CHECK COMPLETE QUESTION BELOW
At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with Xbenzene = 0.580.
Answer:
The total vapor pressure is [tex]81.3 mmHg[/tex]
Explanation:
We will be making use of Dalton and Raoults equation in order to calculate the total pressure,
Which is [tex]PT= (PA × XA) +(PB ×XB)[/tex]
PT= total vapor pressure
From the question
Benene's Mole fraction = 0.580
then to get Mole fraction of toluene we will substract the one of benzene from 1. because total mole fraction is always 1.
= (1 - 0.580) = 0.420
Vapor pressure of benzene given = 183 mmHg
Vapor pressure of toluene given= 59.2 mmHg
If we substitute those value into above equation, we have
PT=(183×0.580)+(59.2×0.420)
=81.3mmHg
Therefore,, the total vapor pressure of the solution is 81.3 mmHg
If we represent the equilibrium as:...N2O4(g) 2 NO2(g) We can conclude that: 1. This reaction is: A. Exothermic B. Endothermic C. Neutral D. More information is needed to answer this question. 2. When the temperature is increased the equilibrium constant, K: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question. 3. When the temperature is increased the equilibrium concentration of NO2: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question.
Answer:
1. This reaction is: B. Endothermic.
2. When the temperature is increased the equilibrium constant, K: A. Increases.
3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.
Explanation:
Hello,
In this case, considering the images, we can state that the red color at high temperature is due to the presence of nitrogen dioxide (product) and the lower coloring is due to the presence of dinitrogen tetroxide (reactant) at low temperature.
With the aforementioned, we can conclude that the chemical reaction:
[tex]N_2O_4(g) \rightleftharpoons 2 NO_2(g)[/tex]
Is endothermic since high temperatures favor the formation of the product and the low temperatures favor the consumption of the the reactant. thereby:
1. This reaction is: B. Endothermic.
2. When the temperature is increased the equilibrium constant, K: A. Increases. In this particular case, since the dinitrogen tetroxide has 1 molecule and nitrogen dioxide two molecules in the chemical reaction, the entropy change should be positive, therefore, by increasing the T, the Gibbs free energy of reaction becomes more negative:
[tex]G=H-TS[/tex]
As Gibbs free energy becomes more negative, the equilibrium constant becomes bigger given their relationship:
[tex]K=exp(-\frac{\Delta G}{RT} )[/tex]
3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.
Regards.
g When considering the effects of temperature on spontaneity, if both ΔH and ΔS are positive, _______. Select the correct answer below: the process is spontaneous at all temperatures
Explanation:
The spontaneity of a system is deduced by the sign of the gibbs free energy value. If it is negative, it means the process / reaction is spontaneous however a positive value indicates the such process is not spontaneous.
Gibbs free energy, enthalpy and entropy are related by the following equation;
ΔG = ΔH - TΔS
A positive value of enthalpy, H and entropy, S means that G would always be a negative value at all temperatures.
What is the frequency of a photon having an energy of 4.91 × 10–17 ? (c = 3.00 × 108 m/s, h = 6.63 × 10–34 J · s)
Answer:
The frequency of the photon is 7.41*10¹⁶ Hz
Explanation:
Planck states that light is made up of photons, whose energy is directly proportional to the frequency of radiation, according to a constant of proportionality, h, which is called Planck's constant. This is expressed by:
E = h*v
where E is the energy, h the Planck constant (whose value is 6.63*10⁻³⁴ J.s) and v the frequency (Hz or s⁻¹).
So the frequency will be:
[tex]v=\frac{E}{h}[/tex]
Being E= 4.91*10⁻¹⁷ J and replacing:
[tex]v=\frac{4.91*10^{-17} J}{6.63*10^{-34} J.s}[/tex]
You can get:
v= 7.41*10¹⁶ [tex]\frac{1}{s}[/tex]= 7.41*10¹⁶ Hz
The frequency of the photon is 7.41*10¹⁶ Hz
How many equivalent resonance structures can be drawn for the molecule of SO3 without having to violate the octet rule on the sulfur atom
Answer:
3
Explanation:
Resonance is a valence bond concept put forward by Linus Pauling to explain the fact that the observed properties of a molecule may be as a result of the fact that its actual structure lie somewhere between a given number of structural extremes called canonical structures or resonance structures.
There are three resonance structures for SO3 that obey the octet rule. All the S-O bonds in SO3 are equivalent in these resonance structures.
Seven equivalent resonance structures for the molecular of SO3 can be drawn without breaking the octet rule.
We can arrive at this answer because:
The octet rule is a rule that states that an atom must reach stability when it has eight electrons in the valence layer.This means that in bonds that cause the donation or sharing of electrons between atoms, each atom has eight electrons in the valence layer.In chemistry, resonance is a term that refers to structures created to represent the donation or sharing of electrons between the atoms of a molecule.These structures can be arranged in different ways, as long as they respect the octet rule.In an SO3 molecule, electrons are shared between atoms. This sharing can be done with seven resonance structures.
These structures are shown in the figure below.
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4. Given that the enthalpy of reaction for a system at 298 K is -292 kJ/mol and the entropy for that system is 224 J/mol K, what's the free energy for the system?
A.-87,793 kJ
B.-358 kJ
C.-225 kJ
D. -66,751 kJ
Answer:
[tex]\Delta G=-359\frac{kJ}{mol}[/tex]
Explanation:
Hello,
In this case, we must remember that the Gibbs free energy is defined in terms of the enthalpy, temperature and entropy as shown below:
[tex]\Delta G=\Delta H -T\Delta S\\[/tex]
In such a way, for the given data, we obtain it, considering the conversion from J to kJ for the entropy in order to conserve the proper units:
[tex]\Delta G=-292\frac{kJ}{mol} -(298)(224\frac{J}{mol}*\frac{1kJ}{1000J} )\\\\\Delta G=-359\frac{kJ}{mol}[/tex]
Best regards.
Answer:
B- 358 kj
Explanation: I took the test
Which option draws the correct conclusion from the following case study?
A patient with sickle-cell anemia and a fever goes to the emergency room and is given Tylenol to reduce
the fever. The patient has seizures and dies after taking the Tylenol. The physician writes up this case as
an interesting outcome for a patient with sickle-cell anemia.
The case study's validity is obvious because it describes a real-life situation.
The case study was influenced by bias, and led to incorrect conclusions being drawn
The case study was not intended to produce a generalized conclusion about treatment
Upon reading this case study, physicians should stop treating sickle cell patients with fevers using Tylenol
Answer:
I believe the answer The case study was influenced by bias, and led to incorrect conclusions being drawn. plz correct me if I am wrong
Explanation:
Answer: options B
Explanation:
the pain reliever codeine is a weak base with a kb equal to 1.6 x 10^-6. what is the ph of a 0.05 m aqueous codeine solution
Answer:
[tex]pH=10.45[/tex]
Explanation:
Hello,
In this case, for the dissociation of the given base, we have:
[tex]base\rightleftharpoons OH^-+CA[/tex]
Whereas CA accounts for conjugated acid and OH⁻ for the conjugated base. In such a way, equilibrium expression is:
[tex]Kb=\frac{[OH^-][CA^+]}{[base]}[/tex]
And in terms of the reaction extent [tex]x[/tex] we can write:
[tex]1.6x10^{-6}=\frac{x*x}{0.05M-x}[/tex]
For which the roots are:
[tex]x_1=-0.000284M\\x_2=0.000282M[/tex]
For which clearly the result is the positive root which also equals the concentration of hydroxyl ions and we can compute the pOH:
[tex]pOH=-log([OH^-])=-log(0.000282)\\\\pOH=3.55[/tex]
And the pH:
[tex]pH=14-pOH=14-3.55\\\\pH=10.45[/tex]
Regards.
The pH of the solution is 10.45.
Let us represent codeine with the generic formula BH. We can set up the ICE table as follows;
:B(aq) + H2O(l) ⇄ BH(aq) + OH^-(aq)
I 0.05 0 0
C -x +x +x
E 0.05 - x x x
We know that the Kb of codeine is 1.6 x 10^-6, Hence;
1.6 x 10^-6 = x^2/0.05 - x
1.6 x 10^-6 (0.05 - x ) = x^2
8 x 10^-8 - 1.6 x 10^-6x = x^2
x^2 + 1.6 x 10^-6x - 8 x 10^-8 = 0
x = 0.00028 M
The concentration of hydroxide ions = 0.00028 M
Given that pOH = - log[0.00028 M]
pOH = 3.55
pH + pOH = 14
pH = 14 - 3.55
pH = 10.45
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A laboratory technician combines 35.9 mL of 0.258 M chromium(II) chloride with 35.8 mL 0.338 M potassium hydroxide. How many grams of chromium(II) hydroxide can precipitate
Answer:
0.52 g of chromium(II) hydroxide, Cr(OH)2.
Explanation:
We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.
This can be obtained as follow:
Molarity of CrCl2 = 0.258 M
Volume = 35.9 mL = 35.9/1000 = 0.0359 L
Mole of CrCl2 =?
Molarity = mole /Volume
0.258 = mole of CrCl2 /0.0359
Cross multiply
Mole of CrCl2 = 0.258 x 0.0359
Mole of CrCl2 = 0.0093 mole
Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.
This can be obtained as follow:
Molarity of KOH = 0.338 M
Volume = 35.8 mL = 35.8/1000 = 0.0358 L
Mole of KOH =.?
Molarity = mole /Volume
0.338 = mole of KOH /0.0358
Cross multiply
Mole of KOH = 0.338 x 0.0358
Mole of KOH = 0.0121 mole.
Next, we shall write the balanced equation for the reaction. This is given below:
2KOH + CrCl2 → Cr(OH)2 + 2KCl
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2.
Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.
From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.
Therefore, KOH is the limiting reactant.
Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.
In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.
The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:
From the balanced equation above,
2 mole of KOH reacted to produce 1 mole of Cr(OH)2.
Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.
Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.
This is illustrated below:
Mole of Cr(OH)2 = 0.00605 mole
Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol
Mass of Cr(OH)2 =..?
Mole = mass /Molar mass
0.00605 = mass of Cr(OH)2/86
Cross multiply
Mass of Cr(OH)2 = 0.00605 x 86
Mass of Cr(OH)2 = 0.52 g
Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.
Determine whether each phrase describes carboxylic acids or esters.a. Do not form hydrogen bonds amongst themselves and have higher vapor pressureb. Form hydrogen bonds amongst themselves and have lower vapor pressurec. Notable for their pleasant fragrancesd. Their reactions with base are kn. own as saponificationse. Usually have a sour odorf. Their reactions with base are known as neutralizations
Explanation:
a. Esters
b. Carboxylic acids
c. Esters (ethyl hexanoate smells like pineapple)
d. Carboxylic acids (produces a carboxylic salt)
For the given phrases the following description is better.
a. Esters
b. Carboxylic acids
c. Esters (ethyl hexanoate smells like pineapple)
d. Carboxylic acids (produces a carboxylic salt)
Esters and carboxylic acids:An ester is a synthetic compound got from a corrosive in which somewhere around one - OH hydroxyl bunch is supplanted by an - O-alkyl (alkoxy) bunch, as in the replacement response of a carboxylic acid and a liquor.
Carboxylic acid is any of a class of natural mixtures in which a carbon (C) particle is clung to an oxygen (O) molecule by a twofold bond and to a hydroxyl bunch (―OH) by a solitary bond.
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Which state of matter does this image represent? Image of water Solid Liquid Gas Plasma
Answer:Liquid
Explanation:
Assume you dissolve 0.235 g of the weak benzoic acid, C6H5CO2H in enough water to make 100.0 mL of the solution and then titrate the solution with 0.108 M NaOH. Benzoic acid is a monoprotic acid.
1. What is the pH of the original benzoic acid solution before the titration is started?
2. What is the pH when 7.00 mL of the base is added? (Hint: This is in the buffer region.)
3. What is the pH at the equivalence point?
Answer:
1. pH = 2.98
2. pH = 4.02
3. pH = 8.12
Explanation:
1. Initial molarity of benzoic acid (Molar mass: 122.12g/mol; Ka = 6.14x10⁻⁵) is:
0.235 ₓ (1mol / 122.12g) = 1.92x10⁻³ moles / 0.100L = 0.01924M
The equilibrium of benzoic acid with water is:
C6H5CO2H(aq) + H2O(l) → C6H5O-(aq) + H3O+(aq)
And Ka is defined as the ratio between equilibrium concentrations of products over reactants, thus:
Ka = 6.14x10⁻⁵ = [C6H5O⁻] [H3O⁺] / [C6H5CO2H]
The benzoic acid will react with water until reach equilibrium. And equilibrium concentrations will be:
[C6H5CO2H] = 0.01924 - X
[C6H5O⁻] = X
[H3O⁺] = X
Replacing in Ka:
6.14x10⁻⁵ = [X] [X] / [0.01924 - X]
1.1815x10⁻⁶ - 6.14x10⁻⁵X = X²
1.1815x10⁻⁶ - 6.14x10⁻⁵X - X² = 0
Solving for X:
X = -0.0010→ False solution. There is no negative concentrations
X = 0.0010567M → Right solution.
pH = - log [H3O⁺] and as [H3O⁺] = X:
pH = - log [0.0010567M]
pH = 2.982.
pH of a buffer is determined using H-H equation (For benzoic acid:
pH = pka + log [C6H5O⁻] / [C6H5OH]
pKa = -log Ka = 4.21 and [] could be understood as moles of each chemical
The benzoic acid reacts with NaOH as follows:
C6H5OH + NaOH → C6H5O⁻ + Na⁺ + H₂O
That means NaOH added = Moles C6H5O⁻ And C6H5OH = Initial moles (1.92x10⁻³ moles - Moles NaOH added)
7.00mL of NaOH 0.108M are:
7x10⁻³L ₓ (0.108 mol / L) = 7.56x10⁻⁴ moles NaOH = Moles C₆H₅O⁻
And moles C6H5OH = 1.92x10⁻³ moles - 7.56x10⁻⁴ moles = 1.164x10⁻³ moles C₆H₅OH
Replacing in H-H equation:
pH = 4.21 + log [7.56x10⁻⁴ moles] / [ 1.164x10⁻³ moles]
pH = 4.023. At equivalence point, all C6H5OH reacts producing C6H5O⁻. The moles are 1.164x10⁻³ moles
Volume of NaOH to reach equivalence point:
1.164x10⁻³ moles ₓ (1L / 0.108mol) = 0.011L. As initial volume was 0.100L, In equivalence point volume is 0.111L and concentration of C₆H₅O⁻ is:
1.164x10⁻³ moles / 0.111L = 0.01049M
Equilibrium of C₆H₅O⁻ with water is:
C₆H₅O⁻(aq) + H₂O(l) ⇄ C₆H₅OH(aq) + OH⁻(aq)
Kb = [C₆H₅OH] [OH⁻]/ [C₆H₅O⁻]
Kb = kw / Ka = 1x10⁻¹⁴ / 6.14x10⁻⁵ = 1.63x10⁻¹⁰
Equilibrium concentrations of the species are:
C₆H₅O⁻ = 0.01049M - X
C₆H₅OH = X
OH⁻ = X
Replacing in Kb expression:
1.63x10⁻¹⁰ = X² / 0.01049- X
1.71x10⁻¹² - 1.63x10⁻¹⁰X - X² = 0
Solving for X:
X = -1.3x10⁻⁶ → False solution
X = 1.3076x10⁻⁶ → Right solution
[OH⁻] = 1.3076x10⁻⁶
as pOH = -log [OH⁻]
pOH = 5.88
And pH = 14 - pOH
pH = 8.12Draw the structure of beeswax.beeswax is made from the esterfication of a saturated 16-carbon fatty acid and a 30 carbon straight chain primary alcohol.
Answer:
Triacontyl palmitate
Explanation:
In this case, we have a reaction between an acid and an alcohol. When we put together these kind of compounds an ester is produced. This reaction is called "esterification".
In our case, the alcohol is a structure with 30 carbon in which the "OH" group is bonded on carbon 1. The name of this compound is "n-triacontanol". The acid is a structure in which we have 16 carbon in which the "COOH" group is placed on carbon 1. The name of this compound is "palmitic acid". The ester produced by the acid and the alcohol is "Triacontyl palmitate".
See figure 1.
I hope it helps!
Suppose a student completes an experiment with an average value of 2.9 mL and a calculated standard deviation of 0.71 mL. What is the minimum value within a 1 SD range of the average
Answer:
The correct answer is 2.2 mL.
Explanation:
Given:
Average: 2.9 mL
SD: 0.71 mL
We can define a 1 SD range in which the value of volume (in mL) will be comprised:
Volume (mL) = Average ± SD = (2.9 ± 0.7) mL
Maximum value= Average + SD= 2.9 + 0.7 mL = 3.6 mL
Minimum value= Average - SD = 2.9 - 0.7 mL = 2.2 mL
Thus, the minimum value within a 1 SD range of the average is 2.2 mL
The minimum value within 1 SD is 2.19 mL
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x\ is\ raw\ score, \mu=mean,\sigma=standard\ deviation[/tex]
Given that μ = 2.9 mL, σ = 0.71 mL; hence:
The minimum value within 1 SD range = μ ± σ = 2.9 ± 0.71 = (2.19, 3.61)
Therefore the minimum value within 1 SD is 2.19 mL
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Which of the following do we need to know in order to calculate pH during an acid-base titration of a strong monoprotic acid with a strong monoprotic base? Select all that apply
a. the concentration of the acid
b. the concentration of the base titrant
c. the initial volume of the acid solution
d. the volume of the titrant used
Answer:
the volume of the titrant used
Explanation:
Acid-base titrations are usually depicted on special graphs referred to as titration curve. A titration curve is a graph that contains a plot of the volume of the titrant as the independent variable and the pH of the system as the dependent variable.
Hence, a titration curve is a graphical plot showing the pH of the analyte solution plotted against the volume of the titrant as the reaction is in progress. The titration curve is drawn by plotting data obtained during a titration, that is, volume of the titrant added (plotted on the x-axis) and pH of the system (plotted on the y-axis).
Using these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)/Cu(s) -0.40 V -0.76 V ‑0.25 V +0.34 V Calculate the standard cell potential for the cell whose reaction is Ni2+(aq) + Zn(s) →Zn2+(aq)+ Ni(s)
Answer: The standard cell potential for the cell is +0.51 V
Explanation:
Given : [tex]E^0_{Ni^{2+}/Ni}=-0.25V[/tex]
[tex]E^0_{Zn^{2+}/Zn}=-0.76V[/tex]
The given reaction is:
[tex]Ni^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Ni(s)[/tex]
As nickel is undergoing reduction, it acts as cathode and Zinc is undergoing oxidation, so it acts as anode.
[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]
where both [tex]E^0[/tex] are standard reduction potentials.
Thus putting the values we get:
[tex]E^0_{cell}=-0.25-(-0.76)[/tex]
[tex]E^0_{cell}=0.51V[/tex]
Thus the standard cell potential for the cell is +0.51 V
explain how the liquid in a thermometer changes so that it can be used to measure a temprature
Answer:
The liquid that is often used in thermometers is chrome.
It is khwon for raising its volule when the temperature raises and vice-versa. ● the temperature and the volume are proprtional to each other so using Mathematics, scientists have figured out a way to benefit from it to make a thermometer.
Calculate the enthalpy change (∆H) for the reaction- N2(g) + 3 F2(g) –––> 2 NF3(g) given the following bond enthalpies: N≡N 945 kJ/mol F–F 155 kJ/mol N–F 283 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Explanation:
If each NADHNADH generates 3 ATPATP molecules and each FADH2FADH2 generates 2 ATPATP molecules, calculate the number of ATPATP molecules generated from one saturated 18 ‑carbon fatty acid.
Answer:
[tex]128~ATP[/tex]
Explanation:
The metabolic pathway by which energy can be obtained from a fatty acid is called "beta-oxidation". In this route, acetyl-Coa is produced by removing 2 carbons from the fatty acid for each acetyl-Coa produced. In other words, for each round, 1 acetyl Coa is produced and for each round 2 carbons are removed from the initial fatty acid. Therefore, the first step is to calculate the number of rounds that will take place for an 18-carbon fatty acid using the following equation:
[tex]Number~of~Rounds=\frac{n}{2}-1[/tex]
Where "n" is the number of carbons, in this case "18", so:
[tex]Number~of~Rounds=\frac{18}{2}-1~=~8[/tex]
We also have to calculate the amount of Acetyl-Coa produced:
[tex]Number~of~Acetyl-Coa=\frac{18}{2}~=~9[/tex]
Now, we have to keep in mind that in each round in the beta-oxidation we will have the production of 1 [tex]FADH_2[/tex] and 1 [tex]NADH[/tex]. So, if we have 8 rounds we will have 8 [tex]FADH_2[/tex] and 8 [tex]NADH[/tex].
Finally, for the total calculation of ATP. We have to remember the yield for each compound:
-) [tex]1~FADH_2~=~2~ATP[/tex]
-) [tex]1~NADH~=~3~ATP[/tex]
-) [tex]Acetyl~CoA~=~10~ATP[/tex]
Now we can do the total calculation:
[tex](8*2)~+~(8*3)~+~(9*10)=130~ATP[/tex]
We have to subtract "2 ATP" molecules that correspond to the activation of the fatty acid, so:
[tex]130-2=128~ATP[/tex]
In total, we will have 128 ATP.
I hope it helps!
A sample of a hydrocarbon is found to contain 7.99g carbon and 2.01g hydrogen. What is the empirical formula for this compound
Answer:
The empirical formulae for the compound is CH3.
Identify the term that matches each electrochemistry definition. The electrode where oxidation occurs Cathode The electrode where reduction occurs Choose... An electrochemical cell powered by a spontaneous redox reaction Choose... An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction Choose... A chemical equation showing either oxidation or reduction Choose...
Answer:
An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction
Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb 2+ and 0.0050 M Ag +. [K sp (PbI 2) = 1.4 × 10 –8; K sp (AgI) = 8.3 × 10 –17] Calculate the Ag + concentration when PbI 2 just begins to precipitate.
Answer:
[Ag⁺] = 5.0x10⁻¹⁴M
Explanation:
The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:
Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸
Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷
The PbI₂ just begin to precipitate when the product [Pb²⁺] [I⁻]² = 1.4x10⁻⁸
As the initial [Pb²⁺] = 0.0050M:
[Pb²⁺] [I⁻]² = 1.4x10⁻⁸
[0.0050] [I⁻]² = 1.4x10⁻⁸
[I⁻]² = 1.4x10⁻⁸ / 0.0050
[I⁻]² = 2.8x10⁻⁶
[I⁻] = 1.67x10⁻³So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:
[Ag⁺] [I⁻] = 8.3x10⁻¹⁷
[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷
[Ag⁺] = 5.0x10⁻¹⁴MOne hundred fifty joules of heat are removed from a heat reservoir at a temperature of 150 K. What is the entropy change of the reservoir (in J/K)?
Answer:
ΔS surrounding (entropy change of the reservoir) = -1 J/K
Explanation:
Given:
Change in heat (ΔH) = 150 joules
Temperature (T) = 150 K
Find:
ΔS surrounding (entropy change of the reservoir)
Computation:
ΔS surrounding (entropy change of the reservoir) = - ΔH / T
ΔS surrounding (entropy change of the reservoir) = - 150 / 150
ΔS surrounding (entropy change of the reservoir) = -1 J/K
PV = nRT. If P = 1 atm, V = 5.0 liter, R = 0.0821 L.atm/mol.K, and T = 293 K; what is the value of n?
Answer:
n = 0.207 mole
Explanation:
We have,
P = 1 atm
V = 5 liter
R = 0.0821 L.atm/mol.K
T = 293 K
We need to find the value of n. The relation is as follows :
PV = nRT
Solving for n,
[tex]n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 5}{0.0821 \times 293}\\\\n=0.207\ \text{mol}[/tex]
So, the value of n is 0.207 mol.
Divers often inflate heavy duty balloons attached to salvage items on the sea floor. If a balloon is filled to a volume of 1.20 L at a pressure of 6.25 atm, what is the volume of the balloon when it reaches the surface?
Answer:
7.50 L
Explanation:
The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 6.25 atm × 1.20 L / 1.00 atm
V₂ = 7.50 L
Which of the following pairs of chemical reactions are inverses of each other? Answer options: a. Hydrogenation and alkylation b.Halogenation and hydrolysis c. Ammoniation and alkylation d. Oxidation and reduction
Answer:
d. Oxidation and reduction
Explanation:
For this question we have to remember the definition of each type of reaction:
-) Hydrogenation
In this reaction, we have the addition of hydrogen to a molecule. Usually, an alkene or alkyne. In the example, molecular hydrogen is added to a double bond to produce an alkane.
-) Alkylation
In this reaction, we have the addition of a chain of carbon to another molecule. In the example, an ethyl group is added to a benzene ring.
-) Hydrolysis
In this reaction, we have the breaking of a bond by the action of water. In the example, a water molecule can break the C-O bond in the ester molecule.
-) Halogenation
In this reaction, we have the addition of a halogen (atoms on the VIIIA group). In the example, "Cl" is added to the butene.
-) Ammoniation
In this reaction, we have the addition of the ammonium ion ([tex]NH_4^+[/tex]). In the example, the ammonium ion is added to an acid.
-) Oxidation and reduction
In this reaction, we have opposite reactions. The oxidation is the loss of electrons and the reduction is the gain of electrons. For example:
[tex]Ag^+~+~e^-~->~Ag[/tex] Reduction
[tex]Al~->~Al^+^3~+~3e^-[/tex] Oxidation
"What is the difference between the revertible and nonrevertible rII mutants that Benzer generated?"
If the lead concentration in water is 1 ppm, then we should be able to recover 1 mg of lead from _____ L of water.
Answer:
1 L
Explanation:
ppm means parts per million. Generally the relationship between mass and litre is given as;
1 ppm = 1 mg/L
This means that 1 ppm is equivalent to 1 mg of a substance dissolved in 1 L of water.
For the reaction CO2(g) + H2(g)CO(g) + H20(g)
∆H°=41.2 kJ and ∆S°=42.1 J/K
The standard free energy change for the reaction of 1.96 moles of Co2(g) at 289 K, 1 atm would be_________KJ.
This reaction is (reactant, product)___________ favored under standard conditions at 289 K.
Assume that ∆H° and ∆S° are independent of temperature.
Answer:
The ΔG° is 29 kJ and the reaction is favored towards reactant.
Explanation:
Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,
ΔG° = ΔH°rxn - TΔS°rxn
= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K
= 41.2 kJ - 12.2 kJ
= 29 kJ
As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.
o prepare vegetables for finishing by grilling, sautéing, pan frying, deep frying, or stewing, you should _______ them to cook them to partial doneness
Answer:
To prepare vegetables for finishing by grilling, sautéing, pan frying, deep frying, or stewing, you should parboil them to cook them to partial doneness.
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as
described by the chemical equation
MnO,(s) + 4 HCl(aq)
MnCl(aq) + 2 H2O(l) + Cl (8)
How much MnO(s) should be added to excess HCl(aq) to obtain 175 mL C12(g) at 25 °C and 715 Torr?
mass of MnO2
Answer:
Explanation:
MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂
87 g 22.4 x 10³ mL
volume of given chlorine gas at NTP or at 760 Torr and 273 K
= 175 x ( 273 + 25 ) x 715 / (273 x 760 )
= 179.71 mL
22.4 x 10³ mL of chlorine requires 87 g of MnO₂
179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g
= 696.77 x 10⁻³ g
= 696.77 mg .
