Answers
Answer:
The answer is 3-Phenylpropanoic acid (see attached structure)
Explanation:
From spectral data:
3005 cm-1 ⇒ carboxylic acid (broad band)
1670 cm-1 ⇒ C=C
1603 cm-1 ⇒ Aromatic C-C bond
H NMR frequency at 2.6 ppm, singlet, ⇒ OH with no surrounding protons, possible deshielding (clearer investigation of spectrum would be expedient).
Hence, our C9H10O2 compound has an aromatic ring and carboxylic acid group attached to it.
The compound is probably a molecule with an aromatic ring (benzene) with an attached ester group (COOCH3) that satisfies all the given spectral data.
Explanation:The spectral data given corresponds to a compound with molecular formula C9H10O2. From the IR data, the bands at 3005 cm-1 suggests C-H sp2 bond (alkene or aromatic), at 1676 cm-1 indicates carbonyl group (C=O), and 1603 cm-1 suggests a carbon-carbon double bond (C=C) which might be in an aromatic ring. The 1H NMR data: 2.6 ppm singlet signifies the protons of a methyl group (CH3) attached to an electronegative atom like a carbonyl carbon.
Based on these data, a probable structure for the compound could be a molecule, which is an aromatic ring (benzene) with an attached ester group (COOCH3). That gives the right molecular formula, the required carbonyl, alkene and sp2 hybridized C-H bonds for the IR, and the singlet in the NMR for the methyl group (CH3) of the ester.
Learn more about Structure of Compound here:https://brainly.com/question/31468426
#SPJ3
Related Questions
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 7.50 l tank with 29.0 mol of ammonia gas at 35.0 °C. She then raises the temperature, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 13.0 mol.
Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to significant digits.
Answers
Answer: The equilibrium constant for the reaction is 2.47
Explanation:
We are given:
Initial moles of ammonia gas = 29.0 moles
Equilibrium moles of nitrogen gas = 13.0 moles
Volume of the tank = 7.50 L
Molarity is calculated by using the formula:
[tex]\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of tank}}[/tex]
[tex]\text{Initial molarity of ammonia}=\frac{29.0}{7.50}=3.87M[/tex]
[tex]\text{Equilibrium molarity of nitrogen gas}=\frac{13.0}{7.50}=1.73M[/tex]
The chemical equation for the decomposition of ammonia follows:
[tex]2NH_3\rightleftharpoons N_2+3H_2[/tex]
Initial: 3.87
At eqllm: 3.87-2x x 3x
Evaluating the value of 'x'
[tex]\Rightarrow 3x=1.73\\\\x=\frac{1.73}{3}=0.577[/tex]
So, equilibrium concentration of ammonia = (3.87 - 2x) = [3.87 - 2(0.577)] = 2.716 M
Equilibrium concentration of nitrogen gas = x = 0.577 M
The expression of [tex]K_{eq}[/tex] for above equation follows:
[tex]K_{eq}=\frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]
Putting values in above equation, we get:
[tex]K_{eq}=\frac{(2.716)^2}{0.577\times (1.73)^3}\\\\K_{eq}=2.47[/tex]
Hence, the equilibrium constant for the reaction is 2.47
A sucrose solution is prepared to a final concentration of 0.210 MM . Convert this value into terms of g/Lg/L, molality, and mass %%. (Use the following values: molecular weight MWsucroseMWsucrose = 342.296 g/molg/mol ; density rhosol′nrhosol′n = 1.02 g/mLg/mL ; and mass of water, mwatmwat = 948.1 gg ). Note that the mass of solute is included in the density of the solution.
Answers
Answer:
1) 71.9 g/L
2) 0.221 m olal
3) 7.05% by mass
Explanation:
Step 1: Data given
Concentration of sucrose = 0.210 M
Molar weight of sucrose = 342.3 g/mol
Density of solution = 1.02 g/mL
Mass of water = 948.1 grams
Step 2: Convert this value into terms of g/L
(0.210 mol/L) * (342.3 g/mol) = 71.9 g/L
Calculate the molality
Step 1: Calculate mass water
Suppose we have a volume of 1.00L
Mass of the solution = 1000 mL * 1.02 g/mL = 1020 g solution
We know that there are 71.9 g of solute in a liter of solution from the first calculation. This means
(1020 grams solution) - (71.9 g solute) = 948.1 g = 0.9481 kg water
Step 2: Calculate molality
Molality = moles sucrose / mass water
(0.210 mol) / (0.9481 kg) = 0.221 mol/kg = 0.221 m olal
Mass %
% MAss = (mass solute / mass solution)*100%
(71.9 g) / (1020 g) *100% = 7.05% by mass
In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class to allow the gas to reach room temperature. If the pressure gauge reads 173 psi and the lecture hall is 20°C, how many moles of sulfur dioxide are in the bottle? In order to solve this problem, you will first need to calculate the pressure of the gas. Hint: The gauge reads zero when 14.7 psi of gas remains.
Answers
Answer:
0.66 mol
Explanation:
Zero Gauge pressure = 14.7 psi
Pressure read = 173 psi
Actual pressure = 173 psi - 14.7 psi = 158.3 psi
P (psi) = 1/14.696 P(atm)
So, Pressure = 10.77 atm
Given that:
Temperature = 20 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (20+ 273.15) K = 298.15 K
V = 1.50 L
Using ideal gas equation as:
[tex]PV=nRT[/tex]
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
10.77 atm × 1.50 L = n ×0.0821 L atm/ K mol × 298.15 K
⇒n = 0.66 mol
Molten solder flows between the two base metals being soldered because ____. A. the atmospheric pressure pushes it in B. of capillary action C. there are no surface pores in the base metal D. the tubing surface melts and the pressure differential forces it in
Answers
Answer:
Molten solder flows between the two base metals being soldered because of capillary action.
Explanation:
Capillary molten solder takes place when a metal and a melt is brought into contact with the tube and an accessory after heating. Due to the phenomenon of capillary action, the molten metal rises and extends in any direction, due to the small space that remains between the wall of the tube and that of the fitting; With this, when cooling, a completely hermetic union is achieved.
Molten solder flows between two base metals due to capillary action, which allows the solder to spread evenly and adhere to the bases, metal ensuring a strong bond.
Explanation:Molten solder flows between the two base metals being soldered mainly because of a process known as capillary action. Capillary action is a natural occurrence where a liquid, in this case the molten solder, moves along a narrow space, such as between the two pieces of metal, against the force of gravity. The solder is drawn into the space and spread evenly between the two base metals because of the adhesive forces between the liquid and the surrounding materials are stronger than the cohesive forces within the liquid.
This phenomenon plays a fundamental role in the soldering process, as it allows the solder to spread evenly and adhere to the base metals, ensuring a strong and stable bond between them. This principle is used not only in soldering but also in various other fields such as plant biology, painting, and inkjet printing.
Learn more about Capillary Action:https://brainly.com/question/14451466
#SPJ3
A chemist adds 55.0mL of a ×1.710−5/mmolL mercury(II) iodide solution to a reaction flask. Calculate the micromoles of mercury(II) iodide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
Answers
Answer:
The answer is 0.017 μmol of mercury iodide
Explanation:
To know the micromoles we must use the following conversion, 1 mmol = 1000μmol. we use a rule of three for the calculation of micromoles:
1mmol------------------1000μmol
1.7x10^-5mmol------- Xμmol
Clearing the X, we have:
X μmol = (1.7x10^-5x1000)/1 = 0.017μmol
how many moles of solute are in 300 mL of 1.5 M CaCl2? How many grams fo CaCl2 is this?
Answers
Answer:
1. 0.45 mole
2. 49.95g
Explanation:
The following were obtained from the question:
Volume of solution = 300mL = 300/1000 = 0.3L
Molarity = 1.5 M
Mole of CaCl2 =?
1. We can obtain the mole of the solute as follow:
Molarity = mole of solute /Volume of solution
1.5 = mole of solute/0.3
Mole of solute = 1.5 x 0.3
Mole of solute = 0.45 mole
2. The grams in 0.45 mole of CaCl2 can be obtained as follow:
Molar Mass of CaCl2 = 40 + (35.5 x 2) = 40 + 71 = 111g/mol
Mole of CaCl2 = 0.45 mole
Mass of CaCl2 =?
Mass = number of mole x molar Mass
Mass of CaCl2 = 0.45 x 111
Mass of CaCl2 = 49.95g
Answer:
We have 0.45 moles CaCl2 in this 1.5 M solution
This 49.9 grams of CaCl2
Explanation:
Step 1: data given
Volume of the CaCl2 solution = 300 mL = 0.300 L
Molarity of the CaCl2 solution = 1.5 M
Molar mass CaCl2 = 110.98 g/mol
Step 2: Calculate number of moles in the solution
Moles CaCl2 = molarity solution * volume of solution
Moles CaCl2 = 1.5 M * 0.300 L
Moles CaCl2 = 0.45 moles
Step 3: Calculate mass CaCl2
Mass CaCl2 = moles CaCl2 * molar mass CaCl2
Mass CaCl2 = 0.45 moles * 110.98 g/mol
Mass CaCl2 = 49.9 grams CaCl2
We have 0.45 moles CaCl2 in this 1.5 M solution
This 49.9 grams of CaCl2
Select the statements that are correct with respect to waste disposal in the Electrochemical Cells Experiment. (Select all that apply.)
a. Dilute copper and ascorbic acid solutions should be disposed of in the waste container in the hood.
b. Rinsings from the half-cell module can be flushed down the sink.
c. Metal and graphite electrodes should be rinsed with water, dried, and returned to their positions on the lab bench.
d. The contents of the half-cell module should be disposed of in the waste container in the hood.
e. No waste will be generated during this experiment.
f. The electrodes should be discarded in the proper jar.
Answers
Answer:A. Dilute copper and ascorbic acid solutions should be disposed of in the waste container in the hood.
C. Metal and graphite electrodes should be rinsed with water, dried, and returned to their positions on the lab bench.
D. The contents of the half-cell module should be disposed of in the waste container in the hood.
Explanation: An electrochemical cell is a cell that has the capability of producing Electric energy from chemical reaction (voltaic cells) or using Electric energy to make chemical reactions to take place(electrolytic cell). For a proper or effective waste disposal in an electrochemical cell the following options are correct.
Dilute copper and ascorbic acid solutions should be disposed of in the waste container in the hood.
Metal and graphite electrodes should be rinsed with water, dried, and returned to their positions on the lab bench.
The contents of the half-cell module should be disposed of in the waste container in the hood.
In the Electrochemical Cells Experiment, the correct statements regarding waste disposal include disposing copper and ascorbic acid solutions in the waste container, flushing rinsings from the half-cell module down the sink, and rinsing and drying the electrodes.
Explanation:The correct statements with respect to waste disposal in the Electrochemical Cells Experiment are:
Dilute copper and ascorbic acid solutions should be disposed of in the waste container in the hood.Rinsings from the half-cell module can be flushed down the sink.Metal and graphite electrodes should be rinsed with water, dried, and returned to their positions on the lab bench.The contents of the half-cell module should be disposed of in the waste container in the hood.No waste will be generated during this experiment.The electrodes should be discarded in the proper jar.he rate constant for this zero‑order reaction is 0.0130 M ⋅ s − 1 at 300 ∘ C. A ⟶ products How long (in seconds) would it take for the concentration of A to decrease from 0.890 M to 0.280 M?
Answers
Answer:
188 s
Explanation:
We are told the reaction is second order respect to A so we know the expression for the rate law is
rate = - Δ[A]/Δt = k[A]²
where the symbol Δ stands for change, [A] is the concentration of A, and k is the rate constant.
The integrated rate law for this equation from calculus is
1 / [A]t = kt + 1/[A]₀
where [A]t is the concentration of A at time t, k is the rate constant, and [A]₀ is the initial concentration.
Since we have all the information required to solve this equation lets plug our values
1 / 0.280= 0.0130x t + 1 / 0.890
( 1 / 0.280 - 1 / 0.890)M⁻¹ = 0.0130 M⁻¹ ·s⁻¹t
t = 188 s
Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.90 moles of N2(g) react at standard conditions. S°surroundings =
Answers
Final answer:
To calculate the entropy change for the surroundings during the reaction of N2(g) and O2(g) to form NO2(g), the standard enthalpy change of the reaction must be known. This value can be used with the equation ΔS° = -ΔH°/T to find the entropy change of the surroundings.
Explanation:
The student is asking to calculate the entropy change for the surroundings (ΔS°surroundings) when 1.90 moles of N2(g) react with O2(g) to form NO2(g) according to the reaction N2(g) + 2O2(g) → 2NO2(g) at standard conditions of 298 K.
To find this, we'll first need the standard enthalpy change (ΔH°) for the reaction, which can be obtained from standard thermodynamic tables. We then apply the equation ΔS° = -ΔH°/T, which relates the entropy change of the surroundings to the enthalpy change of the system at a constant temperature (T).
Given that the standard enthalpy change for the formation of NO2(g) is 33.2 kJ/mol, and the reaction produces 2 moles of NO2 for 1 mole of N2, the standard enthalpy change for the reaction when 1.90 moles of N2 react is (1.90 moles * 33.2 kJ/mol * 2). We'll convert kJ to J by multiplying by 1,000 and then calculate ΔS°surroundings.
Learn more about Entropy Change for the Surroundings here:
https://brainly.com/question/28244712
#SPJ11
Final answer:
The standard entropy change for the reaction N2(g) + 2O2(g) → 2NO2(g) at 298K can be calculated using the equation: ΔS° = 2∙S°(NO2) - [S°(N2) + 2∙S°(O2)]. The entropy change is -198.3 J/mol K.
Explanation:
The standard entropy change for the reaction N2(g) + 2O2(g) → 2NO2(g) at 298K can be calculated using the equation: ΔS° = 2∙S°(NO2) - [S°(N2) + 2∙S°(O2)].
Using the standard entropy values at 298K from the reference table, we can substitute the values and calculate the entropy change:
ΔS° = 2∙192.5 - [191.5 + 2∙130.6]
ΔS° = -198.3 J/mol K
Arrange the following aqueous solutions, all at 25 ∘C, in order of decreasing acidity. Rank from most acidic to most basic. To rank items as equivalent, overlap them.
Answers
Answer:
Rank from most acidic to most basic is
0.0023 M HClpH = 5.45 and pOH = 8.550.0018 M KOHExplanation:
This question is missing options.
Options are
pOH = 8.55 pH= 5.45 0.0023 M HCl 0.0018 M KOHTo rank these solutions first calculate either pH or pOH of these solutions.
We will use pH as an indicator to rank these.
Relation between pH and pOH is
pH + pOH = 14 ................... Eq (A)
For pOH = 8.55
use equation A
pH + 8.55 = 14
pH = 5.45
For pH = 5.45
pH = 5.45
For 0.0023 M HCl
HCl is a strong acid. It will ionize 100% in aqueous solution and produce Hydronium Ion. Formula to calculate pH of 0.0023 M HCl is,
pH = - log (Molarity)
pH = - log (0.0023)
pH = - (-2.64)
pH = 2.64
For 0.0018 M KOH
KOH is a base. It will produce Hydroxide Ion in aqueous solution. First calculate pOH and convert it into pH.
pOH = - log (Molarity)
pOH = - log (0.0018)
pOH = 2.74
use equation A
pH + 2.74 = 14
pH = 11.26
Lower the pH more acidic the solution is. Rank from most acidic to most basic is
0.0023 M HClpH = 5.45 and pOH = 8.550.0018 M KOH
The solutions, arranged from most acidic to most basic, are H₂SO₄, HCl, NH₄NO₃, NaCl, NaOH, and NaCN.
Explanation:The arrangement of the given solutions in order of decreasing acidity is as follows:
H₂SO₄ (Strong acid)HCl (Strong acid)NH₄NO₃ (Neutral salt)NaCl (Neutral salt)NaOH (Strong base)NaCN (Weak base)Strong acids completely dissociate in water, producing a large number of hydronium ions (H₃O⁺) and making the solution highly acidic. Neutral salts, such as NH₄NO₃ and NaCl, do not affect the pH of the solution. Strong bases, like NaOH, ionize completely and produce a high concentration of hydroxide ions (OH⁻), resulting in a basic solution. Weak bases, such as NaCN, only partially ionize, resulting in a lower concentration of OH⁻ ions and a slightly basic solution.
Learn more about pH and Acid-Base Solutions here:https://brainly.com/question/15330062
#SPJ3
The complete question is here:
Arrange the following aqueous solutions, all at 25 ∘C, in order of decreasing acidity. Rank from most acidic to most basic. To rank items as equivalent, overlap them.
H₂SO₄
HCl
NH₄NO₃
NaCl
NaOH
NaCN
If 50.0 g of N2O4 is introduced into an empty 2.12 L container, what are the partial pressures of NO2 and N2O4 after equilibrium has been achieved at 45∘C?
Answers
Answer:
p(N2O4) = 0.318 atm
p(NO2) = 7.17 atm
Explanation:
Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, what are the partial pressures of NO2 and N2O4 after equilibrium has been achieved at 45 degrees C?
Step 1: Data given
Kc = 0.619
Temperature = 45.0 °C
Mass of N2O4 = 50.0 grams
Volume = 2.10 L
Molar mass N2O4 = 92.01 g/mol
Step 2: The balanced equation
N2O4 ⇔ 2NO2
Step 3: Calculate moles N2O4
Moles N2O4 = 50.0 grams / 92.01 g/mol
Moles N2O4 = 0.543 moles
Step 4: The initial concentration
[N2O4] = 0.543 moles/2.10 L = 0.259 M
[NO2]= 0 M
Step 5: Calculate concentration at the equilibrium
For 1 mol N2O4 we'll have 2 moles NO2
[N2O4] = (0.259 -x)M
[NO2]= 2x
Step 6: Calculate Kc
Kc = 0.619= [NO2]² / [N2O4]
0.619 = (2x)² / (0.259-x)
0.619 = 4x² / (0.259 -x)
x = 0.1373
Step 7: Calculate concentrations
[N2O4] = (0.259 -x)M = 0.1217 M
[NO2]= 2x = 0.2746 M
Step 8: The moles
Moles = molarity * volume
Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles
Moles NO2 = 0.2746 M * 2.10 = 0.577 moles
Step 9: Calculate partial pressure
p*V = n*R*T
⇒ with p = the partial pressure
⇒ with V = the volume = 2.10 L
⇒ with n = the number of moles
⇒ with R = the gas constant = 0.08206 L*atm/mol*K
⇒ with T = the temperature = 45 °C = 318 K
p = (nRT)/V
p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10
p(N2O4) = 0.318 atm
p(NO2) = (0.577 *0.08206 * 318)/ 2.10
p(NO2) = 7.17 atm
At a given temperature, 4.92 atm of Cl2 and 4.65 atm of Br2 are mixed and allowed to come to equilibrium. The equilibrium pressure of BrCl is found to be 1.597 atm. Calculate Kp for the reaction at this temperature. Cl2(g) + Br2(g) <=> 2 BrCl(g). Give answer to 2 decimal places.
Answers
Answer:
Kp = 0.16
Explanation:
Step 1: Data given
Initial pressure of Cl2 = 4.92 atm
Initial pressure of Br2 = 4.65 atm
The equilibrium pressure of BrCl is found to be 1.597 atm
Step 2: The balanced equation
Cl2(g) + Br2(g) ⇔ 2 BrCl(g
Step 3: The initial pressures
pCl2 = 4.92 atm
pBr2 = 4.65 atm
pBrCl = 0 atm
Step 4: The pressure at the equilibrium
For 1 mol Cl2 we need 1 mol Br2 to produce 2 moles BrCl
pCl2 = 4.92 - X atm
pBr2 = 4.65 - Xatm
pBrCl = 2X atm = 1.597 atm
X = 1.597/2 = 0.7985 atm
pCl2 = 4.92 - X atm = 4.92 - 0.7985 = 4.1215 atm
pBr2 = 4.65 - Xatm = 3.8515 atm
Step 5: Calculate Kp
Kp = (BrCl)² / (Cl2)*(Br2)
Kp = 1.597² / (4.1215*3.8515)
Kp = 0.16
Final answer:
To calculate Kp for the given reaction, start with the initial pressures of Cl2 and Br2, calculate the change in pressure due to the formation of BrCl, and apply these values in the Kp equation. The result is Kp = 0.084.
Explanation:
The student asked how to calculate Kp for the reaction Cl2(g) + Br2(g) ⇌ 2 BrCl(g) at a given temperature, given initial pressures for Cl2 and Br2, and the equilibrium pressure of BrCl. To solve for Kp, we utilize the change in concentration according to the reaction's stoichiometry and apply it to the equation for the equilibrium constant in terms of pressure (Kp).
Initial pressures: Cl2 = 4.92 atm, Br2 = 4.65 atm. At equilibrium, BrCl = 1.597 atm. The change in pressure for Cl2 and Br2 to form 2 BrCl is equal to the pressure of BrCl divided by 2, since the stoichiometry of the reaction dictates twice the amount of BrCl for each reactant consumed. Thus, the change (δ) is 1.597 atm / 2 = 0.7985 atm. Therefore, the pressures of Cl2 and Br2 at equilibrium are 4.92 - 0.7985 atm and 4.65 - 0.7985 atm, respectively.
To find Kp, the equation is Kp = (PBrCl)2 / (PCl2 × PBr2). Substituting the equilibrium pressures into this equation gives Kp = (1.5972) / ((4.92 - 0.7985) × (4.65 - 0.7985)). Solving this yields Kp = 0.084 (to two decimal places).
A solution contains 4.08 g of chloroform (CHCl3) and 9.29 g of acetone (CH3COCH3). The vapor pressures at 35 ∘C of pure chloroform and pure acetone are 295 torr and 332 torr, respectively. Assuming ideal behavior, calculate the vapor pressures of each of the components and the total vapor pressure above the solution.
Answers
The vapor pressures of chloroform and acetone in the solution are 52.02 torr and 273.50 torr, respectively, and the total vapor pressure above the solution is 325.52 torr.
Explanation:To calculate the vapor pressures of chloroform and acetone above the solution, we should make use of Raoult's law. According to Raoult's law, the partial pressure of each component of a mixture is the product of the vapor pressure of the pure component and its mole fraction in the mixture.
First, calculate the mole fractions of each substance. The molar mass of chloroform (CHCl3) is about 119.38 g/mol, and the molar mass of acetone (CH3COCH3) is about 58.08 g/mol. Therefore, the mole fractions of chloroform and acetone are 4.08 g CHCl3 * (1 mol / 119.38 g) = 0.0342 mol and 9.29 g CH3COCH3 * (1 mol / 58.08 g) = 0.1599 mol, respectively. The total moles are 0.0342 mol + 0.1599 mol = 0.1941 mol, so the mole fraction of chloroform is 0.0342/0.1941 = 0.176 and that of acetone is 0.1599/0.1941 = 0.824.
Next, apply Raoult's Law to find the partial pressure of each component in the mixture. The vapor pressure of chloroform is 0.176 * 295 torr = 52.02 torr, and the vapor pressure of acetone is 0.824 * 332 torr = 273.50 torr.
The total vapor pressure above the solution is the sum of the vapor pressures of chloroform and acetone, that is, 52.02 torr + 273.50 torr = 325.52 torr.
Learn more about Vapor Pressure here:https://brainly.com/question/34135527
#SPJ3
for the hypothetical reaction 2A B -> 2C D, the initial rate of disappearance of A is 1.6x10^-1 mol/(L s) . what is the initial rate of disappearance of B
Answers
Answer:
0.8 x10^-1 mol/(L s)
Explanation:
2A + B -> 2C + D
For the reaction above, the differential rate is usually expressed as;
Rate = - (1 / 2) Δ[A] / Δt = - Δ[B] / Δt
The negative sign denotes disappearance.
Upon comparing ; (1 / 2) Δ[A] / Δt = Δ[B] / Δt
If initial rate of disappearance of A is 1.6x10^-1 mol/(L s);
That means
Δ[B] / Δt = 1.6x10^-1 mol/(L s) / 2
Δ[B] / Δt = 0.8 x10^-1 mol/(L s)
The initial rate of disappearance of B = 1.6x10^-1 mol/(L s)
Insert the IF statement in cell I9 to determine if the % Down is greater than or equal to 20% Down Pmt Rate. If true, the PMI is 0. If false, the PMI is calculated by multiplying the Amount Financed by the PMI Rate divided by the # of Pmts Per Year. The function should be =IF(E9>=$B$7,0,D9*$B$6/$B$5).
Answers
To address a conditional situation in Excel, you would use the IF function. In this case, the asked function is =IF(E9>=$B$7, 0, D9*$B$6/$B$5), wherein a condition is evaluated and if true, returns 0, if false, conducts a specified calculation.
Explanation:The question asked is about using the IF statement in Excel, which falls under the Computers and Technology topic. Excel's IF function is used to create conditional statements, where different calculations can be performed depending on whether a particular condition is met.
For the given question, you are asked to insert the IF function in cell I9. The condition to test is if the value in cell E9 is greater than or equal to the value in cell B7. If this condition is true, the function will return 0. If the condition is not met, the function will execute a calculation: it multiplies the values in cells D9 and B6, then divides the result by the value in cell B5.
Your final function should look like this: =IF(E9>=$B$7,0,D9*$B$6/$B$5).
Learn more about IF statement in Excel here:
https://brainly.com/question/33454350
#SPJ12
The heat of fusion ΔHf of ethyl acetate C4H8O2 is 10.5 /kJmol. Calculate the change in entropy ΔS when 95.g of ethyl acetate freezes at −84.0°C. Be sure your answer contains a unit symbol. Round your answer to 2 significant digits.
Answers
The change in entropy when 95 g of ethyl acetate freezes at -84.0°C is -59.8 J/K, rounded to two significant digits.
The question involves calculating the change in entropy when 95 g of ethyl acetate freezes at -84.0℃. Given the heat of fusion ( Hf) for ethyl acetate is 10.5 kJ/mol, we first need to convert grams to moles using the molar mass of ethyl acetate (C4H8O2) which is 88.11 g/mol.
Next, we use the formula:
moles = mass (g) / molar mass (g/mol)
For 95 g ethyl acetate, moles = 95 g / 88.11 g/mol = 1.078 moles.
The change in entropy ( S) can be found using the formula:
S = H / T
where T is the temperature in Kelvin. To convert -84.0℃ to Kelvin, we add 273.15:
T = -84.0 + 273.15 = 189.15 K
The entropy change then is:
S = 10.5 kJ/mol / 189.15 K
S per mole = 0.0555 kJ/K∙mol
Finally, for 1.078 moles, the total entropy change is:
S total = 0.0555 kJ/K∙mol × 1.078 mol
S total = 0.0598 kJ/K
Expressed in J/K (as 1 kJ = 1000 J),
S total = 59.8 J/K.
Since we are calculating the entropy change for freezing, the sign should be negative because entropy decreases during freezing.
Therefore, the change in entropy when 95 g of ethyl acetate freezes at -84.0℃ is -59.8 J/K, rounded to two significant digits.
In the laboratory, a general chemistry student measured the pH of a 0.329 M aqueous solution of benzoic acid, C6H5COOH to be 2.327. Use the information she obtained to determine the Ka for this acid. Ka(experiment) =
Answers
Answer:
The dissociation constant for the acid ( experimental ) is 1.45 lit/mol
Explanation:
The value of dissociation constant can be calculated as,
[tex]K_{a}[/tex] = C × ∝²
Where, C = concentration of the solution = 0.329M
∝ = Degree of dissociation
again , Degree of dissociation can be obtained form :
[tex]p_{H}[/tex] = C × ∝
∝ = [tex]\frac{p_{H} }{C}[/tex]
∝ = [tex]\frac{2.327}{0.329}[/tex] = 7.072
So, now [tex]K_{a}[/tex] = C × ∝²
= 0.329 ×( 7.072)²
= 1.45 lit/ mol
Answer:
The Ka = 6.74 * 10^-5
Explanation:
Step 1: Data given
Concentration of benzoic acid = 0.329 M
pH = 2.327
Step 2: Calculate the Ka
pH = -log (√([HA]*Ka))
2.327 = -log (√(0.329*Ka))
10 ^ - 2.327 = √(0.329*Ka))
0.0047098 = √(0.329*Ka))
2.218 * 10^-5 = 0.329 * Ka
Ka = 6.74 * 10^-5
The Ka = 6.74 * 10^-5
If a 32.4 gram sample of sodium sulfate (Na2SO4) reacts with a 65.3 gram sample of barium chloride (BaCl2) according to the reaction below: Na2SO4 (aq) + BaCl2 (aq) → BaSO4 (s) + 2NaCl (aq) What is the theoretical yield of barium sulfate (BaSO4) in grams?
Answers
Answer: The theoretical yield of barium sulfate is 50.9 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For sodium sulfate:Given mass of sodium sulfate = 32.4 g
Molar mass of sodium sulfate = 142 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of sodium sulfate}=\frac{32.4g}{142g/mol}=0.228mol[/tex]
For barium chloride:Given mass of barium chloride = 65.3 g
Molar mass of barium chloride = 208.23 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of barium chloride}=\frac{65.3g}{208.23g/mol}=0.314mol[/tex]
The chemical equation for the reaction of barium chloride and sodium sulfate follows:
[tex]Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl[/tex]
By Stoichiometry of the reaction:
1 mole of sodium sulfate reacts with 1 mole of barium chloride
So, 0.228 moles of sodium sulfate will react with = [tex]\frac{1}{1}\times 0.228=0.228mol[/tex] of barium chloride
As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.
Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of sodium sulfate produces 1 mole of barium sulfate.
So, 0.228 moles of sodium sulfate will produce = [tex]\frac{1}{1}\times 0.228=0.228moles[/tex] of barium sulfate
Now, calculating the mass of barium sulfate from equation 1, we get:
Molar mass of barium sulfate = 233.4 g/mol
Moles of barium sulfate = 0.228 moles
Putting values in equation 1, we get:
[tex]0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol\times 223.4g/mol)=50.9g[/tex]
Hence, the theoretical yield of barium sulfate is 50.9 grams
When a diprotic acid, , is titrated with , the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape: Notice that the plot has essentially two titration curves. If the first equivalence point occurs at 100.0 mL added, what volume of added corresponds to the second equivalence point? Volume = mL For the following volumes of added, list the major species present after the reacts completely.
Answers
In a titration involving a diprotic acid, the equivalence points occur when the acid has lost its protons. If the first equivalence point occurs at 100 mL, the second should occur at 200 mL. resultant species in the solution will vary based on the volume of NaOH added.
Explanation:A diprotic acid has two protons to donate in a reaction. During titration, protons are removed one at a time, thus presenting two titration curves or equivalence points. If the first equivalence point occurs at 100.0 mL, the second equivalence point typically occurs at twice that volume because the second proton is just as readily removed as the first. Therefore, the second equivalence point will be at 200.0 mL.
Different volumes of added NaOH will result in different major species present. For example, before the equivalence point is reached (0 mL < V< 25 mL), the pH increases gradually as the diprotic acid reacts with the added NaOH to form its conjugate base. At the equivalence point (V = 25 mL), pH increases abruptly as the reaction transitions from acidic to either neutral or basic, depending on whether the diprotic acid is strong or weak, respectively. After the equivalence point (V > 25 mL), the pH is determined by the amount of added NaOH.
Learn more about Titration here:https://brainly.com/question/31271061
#SPJ12
A student dissolves 14.g of benzoic acid C7H6O2 in 425.mL of a solvent with a density of 0.92 g/mL. The student notices that the volume of the solvent does not change when the benzoic acid dissolves in it.Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.
Answers
The molarity and molality of a solution of 14. g of benzoic acid dissolved in 425.mL of a solvent with a density of 0.92 g/mL are 0.26 M and 0.28 m respectively.
Explanation:To calculate the molarity and molality of the solution, we'll first need to know how many moles of benzoic acid (C7H6O2) are present. Simply converting grams to moles using the molar mass of benzoic acid (122.12 g/mol), we get 14.g ÷ 122.12 g/mol = 0.11 mol.
Molarity is defined as moles of solute divided by liters of solution. Therefore, the molarity of the solution would be 0.11 mol ÷ 0.425 L = 0.26 M.
On the other hand, molality is calculated as moles of solute divided by kilograms of solvent. To obtain the mass of the solvent in kg, we need to multiply the volume by the density, 425.mL x 0.92 g/mL = 391 g = 0.391 kg. Consequently, the molality of the solution would be 0.11 mol ÷ 0.391 kg = 0.28 m.
Learn more about Solution Concentrations here:https://brainly.com/question/34412339
#SPJ3
Quantum numbers arise naturally from the mathematics used to describe the possible states of an electron in an atom. The four quantum numbers, the principal quantum number (n), the angular momentum quantum number (????), the magnetic quantum number (m????), and the spin quantum number (ms) have strict rules which govern the possible values. Identify all allowable combinations of quantum numbers for an electron.
Select all that apply.
a) n= 3 l= -2 ml=-1 ms= +1/2
b) n=3 l= 2 ml= -1 ms= -1/2
c) n= 4 l=4 ml=-1 ms= +1/2
d) n=2 l=1 ml=-1 ms=1
e) n=4 l=1 ml=2 ms=-1/2
f) n=5 l=4 ml=4 ms=+1/2
Answers
Answer:
a), b), c) & f)
Explanation:
d) does not apply because Ms value can be either +½ or -½
e) does not apply because Ml - values range from -l to +l, hence l= 2 doesn't exist when l= 1
The four quantum numbers that govern the possible states of an electron in an atom are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (ml), and the spin quantum number (ms). Allowable combinations of quantum numbers follow specific rules.
Explanation:The four quantum numbers that govern the possible states of an electron in an atom are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (ml), and the spin quantum number (ms).
Allowable combinations of quantum numbers for an electron can be determined according to the following rules:
The principal quantum number (n) can have values of 1, 2, 3, etc.The angular momentum quantum number (l) can have values from 0 to (n - 1).The magnetic quantum number (ml) can have values from -l to +l, including zero.The spin quantum number (ms) can have values of +1/2 or -1/2.Two important indicators of stream pollution are high biological oxygen demand (BOD) and low pH. Of the more than 250 streams draining into a large lake, 30% have high BOD and 20% have low pH levels, with 10% having both characteristics.
Answers
Answer:
a) The Venn diagram is presented in the attached image to this answer.
b) Check Explanation.
c) 0.3333
d) 0.1429
e) 0.6
Explanation:
Let the probability of a lake having high BOD be P(B) = 30% = 0.3
Probability of a lake having low pH = P(P) = 20% = 0.2
Probability that a lake has high BOD and low pH = P(B n P) = 10% = 0.1
Then, probability that a lake has normal BOD = P(B') = 1 - P(B) = 1 - 0.3 = 0.7
Probability that a lake has normal pH = P(P') = 1 - P(P) = 1 - 0.2 = 0.8
Total probability = P(U) = 100% = 1
a) The Venn diagram is presented in the attached image to this answer.
b) Two events are independent if and only if, P(A|B) = P(A) or P(B|A) = P(B).
For this question,
P(B|P) = P(B n P)/P(P) = 0.1/0.2 = 0.5 ≠ P(B) (which is 0.3)
And P(P|B) = P(B n P)/P(B) = 0.1/0.3 = 0.333 ≠ P(P) (which is 0.2).
It is evident that the two events aren't independent of each other.
c) If a stream has high BOD, what is the probability it will also have low pH?
This probability is given as P(P|B) meaning that, the probability of a lake having low pH given that it has high BOD.
Mathematically, this conditional probability is given by
P(P|B) = P(B n P)/P(B) = 0.1/0.3 = (1/3) = 0.3333
d) If a stream has normal levels of BOD, what is the probability it will also have low pH.
This probability is given as P(P|B'); that is, the probability of a lake having low pH given that it has normal BOD.
Mathematically,
P(P|B') = P(B' n P)/P(B')
P(B') = 0.7 (already found above)
But P(B' n P) = ?
Mathematically,
P(B' n P) = P(P) - P(B n P) = 0.2 - 0.1 = 0.1
P(P|B') = 0.1/0.7 = 0.1429
e) What is the probability that a stream will not exhibit either pollution indicator, i.e., will have normal BOD and pH levels?
This is given as P(B' n P')
Mathematically, this represents the region in the Venn diagram outside of the circles representing P(B) and P(P) and it's given mathematically as,
P(B' n P') = P(U) - [P(B n P') + P(B' n P) + P(B n P)] = 1 - (0.2 + 0.1 + 0.1) = 1 - 0.4 = 0.6 or 60%
When the stream has high BOD the probability of low pH is 0.33 and 0.142 when it has normal BOD. 0.6 is the probability that a stream will exhibit neither indicator.
What is BOD?BOD is the biological oxidation demand, that tells about the dissolved oxygen amount in the water body. BOD along with pH are the indicator of pollution.
The Venn diagram is attached in the image below.
The two indicators, high BOD and low pH are dependent on each other. It can be shown as:
[tex]\rm P(A|B) = P(A) \;or \;P(B|A) = P(B)[/tex]
But,
[tex]\begin{aligned}\rm P(B|P) &= \rm \dfrac{P(B \;n \;P)}{P(P)} \\\\&= \dfrac{0.1}{0.2} \\\\&= 0.5 \neq \rm P(B)\end{aligned}[/tex]
And
[tex]\begin{aligned}\rm P(P|B) &=\rm \dfrac{P(B \;n \;P)}{P(B)} \\\\&= \dfrac{0.1}{0.3} \\\\&= 0.333 \neq \rm P(P) \end{aligned}[/tex]
Hence they are not independent of each other.
The probability of low pH at high BOD is given as P(P|B).
[tex]\begin{aligned}\rm P(P|B) &= \rm \dfrac{P(B \;n \;P)}{P(B)}\\\\ &= \dfrac{0.1}{0.3}\\\\&= 0.3333\end{aligned}[/tex]
Hence, 0.33 is the probability of low pH at high BOD.
The probability of low pH at normal BOD is given as P(P|B').
[tex]\begin{aligned}\rm P(P|B') &= \rm \dfrac{P(B' \;n\; P)}{P(B')}\\\\\rm P(P|B') &= \dfrac{0.1}{0.7} \\\\&= 0.1429\end{aligned}[/tex]
Hence, 0.1429 is the probability of low pH at normal BOD.
The probability that a stream will not exhibit any of the indicators is given by, P(B' n P').
[tex]\begin{aligned}\rm P(B' n P') &= \rm P(U) - [P(B n P') + P(B' n P) + P(B n P)]\\\\&= 1 - (0.2 + 0.1 + 0.1) \\\\&= 0.6\end{aligned}[/tex]
Hence, 0.6 is the probability that neither of the indicators will be expressed.
Learn more about BOD and low pH here:
https://brainly.com/question/9866894
A mixture of 0.438 M H2, 0.444 M I2 , and 0.895 M HI is enclosed in a vessel and heated to 430 °C. H2 (g) + I2 (g) <-----> 2 HI (g) Kc = 54.3 at 430∘C Calculate the equilibrium concentrations of each gas at 430∘C.
Answers
Answer:
[H₂] = 0.178 M
[I₂] = 0.184 M
[HI] = 1.415 M
Explanation:
For the equilibrium:
H₂(g) + I₂(g) ⇄ 2 HI(g)
the equilibrium constant is given by the equation:
Kc = [ HI]² / [H₂][I₂]
Lets use first the reaction quotient which has the same expression as the equilibrium constant to predict the direction the reaction will take, i.e towards reactants or product side.
Q =( 0.895)²/(0.438)(0.444) = 4.12
Q is less than Kc so the reaction will favor the product side.
We can set up the following table to account for all the species at equilibrium:
H₂ I₂ HI
initial 0.438 0.444 0.895
change -x -x +2x
equilibrium 0.438 - x 0.444 - x 0.895 + 2x
Now we are in position to express these concentrations in terms of the equilibrium conctant, Kc
54.3 = (0.895 + 2x)² / (0.438 -x)(0.444 - x)
performing the calculatiopns will result in a quadratic equation:
0.801 + 3.580x +4x² = (0.194 - 0.882x + x²)x 54.3
Upon rearrangement and some algebra, we have
0.801 + 3.580 x + 4x² = 10.534 - 47.893x + 54.3 x²
0 = 9.733 - 51.473 x + 54.3 x²
This equation has two roots X₁ = 0.687 and X₂ = 0.26
The first is physically impossible since it will imply that more 0.687 will make the quantity at equilibrium for both H₂ and I₂ negative.
Therefore the concentrations at equilibrium of each gas are:
[H₂] = (0.438 - 0.260) = 0.178 M
[I₂] = (0.444 - 0.260) M = 0.184 M
[HI] = [0.895 + 2x(0.260)] M = 1.415 M
Note if we plug these values into the equilibrium expression we get 61 which is due to the rounding errors propagating in the quadratic equation.
The equilibrium concentrations of H2, I2, and HI at 430 °C are calculated using an expression derived from the reaction quotient equation, plugged into the Kc equation, which is then solved for 'x'. The solutions found are the changes in molarities which applied to the initial molarities give the equilibrium concentrations
Explanation:Let's denote the change in molarity of H2, I2, and HI as 'x'. At equilibrium, the molarities of H2, I2, and HI will be 0.438+x, 0.444+x, and 0.895-2x respectively. We know the equilibrium constant, Kc = 54.3. Thus, (0.895-2x)2/(0.438+x)(0.444+x) = 54.3. This is a quadratic equation in 'x' and needs to be solved to get the value of 'x'.
After finding 'x', put this value back into the equilibrium concentrations of the gases, i.e., 0.438+x for H2, 0.444+x for I2, and 0.895-2x for HI. These will give you the equilibrium concentrations of the gases at 430 °C.
Learn more about Equilibrium Concentrations here:https://brainly.com/question/16645766
#SPJ3
For this heterogeneous system 2 A ( aq ) + 3 B ( g ) + C ( l ) − ⇀ ↽ − 2 D ( s ) + 3 E ( g ) the concentrations and pressures at equilibrium are [ A ] = 9.68 × 10 − 2 M , P B = 9.54 × 10 3 Pa , [ C ] = 14.64 M , [ D ] = 10.11 M , and P E = 9.56 × 10 4 torr . Calculate the thermodynamic equilibrium constant, K.
Answers
Answer:
[tex]2.55*10^{11[/tex]
Explanation:
Equation for the heterogeneous system is given as:
[tex]2A_{(aq)} + 3 B_{(g)} + C_{(l)}[/tex] ⇄ [tex]2D_{(s)}[/tex] [tex]+[/tex] [tex]3E_{(g)}[/tex]
The concentrations and pressures at equilibrium are:
[tex][A] = 9.68*10^{-2}M[/tex]
[tex]P_B = 9.54*10^3Pa[/tex]
[tex][C]=14.64M[/tex]
[tex][D]=10.11M[/tex]
[tex]P_E=9.56*10^4torr[/tex]
If we convert both pressure into bar; we have:
[tex]P_B = 9.54*10^3Pa[/tex]
[tex]P_B = (9.54*10^3)*\frac{1}{10^5} bar[/tex]
[tex]P_B=9.54*10^{-2}bar[/tex]
[tex]P_E=9.56*10^4torr[/tex]
1 torr = 0.001333 bar
[tex]9.54*10^4 *0.001333 = 127.5 bar[/tex]
[tex]K=\frac{[P_E]^3}{[A]^2[P_B]^3}[/tex]
[tex]K=\frac{(127.5)^3}{(9.68*10^{-2})^2(9.54*10^{-2})^3}[/tex]
[tex]K=2.55*10^{11[/tex]
Methylamine, CH3NH2, is a weak base and one of several substances that give herring brine its pungent odor. In .100 M CH3NH2, only 6.4 percent of the base has undergone ionization. What are Kb and pKb of methylamine
Answers
Answer:
[tex]K_{b}[/tex] is 0.000438 and [tex]pK_{b}[/tex] is 3.36
Explanation:
Methylamine is a monoprotic base.
For a monoprotic base, [tex]K_{b}=\frac{ca^{2}}{(1-a)}[/tex]
where, c is concentration of base in molarity and a is it's degree ionization
Here [tex]a=\frac{6.4}{100}=0.064[/tex] and c = 0.100 M
So, [tex]K_{b}=\frac{(0.100)\times (0.064)^{2}}{(1-0.064)}=0.000438[/tex]
We know, [tex]pK_{b}=-logK_{b}[/tex]
Hence, [tex]pK_{b}=-log(0.000438)=3.36[/tex]
A volume of 90.0 mL mL of a 0.590 M M HN O 3 HNO3 solution is titrated with 0.350 M M KOH KOH . Calculate the volume of KOH KOH required to reach the equivalence point. Express your answer to three significant figures and include the appropriate units.
Answers
Answer:
152 mL is the volume of KOH required to reach the equivalence point.
Explanation:
[tex]HNO_3(aq)+KOH(aq)\rightarrow KNO_3(aq)+H_2O(l)[/tex]
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HNO_3[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.
We are given:
[tex]n_1=1\\M_1=0.590 M\\V_1=90.0 mL\\n_2=1\\M_2=0.350 M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]1\times 0.590 M\times 90.00=1\times 0.350 M\times V_2[/tex]
[tex]V_2=\frac{1\times 0.590 M\times 90.0 mL}{1\times 0.350 M}=151 .7 mL\approx 152 mL[/tex]
152 mL is the volume of KOH required to reach the equivalence point.
Answer:
152 ml.
Explanation:
Given:
Volume of HNO3 = 90 ml
Molar concentration of HNO3 = 0.59 M
Molar concentration of KOH = 0.35 M
Equation of the reaction
KOH + HNO3 --> KNO3 + H2O
Number of moles of HNO3 = molar concentration × volume
= 0.59 × 0.09
= 0.0531 moles.
By stoichiometry, 1 mole of HNO3 reacts with 1 mole of KOH. Therefore,
Number of moles of KOH = 0.0531 moles.
Volume = 0.0531 ÷ 0.350
= 0.152 l
= 152 ml.
Write the balanced equation for the combustion of isooctane (C8H18) to produce carbon dioxide and water. Use the smallest possible integers to balance the equation. Also, separate the sign with 1 space and enter the reaction arrow as hyphen greater than sign: ->
Answers
The balanced equation for the combustion of isooctane (C8H18) is C8H18 + 12.5O2 -> 8CO2 + 9H2O.
Explanation:The balanced equation for the combustion of isooctane (C8H18) to produce carbon dioxide (CO2) and water (H2O) is:
C8H18 + 12.5O2 -> 8CO2 + 9H2O
In this equation, the coefficients have been adjusted to balance the number of atoms on both sides of the equation. There are 8 carbon atoms, 18 hydrogen atoms, and 25 oxygen atoms on both sides of the equation.
Final answer:
The balanced chemical equation for the complete combustion of isooctane (C₈H₁₈) is 2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g), comprising a step-by-step process of first balancing the carbon, then hydrogen, and finally the oxygen atoms.
Explanation:
The balanced chemical equation for the combustion of isooctane (C₈H₁₈) to produce carbon dioxide (CO₂) and water (H₂O) can be represented as follows:
2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g)
Here is a step-by-step explanation:
Start by writing the unbalanced equation: C₈H₁₈ + O₂ → CO₂ + H₂O.Balance the carbon (C) atoms first: As there are 8 C atoms in isooctane, you need to have 8 CO₂ molecules to balance the carbons.Next, balance the hydrogen (H) atoms: With 18 H atoms in isooctane, you will need 9 H₂O molecules.Finally, balance the oxygen (O) atoms: There are 16 O atoms in 8 CO₂ molecules and 18 O atoms in 9 H₂O molecules, giving a total of 25 O₂ molecules needed.After balancing, you end up with the smallest possible integers that balance the equation.
Using your data above, draw conclusions about the d-splitting for each ligand (H2O, en, phen). Order the complexes from least to greatest d-splitting, and discuss the reason for your ordering.
Answers
Answer:
H2O<en<phen
Explanation:
The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.
The d-splitting for each ligand is as follows:
H2O (weak field ligand): smallest d-splitting
en (stronger field ligand): medium d-splitting
phen (strongest field ligand): largest d-splitting
The complexes from least to greatest d-splitting are as follows:
[Co(H2O)6]3+
[Co(en)3]3+
[Co(phen)3]3+
The reason for this ordering is that stronger field ligands cause a larger d-splitting. This is because stronger field ligands interact more strongly with the metal's d orbitals, which splits the d orbitals into two sets of orbitals with different energies.
The d-splitting of a transition metal complex is the energy difference between the two sets of d orbitals that are split by the ligand field. The magnitude of the d-splitting depends on the strength of the ligand field. Stronger field ligands cause a larger d-splitting.
The following table shows the spectrochemical series of ligands, which is a ranking of ligands from weakest to strongest field:
| Ligand | Spectrochemical series |
|---|---|---|
| H2O | Weak field |
| en | Medium field |
| phen | Strong field |
Based on the spectrochemical series, we can predict that the d-splitting for each ligand is as follows:
H2O (weak field ligand): smallest d-splitting
en (stronger field ligand): medium d-splitting
phen (strongest field ligand): largest d-splitting
The following table shows the expected d-splitting of the [Co(H2O)6]3+, [Co(en)3]3+, and [Co(phen)3]3+ complexes:
Complex Ligand D-splitting
[Co(H2O)6]3+ H2O Smallest
[Co(en)3]3+ en Medium
[Co(phen)3]3+ phen Largest
The d-splitting of a transition metal complex affects the color of the complex. Complexes with a larger d-splitting absorb higher energy light, which is in the visible region of the spectrum. This is why complexes with strong field ligands tend to be colored.
The following table shows the observed colors of the [Co(H2O)6]3+, [Co(en)3]3+, and [Co(phen)3]3+ complexes:
Complex Ligand Color
[Co(H2O)6]3+ H2O Purple
[Co(en)3]3+ en Purple-pink
[Co(phen)3]3+ phen Pink
The observed colors of the complexes are consistent with the predicted d-splitting. The [Co(phen)3]3+ complex has the largest d-splitting and absorbs the highest energy light, which is why it is pink. The [Co(H2O)6]3+ complex has the smallest d-splitting and absorbs the lowest energy light, which is why it is purple.
For more such information on: ligand
https://brainly.com/question/1869211
#SPJ6
Click in the answer box to activate the palette. List the following molecules in order of increasing dipole moment: H2O, CBr4, H2S, HF, NH3, CO2 < < < < Electronegativities H 2.1 C 2.5 N 3.0 O 3.5 F 4.0 S 2.5 Br 2.8
Answers
Answer:
HF > H2O > NH3 > H2S > CBr4=CO2=0
Explanation:
Dipole moment is a vector quantity. Its a measure of polarity of a bond in a molecule and also a meaure of separation of positive and negative charge in a system.It occurs due to electronegativity difference between the atoms in a molecule.
In order for a molecule to have dipole moment, a molecule must exhibit high electronegativity difference and the shape of the moloecule must be asymmetry
HF has the highest electronegativity difference among all the molecules listed above hence its dipole moment is the greatest.
[tex]H_{2}O[/tex] has a bent structure. There are two O-H bonds hence more charge dipoles. The dipole moment is less than HF molecule because of the net dipole moments of two O-H bonds.
[tex]CO_{2}[/tex] is a linear molecule.However it has polar bonds.But because of the shape of the molecule, the two C-O bond dipoles cancel out each other hence the overall dipole moment will be zero.
Similarly in [tex]CBr_{4}[/tex], ,since the molecule is symmetry, the bond dipole cancels each other out hence the overall dipole moment will be zero.
The increasing order of the dipole moment will be:
[tex]\rm CO_2[/tex] < [tex]\rm CBr_4[/tex] < [tex]\rm H_2S[/tex] < [tex]\rm NH_3[/tex] < [tex]\rm H_2O[/tex] < HF.
Dipole moment can be described as the measure of the polarity of the molecule. The higher the electronegativity difference, the more polar the bond.
In the given molecules,
HF: The electronegativity difference is 1.9.[tex]\rm H_2O[/tex] : The molecule is bend, and the difference in electronegativity is 1.4.[tex]\rm CBr_4[/tex] : The molecule is symmetrical, which cancels the dipole. The net dipole is zero.[tex]\rm H_2S[/tex] : The electronegativity difference is 0.4.[tex]\rm NH_3[/tex] : The electronegativity difference is 0.9.[tex]\rm CO_2[/tex]: The molecule has a symmetrical arrangement. Thus the net dipole of the molecule is 0.The increasing order of the dipole moment will be:
[tex]\rm CO_2[/tex] < [tex]\rm CBr_4[/tex] < [tex]\rm H_2S[/tex] < [tex]\rm NH_3[/tex] < [tex]\rm H_2O[/tex] < HF.
The molecule with the highest dipole moment is HF.
For more information about the dipole moment, refer to the link:
https://brainly.com/question/16260427
The beta-ketoester Claisen product is generated from the product of this final step by addition of dilute HCl. Write the final step of the Claisen condensation using curved arrows to show electron reorganization.
Answers
Answer:
This shown on the second uploaded image
Explanation:
What is occurring in this reaction is the further deprotonation of the base and this would now react with HCl to give the final product