Example: Serum containing Na gave a signal of 4.27 mV in an atomic emission analysis. Then, 5 mL of 2.08 M NaCl were added to 95.0 mL of serum. The spiked serum gave a signal of 7.98 mV. Find the original concentration of Nat in serum.

Answer:

a. [tex]4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)[/tex]

b. 146.0 g

Explanation:

Question 1 (a). Just as the problem states, liquid nitroglycerin decomposes into nitrogen gas [tex]N_2[/tex], oxygen gas [tex]O_2[/tex], water vapor [tex]H_2O[/tex] and carbon dioxide [tex]CO_2[/tex]. Let's write the decomposition of nitroglycerin into these 4 components:

[tex]C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + CO_2 (g)[/tex]

Now we need to balance the equation. Firstly, notice we have 3 carbon atoms on the left and 1 on the right, so let's multiply carbon dioxide by 3:

[tex]C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)[/tex]

Now, we have 3 nitrogen atoms on the left and 2 on the right, so let's multiply nitrogen on the right by [tex]\frac{3}{2}[/tex]:

[tex]C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)[/tex]

We have 5 hydrogen atoms on the left, 2 on the right, so let's multiply the right-hand side by [tex]\frac{5}{2}[/tex]:

[tex]C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)[/tex]

Finally, count the oxygen atoms. We have a total of 9 on the left. On the right we have (excluding oxygen molecule):

[tex]\frac{5}{2} + 6 = 8.5[/tex]

This leaves [tex]9 - 8.5 = 0.5 = \frac{1}{2}[/tex] of oxygen. Since oxygen is diatomic, we need to take one fourth of it to get one half in total:

[tex]C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + \frac{1}{4} O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)[/tex]

To make it look neater without fractional coefficients, multiply both sides by 4:

[tex]4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)[/tex]

Question 2 (b). Now we can make use of the balanced chemical equation and apply it for the context of this separate problem. We're given the following variables:

[tex]V_{CO_2} = 41.0 L[/tex]

[tex]T = -14.0^oC + 273.15 K = 259.15 K[/tex]

[tex]p = 1 atm[/tex]

Firstly, we may find moles of carbon dioxide produced using the ideal gas law [tex]pV = nRT[/tex].

Rearranging for moles, that is, dividing both sides by RT (here R is the ideal gas law constant):

[tex]n_{CO_2} = \frac{pV_{CO_2}}{RT} = \frac{1 atm\cdot 41.0 L}{0.08206 \frac{L atm}{mol K}\cdot 259.15 K} = 1.928 mol[/tex]

According to the stoichiometry of the balanced chemical equation:

[tex]4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)[/tex]

4 moles of nitroglycerin (ng) produce 12 moles of carbon dioxide. From here we can find moles o nitroglycerin knowing that:

[tex]\frac{n_{ng}}{4} = \frac{n_{CO_2}}{12} \therefore n_{ng} = \frac{4}{12}n_{CO_2} = \frac{1}{3}\cdot 1.928 mol = 0.6427 mol[/tex]

Multiplying the number of moles of nitroglycerin by its molar mass will yield the mass of nitroglycerin decomposed:

[tex]m_{ng} = n_{ng}\cdot M_{ng} = 0.6427 mol\cdot 227.09 g/mol = 146.0 g[/tex]

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 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.

The correct steps in the activation of PKA, after the adenylyl cyclase reaction, are:

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.

The correct steps in the activation of PKA after the adenylyl cyclase reaction are:

These steps occur in the cAMP second messenger system and are responsible for the effects of cAMP within the cell.

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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°.

The shape, hybridization of the central atom, ideal bond angle(s), and any expected deviations for the molecules are as follows:

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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.

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:

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Answer:

6,04x10⁻³M

Explanation:

For the reaction:

Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)

The precipitate of Cu(s) weights 96,0 mg. In moles:

Moles of Cu(s):

0,096g×(1mol/63,546g) = 1,51x10⁻³ moles of Cu(s). If you see the balanced equation 1 mole of CuSO₄ produce 1 mole of Cu(s). That means moles of CuSO₄ are the same of Cu(s), 1,51x10⁻³ moles of CuSO₄

As volume of the solution is 250 mL, 0,250L, the molar concentration of the original solution is:

1,51x10⁻³ moles of CuSO₄ / 0,250L = 6,04x10⁻³M

I hope it helps!

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

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.

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.

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.

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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Answer:

The correct option are:

1.  (b) current

2. (c) potential

3.  (b) coulombs (C)

4.  (a) amperes (A)

5.  (d) volts (V)

Explanation:

Electric charge is a property of a particle, like electron and proton. The subatomic particles, electrons and protons are negatively and positively charged particles, respectively.

The electric charge of a particle can be measured in coulomb, denoted by C.

Current or electric current is defined as the net flow of the electric charges through a given region, per unit time. Electric current can be measured in ampere, denoted by A.

Potential or Electric potential is described as the work done to displace the electric charges from one point to another. Potential is the driving force for the movement of charges and can be calculated in volt, denoted by V.

The flow of charge per unit time is current, measured in amperes (A), while the driving force for electron flow is electric potential, measured in volts (V), and charge itself is quantified in coulombs (C).

Understanding Electrical Quantities

The amount of charge that passes per unit time is called current, and is measured in amperes (A). The driving force that causes electrons to flow, known as the electric potential, is measured in volts (V). Charge itself is measured in coulombs (C).

Answering the questions provided:

The amount of charge that passes per unit time is called current.

The driving force for the electrons is measured by potential.

Charge is measured in coulombs (C).

Current is measured in amperes (A).

Potential is measured in volts (V).

The concentration cell involving Al3+ has a standard cell potential (E°) of -1.66 V, and the cell potential (E) is approximately -1.54 V. The correct option is D: E° = -1.66 V and E = -1.54 V.

To determine the standard cell potential (E°) and the cell potential (E) for the given concentration cell, we can use the Nernst equation and the standard reduction potential (E°red) for the half-reaction involved.

The shorthand notation for the concentration cell is given as:

[tex]\[ \text{Al(s) | Al}^{3+} (1.0 \times 10^{-5} \, \text{M}) \, || \, \text{Al}^{3+} (0.100 \, \text{M}) \, | \, \text{Al(s)} \][/tex]

Firstly, let's identify the two half-reactions occurring in the cell. The half-reaction for the left side (anode) is [tex]\(\text{Al} \rightarrow \text{Al}^{3+} + 3e^-\)[/tex], and for the right side (cathode), it is [tex]\(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\)[/tex].

The standard cell potential (E°) can be calculated using the standard reduction potentials [tex](\(E°_{\text{red}}\))[/tex] of the half-reactions:

[tex]\[ E° = E°_{\text{cathode}} - E°_{\text{anode}} \][/tex]

Given[tex]\(E°_{\text{red}} = -1.66 \, \text{V}\) for \(\text{Al}^{3+}/\text{Al}\)[/tex], we can substitute this into the equation to find E°. However, we need to consider the fact that the half-reaction for the anode is reversed in the standard reduction potential table. Therefore, \(E°_{\text{anode}} = -(-1.66 \, \text{V}) = 1.66 \, \text{V}\).

[tex]\[ E° = 0 - 1.66 \, \text{V} = -1.66 \, \text{V} \][/tex]

Now, to find the cell potential (E), we use the Nernst equation:

[tex]\[ E = E° - \frac{0.0592}{n} \log\left(\frac{[\text{Al}^{3+}]_{\text{cathode}}}{[\text{Al}^{3+}]_{\text{anode}}}\right) \][/tex]

Here, n is the number of electrons transferred (which is 3 for both half-reactions), and[tex]\([X]_{\text{cathode}}\[/tex] ) and [tex]\([X]_{\text{anode}}\)[/tex]are the concentrations of X at the cathode and anode, respectively.

For the given cell, plugging in the values:

[tex]\[ E = -1.66 - \frac{0.0592}{3} \log\left(\frac{0.100}{1.0 \times 10^{-5}}\right) \][/tex]

Calculating this gives approximately -1.54 V.

Therefore, the correct answer is option D: [tex]\(E° = -1.66 \, \text{V}\) and \(E = -1.54 \, \text{V}\)[/tex].

Complete question :- Given that E°red = -1.66 V for Al3+/ Al at 25°C, find E° and E for the concentration cell expressed using shorthand notation below.

Al(s) | Al3+(1.0 × 10-5 M) || Al3+(0.100 M) | Al(s)

A) E° = 0.00 V and E = +0.24 V

B) E° = 0.00 V and E = +0.12 V

C) E° = -1.66 V and E = -1.42 V

D) E° = -1.66 V and E = -1.54 V

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:

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 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.

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.

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

Answer:

b

Explanation:

bc

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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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.

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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Answer:

the gas it runs on it

Explanation:

Answer:

All the 8.82*10^-5 moles of lead present is removed.

Explanation:

The total amount of lead II ion present is less than the amount of sulphide ion present and they react in 1:1 ratio so all the lead II ions are removed from the solution. See attached image.

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.

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.

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.

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.

To learn more about Thin-layer chromatography

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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.

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.

Find more information about X-rays here:

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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.

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

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:

Therefore, the correct answer is e. CH3CH3 < CH3CH2CH2CH3 < CH3(CH2)8CH3.