJ.
Explanation:The question refers to the heat absorbed in a chemical reaction, specifically how much heat is consumed when 87.1 g of hydrogen gas (H2) is formed in the given reaction: CH3OH (I) → CO (g) + 2H2 (g). The heat change (ΔH°) for this reaction is +128.1 kJ.
First, we need to establish the relation between the heat change and the mole of hydrogen formed. For every 2 moles of H2, 128.1 kJ of heat is absorbed. Now for the calculation, we use the molar mass of hydrogen (H2) which is roughly 2 g/mol. Therefore, the given 87.1 g of hydrogen gas will approximately be 87.1/2 = 43.55 moles.
Since the molar heat is for 2 moles, the heat consumed by forming 1 mole of H2 will be 128.1/2 = 64.05 kJ. So, for 43.55 moles, it will be 64.05*43.55 = 2.788 x 103 kJ, which can be approximated as 2.76 x 103 kJ (rounded to three significant digits).
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50 g of NaCl is mixed with water to make .5 L of salt water. Which is the solute?
Answer:
The correct answer to the following question will be "NaCl".
Explanation:
A solute is being known as the soluble material within a liquid or a solution. The solvent is available in larger quantities than those of the solute among fluid. Concentration seems to be a calculation including its quantity of solvent contained in such a liquid solution, regard to something like the solvent volume.So that NaCl is the right answer.
A coordination complex is made up of _________________________________________. (Choose one) Selected Answer: IncorrectC. a transition metal center and alkaline earth metal ligands surrounding it. Answers: CorrectA. a Lewis acidic metal center with Lewis basic ligands surrounding it. B. an Arrhenius basic metal center with Lewis acidic ligands surrounding it. C. a transition metal center and alkaline earth metal ligands surrounding it. D. a Lewis basic metal center with Lewis acidic ligands surrounding it.
Answer:
a Lewis acidic metal center with Lewis basic ligands surrounding it.
Explanation:
A lewis acid is a chemical specie capable of accepting a lone pair of electrons while a lewis base is a chemical specie capable of donating a lone pair of electrons.
In a coordination compound, a lewis acid(central metal atom/ion) accepts electron pairs from lewis bases (ligands). Any chemical specie called a ligand must possess at least one electron pair available for coordinate covalent bonding.
Coordination compounds are usually formed between transition metal atoms/ions and neutral molecules having lone pairs of electrons or anionc species .
The correct option is: A. A coordination complex consists of a Lewis acidic metal center with Lewis basic ligands surrounding it.
A coordination complex features a central metal atom or ion, often a transition metal, functioning as a Lewis acid due to its ability to accept electron pairs.
Surrounding this central metal are ligands, which are molecules or ions that act as Lewis bases by donating electron pairs to the metal. This donor-acceptor interaction leads to the formation of coordinate covalent bonds, creating a stable structure.
The unique properties of the metal and ligands, as well as their interactions, shape the geometry and characteristics of the complex. These properties are crucial for various chemical reactions and biological processes, highlighting the importance of coordination complexes in science.
Complete Question: -
"A coordination complex is made up of ____________."
Then the options should be listed below:
A. a Lewis acidic metal center with Lewis basic ligands surrounding it.
B. an Arrhenius basic metal center with Lewis acidic ligands surrounding it.
C. a transition metal center and alkaline earth metal ligands surrounding it.
D. a Lewis basic metal center with Lewis acidic ligands surrounding it.
Consider the following data concerning the equation: H2O2 + 3I– + 2H+ → I3– + 2H2O [H2O2] [I–] [H+] rate I 0.100 M 5.00 × 10–4 M 1.00 × 10–2 M 0.137 M/sec II. 0.100 M 1.00 × 10–3 M 1.00 × 10–2 M 0.268 M/sec III. 0.200 M 1.00 × 10–3 M 1.00 × 10–2 M 0.542 M/sec IV. 0.400 M 1.00 × 10–3 M 2.00 × 10–2 M 1.084 M/sec Two mechanisms are proposed: I. II. Which mechanism and which step as the rate determining step would best fit the data?
Based on the provided data, mechanism I and the decomposition of H2O2 as the rate-determining step best fit the experimental observations.
Explanation:To determine which mechanism and which step is the rate-determining step, we need to analyze the data provided. If we compare the rates of the reactions in experiments I, II, III, and IV, we can see that the rate of the reaction doubles when the concentration of H2O2 doubles, suggesting that the reaction is first-order with respect to H2O2. This information aligns with mechanism I, where the rate-determining step involves the decomposition of H2O2. Therefore, mechanism I and the decomposition of H2O2 as the rate-determining step best fit the data.
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The mechanism and the step as the rate determining step that best fits the data is Mechanism B, with the first step rate determining step. The correct option is c.
To determine which mechanism and rate-determining step best fit the given data, we need to examine the rate laws implied by each mechanism and compare them to the experimental data.
Experimental Data:
[tex]Trial \ I: [H_2O_2] = 0.100 \, M, [I^-] = 5.00 \times 10^{-4} \, M, [H^+] = 1.00 \times 10^{-2} \, M, Rate = 0.137 \, M/s \\\\Trial \ II: [H_2O_2] = 0.100 \, M, [I^-] = 1.00 \times 10^{-3} \, M, [H^+] = 1.00 \times 10^{-2} \, M, Rate = 0.268 \, M/s \\\\Trial \ III: [H_2O_2] = 0.200 \, M, [I^-] = 1.00 \times 10^{-3} \, M, [H^+] = 1.00 \times 10^{-2} \, M, Rate = 0.542 \, M/s \\\\Trial \ IV: [H_2O_2] = 0.400 \, M, [I^-] = 1.00 \times 10^{-3} \, M, [H^+] = 2.00 \times 10^{-2} \, M, Rate = 1.084 \, M/s[/tex]
Observations:
1. Comparing Trials I and II, the concentration of [I⁻] doubles while other concentrations remain constant, and the rate approximately doubles (0.137 [tex]\rightarrow[/tex] 0.268). This suggests that the rate is first-order in [I⁻].
2. Comparing Trials II and III, the concentration of [H₂O₂] doubles while other concentrations remain constant, and the rate approximately doubles (0.268 [tex]\rightarrow[/tex] 0.542). This suggests that the rate is first-order in [H₂O₂].
3. Comparing Trials III and IV, the concentration of [H₂O₂] doubles and [H⁺] also doubles. The rate also approximately doubles (0.542 [tex]\rightarrow[/tex] 1.084). This suggests that the rate is first-order in [H⁺].
Rate Law:
Based on these observations, the rate law can be inferred as:
[tex]\text{Rate} = k[H_2O_2][I^-][H^+][/tex]
Mechanism Analysis:
Let's analyze the proposed mechanisms to see which one fits the observed rate law.
Mechanism A:
[tex]1. \ H_2O_2 + I^- \rightarrow H_2O + OI^- \\\\2. \ OI^- + H^+ \rightarrow HOI \\\\3. \ HOI + I^- + H^+ \rightarrow I_2 + H_2O \\\\4. \ I_2 + I^- \rightarrow I_3^-[/tex]
Mechanism B:
[tex]1. \ H_2O_2 + I^- + H^+ \rightarrow H_2O + HOI \\\\2. \ HOI + I^- + H^+ \rightarrow I_2 + H_2O \\\\3. \ I_2 + I^- \rightarrow I_3^-[/tex]
Rate Determining Step Analysis:
Mechanism A: If the first step [tex](H_2O_2 + I^- \rightarrow H_2O + OI^-)[/tex] is rate-determining, the rate law would be [tex]\text{Rate} = k[H_2O_2][I^-][/tex]. This does not include [H⁺] , which does not match the experimental data. If the second step [tex](OI^- + H^+ \rightarrow HOI)[/tex] is rate-determining, it would still not fit as it involves OI⁻ and H⁺, not [H₂O₂] and [I⁻] directly.Mechanism B: If the first step [tex](H_2O_2 + I^- + H^+ \rightarrow H_2O + HOI)[/tex] is rate-determining, the rate law would be [tex]\text{Rate} = k[H_2O_2][I^-][H^+][/tex]. This fits the experimental rate law perfectly.Thus, Mechanism B with the first step as the rate-determining step matches the observed rate law, c. Mechanism B, with the first step rate determining step.
The complete question is:
Consider the following data concerning the equation:
H₂O₂ + 3I⁻ + 2H⁺ → I₃⁻ + 2H₂O
[H₂O₂] [I⁻] [H⁺] rate
I 0.100 M 5.00 x 10⁻⁴M 1.00 x 10⁻²M 0.137 M/sec
II. 0.100 M 1.00 x 10⁻³M 1.00 x 10⁻²M 0.268 M/sec
III 0.200 M 1.00 x 10⁻³M 1.00 x 10⁻²M 0.542 M/sec
IV. 0.400 M 1.00 x 10⁻³M 2.00 x 10⁻²M 1.084 M/sec
Two mechanisms are proposed, A, the first four equation lines below, while B is the next three lines:
A.
H₂O₂ + I⁻ → H₂O + OI⁻
OI⁻ + H⁺ → HOI
HOI + I⁻ + H⁺ → I₂ + H₂O
I₂ + I⁻ → I₃⁻
B.
H₂O₂ + I⁻ + H⁺ → H₂O + HOI
HOI + I⁻ + H⁺ → I₂ + H₂O
I₂ + I⁻ → I₃⁻
Which mechanism and which step as the rate determining step would best fit the data?
a. Mechanism B, with the second step rate determining step
b. Mechanism A, with the second step rate determining step
c. Mechanism B, with the first step rate determining step
d. Mechanism A, with the first step rate determining step
e. None of the above
Determine the identity of the element that contains exactly three 4p electrons in the ground state
Answer: Arsenic
Explanation: The element that contains exactly three 4p electrons at ground state is
Arsenic, a group 15 and period 4 solid (20°C) element with
atomic weight 74.9216g/mol whose Electronic configuration is
1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p³ with the condensed electronic configuration as [Ar] 3d¹⁰ 4s² 4p³.
The element that contains exactly three 4p electrons in the ground state is Arsenic (As).
Ground state is the lowest energy level of a physical system (such as an atomic nucleus or an atom).
In this case, [tex]\rm p^3[/tex] means nitrogen family. Nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), and bismuth (Bi) are all members of the nitrogen family. Among these elements only Arsenic has 3 electrons in the 4p orbital in the ground energy level.
The electronic configuration of Arsenic is represented as: [Ar] 3d¹⁰ 4s² 4p³.
Therefore, the correct answer is Arsenic is the element that has 3 electrons in the 4p in ground state.
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Question 63 of 70 1.5 Points When 2-bromo-2-methylbutane is treated with a base, a mixture of 2-methyl-2-butene and 2-methyl-1-butene is produced. When potassium hydroxide is the base, 2-methyl-1-butene accounts for 45% of the product mixture. However, when potassium tert-butoxide is the base, 2-methyl-1-butene accounts for 70% of the product mixture. What percent of 2-methyl-1-butene would be in the mixture if potassium propoxide were the base
Answer: between 45% and 70%
Explanation:
taking this step by step, let us analyze this question carefully.
From the original image in the question, as can be seen in the second uploaded image, we observe that;
potassium hydroxide having being less bulky produces about 45% of 2-methyl-butane whereas potassium tert-butoxide having being bulky produces 70% of this product.
Given the potassium propoxide as the base and intermediate to that of potassium hydroxide and potassium tert-butoxide, it produces 2-methyl-1-butane in between 45% to 70%.
Cheers i hope this helped!!!!
Final answer:
Without additional information, it is not possible to determine the exact percentage of 2-methyl-1-butene produced using potassium propoxide as the base for the reaction of 2-bromo-2-methylbutane.
Explanation:
When 2-bromo-2-methylbutane is treated with a base to form a mixture of 2-methyl-2-butene and 2-methyl-1-butene, the specific base used influences the proportion of each product. The use of potassium hydroxide results in 45% yield of 2-methyl-1-butene, while potassium tert-butoxide increases this yield to 70%. If potassium propoxide were used as the base, without further experimental data or specific trends indicated in the question, it is not possible to accurately predict the exact percentage of 2-methyl-1-butene in the mixture. Generally, the bulkiness of the base can influence the product distribution in elimination reactions due to steric hindrance, but without additional information about the reaction conditions or the specific mechanism with potassium propoxide, we can only hypothesize about the potential outcome based on trends we see with other bases.
What would be an equivalent two-dimensional point
lattice for the area-centered hexagon?
Diagram is attached
Answer:
The equivalent two-dimensional point lattice for the area-centered hexagon is a rhombus.
Explanation:
The area centered hexagon is illustrated with a centered figure that has dotted center and many other dots around it that connect each other.
In this case, we need to draw the area centered hexagon first.
After drawing, we then connect the centered atoms of the hexagon. This connected centered atoms now form a rhombus like shape
Note: The shape of a rhombus is said to be flat, and it has 4 equal sides.
A 25.0 mL aliquot of 0.0430 0.0430 M EDTA EDTA was added to a 56.0 56.0 mL solution containing an unknown concentration of V 3 + V3+ . All of the V 3 + V3+ present in the solution formed a complex with EDTA EDTA , leaving an excess of EDTA EDTA in solution. This solution was back-titrated with a 0.0490 0.0490 M Ga 3 + Ga3+ solution until all of the EDTA EDTA reacted, requiring 13.0 13.0 mL of the Ga 3 + Ga3+ solution. What was the original concentration of the V 3 + V3+ solution?
Answer:
Check the explanation
Explanation:
As we know the reaction of EDTA and [tex]Ga^3[/tex]+ and EDTA and [tex]V^3[/tex]+
Let us say that the ratio is 1:1
Therefore, the number of moles of [tex]Ga^3[/tex]+ = molarity * volume
= 0.0400M * 0.011L
= 0.00044 moles
Therefore excess EDTA moles = 0.00044 moles
Given , initial moles of EDTA = 0.0430 M * 0.025 L
= 0.001075
Therefore reacting moles of EDTA with [tex]V^3+[/tex] = 0.001075 - 0.00044 = 0.000675 moles
Let us say that the ratio between [tex]V^3+[/tex] and EDTA is 1:1
Therefore moles of [tex]V^3+[/tex] = 0.000675 moles
Molarity = moles / volume
= 0.000675 moles / 0.057 L
= 0.011 M (answer).
What is the identity of the isotope?
Answer:
Explanation:
The fact that each isotope has one proton makes them all variants of hydrogen: the identity of the isotope is given by the number of protons and neutrons. From left to right, the isotopes are protium (1H) with zero neutrons, deuterium (2H) with one neutron, and tritium (3H) with two neutrons.
Which organelles are common to BOTH plant and animal cells? (Choose all 1 point
that apply.)
cell wall
cell membrane
nucleus
chloroplasts
u
mitochondria
vacuole
cytoplasm
O
cuestion
wering pants and huma
Answer:
The following organelles are common to both plant and animal cells:
Cell membraneCell NucleusMitochondriaCytoplasm(N.B: the other options are peculiar to plant cells only)
Explanation:
Plant and animal cells share similarities in components due to the fact that the are both eukaryotic cells ( having DNA as genetic composition).
The organelles common to both plant and animal cells are the cell membrane, nucleus, mitochondria, and cytoplasm.
Explanation:The organelles that are common to both plant and animal cells are the cell membrane, nucleus, mitochondria, and cytoplasm. Both plant and animal cells have a cell membrane, which controls the movement of substances in and out of the cell. The nucleus is the control center of the cell, responsible for storing and processing genetic information. Mitochondria are the powerhouse of the cell, producing energy through cellular respiration. Cytoplasm is a jelly-like substance that fills the cell and houses the organelles.
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6. A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCI. Calculate the pH
at each volume of added acid: 0 mL, 10 mL, 20 mL
The pH at each volume of added acid is calculated using the Henderson-Hasselbalch equation.
Explanation:The pH of a solution can be calculated using the equation:
pH = -log[H+]
Using the Henderson-Hasselbalch equation, pH can be calculated at each volume of added acid:
Volume of added acid = 0 mL
The initial concentration of pyridine is 0.125 M. Since pyridine is a weak base, [H+] can be calculated using the equation:
Kw / Kb = [H+]
where Kw is the equilibrium constant for water (1.0 x 10^-14) and Kb is the base dissociation constant for pyridine (1.7 x 10^-9). Substituting the values, we get:
[H+] = (1.0 x 10^-14) / (1.7 x 10^-9) ≈ 5.88 x 10^-6 M
Now, we can calculate the pH:
pH = -log(5.88 x 10^-6) ≈ 5.23
Volume of added acid = 10 mL
Using stoichiometry, we can determine the moles of HCI added:
moles of HCI = concentration of HCI x volume of HCI
moles of HCI = (0.100 M) x (0.010 L) = 0.001 mol
Since the acid HCl is strong, all of it will react with the pyridine:
moles of pyridine reacted = moles of HCI added = 0.001 mol
The final volume of the solution is 25.0 mL + 10 mL = 35.0 mL. The concentration of pyridine after the reaction is:
concentration of pyridine after reaction = moles of pyridine / final volume of solution
concentration of pyridine after reaction = (0.125 M)(25.0 mL) / (35.0 mL) ≈ 0.0893 M
Using the Henderson-Hasselbalch equation, we can calculate the pH:
pH = pKa + log ([A-]/[HA])
where pKa is the negative logarithm of the acid dissociation constant for pyridine (9.27) and [A-]/[HA] is the ratio of the conjugate base to the weak acid, which is equal to the ratio of the concentration of pyridine after reaction to the initial concentration of pyridine:
pH = 9.27 + log (0.0893 / 0.125) ≈ 9.07
Volume of added acid = 20 mL
Using the same approach, we can calculate the concentration of pyridine after the reaction:
concentration of pyridine after reaction = (0.125 M)(25.0 mL) / (45.0 mL) ≈ 0.0694 M
Using the Henderson-Hasselbalch equation, we can calculate the pH:
pH = 9.27 + log (0.0694 / 0.125) ≈ 8.88
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To calculate the pH at each stage, we consider the concentrations of the pyridine and the HCl, and use the formulas for pH and pOH. At 0 mL of HCl, we would calculate pH from the pKa of pyridine. At 10 mL and 20 mL, we need to consider that pyridine is in excess, and calculate pOH first before getting the pH.
Explanation:The subject of this question is acid-base titration, specifically the titration of pyridine with hydrochloric acid (HCl). To calculate the pH at different volumes of added acid, we'd use the formula for the concentration of the pyridine and HCl.
1. At 0 mL of HCl, the solution is just the pyridine, which is a weak base. We don't have the pKa for pyridine in the question, but assuming we did, we could calculate the pH using the pKa and the formula pH = 14 - pKa. Let's say it's around 5.2.
2. At 10 mL of 0.100 M HCl (1.0 mmol), and 25 mL of 0.125 M pyridine (3.125 mmol), we have more base than acid, so pyridine is in excess. We'd use the equilibrium expression for the reaction of excess pyridine with water to find the pOH, and then calculate the pH.
3. At 20 mL of 0.100 M HCl (2.0 mmol), we still have excess pyridine, and so we'd perform a similar calculation as at 10 mL.
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