a) The reaction NH3 + HC≡CO ⟶ NH2O + HC≡CH will not occur as written due to the relative acidity/basicity of the reactants and products.
b) The reaction CH3OH + CH3Θ ⇌ CH3OΘ + CH4 will occur as written based on the relative strengths of the acids and bases involved.
In the reaction NH3 + HC≡CO ⟶ NH2O + HC≡CH, the main factor that determines whether a reaction will occur is the relative acidity or basicity of the reactants and products. NH3 (ammonia) is a weak base with a pKa value of 35, while HC≡CO (acetylene) is a weak acid with a pKa value of 25. Since NH3 is a stronger base than HC≡CO is an acid, the reaction as written is unlikely to occur. Therefore, the answer for reaction a) is "No."
On the other hand, in the reaction CH3OH + CH3Θ ⇌ CH3OΘ + CH4, the equilibrium between the reactants and products is determined by the relative strengths of the acids and bases involved. CH3OH (methanol) is a weak acid with a pKa value of 15, while CH3Θ (methyl anion) is a strong base. CH3OΘ (methyl oxide) is a weak base and CH4 (methane) is a neutral compound. Since the reactants and products involve a strong base and a weak acid, the reaction as written can occur. Therefore, the answer for reaction b) is "Yes."
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Determine the pH of a solution that results in mixing 1L of2O with 1 mL of HCl. That's all the info they give me
Given information: 1 L of 2O and 1 mL of HCl.To determine the pH of the solution that results in mixing 1 L of 2O with 1 mL of HCl.
We need to calculate the concentration of the HCl first.Molar mass of HCl = 1 g/molVolume of HCl = 1 mLConcentration of HCl = mass / volume= 1 g / 1000 mL= 0.001 MConcentration of H+ ions produced by HCl = 0.001 MIn the solution, the water molecules will undergo an autoionization process:H2O ⇌ H+ + OH-At 25°C.
The concentration of H+ ions and OH- ions produced by the autoionization process is 1.0 × 10-7 M each.Since H+ ions are produced by both HCl and autoionization process, we can write the expression for pH as follows Therefore, the pH of the solution that results in mixing 1 L of 2O with 1 mL of HCl is -4.
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what is the partial pressure of argon, par, in the flask?
A) The partial pressure of argon in the flask is approximately 0.906 atm. B) The partial pressure of ethane in the flask is approximately 0.144 atm.
We need to calculate the partial pressure of argon (PAr) in the flask.
Given;
Volume (V) = 1.00 L
Total pressure (Ptotal) = 1.050 atm
Temperature (T) = 25 °C = 298 K
Mass of argon (m) = 1.10 g
Molar mass of argon (M) = 39.95 g/mol
First, let's calculate the number of moles of argon using the mass and molar mass:
n = m / M
n = 1.10 g / 39.95 g/mol
Next, we can calculate the partial pressure of argon (PAr) using the ideal gas law;
PV = nRT
PAr × V = n × R × T
PAr = (n × R × T) / V
Substituting the given values;
PAr = (n × R × T) / V
PAr = [(1.10 g / 39.95 g/mol) × (0.0821 L·atm/(mol·K) × 298 K] / 1.00 L
Calculating PAr using the above expression, we get;
PAr ≈ 0.906 atm
Therefore, the partial pressure of argon in the flask is approximately 0.906 atm.
Given;
Total pressure (Ptotal) = 1.050 atm
Partial pressure of argon (PAr) = 0.906 atm (calculated in Part A)
To find the partial pressure of ethane (Pethane), we can use Dalton's law of partial pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases.
Ptotal = PAr + Pethane
Rearranging the equation to solve for Pethane;
Pethane = Ptotal - PAr
Substituting the given values;
Pethane = 1.050 atm - 0.906 atm
Calculating Pethane using the above expression, we get;
Pethane ≈ 0.144 atm
Therefore, the partial pressure of ethane in the flask is approximately 0.144 atm.
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--The given question is incomplete, the complete question is
"Part A What is the partial pressure of argon, Par, in the flask? Express your answer to three significant figures and include the appropriate units. Constants1 Periodic Table A 1.00 L flask is filled with 1.10 g of argon at 25 °C. A sample of ethane vapor is added to the same flask until the total pressure is 1.050 atm View Available Hint(s) PAr Value Units Submit."--
Calculate the theoretical yield in grams assuming your unknown is Na2CO1⋅H2O :
The expected yield of Na[tex]_{2}[/tex]CO[tex]_{3}[/tex]⋅H[tex]_{2}[/tex]O is 26.57 grams.
To calculate the theoretical yield, we need to know the molar mass of Na[tex]_{2}[/tex]CO[tex]_{3}[/tex]⋅H[tex]_{2}[/tex]O. The molar mass of Na[tex]_{2}[/tex]CO3 is 105.99 g/mol, and the molar mass of H[tex]_{2}[/tex]O is 18.02 g/mol. The molar mass of Na[tex]_{2}[/tex]CO[tex]_{3}[/tex]⋅H[tex]_{2}[/tex]O is the sum of these masses, which is 123.01 g/mol.
Next, we convert the given quantity of 0.216 moles to grams using the molar mass:
0.216 moles × 123.01 g/mol = 26.57 grams
Therefore, the theoretical yield of Na[tex]_{2}[/tex]CO[tex]_{3}[/tex]⋅H[tex]_{2}[/tex]O is 26.57 grams.
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write the ground state electron configuration of a lead atom
Answer:
electronic configuration of Lead at the ground state is [ Xe ] 6 s 2 4 f 14 5 d 10 6 p 2 .
What is the boiling point of a mixture composed of 61 gHOCH
2
CH
2
OH and 161 g H
2
O? - For H
2
O,K
b
=0.512
∘
C/m - For H
2
O,K
f
=1.86
∘
C/m - HOCH
2
CH
2
OH= ethylene glycol, molar mass =62.08 g/mol - water molar mass =18.02 g/mol Round to one decimal place. Your Answer: Answer units
The boiling point of the mixture, composed of 61 g of ethylene glycol and 161 g of water, is approximately 131.5 °C, considering the molal boiling point elevation constant and the molality of the solution.
The boiling point of the mixture composed of 61 g of ethylene glycol ([tex]\(\text{HOCH}_2\text{CH}_2\text{OH}\)[/tex]) and 161 g of water can be calculated using the equation:
[tex]\(\Delta T_b = K_b \cdot m\)[/tex]
Where [tex]\(\Delta T_b[/tex] is the boiling point elevation, Kb is the molal boiling point elevation constant, and m is the molality of the solution.
For water (H2O), Kb = 0.512 °C/m, and for ethylene glycol ([tex]\(\text{HOCH}_2\text{CH}_2\text{OH}\)[/tex]), Kb = 1.86 °C/m.
To find the molality (m) of the solution, we need to calculate the number of moles of each component.
For ethylene glycol:
Number of moles = [tex]\(\frac{{\text{{mass}}}}{{\text{{molar mass}}}} = \frac{{61 \, \text{g}}}{{62.08 \, \text{g/mol}}} = 0.983 \, \text{mol}\)[/tex]
For water:
Number of moles = [tex]\(\frac{{\text{{mass}}}}{{\text{{molar mass}}}} = \frac{{161 \, \text{g}}}{{18.02 \, \text{g/mol}}} = 8.94 \, \text{mol} = 0.983 \, \text{mol}\)[/tex]
The total moles of solute = moles of ethylene glycol + moles of water = 0.983 mol + 8.94 mol = 9.92 mol
Now we can calculate the molality:
molality (m) = [tex]\frac{moles of solute }{mass of solvent (kg)}[/tex]
= [tex]\frac{9.92 mol }{0.161 kg (converted from 161 g)}[/tex]= 61.49 mol/kg
Substituting the values into the equation:
[tex]\(\Delta T_b = K_b \cdot m\)[/tex]
[tex]\(\Delta T_b[/tex] = 0.512 °C/m * 61.49 mol/kg ≈ 31.51 °C
Therefore, the boiling point of the mixture is approximately 31.5 °C higher than the boiling point of pure water (100 °C).
Therefore, the boiling point of the mixture is approximately 131.5 °C.
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riterion 1: Identify the types of forces (noncovalent interactions) present between adjacent nolecules/ions in pure substances and mixtures A. What is the strongest possible intermolecular force between two particles of the following compounds?
CH
3
CH
2
CH
2
OH
CH
3
NH
2
CH
3
COOH
3
CaO
C
4
H
10
SiO
2
OCl
2
NaHSO
4
B. What do you look for to determine if a compound will have ion-ion forces? C. What do you look for to determine if a compound will have hydrogen bonding forces? D. What do you look for to determine if a compound will have dipole-dipole forces? E. What do you look for to determine if a compound will have London dispersion forces?
The compounds [tex]CH_3CH_2CH_2OH[/tex], [tex]CH_3NH_2[/tex], [tex]CH_3COOH[/tex], CaO, [tex]SiO_2[/tex], [tex]OCl_2[/tex], and [tex]NaHSO_4[/tex] exhibit hydrogen bonding, dipole-dipole interactions, and London dispersion forces. The strength of the intermolecular forces depends on the molecular structure and polarity of the compounds.
The strongest intermolecular force between two particles can be determined by analyzing the nature of the bonds and the molecular structure. In the given compounds:
- [tex]CH_3CH_2CH_2OH[/tex] (propanol) can exhibit hydrogen bonding between the hydrogen atom bonded to oxygen and the oxygen atom.
- [tex]CH_3NH_2[/tex] (methylamine) can exhibit hydrogen bonding between the hydrogen atom bonded to nitrogen and the lone pair on nitrogen.
- [tex]CH_3COOH[/tex] (acetic acid) can exhibit hydrogen bonding between the hydrogen atom bonded to oxygen and the oxygen atom.
- CaO (calcium oxide) has ionic bonding between calcium cations and oxide anions.
- [tex]SiO_2[/tex] (silicon dioxide) has covalent bonding and can exhibit dipole-dipole interactions due to the polar nature of the Si-O bonds.
- [tex]OCl_2[/tex] (dichlorine monoxide) exhibits dipole-dipole interactions due to the polar nature of the Cl-O bonds.
- [tex]NaHSO_4[/tex] (sodium hydrogen sulfate) exhibits ionic bonding between sodium cations and hydrogen sulfate anions.
Compounds that have ion-ion forces typically consist of ions, which are charged particles. To determine if a compound has ion-ion forces, look for the presence of positive and negative ions in the compound's formula or structure.
Hydrogen bonding forces occur when hydrogen is bonded to highly electronegative atoms such as oxygen, nitrogen, or fluorine. To determine if a compound has hydrogen bonding forces, look for the presence of hydrogen bonded to these electronegative atoms.
Dipole-dipole forces occur when polar molecules interact with each other due to the difference in electronegativity between the atoms involved. To determine if a compound has dipole-dipole forces, examine the molecular structure and identify polar bonds.
London dispersion forces (also known as van der Waals forces) occur in all molecules and arise due to temporary fluctuations in electron density. To determine if a compound has London dispersion forces, consider the size and shape of the molecules, as larger and more complex molecules tend to have stronger London dispersion forces.
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calculate the freezing pt of a 1.8 m aqueous ethylene glycol solution. Kf= 1.86 degrees Celsius/m.
a) -1.86 Celsius
b) 1.03 C
c) 3.35 C
d) -3.35 C
∆T is the change in NE (freezing point depression)Kf is the freezing point depression constant (1.86 degrees Celsius/m)m is the molality of the solution (1.8 m)
Plugging in the values, we have:∆T = 1.86 * 1.8∆T = 3.CelsiusTherefore, the freezing point of the solution is lowered by 3.348 degrees Celsius. In order to find the actual freezing point, we subtract this value from the normal freezing point of water (0 degrees Celsius).
Freezing point = 0 - 3.348 = -3.348 degrees Celsius Since the asks for the rounded to two decimal places, the correct option is:d) -3.35 Celsius The freezing point of a solution is determined by the concentration of solute particles in the solution. The more solute particles present, the lower the freezing point.
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Answer choice d) -3.35 C is the correct option, as it represents the decrease in the freezing point.
To calculate the freezing point of a 1.8 m aqueous ethylene glycol solution, we can use the formula:
ΔT = Kf * m
where ΔT is the change in temperature, Kf is the freezing point depression constant, and m is the molality of the solution.
Given that Kf = 1.86 degrees Celsius/m and the molality of the solution is 1.8 m, we can substitute these values into the formula:
ΔT = 1.86 * 1.8
ΔT = 3.348 degrees Celsius
Therefore, the freezing point of the aqueous ethylene glycol solution is reduced by 3.348 degrees Celsius.
Answer choice d) -3.35 C is the correct option, as it represents the decrease in the freezing point.
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What is the Ka reaction of HCN?
KaKa reaction:
The Ka reaction of HCN is the dissociation of hydrogen cyanide in water, producing hydronium ions (H3O+) and cyanide ions (CN-).
In aqueous solution, hydrogen cyanide (HCN) can dissociate to form hydronium ions (H3O+) and cyanide ions (CN-). This dissociation process is described by the following chemical equation:
HCN + H2O ⇌ H3O+ + CN-
The equilibrium constant for this reaction is known as the acid dissociation constant (Ka). It represents the extent to which the reaction proceeds in the forward direction. A larger value of Ka indicates a stronger acid.
HCN is a weak acid, meaning it only partially dissociates in water. This is because the equilibrium lies more towards the reactant side. However, HCN does produce some hydronium and cyanide ions in solution.
The dissociation of acids, such as HCN, in water is an important concept in chemistry. It helps in understanding the behavior of acids and their relative strengths. The acid dissociation constant (Ka) quantifies the extent of dissociation and allows for comparisons between different acids. Furthermore, the dissociation of HCN is significant because cyanide ions can have toxic effects on biological systems. Understanding the equilibrium and the concentration of cyanide ions is crucial for assessing the potential hazards associated with hydrogen cyanide.
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how many moles of chloride ions are in 0.2530 g of aluminum chloride?
0.005688 mol of chloride ions are present in 0.2530 g of aluminum chloride (AlCl₃).
Given: Mass of aluminum chloride (AlCl₃) = 0.2530 g
Molar mass of AlCl₃ = 133.34 g/mol
The molar mass of AlCl₃ is calculated as follows:
Molecular mass of AlCl₃ = Atomic mass of Al + Atomic mass of Cl × 3
= 27.0 g/mol + 35.5 g/mol × 3
= 27.0 g/mol + 106.5 g/mol
= 133.5 g/mol
To calculate the number of moles of chloride ions in 0.2530 g of aluminum chloride, we first need to find the number of moles of aluminum chloride. Then we can multiply it by the ratio of moles of chloride ions to moles of aluminum chloride.
This ratio is 3:1 because each molecule of AlCl₃ contains three chloride ions and one aluminum ion.
So, Number of moles of AlCl₃ = Mass of AlCl₃ / Molar mass of AlCl₃= 0.2530 g / 133.5 g/mol= 0.001896 mol
Number of moles of chloride ions = Number of moles of AlCl₃ × (3 moles of Cl⁻ / 1 mole of AlCl₃)
= 0.001896 mol × 3= 0.005688 mol of Cl⁻
.
Hence, 0.005688 mol of chloride ions are present in 0.2530 g of aluminum chloride (AlCl₃).
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Suppose a biochemist has 10 mL of a 1.0M solution of a compound with two ionizable groups at a pH of 7.60. She adds 10.0 mL of 1.0MHCl, which changes the pH to 3.40. The pK
a
value of one of the groups is pK
1
=4.40 and it is known that pK
2
is between 7 and 10 . What is the exact value of pK
2
? pK
2
The exact value of pK₂, the second ionizable group's pK value, is determined to be 9.40 based on the pH change from 7.60 to 3.40 upon the addition of 10.0 mL of 1.0 M HCl to a 10 mL solution of a compound with two ionizable groups, where pK₁ is known to be 4.40.
When the biochemist adds 10.0 mL of 1.0 M HCl to the 10 mL solution with a pH of 7.60, the pH decreases to 3.40. This indicates that the compound's ionizable group with pK₁ = 4.40 is completely protonated at pH 3.40.
The difference between the initial pH and the final pH gives us insight into the protonation state of the second ionizable group. Since pK₂ is between 7 and 10, and the pH decreased by 4 units (from 7.60 to 3.40), it implies that the second ionizable group was only partially protonated at pH 7.60.
To determine the exact value of pK₂, we can use the Henderson-Hasselbalch equation, which relates pH, pK, and the ratio of the concentrations of the ionized and un-ionized forms of the compound. The Henderson-Hasselbalch equation can be written as follows:
pH = pK + log([A-]/[HA])
In this case, [A-] represents the concentration of the ionized form, and [HA] represents the concentration of the un-ionized form. Since the concentration of the un-ionized form is much larger than the ionized form (as the second ionizable group was only partially protonated at pH 7.60), we can simplify the equation to:
pH ≈ pK + log([A-]/[HA]) ≈ pK
By substituting the pH value of 7.60 into the equation, we find:
7.60 ≈ pK
Therefore, the approximate value of pK₂ is 7.60.
However, to determine the exact value of pK₂, we need to consider the change in pH when 10.0 mL of 1.0 M HCl is added. The decrease in pH from 7.60 to 3.40 is 4.20 units. Since pK₂ is greater than 7, it means that the second ionizable group was not fully protonated at pH 7.60. Therefore, the 4.20 unit decrease in pH indicates that the second ionizable group has undergone partial protonation.
To find the exact value of pK₂, we subtract the decrease in pH (4.20) from the initial pH (7.60):
pK₂ = 7.60 - 4.20 = 3.40
Hence, the exact value of pK₂ is 9.40.
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Consider the following fictitious balanced chemical equation below. In the first 580.0 seconds of the resction, the concentration of C increases from 0.1602M to 0.2418M. What is the average rate of this resction? 5 A( g)+4 B( g)→3C(g)+7D(g)4.6 S×10−5M/s1.407×10−4M/s2.470×10−4M/s6.223×10−5M/s4.221×10−4M/s QUESTION 3 Given the rate of change of the concentration of B, deterrine the rate of change of the concentration of A. 5 A( g)+2 B( g)→3C(g)+7D(g)Δ[B]Δtt=−9.652×10−2.413×10−4M/s−6.520×10−5M/s−9.652×10−5M/s−1.566×10−4M/s−3.861×10−5M/s
QUESTION 4 Consider the fictitious reaction and the rate law below. What is the overall order of the reaction? What happens to the rate of the reaction when the concentration of X stays the same and the concentration of Y triples? X(g)+Y⟨g⟩→XY(g) rate =k[X∣2[Y] The overall order of the reaction is third order. The rate increases by a tactor ot 3. The overall order of the reaction is second order. The rate increases by a factor of 3 . The overall order of the reaction is third order. The rate increases by a factor of 9 . The overall order of the reaction is second order. The rate increases by a factor of 9
Option (c) is the correct answer.Consider the following fictitious balanced chemical equation below. In the first 580.0 seconds of the reaction, the concentration of C increases from 0.1602M to 0.2418M.
5 A( g)+4 B( g)→3C(g)+7D(g)To calculate the average rate of a reaction, you can use the formula given below:Average rate=change in concentration of reactant or producttime intervalThe change in concentration of C=0.2418 M - 0.1602 M = 0.0816 MGiven that the time interval is 580.0 s.Average rate = (0.0816 M)/ (580.0 s) = 1.407×10−4 M/s.The average rate of the reaction is 1.407×10−4 M/s. Therefore, the option (b) is the correct answer.QUESTION 4:X(g)+Y⟨g⟩→XY(g)rate =k[X∣2[Y]The overall order of the reaction is third order. The rate increases by a factor of 9.
If we add the exponents of the concentration terms in the rate law equation, we get the order of the reaction. Here, the order of X is 1, and the order of Y is 2. The sum of the exponents is 1+2 = 3, which is the overall order of the reaction.When the concentration of Y triples, the concentration of Y becomes 3 times its original concentration. Therefore, we can replace [Y] by 3[Y] in the rate law equation. The new rate law becomes:rate =k[X][Y]²= k[X][3Y]² = 9k[X][Y]²So, the rate of the reaction increases by a factor of 9. Therefore, the option (c) is the correct answer.
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Check ALL statements that are TRUE in application to electrophilic aromatic substitution (EAS). The arenium ion intermediate acts as an acid in re-aromatization step The re-aromatization step is the rate-limiting step Benzene ring acts as a nucleophile in the eletcrophilic attack step Electrophilic attack step is exothermic
The arenium ion intermediate acts as a base, abstracting a proton from a neighboring carbon atom, thus creating a double bond and re-aromatizing the benzene ring. Therefore, the first statement is false.
The following statements are TRUE in application to electrophilic aromatic substitution (EAS):
The re-aromatization step is the rate-limiting step.
Electrophilic attack step is exothermic.The electrophilic aromatic substitution (EAS) is an organic reaction in which an atom that is attached to an aromatic system (generally hydrogen) is replaced by an electrophile.
In this reaction, the benzene ring acts as a nucleophile in the electrophilic attack step.
It is because the benzene ring is rich in electrons due to the presence of alternating double bonds. Therefore, it can attack electrophiles.
The arenium ion intermediate acts as a base in the re-aromatization step, not an acid. The re-aromatization step is the step in which the aromaticity is restored in the benzene ring after the addition of an electrophile.
In this step, the arenium ion intermediate acts as a base, abstracting a proton from a neighboring carbon atom, thus creating a double bond and re-aromatizing the benzene ring. Therefore, the first statement is false.
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identify the atoms in the ring portion of the haworth structure of glucose.
The atoms in the ring portion of the Haworth structure of glucose are oxygen, carbon, and hydrogen.
The Haworth structure is a way to represent the cyclic form of a monosaccharide, such as glucose. In this structure, the atoms are shown in a planar representation, with the oxygen and carbon atoms forming a hexagonal ring and the hydrogen atoms attached to the carbon atoms.
The ring is formed when the hydroxyl group on the fifth carbon atom of the linear structure reacts with the aldehyde group on the first carbon atom, forming a hemiacetal bond. The resulting ring structure is known as a pyranose ring. The ring structure of glucose is important because it affects the reactivity of the molecule and its interactions with other molecules in biological systems.
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Electron Configuration Diagram for Carbon Singlet and Triplet Forms
The electronic configuration diagram of Carbon in singlet and triplet forms are given below: Singlet State:It has the electron configuration 1s2 2s2 2p2.
It has two unpaired electrons in its outermost shell. The ground state electronic configuration of carbon in the singlet state is shown below. Triplet State:It has the electron configuration 1s2 2s2 2p2. It has two unpaired electrons in its outermost shell. Carbon is a chemical element having an atomic number of 6. Its electronic configuration is 1s2 2s2 2p2.
The electronic configuration of carbon in singlet and triplet forms are shown below. Both the states have two unpaired electrons in its outermost shell.The singlet state of carbon is represented as 1s2 2s2 2p2 with two unpaired electrons in its outermost shell. The triplet state of carbon is represented as 1s2 2s2 2p2 with two unpaired electrons in its outermost shell.
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what's the boiling point of dimethyl disulfide ?
Answer:
109.72°C or 110°C
Explanation:
The boiling point of dimethyl disulfide is 109.72 °C (229.5 °F).
Some additional information about dimethyl disulfide:
Molecular formula:[tex]\tt{ CH_3SSCH_3}[/tex]
Molar mass: 94.19 g/mol
Density: 1.06 g/cm3
Melting point: -85 °C (-121 °F)
Flash point: 10 °C (50 °F)
Solubility in water: 2.5 g/L (20 °C)
State: colorless, flammable liquid with a strong garlic-like odor
Hazards: Flammable, toxic, irritant
Uses: Production of plastics, rubber, pesticides, solvent, reagent.
Answer and Explanation:
The boiling point of dimethyl disulfide is approximately 109-110 degrees Celsius (228-230 degrees Fahrenheit). This is the temperature at which the substance changes from a liquid to a gas under standard atmospheric pressure. It's important to note that boiling points can vary slightly depending on factors such as impurities or the specific conditions under which the measurement is taken. The boiling point of dimethyl disulfide is within a relatively narrow range, allowing it to be easily vaporized for various industrial and chemical processes.
How do I find the solution to this question?
1.) A hydrogen atom absorbs energy and an electron is excited to the =4n=4 shell. When the electron relaxes back to the ground state, =1,n=1, it emits light.
Calculate the change in energy of the electron from the excited state back to the ground state. Then, calculate the frequency of the light that was emitted.
2.) Identify the region on the electromagnetic spectrum where the light that was emitted is located.
a. ultra-violet (UV)
b. visible
c. x rays
d. infrared (IR)
(1) We can use the energy level equation and the relationship between energy and frequency. The change in energy can be calculated by subtracting the energy of the ground state from the energy of the excited state. (2) The light that was emitted when the electron relaxed from the excited state (n=4) to the ground state (n=1) is located in the visible region of the electromagnetic spectrum. (b) visible
The change in energy of the electron can be calculated using the energy level equation: ΔE = E_final - E_initial. In this case, the excited state (n=4) corresponds to the final state, and the ground state (n=1) corresponds to the initial state. The energy of an electron in a hydrogen atom is given by the equation E = -13.6 eV/[tex]n^{2}[/tex], where eV represents electron volts. Plugging in the values for the excited state (n=4) and the ground state (n=1), we can calculate the change in energy.
Once the change in energy is determined, we can use the equation E = hν, where E is the energy of a photon, h is Planck's constant (6.626 x 10^-34 J·s), and ν is the frequency of the emitted light. Rearranging the equation, we can solve for the frequency ν.
Based on the frequency or wavelength of the emitted light, we can identify the region on the electromagnetic spectrum where it is located. In this case, the emitted light corresponds to the transition from the fourth energy level (n=4) to the first energy level (n=1), which corresponds to a transition in the visible region of the electromagnetic spectrum.
Therefore, the change in energy can be calculated using the energy level equation, the frequency of the emitted light can be determined using the energy-frequency relationship, and the region on the electromagnetic spectrum can be identified based on the wavelength or frequency of the emitted light.
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Which of following is the actual amount of ammonia gas if you mix one mole of nitrogen gas and two moles of hydrogen gas with 100% yield? 4/3 moles Two moles 2/3 moles One mole 3/4 moles
If you mix one mole of nitrogen gas and two moles of hydrogen gas with 100% yield, the actual amount of ammonia gas produced would be two moles. The correct answer is: Two moles.
The balanced chemical equation for the reaction between nitrogen gas (N2) and hydrogen gas (H2) to form ammonia gas (NH3) is:
N2 + 3H2 → 2NH3
According to the stoichiometry of the equation, for every one mole of nitrogen gas (N2), three moles of hydrogen gas (H2) are required to produce two moles of ammonia gas (NH3).
Therefore, if you mix one mole of nitrogen gas and two moles of hydrogen gas with 100% yield, the actual amount of ammonia gas produced would be two moles.
Hence, the correct answer is: Two moles.
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The actual amount of ammonia gas produced when one mole of nitrogen gas and two moles of hydrogen gas react with 100% yield is two moles.
Explanation:The balanced equation for the reaction between nitrogen gas (N₂) and hydrogen gas (H₂) to produce ammonia gas (NH₃) is N₂(g) + 3H₂(g) → 2NH₃(g).
Based on the stoichiometry of the reaction, for every 1 mole of nitrogen gas and 2 moles of hydrogen gas, we can obtain 2 moles of ammonia gas with 100% yield.
Therefore, the actual amount of ammonia gas produced when one mole of nitrogen gas and two moles of hydrogen gas react with 100% yield is Two moles.
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Consider the reaction BaCl2+Na2SO4 --> BaSO4 +2NaCl How many grams of solid BaSO4 will be produced if 40.0mL of 0.30M BaCl2 are mixed with 25.0mL of 0.50M Na2SO4?
2.80 grams of BaSO4 will be produced if 40.0mL of 0.30M BaCl2 are mixed with 25.0mL of 0.50M Na2SO4
The balanced chemical equation for the given reaction is;BaCl2 + Na2SO4 → BaSO4 + 2NaClNumber of moles of BaCl2 in 40.0 mL of 0.30M solution;Molarity of BaCl2 = 0.30 MVolume of solution = 40.0 mL = 0.0400 LNumber of moles of BaCl2 = Molarity x Volume= 0.30 mol/L x 0.0400 L= 0.0120 moles of BaCl2.
Similarly, Number of moles of Na2SO4 in 25.0 mL of 0.50M solution;Molarity of Na2SO4
= 0.50 MVolume of solution
= 25.0 mL
= 0.0250 L
Number of moles of Na2SO4 = Molarity x Volume
= 0.50 mol/L x 0.0250 L
= 0.0125 moles of Na2SO4. In the reaction, BaCl2 and Na2SO4 react in 1:1 molar ratio.Therefore, 0.0120 moles of BaCl2 react with 0.0120 moles of Na2SO4.In the reaction, BaSO4 is formed which means the amount of BaSO4 produced is equal to the amount of BaCl2 reacted.0.0120 moles of BaCl2 produces 0.0120 moles of BaSO4.Molar mass of BaSO4 = 233.39 g/mol. Number of grams of BaSO4 produced;
Mass = Number of moles x Molar mass
= 0.0120 moles x 233.39 g/mol
= 2.80 g. Hence, 2.80 grams of BaSO4 will be produced if 40.0mL of 0.30M BaCl2 are mixed with 25.0mL of 0.50M Na2SO4
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Sulfur tetrefluoride gas is collected at 31.0 C in an evacuated fask with a measured volume of 10.0.1. When all the gas has been collected, the pressure in the faskis measured to be 0.360 atm. Calculate the mass and number of moles of sulfur tetrafluoride gas that were collected. Round your answer to 3 significant digits.
The mass of sulfur tetrafluoride gas that was collected is 21.09 g. The number of moles of sulfur tetrafluoride gas that was collected is 0.144 mol.
Here are the steps to calculate the mass and number of moles of sulfur tetrafluoride gas that were collected:
Calculate the number of moles of sulfur tetrafluoride gas using the ideal gas law:
PV = nRT
where:
P is the pressure of the gas (0.360 atm)
V is the volume of the gas (10.01 L)
R is the ideal gas constant (0.08206 L * atm / mol * K)
T is the temperature of the gas (31.0 + 273.15 = 304.15 K)
n = PV / RT = (0.360 atm) * (10.01 L) / (0.08206 L * atm / mol * K) * 304.15 K = 0.144 mol
Calculate the mass of sulfur tetrafluoride gas using the molar mass of sulfur tetrafluoride (146.06 g/mol):
m = n * M = 0.144 mol * 146.06 g/mol = 21.09 g
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Aqueous sulfuric acid (H
2
SO
4
) will react with solid sodium tydroxide (NaOH) to produce aqueous sodium suifate (Na2SO ) and Hquid water (H
2
O). Suppose 72.69 of sulfuric acid is mixed with 37.9 of sodium hydroxide. Calculate the minimum mass of sulfuric acid that could be left over by the chemical reaction. found your answer to 2 significant digits.
We need to determine the limiting reagent and compare the amount of sulfuric acid used to the initial mass of sulfuric acid. The minimum mass of sulfuric acid that could be left over by the chemical reaction is approximately 26.2 grams (rounded to two significant digits).
To calculate the minimum mass of sulfuric acid (H2SO4) that could be left over by the chemical reaction with sodium hydroxide (NaOH), we need to determine the limiting reagent and compare the amount of sulfuric acid used to the initial mass of sulfuric acid.
First, we need to find the limiting reagent by comparing the number of moles of each reactant.
Molar mass of H2SO4 = 2 * (1.01 g/mol (hydrogen) + 32.07 g/mol (sulfur) + 4 * 16.00 g/mol (oxygen)) = 98.09 g/mol
Molar mass of NaOH = 22.99 g/mol (sodium) + 16.00 g/mol (oxygen) + 1.01 g/mol (hydrogen) = 39.99 g/mol
Moles of H2SO4 = 72.69 g / 98.09 g/mol ≈ 0.741 moles
Moles of NaOH = 37.9 g / 39.99 g/mol ≈ 0.948 moles
From the balanced equation: H2SO4 + 2NaOH -> Na2SO4 + 2H2O, we can see that the stoichiometric ratio between H2SO4 and NaOH is 1:2.
Since the number of moles of H2SO4 (0.741 moles) is less than half the number of moles of NaOH (0.948 moles), H2SO4 is the limiting reagent.
To determine the minimum mass of H2SO4 left over, we use the stoichiometric ratio between H2SO4 and NaOH.
Moles of H2SO4 used = 0.741 moles
Moles of H2SO4 left over = Moles of H2SO4 initially - Moles of H2SO4 used
= 0.741 moles - 0.5 * Moles of NaOH
= 0.741 moles - 0.5 * 0.948 moles
≈ 0.741 moles - 0.474 moles
≈ 0.267 moles
Mass of H2SO4 left over = Moles of H2SO4 left over * Molar mass of H2SO4
= 0.267 moles * 98.09 g/mol
≈ 26.2 g
Therefore, the minimum mass of sulfuric acid that could be left over by the chemical reaction is approximately 26.2 grams (rounded to two significant digits).
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Galvanic corrosion is the degradation of one metal near a joint or juncture that occurs when two electrochemically dissimilar metals are in electrical contact in an electrolytic environment. On the other hand, microbiologically influenced corrosion (MIC) is metal deterioration as a result of the metabolic activity of various microorganisms. a. Evaluate FiVE (5) strategies that can be adapted to minimize the impact of galvanic corrosion of metalic parts that are submerged in a controlled liquid environment. (CO2:PO4 - 10 Marks) b. Select THREE (3) anti-corrosion strategies to be applied at surfaces that are prone to MIC during prolonged submergence in seawater. (CO3:PO9 - 15 Marks)
Galvanic corrosion is the degradation of one metal near a joint or juncture that occurs when two electrochemically dissimilar metals are in electrical contact in an electrolytic environment.
Here are five strategies that can be adopted to minimize the impact of galvanic corrosion of metallic parts that are submerged in a controlled liquid environment:
1. Selecting electrochemically similar metals: Choosing metals that have a similar electrochemical reaction will minimize the likelihood of galvanic corrosion.
2. Isolation of the metals: Placing an insulator such as plastic, rubber or ceramic between the two metals to avoid electrical contact is another way to prevent galvanic corrosion.
3. Removing electrolytes: This method can be used to stop the flow of electricity by removing the electrolytes from the environment in which the metals are placed.
4. Coating or plating: Applying an electrochemically neutral coating or plating on the metal surfaces, which acts as a barrier, can prevent corrosion.
5. Cathodic protection: In this method, a cathode is connected to the metal, which reduces the anodic current. It involves providing an external source of electrons, which protects the anode surface. Three anti-corrosion strategies that can be applied at surfaces that are prone to microbiologically influenced corrosion (MIC) during prolonged submergence in seawater are as follows:
1. Biocide treatment: Adding biocides such as chlorine, bromine, iodine, or oxidizing agents such as hydrogen peroxide to the water can kill the microorganisms responsible for corrosion.
2. Cathodic protection: A sacrificial anode, such as zinc or magnesium, can be attached to the metal. As a result, the anode is consumed instead of the metal surface, preventing corrosion.
3. Coating: Applying coatings such as epoxy or vinyl ester to the metal surface, which prevents the microorganisms from contacting the metal surface, is also an effective method of preventing corrosion.
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Show which are necessarily Hermitian operators or non-Hermitian operators, where
A
^
is an arbitrary operator: (i)
A
^
+
A
^
+
(ii)
A
^
A
^
+
(iii)
A
^
−
A
^
†
(iv) exp(
A
^
−
A
^
+
) (v) exp(
A
^
+
A
^
†
) 0.5×5=2.5
In summary:
- (i) A^+ A^ is necessarily Hermitian.
- (ii) A^ A^+ is non-Hermitian in general.
- (iii) A^ - A^† is non-Hermitian in general.
- (iv) exp(A^ - A^+) is non-Hermitian in general.
- (v) exp(A^ + A^†) is necessarily Hermitian.
To determine whether the given operators are necessarily Hermitian or non-Hermitian, we need to examine their properties.
(i) A^+ A^: This operator is necessarily Hermitian. The Hermitian conjugate of the operator (A^+ A^) is (A^+ A^)† = A^† (A^+)† = A^† A^, which is equal to the original operator.
(ii) A^ A^+: This operator is non-Hermitian in general. The Hermitian conjugate of the operator (A^ A^+) is (A^ A^+)† = (A^+)† A^† = A^ A^+, which is not equal to the original operator unless A^ commutes with A^+.
(iii) A^ - A^†: This operator is non-Hermitian in general. The Hermitian conjugate of the operator (A^ - A^†) is (A^ - A^†)† = (A^†)† - (A^)† = A^ - A^†, which is not equal to the original operator unless A^ is Hermitian.
(iv) exp(A^ - A^+): This operator is non-Hermitian in general. The Hermitian conjugate of the operator exp(A^ - A^+) is (exp(A^ - A^+))† = exp((A^ - A^+)†) = exp(A^ - A^†), which is not equal to the original operator unless A^ is Hermitian.
(v) exp(A^ + A^†): This operator is necessarily Hermitian. The Hermitian conjugate of the operator exp(A^ + A^†) is (exp(A^ + A^†))† = exp((A^ + A^†)†) = exp(A^† + A^), which is equal to the original operator.
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Which of the following has the greatest London dispersion forces acting between molecules of the same kind? A) B) CH
3
CH
2
CH
2
CH
3
C) D) CH
3
CH
2
CH
2
CH
2
CH
3
8) Answer E)
The molecules with the greatest London dispersion forces are A) butane (CH3CH2CH2CH3), C) pentane (CH3CH2CH2CH2CH3), and D) hexane (CH3CH2CH2CH2CH2CH3).
To determine which molecule has the greatest London dispersion forces, we need to consider the molecular size and shape.
A) CH3CH2CH2CH3 (butane): Butane is a straight-chain alkane with four carbon atoms. The molecule is relatively large and has a linear shape, allowing for a larger surface area of contact between molecules. This leads to stronger London dispersion forces.
B) CH3CH2CH2CH2 (butylamine): Butylamine is also a straight-chain molecule with four carbon atoms. However, it has an amino group (-NH2) attached to the end, which adds some polarity to the molecule. While the presence of the amino group can introduce some dipole-dipole interactions, the London dispersion forces will still be significant due to the size and shape of the molecule.
C) CH3CH2CH2CH2CH3 (pentane): Pentane is a straight-chain alkane with five carbon atoms. Like butane, it is relatively large and has a linear shape, resulting in strong London dispersion forces.
D) CH3CH2CH2CH2CH2CH3 (hexane): Hexane is a straight-chain alkane with six carbon atoms. Again, it is relatively large and linear, leading to strong London dispersion forces.
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he following compounds are in their chair conformations. Label the each cyclohexane below as the cis or trans isomer. 7. Label each carbon on the following compound as 1
∘
,2
∘
,3
∘
, or 4
∘
.
Cyclohexane with methyl groups is a cis isomer of cyclohexane.
Cyclohexane with ethyl groups is a trans isomer of cyclohexane.
The carbons are labeled as 1°, 2°, 3°, and 4°.
In the given question, the following compounds are in their chair conformations.
We have to label each cyclohexane below as cis or trans isomer, and we have to label each carbon on the following compound as 1 ∘, 2 ∘, 3 ∘, or 4 ∘.
The given structures are shown below:
Cyclohexane with methyl groups:
Here, each of the methyl groups is equatorial because it can't fit axial position without making 1,3-diaxial interactions. This compound is a cis isomer of cyclohexane.
It is because cis-isomers have similar substituents on the same side.
Cyclohexane with ethyl groups:
Here, one of the ethyl groups is equatorial and the other axial.
This compound is a trans isomer of cyclohexane.
It is because trans-isomers have similar substituents on opposite sides.
Labeling each carbon:
In the given compound, there are four carbons present.
The carbon atoms connected directly to two other carbon atoms are sp3 hybridized and called quaternary carbons.
These quaternary carbons are numbered as 1°.
The carbon atoms connected directly to one other carbon atom and two hydrogen atoms are sp3 hybridized and called tertiary carbons.
These tertiary carbons are numbered as 2°.
The carbon atoms connected directly to one other carbon atom and one hydrogen atom are sp3 hybridized and called secondary carbons.
These secondary carbons are numbered as 3°.
The carbon atom connected directly to two hydrogen atoms is sp3 hybridized and called primary carbon.
This primary carbon is numbered as 4°.
The labeling of each carbon is shown below:
Therefore, the correct answers are:
Cyclohexane with methyl groups is a cis isomer of cyclohexane.
Cyclohexane with ethyl groups is a trans isomer of cyclohexane.
The carbons are labeled as 1°, 2°, 3°, and 4°.
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what is the total number of valence electrons in one molecule of n2o3?
The total number of valence electrons in one molecule of N2O3 is 24.
Nitrogen (N) and Oxygen (O) are both nonmetals with varying numbers of valence electrons.
So, you need to find out the number of valence electrons for each atom in N2O3.
Valence electrons of each atom: N = 5 valence electrons O = 6 valence electrons
So, for N2O3:
N2O3 has two nitrogen atoms (2 x 5 valence electrons) + three oxygen atoms (3 x 6 valence electrons)
= 10 + 18 = 28 valence electrons
The total number of valence electrons in one molecule of N2O3 is 28.
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Which of the following is the correctly charged balanced equation for the reaction below, and has the correct standard cell potential? Cr(s)+Zn
2+
(aq)→Zn(s)+Cr
3+
(aq)
2Cr(s)+3Zn
2+
(aq)→3Zn(s)+2Cr
3+
(aq),E
0
=−1.50 V
2Cr(s)+3Zn
2+
(aq)→3Zn(s)+2Cr
3+
(aq),E
0
=−0.02 V
Cr(s)+Zn
2+
(aq)→Zn(s)+Cr
3+
(aq),E
0
=0.02 V
Zn(s)+Cr
3+
(aq)→Cr(s)+Zn
2+
(aq), E
0
=−0.02 V
Consider the following unbalanced redox equation: M
+
(aq)+X(s)⟶M(s)+X
3+
(aq)K=7×10
37
Calculate the E
cell
0
for this reaction (NB: Make sure to balance the reaction first!). Consider the following half reactions:
M
1+
(aq)+1e
−
(aq)⟶M(s)
X
1+
(aq)+1e
−
(aq)⟶X(s)
E
0
=0.2
E
0
=0.5
Calculate the Gibbs free energy for the overall spontaneous redox reaction. Give your answer to one decimal place and in kJ. Consider the following half reactions:
M
2+
(aq)+2e
−
(aq)⟶M(s)
X
2+
(aq)+2e
−
(aq)⟶X(s)
E
0
=1.22
E
0
=0.59
Calculate the spontaneous cell potential when [M
2+
]=0.8M and [X
2+
]=1.1M
The correctly charged balanced equation for the given reaction is:
Cr(s) + Zn2+(aq) → Zn(s) + Cr3+(aq)
The correct standard cell potential for this reaction is:E0 = -0.02 VTo calculate the Ecell0 for the unbalanced redox equation:M+(aq) + X(s) ⟶ M(s) + X3+(aq)First, we need to balance the equation:
Ecell = Ecell0 - (0.0592/n) * log([X2+]/[M2+])Plugging in the values:The correctly charged balanced equation for the given reaction is:
Cr(s) + Zn2+(aq) → Zn(s) + Cr3+(aq)
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To calculate the spontaneous cell potential when [M2+] = 0.8 M and [X2+] = 1.1 M, we can use the Nernst equation
Ecell = Ecell0 - (0.0592 V/n) * log([Zn2+]^3/[Cr3+]^2)
The correctly charged balanced equation for the reaction is: 2Cr(s) + 3Zn2+(aq) → 3Zn(s) + 2Cr3+(aq). The correct standard cell potential for this reaction is E0 = -1.50 V.
To calculate the Ecell0 for the given unbalanced redox equation: M+(aq) + X(s) → M(s) + X3+(aq), we need to balance the reaction first.
Balanced equation: 6M+(aq) + 2X(s) → 2M(s) + 3X3+(aq)
The Ecell0 for this reaction can be calculated by summing the half-cell potentials of the half reactions involved. The half reactions are:
M1+(aq) + 1e- → M(s) E0 = 0.2 V
X1+(aq) + 1e- → X(s) E0 = 0.5 V
The Gibbs free energy (∆G) for the overall spontaneous redox reaction can be calculated using the formula ∆G = -nFE, where n is the number of electrons transferred and F is the Faraday constant. Here, n = 2.
∆G = -2(96.485 C/mol)(Ecell0)
To calculate the spontaneous cell potential when [M2+] = 0.8 M and [X2+] = 1.1 M, we can use the Nernst equation:
Ecell = Ecell0 - (0.0592 V/n) * log([Zn2+]^3/[Cr3+]^2)
Substitute the values and calculate the Ecell.
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Which of the following factors does NOT influence the stability of a resonance form?
A) The possibility of gaining or losing aromaticity depending on the location of electrons
B) The electronegativity of the atoms bearing charges
C) The number of heteroatom (non-carbon atoms) included in the structure
D) The number of formal charges the structure has
The factor that does NOT influence the stability of a resonance form is the number of heteroatoms (non-carbon atoms) included in the structure. Thus, option (C) is correct.
Resonance stability is primarily determined by factors such as delocalization of electrons, distribution of charges, and aromaticity. The presence of heteroatoms in the structure can affect the polarity or reactivity of the molecule but does not directly impact the stability of a resonance form.
Therefore, the number of heteroatoms is not a significant factor in assessing the stability of resonance forms.
In conclusion, the factor that does NOT influence the stability of a resonance form is the number of heteroatom (non-carbon atoms) included in the structure.
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A clean, dry container was placed on a balance and the mass recorded as 25.0678 g. Using a standard burette, a sample of a clear liquid was placed inside the container and the new mass recorded as 40.3454 g. The initial volume of the liquid in the burette was recorded as 0.00mI and the final volume as 15.30ml. Calculate the density of the liquid using the proper number of sib figs. 5 points 2. Which law does the picture below describe? Who was key in the development? 2 points Novemanc num mannumnnim 3. Match the following (can have more than one answer and not all answers are used) by writing the numbers in the blanks A) JJ Thompson B) Robert Millikan C) Ernest Rutherford 1)Oil Drops experiment 2)electron discovery 3) neutron discovery 4) gold foll 5) law of conservation of nass 6) proton discovery 7) cathode ray experiment 8) Saturn model 9) nuclear model 10) plumb udding model
The density of the given liquid is 0.9985 g/mL. The scientist are matched with their discoveries.
To determine the density, we divide the mass by the volume, considering the appropriate significant figures.The density of a substance is calculated by dividing its mass by its volume. In this scenario, the mass of the liquid is the difference between the initial and final masses of the container, which is 40.3454 g - 25.0678 g = 15.2776 g. The volume is the difference between the initial and final volumes of the liquid in the burette, which is 15.30 mL - 0.00 mL = 15.30 mL.
To calculate the density, we divide the mass by the volume: density = mass/volume. The result is 15.2776 g / 15.30 mL = 0.9985 g/mL. The density of the liquid, considering the proper number of significant figures, is approximately 0.9985 g/mL.
Moving on to the second question, the picture mentioned likely describes the "Oil Drops experiment," which was key in the discovery of the charge of an electron. This experiment was conducted by Robert Millikan.
For the third question, the matches between the scientists and their respective experiments/models are as follows:
A) JJ Thompson: 7) cathode ray experiment, 2) electron discovery.
B) Robert Millikan: 1) Oil Drops experiment.
C) Ernest Rutherford: 4) gold foil experiment, 6) proton discovery, 9) nuclear model.
These matches highlight the contributions of these scientists to the understanding of atomic structure and the discovery of key particles such as electrons, protons, and the nuclear model of the atom.
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Using chemical equations, show how the triprotic acid H
3
PO
4
ionizes in water. Phases are optional.
The triprotic acid H[tex]_3[/tex]PO[tex]_4[/tex] ionizes in water to produce H+ ions and phosphate ([tex]\[ \text{PO}_4^{3-} \][/tex]) ions through three successive ionization reactions: [tex]\[H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^-\][/tex], [tex]\[ \text{H}_2\text{PO}_4^- \rightleftharpoons \text{H}^+ + \text{HPO}_4^{2-} \][/tex], and [tex]\[ \text{HPO}_4^{2-} \rightleftharpoons \text{H}^+ + \text{PO}_4^{3-} \][/tex].
Triprotic acids, such as H[tex]_3[/tex]PO[tex]_4[/tex], have three ionizable hydrogen atoms. When H[tex]_3[/tex]PO[tex]_4[/tex] is dissolved in water, it undergoes ionization through a stepwise process. In the first ionization step, one hydrogen atom is released as an H+ ion, resulting in the formation of the dihydrogen phosphate ion ([tex]\[ \text{H}_2\text{PO}_4^-[/tex]). The equation for this reaction is:
[tex]\[H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^-\][/tex]
In the second ionization step, another hydrogen atom is released as an H+ ion, forming the hydrogen phosphate ion ([tex]\(\text{HPO}_4^{2-}\)[/tex]). The equation for this reaction is:
[tex]\[ \text{H}_2\text{PO}_4^- \rightleftharpoons \text{H}^+ + \text{HPO}_4^{2-} \][/tex]
Finally, in the third ionization step, the last hydrogen atom is released as an H+ ion, resulting in the formation of the phosphate ion ([tex]\[ \text{PO}_4^{3-} \][/tex]). The equation for this reaction is:
[tex]\[ \text{HPO}_4^{2-} \rightleftharpoons \text{H}^+ + \text{PO}_4^{3-} \][/tex]
These ionization reactions progressively release H+ ions and form negatively charged phosphate ions. The extent of ionization at each step depends on factors such as acid concentration, pH, and equilibrium constants.
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Vaper Pressuee of tiguid? DATA AND CALCULATIONS Do the folbwing: 1. Total Pressure = Air pressure + Vapor pressure 2. Vapor pressure = Total pressure - Air pressure 3. Also find: Δ Hvap using Clausius-Clapeyron Equation lnP=RΔHvap⋅T1+B y=mx+c
the vapor pressure of tiguid, you can use the following steps and equations Total Pressure = Air pressure + Vapor pressure Vapor the total pressure by adding the air pressure and the vapor pressure.
This will give you the total pressure in the system. Subtract the air pressure from the total pressure to find the vapor pressure. This will give you the partial pressure of the vapor in the system. To find ΔHvap, you can use the Clausius-Clapeyron equation. The equation ln(P) = R * ΔHvap / T + B relates the natural logarithm of the vapor pressure (P) to the enthalpy of vaporization (ΔHvap), the temperature (T), and a constant (B). R is the gas constant.
Rearrange the equation to solve for ΔHvap: ΔHvap = (ln(P) - B) * T / R. Plug in the values for ln(P), B, T, and R to find the enthalpy of vaporization.
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