Upon the complete reaction of 29.6 grams of phosphoric acid with excess potassium hydroxide, approximately 64 grams of potassium phosphate will be produced.
Explanation:To determine the amount of potassium phosphate that will be produced, we first need to find the molar mass of phosphoric acid (H3PO4), which is approximately 98 g/mol. So, 29.6 grams of phosphoric acid is equal to 0.302 moles (29.6g / 98g/mol).
The balanced equation shows that each mole of phosphoric acid will produce one mole of potassium phosphate (K3PO4) in the reaction. Therefore, 0.302 moles of phosphoric acid will produce 0.302 moles of potassium phosphate.
Next, we need to convert this amount from moles to grams. The molar mass of potassium phosphate is roughly 212 g/mol. So, 0.302 moles of K3PO4 is equal to approximately 64 grams (0.302 moles * 212 g / mol).
Therefore, upon the complete reaction of 29.6 grams of phosphoric acid with excess potassium hydroxide, approximately 64 grams of potassium phosphate will be formed.
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Which of the following sequences ranks the structures below in order of increasing acidity?
A) 1 < 2 < 3
B) 2 < 3 < 1
C) 3 < 1 < 2
D) 2 < 1 < 3
Answer:
Hi, no structure in the question was given to the answer the question.
Explanation:
Certain molecules or groups, when attached to an organic compound can make it become more acidic or less acidic. These groups are classified into 2
Electron withdrawing groups e.g the halogens F, Cl, Br and I. Fluorine is the best withdrawing group, next to chlorine and the least Iodine. Electron donating groups e.g Alkyls and Phenyl groupsElectron withdrawing groups makes the carboxylic more acidic. Also, the closer the electron withdrawing groups to the proton in the acid also makes the acid more acidic. This electron withdrawing group allows the electron density around the oxygen atom to be decreased.
Electron donating group makes the carboxylic acid less acidic. The more alkyl groups in a carboxylic acid, the less acidic the acid becomes. This allows the electron density around the oxygen atom in to be increased.
The human body obtains 1062 kJ from a candy bar.If this energy were used to vaporize water at 100 ∘C, how much water in liters could be vaporized? (Assume that the density of water is 1.0 g/mL.) The heat of vaporization of water at 100 ∘C is 40.7 kJ/mole.
Answer:
The Answer Of this question is 0.4052 L ...
In an ionic compound, the size of the ions affects the internuclear distance (the distance between the centers of adjacent ions), which affects lattice energy (a measure of the attractive force holding those ions together). Based on ion sizes, rank these compounds of their expected lattice energy..
Note: Many sources define lattice energies as negative values. Please rank by magnitude and ignore the sign. |Lattice energy| = absolute value of the lattice energy.
a. RbCl ,b. RbBr ,c. Rbl ,d. RbF
Answer:
b
Explanation:
bc
Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an electrolytic cell. How many hours will it take to plate 15.0 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A ? Assume the copper in the electrolytic solution is present as Cu 2 + .
Final answer:
To electroplate 15.0 kg of copper onto the cathode using 34.5 A, first calculate the moles of copper needed then convert this to the required charge using Faraday's constant. Finally, divide the total charge by the current to find the time in seconds and convert to hours, resulting in approximately 367 hours.
Explanation:
To calculate the time required to electroplate 15.0 kg of copper using a current of 34.5 A, we need to use Faraday's laws of electrolysis. First, we need to determine the number of moles of copper to be plated. The molar mass of copper is approximately 63.55 g/mol, so:
15,000 g / 63.55 g/mol = 236.025 mol Cu
Since each copper ion (Cu2+) requires two electrons to be reduced to copper metal, the number of moles of electrons needed is twice the number of moles of copper:
2 × 236.025 mol = 472.05 mol e-
Each mole of electrons corresponds to a charge of 96,485 coulombs (Faraday's constant), so:
472.05 mol e- × 96,485 C/mol = 45,562,240.25 C
Now, we can calculate the time required to deliver this charge at a rate of 34.5 A (since 1 A = 1 C/s), using:
Time (s) = Total Charge (C) / Current (A)
Time (s) = 45,562,240.25 C / 34.5 A = 1,320,643 s
Converting seconds to hours:
1,320,643 s / (60 s/min) / (60 min/h) ≈ 367 h
Therefore, it will take approximately 367 hours to electroplate 15.0 kg of copper onto the cathode with a constant current of 34.5 A.
For the reaction X2 + Y + Z -> XY + XZ it is found that doubling the concentration of X2 doubles the reaction rate. triplingthe concentration on triples the rate, and doubling the concentration of 2 has no effect, (a) What is the rate law for thisreaction? (b) Why is it that the change in the concentration of 2 has no effect on the rate? (c) Suggest a mechanism for thereaction that is consistent with the rate law.
Answer:
(a) R=k[X2][Y] (b) reaction is zero order wrt Z (c) X2 + Y --- XY + X (slow step)
X + Z --- XZ (fast step)
Explanation:
(a) Suppose the reaction rate R with respect to a component X2 with concentration [X2] is generally expressed as follows:
R = k [X2]^n (1)
Where k is the rate constant and n is the order of reaction with respect to X2.
When the reaction rate is found to double by doubling the concentration of X2, the following equation could be developed:
2R = k *(2[X2])^n (2)
Dividing equation (2) by (1) yields
2R/R= k *(2[X2])^n / k *[X2]^n
2=2^n , n = 1
Thus, the reaction is first order with respect to X2.
In the same manner,
Tripling the concentration of Y triples the rate,
R = k [Y]^m (3)
Where k is the rate constant and m is the order of reaction with respect to Y.
When the reaction rate is found to triple by tripling the concentration of Y, the following equation could be developed:
2R = k *(3[Y])^m (4)
Dividing equation (4) by (3) yields
3R/R= k *(3[Y])^m / k *[Y]^m
3=3^m , m = 1
Thus, the reaction is first order with respect to Y.
Therefore,the rate law for this reaction is R = k [X2] [Y]
(b)
Provided that doubling the concentration of Z has no effect, we perform similar analysis as above with p representing order of reaction wrt Z:
R = k [Z]^p (5)
R = k (2*[Z]^p (6)
R/R= k *(2[Z])^p / k *[Z]^p
1=2^p , p = 0
Thus, the reaction is zero order wrt Z. This explains why the change in concentration has no effect on the rate.
(c) A suggested mechanism for the reaction that is consistent with the rate law will therefore be
X2 + Y ---- XY + X (slow step)
X + Z ---- XZ (fast step)
The rate law is determined by the slow step, and this is very consistent with the experimentally observed data.
Besides, addition of the two step yields the overall reaction, and both steps are reasonable.
Final answer:
The rate law is Rate = k[X2][Y] indicating first order dependencies on X2 and Y. Z has no effect on the rate, suggesting it's not involved in the rate-determining step, which implies a reaction mechanism with an initial slow step not involving Z.
Explanation:
Rate Law and Reaction Mechanism
Based on the given information, we can deduce the rate law for the reaction. As the rate doubles when the concentration of X2 is doubled, and it triples when the concentration of Y is tripled, this suggests first order dependencies with respect to X2 and Y. There is no effect on the rate when the concentration of Z is doubled, which means Z is zero order in the rate law. Therefore, the rate law can be expressed as:
Rate = k[X2][Y]
The change in concentration of Z has no effect on the rate because it is not involved in the rate-determining step of the mechanism. The reaction mechanism likely involves a slow initial step that does not include Z, followed by a rapid step that includes it. Thus, the overall rate of the reaction is controlled by the first, slower step in the mechanism.
Express the equilibrium constant for the following reaction.P(g) + 3/2 Cl2(g) ↔ PCl3(g)K = [PCl3]^2/[P]^2[Cl2]^3K = [PCl3]/[P][Cl2]^3/2K = [P][Cl2]^3/2/[PCl3]K = [P]^1/2[Cl2]^1/3/[PCl3]^1/2K = [PCl3]^2/[P]2[Cl2]3
The correct expression for the equilibrium constant (K) is:
K = [PCl₃]² / [P] × [Cl₂](3/2)
The correct expression for the equilibrium constant (K) of the given reaction is:
K = [PCl₃]² / [P] × [Cl₂](3/2)
In this expression:
[PCl₃] represents the molar concentration of PCl₃ (gaseous product).
[P] represents the molar concentration of P (reactant P in its gaseous state).
[Cl₂] represents the molar concentration of Cl₂ (reactant Cl₂ in its gaseous state).
The exponents in the expression correspond to the stoichiometric coefficients in the balanced chemical equation:
P(g) + 3/2 Cl₂(g) ↔ PCl₃(g)
The stoichiometric coefficients of P and Cl₂ are both 1, and the coefficient of PCl₃ is 1. To make the equation balanced, we need to multiply Cl₂ by 3/2 (1.5).
So, the correct expression for the equilibrium constant (K) is:
K = [PCl₃]² / [P] × [Cl₂](3/2)
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2) You are trying to determine a TLC solvent system which will separate the compounds X, Y, and Z. You ran the compounds on a TLC plate using hexanes/ethyl acetate 95:5 as the eluting solvent and obtained the chromatogram below. How could you change the solvent system to give better separation of these three compounds?
Answer:
Answer is in the explanation.
Explanation:
Thin layer chromatography is a chromatographic technique used to separate the components of a mixture using a thin stationary phase supported by an inert backing and a mobile phase. The separation principle is in the different affinities between the components of the mixture and the stationary or mobile phase.
The affinity in mobile phase could be improved changing the polarity of this phase. In this case, you could change proportion of hexane/ethyl acetate to change polarity of mobile phase and the affinity of the different compounds to mobile or stationary phase.
I hope it helps!
Thin-layer chromatography with its adjustable hexane/ethyl acetate ratio, serves as a precise and versatile tool for optimizing separation conditions and exploring diverse interactions in chromatography.
Thin-layer chromatography (TLC) is a chromatographic method that segregates mixture components via a slender stationary phase on an inert support and a mobile phase. The separation hin-ges on distinct affinities between mixture constituents and the stationary or mobile phase. Modifying the polarity of the mobile phase can enhance its affinity.
By adjusting the hexane/ethyl acetate ratio, the mobile phase's polarity transforms, influencing compounds' interactions with the stationary and mobile phases. This dynamic shift in affinity leads to differential migration rates, facilitating component separation.
TLC stands as a versatile tool in analytical chemistry, where subtle adjustments in solvent composition yield nuanced variations in separation patterns.
Fine-tuning the hexane/ethyl acetate proportions allows for targeted optimization of separation conditions, offering a precise means to explore and exploit the diverse interactions governing the chromatographic process.
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Which of the following are characteristics of reverse-phase chromatography?I The stationary phase is polar.II The mobile phase is more polar than the stationary phase.III Less polar mobile phase has a lower eluent strength.IV More polar mobile phase has a higher eluent strength.V The stationary phase is nonpolar.
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The correct options would be:
The mobile phase is more polar than the stationary phase
The stationary phase is nonpolar
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The characteristics of reversed phase chromatography are The mobile phase is more polar than the stationary phase and The stationary phase is non polar.
Hence, option II and V are correct.
What is RPC?RPC is Reversed Phase Chromatography. In reversed phase chromatography the stationary phase is non polar [hydrophobic] and the mobile phase is very polar [hydrophilic].
Now lets check all the option one by one:
Option (I): The stationary phase is non polar in reverse phase chromatography.
So it is incorrect option.
Option (II): The mobile phase is more polar than the stationary phase because in polar mobile phase has high affinity towards the polar solute.
So, it is correct option.
Option (III): Less polar mobile phase has higher eluent strength not lower eluent strength.
So, it is incorrect option.
Option (IV): More polar mobile phase has less eluent strength.
So, it is incorrect option.
Option (V): The stationary phase is non polar [hydrophobic] is reverse phase chromatography.
So, it is correct option.
Thus, from above conclusion we can say that the The mobile phase is more polar than the stationary phase and The stationary phase is non polar are the characteristics of reverse phase chromatography.
Hence option II and V are correct.
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Part A Find ΔS∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia: 12N2(g)+32H2(g)→NH3(g) Express your answer using one decimal place. ΔS∘Δ S ∘ = J/K Previous Answer Request Answer Incorrect; Try Again; 3 attempts remaining Part B Complete previous part(s) Provide Feedback
Answer:
[tex]\Delta S^{0}[/tex] for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: [tex]\frac{1}{2}N_{2}(g)+\frac{3}{2}H_{2}(g)\rightarrow NH_{3}(g)[/tex]
[tex]\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}][/tex]
where [tex]S^{0}[/tex] represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
[tex]\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K[/tex]
So, [tex]\Delta S^{0}[/tex] for the given reaction is -99.4 J/K
The change in entropy in the reaction forming ammonia from nitrogen and hydrogen gas can be calculated using the stoichiometry of the balanced equation and the entropy values of the reactants and products.
Explanation:The question is, 'Find ΔS∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia: 1/2N2(g) + 3/2H2(g) → NH3(g)' To calculate this, we need to consider the stoichiometry of the balanced chemical equation, which tells us that one mole of N₂ will react with three moles of H₂ to form two moles of NH3. The stoichiometric factors derived from this equation can be used to determine the change in entropy (ΔS∘) for the reaction.
The change in entropy for a reaction can be determined using the equation ΔS∘ = ΣS∘(products) - ΣS∘(reactants). As per the balanced chemical equation N₂(g) + 3H₂(g) → 2NH3(g), ΔS∘ would be equal to the sum of the absolute entropies of 2 moles of NH3 (products) minus the sum of the absolute entropies of 1/2 moles of N2 and 3/2 moles of H2 (reactants). The specific entropy values would depend on data provided in a standard thermodynamic table.
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Compare and contrast the three different types of friction. Explain one way they are all alike, and at least one way they are different.
Answer:
The correct answer is static, sliding and rolling friction.
Explanation:
There are three different types of friction. Static, sliding and rolling friction occur between solid surfaces. Static friction is strongest, followed by sliding friction and then rolling friction, which is the weakest of the three. The equality of these three types of friction is that they produce heat and make movement difficult. The difference is in their magnitude and in the conditions that produce each type of friction. Static is produced when a body at rest begins to move, sliding is produced when this body is already in motion and rolling is produced when a body rolls on a surface, deforming one or both of them.
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What is the conjugate acid of each of the following? What is the conjugate base of each?
(a) OH−
(b) H2O
(c) HCO3 −
(d) NH3
(e) HSO4 −
(f) H2O2
(g) HS−
(h) H5N2+
The specie formed when an acid looses a proton is its conjugate base.
According to the Brownstead - Lowry definition, an acid donates a proton while a base accepts a proton. The specie formed when an acid looses a proton is its conjugate base.
The conjugate acids of the following Brownstead - Lowry bases are;
[tex]OH^-[/tex] ----> [tex]H2O[/tex]
[tex]H2O[/tex]----->[tex]H3O^+[/tex]
[tex]NH3[/tex] -----> [tex]NH4^+[/tex]
[tex]HS^-[/tex]----> [tex]H2S[/tex]
The conjugate bases of the following acids are shown;
[tex]H2O2[/tex] -----> [tex]HO2^-[/tex]
[tex]H5N2+[/tex] ------> [tex]H4N2[/tex]
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This is a high school chemistry question about identifying conjugate acids and bases. The conjugate acid of a base is formed when the base accepts a proton, while a conjugate base is formed when an acid donates a proton.
Explanation:When a base accepts a proton (H+), it becomes a conjugate acid, and when an acid donates a proton, it becomes a conjugate base.
For the given compounds:
(a) OH−: Conjugate acid is H2O; Conjugate base doesn't exist as OH- is a base.(b) H2O: Conjugate acid is H3O+; Conjugate base is OH-.(c) HCO3− : Conjugate acid is H2CO3; Conjugate base is CO3^2-.(d) NH3 : Conjugate acid is NH4+; Conjugate base doesn't exist as NH3 is a base.(e) HSO4− : Conjugate acid is H2SO4; Conjugate base is SO4^2-.(f) H2O2 : Conjugate acid is H3O2+; Conjugate base is HO2-.(g) HS− : Conjugate acid is H2S; Conjugate base is S^2-.(h) H5N2+ : Conjugate acid doesn't exist as H5N2+ is an acid; Conjugate base is H4N2.Learn more about Conjugate Acids and Bases here:https://brainly.com/question/36337476
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A fixed amount of gas at 25.0°C occupies a volume of 10.0 L when the pressure is 667 torr. Use Boyle's law to calculate the pressure (torr) when the volume is reduced to 7.88 L at a constant temperature of 25.0°C.a. 846 torrb. 526 torrc. 0.118 torrd. 1.11 torre. 5.26 × 104 torr
Answer:
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
Explanation:
Step 1: Data given
The temperature of a gas = 25.0°C
AT 25 °C the gas occupies a volume of 10.0L and a pressure of 667 torr.
The volume reduces to 7.88 L but the temperature stays constant.
Step 2: Boyle's law
(P1*V1)/T1 = (P2*V2)/T2
⇒ Since the temperature stays constant, we can simplify to:
P1*V1 = P2*V2
⇒ with P1 = the initial pressure 667 torr
⇒ with V1 = the initial volume = 10.0 L
⇒ with P2 = the final pressure = TO BE DETERMINED
⇒ with V2 = the final volume = 7.88L
P2 = (P1*V1)/V2
P2 = (667*10.0)/7.88
P2 = 846 torr
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
Final answer:
Using Boyle's law, the new pressure of the gas when its volume decreases from 10.0 L to 7.88 L (with initial pressure 667 torr) is calculated to be 846 torr.
Explanation:
To calculate the new pressure of a gas when its volume changes, we can use Boyle's law, which states that for a fixed amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume.
This means if the volume decreases, the pressure increases, and vice versa, as long as the temperature and the amount of gas remain unchanged.
When the volume of the gas decreases from 10.0 L to 7.88 L and the initial pressure is 667 torr, we set up Boyle's law as follows: P1 * V1 = P2 * V2
Substituting the known values gives: 667 torr * 10.0 L = P2 * 7.88 L
Assuming no typographical errors, to solve for P2 (the new pressure), we rearrange the equation:
P2 = (667 torr * 10.0 L) / 7.88 L
After performing the division, we find: P2 = 846 torr.
Therefore, the answer is (a) 846 torr, which aligns with Boyle's expectation that a decrease in volume leads to an increase in pressure.
In the development of atomic models, it was realized that the atom is mostly empty. Consider a model for the hydrogen atom where its nucleus is a sphere with a radius of roughly 10−15 m, and assume the electron orbits in a circle with a radius of roughly 10−10 m. In order to get a better sense for the emptiness of the atom, choose an object and estimate its width. This object will be your "nucleus". How far away would the "electron" be located away from your "nucleus"?
Answer:
See below
Explanation:
Lets choose a baseball ball which has to have a minimum of diameter of 1.43 inches ( 1.43 in x 2.54 cm/in = 3.63 cm )
ratio electron/nucleus = 10−10 m / 10−15 m = 100000
distance "e" = 3.63 cm x 100000 = 363000 cm ( 3630 m or 3.630 km )
Imagine to bat a baseball 3.6 km away !
The van Deemter equation describes plate height in terms of constants A, B, C, and the linear velocity, ux. Which statement is NOT true for the van Deemter equation?
A) A takes into account multiple pathways through the column. The value of A is column specific and independent of linear flow.
B) B takes into account longitudinal diffusion of the analyte in the mobile phase. The value of B is column specific and the impact of B on the plate height is inversely proportional to the linear velocity. x x + Cu u B H » A +
C) C takes into account equilibrium time between the stationary and mobile phase. The value of C is column specific and the impact of C on plate height is proportional to the linear velocity.
D) To minimize plate height, the optimal flow rate is the maxima for the plot of versus flow rate.
E) Linear velocity is the flow rate of the mobile phase.
Answer:
D) To minimize plate height, the optimal flow rate is the maxima for the plot of plate height versus flow rate.
Explanation:
Van Deemter equation in chromatography, relates the variance per unit length of a separation column to the linear mobile phase velocity by considering physical, kinetic, and thermodynamic properties of a separation.
The formula is:
H = A + B/u + C×u
Where:
H = HETP (plate height)
A = Eddy diffusion term .
B = Longitudinal diffusion.
C = Resistance against mass transfer.
u = Linear velocity .
A) A takes into account multiple pathways through the column. The value of A is column specific and independent of linear flow. TRUE. Eddy difussion term depends of column packing
B) B takes into account longitudinal diffusion of the analyte in the mobile phase. The value of B is column specific and the impact of B on the plate height is inversely proportional to the linear velocity. TRUE. Longitudinal difussion is another term that is specific to a column. Also, as the formula is H = A + B/u + C×u, the impact of B on the plate height is inversely proportional to the linear velocity.
C) C takes into account equilibrium time between the stationary and mobile phase. The value of C is column specific and the impact of C on plate height is proportional to the linear velocity. TRUE. C is the time that system needs to equilibirum. Based on the formula, the impact of C on plate height is proportional to the linear velocity
D) To minimize plate height, the optimal flow rate is the maxima for the plot of plate height versus flow rate. FALSE. The optimal flow is the minimum of the graph
E) Linear velocity is the flow rate of the mobile phase. TRUE. The Van Deemter equation describes u as the linear velocity of the mobile phase.
I hope it helps!
Final answer:
The correct statement that is NOT true for the van Deemter equation is option D) To minimize plate height, the optimal flow rate is the maxima for the plot of versus flow rate.
Explanation:
The correct statement that is NOT true for the van Deemter equation is option D) To minimize plate height, the optimal flow rate is the maxima for the plot of versus flow rate.
This means that in order to minimize the plate height, the optimal flow rate is not at the maximum point on the plot of plate height versus flow rate. The optimal flow rate is usually at a point of minimum plate height.
A solid sample of Rb2SO3 weighing 6.24 g reacts with 1.38 L gaseous HBr, measured at 75°C and 0.953 atm pressure. The solid RbBr, extracted from the reaction mixture and purified, has a mass of 7.32 g.
(a) What is the limiting reactant?
(b) What is the theoretical yield of RbBr, assuming com- plete reaction?
(c) What is the actual percentage yield of product?
Answer:
(a) HBr;
(b) 7.61 g;
(c) 96.2 %
Explanation:
Firstly, write the balanced chemical equation:
[tex]Rb_2SO_3 (aq) + 2 HBr (aq)\rightarrow 2 RbBr (aq) + SO_2 (g) + H_2O (l)[/tex]
(a) Find moles of each reactant dividing the mass by the molar mass of rubidium sulfite, then applying the ideal gas law for HBr:
[tex]n_{Rb_2SO_3}=\frac{6.24 g}{251.00 g/mol} = 0.02486 mol[/tex]
[tex]pV_{HBr}=n_{HBr}RT[/tex]
[tex]\therefore n_{HBr} = \frac{pV_{HBr}}{RT} = \frac{0.953 atm\cdot 1.38 L}{0.08206 \frac{L atm}{mol K}\cdot 348.15 K} = 0.04603 mol[/tex]
Find the limiting reactant by dividing each moles by the stoichiometric coefficients and comparing the two numbers:
[tex]eq._{Rb_2SO_3} = \frac{0.02486 mol}{1} = 0.02486 mol[/tex]
[tex]eq._{HBr} = \frac{0.04603 mol}{2} = 0.02302 mol[/tex]
That said, the equivalent of HBr is lower, so it's the limiting reactant.
(b) According to the balanced equation, the moles of HBr are equal to the moles of RbBr, so moles of RbBr theoretically are equal to:
[tex]n_{RbBr} = 0.04603 mol[/tex]
Using the molar mass of RbBr, convert this into mass:
[tex]m_{RbBr} = 0.04603 mol\cdot 165.372 g/mol = 7.61 g[/tex]
(c) To find the percent yield, divide the actual mass produced by the theoretical mass calculated in (b) and multiply by 100 %:
[tex]\%_{yield} =\frac{7.32 g}{7.61 g}\cdot 100\% = 96.2 \%[/tex]
A chemist prepares a solution of copper(II) sulfate CuSO4 by measuring out 79.g of copper(II) sulfate into a 500.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /molL of the chemist's copper(II) sulfate solution. Be sure your answer has the correct number of significant digits.
Answer:
The concentration of the copper(II) sulfate solution is 0.99 M
Explanation:
Step 1: Data given
Mass of copper(II) sulfate = 79 grams
Volume of the flask = 500 mL
Molar mass copper(II) sulfate = 159.61 g/mol
Step 2: Calculate moles copper(II) sulfate
Moles CuSO4 = Mass CuSO4 / molar mass CuSO4
Moles CuSO4 = 79.0 grams / 159.61 g/mol
Moles CuSO4 = 0.495 moles
Step 3: Calculate concentration
Concentration CuSO4 = moles / volume
Concentration CuSO4 = 0.495 moles / 0.5 L
Concentration = 0.99 M
The concentration of the copper(II) sulfate solution is 0.99 M
Predict the shape, state the hybridizationof the central atom,and give the ideal bond angle(s) and any expected deviations ineach of the following.
a) BiF52-
Shape:
Hybridization:
Ideal bond angle(s):
Deviation from ideal angle(s):
b) BrO3-
Shape:
Hybridization:
Ideal bond angle(s):
Deviation from ideal angle(s):
c) IF4+
Shape:
Hybridization:
Ideal bond angle(s):
Deviation from ideal angle(s):
Answer:
BiF5
Shape- trigonal bipyramid
Hybridisation-sp3d
Ideal bond angle-120 and 90
BrO3-
Shape- tetrahedral
Hybridisation-sp3
Ideal bond angle 109o28'
IF4+
Shape- trigonal bipyramid
Hybridisation-sp3d
Ideal bond angle-120o and 90o
Explanation:
The shape adopted by a molecule according to VSEPR is such as minimizes repulsion between electron pairs. The shape of a molecule depends on electron pairs present on the outermost shell of the central atom. The bond angles are such that electrons are positioned as far apart in space as possible, given the number of electron pairs present.
BiF52- has a square pyramidal shape, sp3d2 hybridization, with ideal bond angles of 90° and 120°. BrO3- has a trigonal pyramidal shape, sp3 hybridization, with an ideal bond angle of 109.5°. IF4+ has a square planar shape, sp3d2 hybridization, with ideal bond angles of 90° and 180°.
Explanation:The shape, hybridization of the central atom, ideal bond angle(s), and any expected deviations for the molecules are as follows:
BiF52-: The shape is square pyramidal. The hybridization of the central atom, Bi, is sp3d2. The ideal bond angles are 90° and 120°. Due to the presence of lone pairs, there can be deviations from these angles.BrO3-: This molecule has a trigonal pyramidal shape. The hybridization of the central Br atom is sp3. The ideal bond angle is 109.5°. Since there is one lone pair of electrons, deviations from the ideal angle occur.IF4+: The shape of this molecule is square planar. The central atom I is in sp3d2 hybridization. The ideal bond angles are 90° and 180°. As it is a symmetrical molecule with no lone pairs, there aren't deviations from this angle.Learn more about Molecule Structures here:https://brainly.com/question/31544427
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Concerning the 10.0 mL of 0.50 M NaCl to 100 mL of solution:
Does dilution change the concentration?
Yes
No
Answer:
Yes
Explanation:
Before dilution:
Volume of NaCl = 10 mL
Concentration of NaCl = 0.5 M
Number of moles = Molarity*Volume = 0.5*10 = 5 millimoles.
Note that number of moles of NaCl does not change on dilution as we are only adding water.
After dilution:
Volume of NaCl = 100 mL
Number of moles = 5 millimoles (no change)
New Concentration = Number of moles per volume in litres = [tex]\frac{5\times10^{-3}}{100\times10^{-3}}[/tex]= 0.05 Molar
Hence the concentration became one-tenth of the initial concentration after dilution.
Final answer:
Yes, dilution does change the concentration of a solution. By diluting a 0.50 M NaCl solution to a total volume of 100 mL, the concentration is reduced to 0.050 M, because the same amount of NaCl is spread out in a larger volume of water.
Explanation:
Concerning dilution, yes, it does change the concentration of the solution. When you add more solvent to a solution, the concentration of the solute decreases. For example, if you have a 0.50 M NaCl solution and add enough water to make the total volume 100 mL, the concentration of NaCl changes because the same amount of solute (NaCl) is now dispersed in a greater volume of solvent (water).
Step-by-step Calculation:
Determine the initial amount of NaCl in moles by multiplying the initial volume by the concentration: Moles of NaCl = 0.50 M × 0.010 L = 0.005 moles.
Keep in mind that the amount of NaCl does not change during dilution.
Calculate the final concentration by dividing the moles of NaCl by the final volume of the solution after dilution: Final concentration = 0.005 moles / 0.100 L = 0.050 M NaCl.
The zinc within a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (so that the hydrochloric acid can get to the zinc). The reaction between the acid and the zinc is as follows: 2H+(aq)+Zn(s)→H2(g)+Zn2+(aq). When the zinc in a certain penny dissolves, the total volume of gas collected over water at 25 ∘C was 0.949 L at a total pressure of 758 mmHg .
Answer:
Mass of hydrogen gas evolved is 0.0749 grams.
Explanation:
Total pressure of the gases = p = 758 mmHg
Vapor pressure of water = 23.78 mmHg
Pressure of hydrogen gas ,P = p - 23.78 mmHg = 758 mmHg - 23.78 mmHg
P = 734.22 mmHg = [tex]\frac{734.22}{760} atm=0.966 atm[/tex]
Temperature of of hydrogen gas ,T= 25°C =298.15 K
Volume of hydrogen gas = V = 0.949 L
Moles of hydrogen gas =n
PV = nRT (Ideal gas equation )
[tex]n=\frac{PV}{RT}=\frac{0.966 atm\times 0.949 L}{0.0821 atm L/mol K\times 298.15 K}[/tex]
n = 0.03745 mol
Moles of hydrogen gas = 0.03745 mol
Mass of 0.03745 moles of hydrogen gas = 0.03745 mol × 2 g/mol = 0.0749 g
Mass of hydrogen gas evolved is 0.0749 grams.
reddit The amount of NE released by sympathetic nerve terminals will be most strongly influenced by a change in which of the following? Alpha receptor sensitivity Alpha receptor density Extracellular [Ca2+] COMT activity
Answer: Extracellular [Ca2+]
Explanation:
The sensitivity and density of the alpha receptors serve to enhance the response to the release of norepinephrine (NE) . However, they do not exert a strong influence as the concentration of calcium ions on the amount of norepinephrine (NE) released by sympathic nerve terminals.
The release of neurotransmitters depends more on either an external or internal stimulus.This results in an action potential which on reaching a nerve terminal, results in the opening of Ca²⁺ channels in the neuronal membrane. Because the extracellular concentration of Ca²⁺ is greater than the intracellular Ca²⁺ concentration, Ca²⁺ flows into the nerve terminal. This triggers a series of events that cause the vesicles containing norepinephrine (NE) to fuse with the plasma membrane and release norepinephrine (NE) into the synapse. The higher the action potential, the higher the Ca²⁺ flow into the terminals resulting in higher amount of norepinephrine (NE) into the synapse, and vice versa.
Catechol-O-methyltransferase (COMT) is one of several enzymes that degrade catecholamines such as dopamine, epinephrine, and norepinephrine. It serves a regulatory purpose to lower the concentration of norepinephrine upon its release from nerve terminals.
Place the following compounds in order of increasing strength of intermolecular forces.
a. CH3CH3 CH3(CH2)8CH3 CH3CH2CH2CH3
b. CH3CH3 < CH3(CH2)8CH3 < CH3CH2CH2CH3
c. CH3CH2CH2CH3 < CH3CH3 < CH3(CH2)8CH3
d. CH3(CH2)8CH3 < CH3CH3 < CH3CH2CH2CH3
e. CH3CH3 < CH3CH2CH 2CH3 < CH3(CH2)8CH3
f. CH3(CH2)8CH3 < CH3CH2CH2CH3 < CH3CH3
Answer:
CH3(CH2)8CH3 > CH3CH2CH2CH3 > CH3CH3 (Option f)
Explanation:
Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones. In a larger atom or molecule, the valence electrons are, on average, farther from the nuclei than in a smaller atom or molecule. They are less tightly held and can more easily form temporary dipoles.
CH3CH3 has a molar mass of 30.07 g/mol
CH3(CH2)8CH3 has a molar mass of 142.28 g/mol
CH3CH2CH2CH3 has a molar mass of 58.12 g/mol
CH3(CH2)8CH3 > CH3CH2CH2CH3 > CH3CH3 (Option f)
The compounds should be ordered by increasing strength of their London dispersion forces, which relate to their molecular size. The correct order is CH3CH3 < CH3CH2CH2CH3 < CH3(CH2)8CH3, reflecting the increasing number of electrons and molecule size from ethane, to butane, to decane. Option e) is correct.
When placing the compounds CH3CH3, CH3(CH2)8CH3, and CH3CH2CH2CH3 in order of increasing strength of intermolecular forces (IMFs), it is important to consider the types of intermolecular forces present and the size of the molecules. All three compounds are nonpolar and primarily exhibit London dispersion forces. Because dispersion forces increase with the number of electrons and, thus, with the size of the molecule, the compound with the longest carbon chain will have the strongest intermolecular forces.
Thus, the correct ordering from weakest to strongest intermolecular forces is:
CH3CH3 (Ethane) - smallest molecule with the fewest electrons and hydrocarbonsCH3CH2CH2CH3 (Butane) - larger molecule with more electrons and hydrocarbons than ethaneCH3(CH2)8CH3 (Decane) - largest molecule with the most electrons and hydrocarbons, thus the strongest dispersion forcesTherefore, the correct answer is e. CH3CH3 < CH3CH2CH2CH3 < CH3(CH2)8CH3.
At 10 K Cp,m(Hg(s)) = 4.64 J K−1 mol−1. Between 10 K and the melting point of Hg(s), 234.3 K, heat capacity measurements indicate that the entropy increases by 57.74 J K−1 mol−1. The standard enthalpy of fusion of Hg(s) is 2322 J mol−1 at 234.3 K. Between the melting point and 298.0 K, heat capacity measurements indicate that the entropy increases by 6.85 J K−1 mol−1. Determine the Third-Law standard molar entropy of Hg(l) at 298 K.
Answer:
S°m,298K = 85.184 J/Kmol
Explanation:
∴ T = 10 K ⇒ Cp,m(Hg(s)) = 4.64 J/Kmol
∴ 10 K to 234.3 K ⇒ ΔS = 57.74 J/Kmol
∴ T = 234.3 K ⇒ ΔHf = 2322 J/mol
∴ 234.3 K to 298.0 K ⇒ ΔS = 6.85 J/Kmol
⇒ S°m,298K = S°m,0K + ∫CpdT/T(10K) + ΔS(10-234.3) + ΔHf/T(234.3K) + ΔS(234.3-298)
⇒ S°m,298K = 0 + 10.684 J/Kmol + 57.74 J/Kmol + 9.9104 J/Kmol + 6.85 J/kmol
⇒ S°m,298K = 85.184 J/Kmol
To find the Third-Law standard molar entropy of Hg(l) at 298 K, we sum up the entropy changes between 10K and 234.3 K, the entropy change due to fusion, and then the entropy change between the melting point and 298K. The sum gives us a final value of 74.51 J K−1 mol−1.
Explanation:The Third-Law standard molar entropy of Hg(l) at 298 K can be calculated by summing up the entropy changes that occur from 10K to 298K.
First calculate the entropy up to the melting point from 10 K which can be determined using the equation ΔS = ∫(Cp,mdT)/T.
This integral can be approximated as a rectangle from 10K to 234.3 K, hence ΔS₁ = (234.3-10)(4.64 J K−1 mol−1)/10 = 57.74 J K−1 mol−1.
Next, calculate the entropy change associated with the fusion process using the equation, ΔS = ΔH/T, giving ΔS₂ = 2322 J mol−1 / 234.3 K = 9.92 J K−1 mol−1.
Finally, add the entropy increase from the melting point to 298.0 K, which is given as 6.85 J K−1 mol−1.
Summing these values gives the Third-Law standard molar entropy of Hg(l) at 298 K: 57.74 J K−1 mol−1 + 9.92 J K−1 mol−1 + 6.85 J K−1 mol−1 = 74.51 J K−1 mol−1.
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Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume in (a) an isothermal reversible expansion, (b) an isothermal irreversible expansion against Pext = 0, and (c) an adiabatic reversible expansion.
Answer:
(a) Δ[tex]S_{sys}[/tex] = 2.881 J/K; Δ[tex]S_{sur}[/tex] = -2.881 J/K; total change in entropy = 0
(b)Δ[tex]S_{sys}[/tex] = 2.881 J/K; Δ[tex]S_{sur}[/tex] = 0 ; total change in entropy = 2.881 J/K
(c) Δ[tex]S_{sys}[/tex] = 0 ; Δ[tex]S_{sur}[/tex] = 0 ; total change in entropy = 0
Explanation:
In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:
mass (m) = 14 g
Temperature = 298 K
Pressure = 1.00 bar
Initial volume = [tex]V_{1}[/tex]
Final volume = [tex]V_{2}[/tex] = [tex]2V_{1}[/tex]
(a) Change in entropy of the system Δ[tex]S_{sys}[/tex] = [tex]nRIn\frac{V_{2} }{V_{1} }[/tex]
where R = 8.314 J/(mol*K)
n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles
Δ[tex]S_{sys}[/tex] = 0.5*8.314*ln2 = 2.881 J/K
Change in entropy of the surrounding Δ[tex]S_{sur}[/tex] = -2.881 J/K
Therefore, for a reversible process, the total change in entropy = Δ[tex]S_{sys}[/tex]+Δ[tex]S_{sur}[/tex] = 2.881 - 2.881 = 0
(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ[tex]S_{sys}[/tex] = 2.881 J/K
Since surrounding does not change in this process Δ[tex]S_{sur}[/tex] = 0.
total change in entropy = Δ[tex]S_{sys}[/tex]+Δ[tex]S_{sur}[/tex] = 2.881 - 0 = 2.88 J/K
(c) For an adiabatic reversible expansion, q(rev) = 0, thus:
Δ[tex]S_{sys}[/tex] = 0
Since heat energy is not transferred from the system to the surrounding
Δ[tex]S_{sur}[/tex] = 0
total change in entropy = Δ[tex]S_{sys}[/tex]+Δ[tex]S_{sur}[/tex] = 0
Dinitrogen tetroxide partially decomposes according to the following equilibrium: N2O4 (g) → 2 NO2 (g)
A 1.000-L flask is charged with 3.00 × 10-2 mol of N2O4. At equilibrium, 2.36 × 10-2 mol of N2O4 remains.
Keq for this reaction is __________.
Answer:
Keq for this reaction is 6.94x10⁻³
Explanation:
The equilibrium equation is this one:
N₂O₄ (g) ⇄ 2NO₂ (g)
Initially we have 0.03 moles from the dinitrogen tetroxide and nothing from the dioxide.
In the reaction, some amount of compound (x) has reacted.
As ratio is 1:2, we have double x in products.
Finally in equilibrium we have:
N₂O₄ (g) ⇄ 2NO₂ (g)
0.03 - x 2x
And we know [N₂O₄] in equilibrium so:
0.03 - x = 0.0236
x = 0.03 - 0.0236 → 6.4x10⁻³
As this is the amount that has reacted, in equilibrium I have produced:
6.4x10⁻³ .2 = 0.0128 moles of NO₂
This is the expression for K,
[NO₂] ² / [N₂O₄]
0.0128² / 0.0236 = 6.94x10⁻³
Answer:
6.94x10-3
Explanation:
ethanol is used as fuel for cars. Explain how energy is obtained from ethanol to fuel a car.
Answer:
the gas it runs on it
Explanation:
The percentage of water in an unknown hydrated salt is to bedetermined by weighing a sample of the salt, heating it to driveoff water, cooling it to room temperature, and re-weighing. Whichprocedural mistake would result in determining a percentage ofwater that is too low?
Procedural Mistakes
I. heating the sample in aclosed, rather than an open, container
II. re-weighing thesample before it has cooled to room temperature
A. I only
B. II only
C. both I andII
D. neither I norII
Answer:
B
Explanation:
If the sample is reweighed after it is removed from the furnace without being cooled to room temperature, its percentage water content will be too low as there would be almost no water if crystalization at such high temperature. However, when it is cooled to room temperature, the actual percentage of water contained in the sample can be accurately determined by weighing.
Final answer:
The procedural mistake that would result in determining a percentage of water that is too low is heating the sample in a closed container. Mistake I prevents the water from fully evaporating, which would cause an undercalculation of the water content.
Explanation:
When determining the percentage of water in a hydrated salt through heating, two procedural mistakes could lead to an inaccurate calculation of a lower percentage of water than the actual value. The correct answer to the given options is A. I only.
Heating the sample in a closed container (Mistake I) could prevent all the water from escaping, meaning that some water may remain inside, leading to an underestimation of the percentage of water in the salt.
In contrast, re-weighing the sample before it has cooled to room temperature (Mistake II) would not cause an underestimation but rather an overestimation, as the sample would weigh more due to the warmth. Therefore, Mistake II would not result in a lower percentage but rather potentially a higher percentage if the heat impacted the scale's reading.
vWould radiation with a wavelength 0.91 nm or a frequency of 5.9×1011 s−1 be detected by an X-ray detector? Would radiation with a wavelength 0.91 or a frequency of 5.9×1011 be detected by an X-ray detector? Only the radiation with a wavelength 0.91 nm can be observed by an X-ray detector. Only the radiation with a frequency of 5.9×1011 s−1 can be observed by an X-ray detector. Both types of radiation can be observed by an X-ray detector. Neither type of radiation can be observed by an X-ray detector.
Answer:
Only the radiation with a wavelength 0.91 nm can be observed by an X-ray detector.
Explanation:
To answer this question we need to consult the ranges in which x rays are in the electromagnetic spectrum:
The X radiation in the electromagnetic spectrum fall in the region of:
frequency: 3 x 10¹⁶ Hz to 3 x 10¹⁹ Hz (1Hz = 1s⁻¹)
wavelengt: 1 pm to 10 nm
Comparing the values in our question,
0.91 nm will be detected
5.9 x 10¹¹ Hz will not be detected.
The radiation with a wavelength 0.91 or a frequency of [tex]5.9*10^{11}[/tex] will have:
A. Only the radiation with a wavelength 0.91 nm can be observed by an X-ray detector.
Electromagnetic spectrum:X-rays are both types of high energy (high frequency) electromagnetic radiation. They are packets of energy that have no charge or mass (weight).
The X radiation in the electromagnetic spectrum fall in the region of:
Frequency : 3 x 10¹⁶ Hz to 3 x 10¹⁹ Hz (1Hz = 1s⁻¹)
Wavelength : 1 pm to 10 nm
On comparing to the values given in the question:
0.91 nm will be detected
5.9 x 10¹¹ Hz will not be detected.
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A thermos bottle (Dewar vessel) has an evacuated space between its inner and outer walls to diminish the rate of transfer of thermal energy to or from the bottle’s contents. For good insulation, the mean free path of the residual gas (air; average molecular mass 5 29) should be at least 10 times the distance between the inner and outer walls, which is about 1.0 cm. What should be the maximum residual gas pressure in the evacuated space if T 5 300 K? Take an average diameter of d 5 3.1 × 10210 m for the molecules in the air.
There are quite a bunch of typo errors in the question; here is the correct question below:
A thermos bottle (Dewar vessel) has an evacuated space between its inner and outer walls to diminish the rate of transfer of thermal energy to or from the bottle’s contents. For good insulation, the mean free path of the residual gas (air; average molecular mass = 29) should be at least 10 times the distance between the inner and outer walls, which is about 1.0 cm. What should be the maximum residual gas pressure in the evacuated space if T = 300 K? Take an average diameter of d = 3.1 × 10⁻¹⁰ m for the molecules in the air.
Answer:
9.57 × 10⁻⁷ atm
Explanation:
The mean free path ( λ ) can be illustrated by the equation:
λ = [tex]\frac{1}{\sqrt{2} \pi d{^2}N/V }[/tex] ---------- (1)
N/V = [tex]\frac{N_AP}{RT}[/tex] ------------- (2)
From the above relation, we can deduce that;
P= [tex]\frac{RT}{\sqrt{2}\pi d{^2}N{_A}λ }[/tex] -------------(3)
let I= λ
From the above equations;
d= diameter of the atom
[tex]{N_A}[/tex] = avogadro's constant
P= pressure
R= universal rate constant which is given by 0.08206 L atm mol⁻¹ k⁻¹
T= temperature
From the question, we are given that the mean free path of the residual air molecules ( d = 3.1 × 10⁻¹⁰ m) is equal to 10cm = 0.1m
Therefore, we can determine the pressure using equation (3)
i.e
P= [tex]\frac{RT}{\sqrt{2}\pi d{^2}N{_A}λ }[/tex]
= [tex]\frac{(8.314J{K^-^1)(300K)}}{\sqrt{2}(3.14)(3.1*10^{-10}m)^2(6.022*10^{23}mol^{-1})(0.1m) }[/tex]
=97.06 × 10⁻³ Pa × [tex]\frac{1atm}{1.01325*10^5Pa}[/tex]
=9.57 × 10⁻⁷ atm
Therefore, the maximum residual gas pressure in the calculated space is; 9.57 × 10⁻⁷ atm
My grandmother left me some silver years ago. I opened the box containing the silver and it had tarnished. Using the equation below, if I had 300 g of silver that had tarnished and 175g H2S, which would be the limiting reactant? 2 Ag(s) + H2S(g) → Ag2S(s) + H2(g) A. hydrogen B. silver C. There is no limiting reactant in this case D. hydrogen sulfide E. silver sulfide
Answer:
B
Explanation:
The limiting reagent in a chemical reaction is the one that is consumed completely in the course of the reaction. It dictates the extent to which the reaction would proceed as the completion of the reaction is based solely on it.
There are several ways to determine the limiting reagent. The easiest way to do this that works is to divide the number of moles of each reagent by their stoichiometric coefficient in the balanced equation. The reactant with the least value is the limiting reagent.
Thus, we need to calculate the number of moles of silver and hydrogen sulphide. To do this, we simply divide the masses by the relative atomic masses or molecular masses.
The atomic mass of silver is 108.
The number of moles of silver is =
300/108 = 2.8
We now divide this by the stoichiometric coefficient: 2.8/2 = 1.4
We do same for hydrogen sulphide. The molar mass of hydrogen sulphide is 34g/mol
The number of moles is thus 175/34 = 5.15
We now divide this by stoichiometric coefficient = 5.15/1 = 5.15
We can see that silver has less number of the value and hence it is the limiting reactant.
The second messenger cyclic AMP (cAMP) is synthesized from ATP by the activity of the enzyme adenylyl cyclase. Cyclic AMP, in turn, activates protein kinase A (PKA), also called cAMP‑dependent protein kinase, which is responsible for most of the effects of cAMP within the cell. Determine the correct steps in the activation of PKA, and then place them in the correct order, starting after the adenylyl cyclase reaction.
The correct steps in the activation of PKA after the adenylyl cyclase reaction are: cAMP binding to and activating the regulatory subunits of PKA, the catalytic subunits of PKA phosphorylating target proteins, and PKA being inactivated when cAMP is hydrolyzed by phosphodiesterase.
Explanation:The correct steps in the activation of PKA, after the adenylyl cyclase reaction, are:
cAMP binds to and activates the regulatory subunits of PKA, causing them to release the catalytic subunits.The catalytic subunits of PKA phosphorylate target proteins, leading to their activation or inhibition.PKA is inactivated when cAMP is hydrolyzed by phosphodiesterase, reducing the levels of cAMP in the cell.Overall, the activation of PKA by cAMP allows for the regulation of various cellular processes, including metabolism, gene expression, and cell signaling.
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The correct steps in the activation of PKA after the adenylyl cyclase reaction are: adenylyl cyclase converts ATP to cAMP, cAMP binds to and activates Protein Kinase A (PKA), and activated PKA phosphorylates serine and threonine residues of target proteins, activating them.
Explanation:The correct steps in the activation of PKA after the adenylyl cyclase reaction are:
Adenylyl cyclase converts ATP to cAMP.cAMP binds to and activates Protein Kinase A (PKA).Activated PKA phosphorylates serine and threonine residues of target proteins, activating them.These steps occur in the cAMP second messenger system and are responsible for the effects of cAMP within the cell.
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