Draw the mechanism of the slow step that occurs in both first-order substitution and first-order elimination reactions for (R)-3-bromo-2,3-dimethylpentane in methanol with heat applied. Provide curved arrows in Box 1 to depict the flow of electrons and draw the intermediate in Box 2.

Answer:

Check the explanation

Explanation:

Answer – Given, [tex]H_3PO_4[/tex] acid and there are three Ka values

[tex]K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}[/tex]

The transformation of [tex]H_2PO_4- (aq) to HPO_4^2-(aq)[/tex]is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

Answer:

See explanation below

Explanation:

The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.

For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)

For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.

Answer:

the water evaporates into the carbon dioxide

Explanation:

i just know by heart

Answer:

See explanation

Explanation:

Enamines are nucleophiles, and will react with alkyl halides to give alkylation products. Subsequent treatment with aqueous acid will give ketones.

The mechanism of an enamine reaction is shown in the image attached.

The structure of the carbanion is also shown in the image attached.

Final answer:

The sudden rise in cytosolic Ca²+ concentration in skeletal muscle leads to activation of phosphorylase kinase, not inactivation. This process is essential for muscle contraction, where Ca²+ plays a critical role by binding to proteins to expose actin for myosin attachment.

Explanation:

In response to a sudden elevation of cytosolic Ca²+ concentration in a contracting skeletal muscle, the activation of phosphorylase kinase occurs. This elevation is primarily due to excitation-contraction coupling, where an action potential triggers the release of Ca²+ from the sarcoplasmic reticulum into the cytosol. Immediately following this, calcium ions interact with troponin, altering its configuration and moving tropomyosin off the actin-binding sites. This allows the myosin head to bind to actin, resulting in muscle contraction.

Separately, in the context of ß-adrenergic receptor activation by adrenaline, an increase in cyclic AMP (cAMP) inside the muscle cell activates PKA (protein kinase A), leading to the phosphorylation of enzymes involved in glycogen degradation and inhibition of glycogen synthesis. Specifically, cAMP-dependent protein kinase is activated, causing a ready pool of glucose to be available for muscular activity.

Therefore, the correct answer is that a sudden increase in Ca²+ results in the activation of phosphorylase kinase, which is critical for initiating muscle contraction and metabolic responses.

Answer:

Mixing of O2 and N2 for the production of air.

Explanation:

An irreversible process is defined as any process which cannot return both the system and its surroundings to their original conditions. That is, the system and its surroundings would not return to their original conditions if the process was reversed. Irreversible Processes usually increase the entropy of the universe.

Processes that involve evolution of heat are usually irreversible processes since heat is lost to the surrounding. Hence the answer

Answer: 205.70 mL of acid solution is needed

Explanation: Please see the attachments below

Answer:

Mostly Para

Explanation:

First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).

Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.

Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.

Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.

Answer:

Spices

Explanation:

Herbs are the leaves.

Vegetables are the seeds.

Fruits are the seeds.

So spices are the only option left.

The correct term for the aromatic parts of plants such as bark, roots, seeds, buds, or berries is Option a i.e,  spices. Spices are used for their flavor and aroma and come from various dried parts of plants. Examples include black pepper from the fruit of Piper nigrum and cinnamon from tree bark.

When referring to the aromatic parts of plants such as bark, roots, seeds, buds, or berries, the correct term is spices. Spices are aromatic substances derived from various dried parts of plants including roots, shoots, fruits, bark, and leaves. They are often used in cooking to add flavor and aroma and are sold in forms such as whole spices, ground spices, or seasoning blends. The correct option is a i.e,  Spices.

An example is black pepper, which comes from the fruit of the Piper nigrum plant, and cinnamon, which is derived from the bark of a tree in the Laurales family.

Answer:

12.00

Explanation:

Final answer:

To find the pH of the diluted Ca(OH)₂ solution, calculate the new concentration after dilution, determine the [OH-], find pOH, and then subtract from 14 to get pH. The final pH of the solution after dilution to 1 L will be 12.

Explanation:

The question is about calculating the pH of a diluted solution of Ca(OH)₂ after it has been diluted to 1.00 L. We begin by finding the new concentration of Ca(OH)₂ after dilution, which would be:

Determine the initial number of moles of Ca(OH)₂ : moles = 200 mL times 0.025 M = 0.005 moles.

Calculate the new concentration after dilution: concentration = 0.005 moles / 1.00 L = 0.005 M.

As Ca(OH)₂ dissociates to give 2 OH - per molecule, the [OH-] = 2 times 0.005 M = 0.01 M.

Calculate the pOH: pOH = -log(0.01) = 2.

Finally, calculate the pH: pH = 14 - pOH = 14 - 2 = 12.

The pH of the solution after dilution to 1 L will be 12.

Answer:

The first stage of this reaction involves the 2-propanol reacting with benzophenone which involves the formation of the most stable carbocation, the reaction is then heated under high temperature and pressure to form benzopinacol, on cooling the benzopinacololone crystallizes out

Melting point of benzopinacolol= 184-186°C

Explanation:

Please find attached the detailed step by step mechanism of synthesis of benzopinacololone

The volume of oxygen produced by the decomposition of 129.7 g of BaO2 at 423.0 K and a pressure of 127.4 kPa is 10.37 L.

The decomposition of BaO2 (Barium peroxide) to BaO (Barium oxide) and O2 (Oxygen) is represented by the balanced chemical reaction: 2BaO2(s) → 2BaO(s) + O2(g). Using the molar mass of BaO2 (169.33 g/mol), we can calculate the number of moles of BaO2 in 129.7 g which is 0.766 moles. From the reaction stoichiometry, we can see that 2 moles of BaO2 produces 1 mole of O2. Therefore, 0.766 moles of BaO2 produces 0.766/2 = 0.383 moles of O2. Using the ideal gas law, PV=nRT, we can solve for volume (V) using n=0.383 moles, R=8.314 kPa L/mol K (universal gas constant) and T=423 K (Temperature), and P=127.4 kPa (Pressure) which gives us, V = (nRT)/P , V = (0.383 moles * 8.314 kPa L/mol K * 423 K) / 127.4 kPa = 10.37 L. Thus, the volume of oxygen produced by the decomposition of 129.7 g of BaO2 at 423.0 K and a pressure of 127.4 kPa is 10.37 L.

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The volume of oxygen produced by the decomposition of 129.7 g of BaO2 under the given conditions is approximately 13.2 liters.

The decomposition of barium peroxide (BaO2) to barium oxide (BaO) and oxygen (O2) is a reaction that can be described by the equation:

2 BaO2(s) → 2 BaO(s) + O2(g)

From the molar mass of BaO2, which is 169.34 g/mol, we can calculate the number of moles of BaO2 that the 129.7 g represents:

moles of BaO2 = 129.7 g / 169.34 g/mol ≈ 0.766 moles

According to the balanced equation, 2 moles of BaO2 produce 1 mole of O2, hence:

moles of O2 produced = 0.766 moles BaO2 / 2 ≈ 0.383 moles

Using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/K·mol), and T is the temperature:

We first convert the pressure from kPa to atm: 127.4 kPa = 1.258 atm (using the conversion 101.3 kPa = 1 atm).

Then, we solve for V using the ideal gas law equation:

V = nRT / P = (0.383 moles) × (0.0821 L·atm/K·mol) × (423.0 K) / 1.258 atm ≈ 13.2 L

Therefore, the volume of oxygen produced under the given conditions is approximately 13.2 liters.

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Answer: The number of moles of gas added to the container are 5.82

Explanation:

Avogadro's law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

where,

[tex]V_1[/tex] and [tex]n_1[/tex] are the initial volume and number of moles

[tex]V_2[/tex] and [tex]n_2[/tex] are the final volume and number of moles

Putting values in above equation, we get:

[tex]\frac{6.13}{3.51}=\frac{16.3}{n_2}\\\\n_2=9.33[/tex]

number of moles of gas added to the container = (9.33-3.51) = 5.82

Thus the number of moles of gas added to the container are 5.82

Answer:

It produces water.

Explanation:

H+   +    OH-    produces    H2O.

It is a type of Neutralization reaction.

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

To calculate the final pH of the solution, you need to consider the reaction between acetic acid (CH3COOH) and NaOH. The concentration of OH- ions in the final solution determines the pH, which can be calculated using the formula pH = -log10(H+ concentration). Using the given information, we can calculate the concentration of OH- ions, and then convert that to the concentration of H+ ions to find the pH of the solution.

To calculate the final pH of the solution, we need to consider the reaction between acetic acid (CH3COOH) and NaOH. The reaction between the two compounds forms sodium acetate (CH3COONa) and water.

The balanced equation for the reaction is:

CH3COOH + NaOH → CH3COONa + H2O

Since NaOH is a strong base, it completely dissociates in water to produce OH- ions. These OH- ions react with the acetic acid to form water. Therefore, the concentration of OH- ions will determine the pH of the final solution.

Using the given information, we can calculate the moles of acetic acid and OH- ions:

Moles of acetic acid = volume of acetic acid solution (L) x concentration of acetic acid (M)

= 0.200 L x 0.5000 M = 0.100 moles

Moles of OH- ions = volume of NaOH solution (L) x concentration of NaOH (M)

= 0.100 L x 0.5000 M = 0.050 moles

Since the mole ratio between acetic acid and OH- ions is 1:1, there will be an equal number of moles of OH- ions and acetic acid in the solution after the reaction is complete.

Therefore, the concentration of OH- ions in the final solution is:

Concentration of OH- ions = Moles of OH- ions / volume of solution (L)

= 0.050 moles / 0.300 L = 0.167 M

To calculate the final pH, we can use the formula: pH = -log10(H+ concentration)

Since in water, OH- ions and H+ ions are inversely proportional, we can calculate the concentration of H+ ions using:

H+ concentration = Kw / OH- concentration

Where Kw is the ion product of water and is equal to 1.0 x 10^-14 at 25°C.

Therefore, H+ concentration = 1.0 x 10^-14 / 0.167 M = 5.99 x 10^-14 M

Finally, calculating the pH:

pH = -log10(5.99 x 10^-14) ≈ 13.22

Answer:

false

Explanation:

it is MUCH lower in temperature

Temperature comparison on the Celsius and Fahrenheit scales are being compared, with a focus on -60 °C and -60 °F. -60 °C is colder than -60 °F because the Celsius scale has a smaller degree interval.

The question is about comparing temperatures in different units, specifically Celsius and Fahrenheit. In physics, temperature is measured using a scale known as the Celsius scale.

The Celsius scale is based on the freezing and boiling points of water, with 0 °C being the freezing point and 100 °C being the boiling point at standard atmospheric pressure. On the other hand, the Fahrenheit scale is another temperature scale commonly used in countries like the United States.

The freezing point of water on the Fahrenheit scale is 32 °F, while the boiling point is 212 °F at standard atmospheric pressure.

To answer the question, let's compare -60 °C and -60 °F. Since the Fahrenheit scale has a larger degree interval between freezing and boiling points, it means that each degree on the Fahrenheit scale is smaller than each degree on the Celsius scale.

So, -60 °C is actually colder than -60 °F. This is because -60 °C is closer to the freezing point of water (0 °C) than -60 °F is to the freezing point of water (32 °F).

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The probable question may be:

The temperature –60 °C is higher than –60 °F. Explain.

Answer:

Option B is the correct option- Ethanol increases apparent Km without affecting Vmax.

Explanation:

Vmax remains the same, and Km increases in competitive inhibition. There is an increment in Km because competitive inhibitors interfere with substrate binding to the enzyme. Vmax is not affected because the competitive inhibitor cannot bind to ES and therefore does not alter the catalysis.

Option B is the correct option- Ethanol increases apparent Km without affecting Vmax.

In the attached image, the first picture is the Michaelis Menten Plot for competitive inhibition in which an increase in km but constant Vmax is observed.

In the images,  ............................. represents the case with competitive inhibitor, while _______________ represents the case without competitive inhibitor.

The enthalpy change (ΔH) of the reaction C4H10 ---> C2H6 + C2H4 can be calculated as the difference between the activation energies of the forward and reverse reactions, giving a result of 200 kJ/mol. Activation energy, the minimum energy that reactants need to react, influences the reaction rate.

The question pertains to the activation energy and enthalpy change in the chemical reaction C4H10 ---> C2H6 + C2H4. The activation energy (Ea) is the minimum energy that reactants need to undergo a reaction, and it varies depending on the direction of the reaction. Using the given activation energies, the enthalpy change (ΔH) can be estimated as ΔH = Ea(forward) - Ea(reverse). Substituting the given values, ΔH = 450 kJ/mol - 250 kJ/mol = 200 kJ/mol.

Activation energy is highly significant to the rate of a chemical reaction. If the activation energy is larger than the average kinetic energy of the reactants, the reaction will occur slowly as only a few high-energy molecules can react. Conversely, if the activation energy is smaller, more molecules can react and the reaction rate is higher.

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Answer:

Explanation:

Larger percentage of the cells would be in mitosis, although the differences in percentages within the different stages of mitosis would still look alike, and interphase might likely still maintain the largest percentage, although it might sometimes be 50% and not  88%.

The slides you viewed in this lab are different from the slides, A larger percentage of the meristematic cells would be in mitosis in treated onion tips

The differences in percentages within the different stages of mitosis look the same, and interphase might likely still maintain the largest percentage.

Thus, The slides you viewed in this lab are different from the slides, A larger percentage of the meristematic cells would be in mitosis in treated onion tips

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Answer:

7.5x10⁻¹³M = [H⁺]

Explanation:

When a solution of Na₂CO₃ is dissolved in water, the equilibrium produced is:

Na₂CO₃(aq) + H₂O(l) ⇄ HCO₃⁺(aq) + OH⁻(aq) + 2Na⁺

Where Kb is defined from equilibrium concentrations of reactants, thus:

Kb = [HCO₃⁺][OH⁻] / [Na₂CO₃] (1)

It is possible to obtain Kb value from Ka2 and Kw thus:

Kb = Kw /  Ka2

Kb = 1x10⁻¹⁴ / 5.6x10⁻¹¹

Kb = 1.8x10⁻⁴

Replacing in (1):

1.8x10⁻⁴ = [HCO₃⁺][OH⁻] / [Na₂CO₃]

The equilibrium concentrations are:

[Na₂CO₃] = 1.0M - X

[HCO₃⁺] = X

[OH⁻] = X

Thus:

1.8x10⁻⁴ = [X][X] / [1-X]

1.8x10⁻⁴ -  1.8x10⁻⁴X = X²

X² + 1.8x10⁻⁴X - 1.8x10⁻⁴ = 0

Solving for X:

X = -0.0135 → False answer, there is no negative concentrations

X = 0.0133

As [OH⁻] = X;

[OH⁻] = 0.0133

From Kw:

Kw = [OH⁻] [H⁺]

1x10⁻¹⁴ = 0.0133[H⁺]

7.5x10⁻¹³M = [H⁺]

The concentration of [H+] in a 1.0M solution of Na2CO3 is approximately 1.2 x 10-4M, closest to 6.6 × 10^-4 M

To solve this problem, we will use the dissociation constants (Ka1 and Ka2) provided for the diacid H2CO3, as well as the fact that Na2CO3 will dissociate completely in water to form 2Na+ and CO32-. The CO32- anion can then react with water (H2O) to produce HCO3- and OH-, the latter of which will increase the pH of the solution. However, the HCO3- anion is amphiprotic and can further react with water to produce H2CO3 and OH-, again increasing the pH.

The calculations necessary to solve this question require solving the equilibrium problems for two equations: HCO3- + H2O <=> H2CO3 + OH- and H2CO3 + H2O <=> H3O+ + HCO3-. The concentrations at equilibrium are given as [H2CO3] = 0.033M, [HCO3-] = 1.2 × 10-4 M, [CO32-] = 4.7 x 10-11M, [H3O+] = 1.2 × 10-4M. Hence, [H+] = [H3O+] = 1.2 x 10-4 M. Given the multiple choice options, the closest is 6.6 × 10-4M.

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The bulk modulus of the crystal is -27.3 eV or [tex]\( -4.37 \times 10^{-9} \) J/m³.[/tex]

The cohesive energy per ion in an ionic crystal is given by the expression:

[tex]\[ E = \frac{k}{r^n} \][/tex]

Where:

E  is the cohesive energy per ion.

k  is a constant.

r is the interionic distance (lattice constant).

n  is the exponent.

Given that E = 9.1 eV, r = 1.7 nm, and n = 7 , we can solve for k :

[tex]\[ 9.1 \text{ eV} = \frac{k}{(1.7 \times 10^{-9} \text{ m})^7} \][/tex]

[tex]\[ k = 9.1 \text{ eV} \times (1.7 \times 10^{-9} \text{ m})^7 \][/tex]

Now, the bulk modulus B  of the crystal can be calculated using the relationship between cohesive energy and bulk modulus:

[tex]\[ B = -V \frac{dE}{dV} \][/tex]

Given that [tex]\( V = r^3 \)[/tex], we can find [tex]\( \frac{dE}{dV} \)[/tex] by differentiating  E  with respect to V , then substituting the values:

[tex]\[ B = -3E \][/tex]

[tex]\[ B = -3 \times 9.1 \text{ eV} \][/tex]

[tex]\[ B = -27.3 \text{ eV} \text{ (or } -4.37 \times 10^{-9} \text{ J/m}^3 \text{)} \][/tex]

Thus, the bulk modulus of the crystal is -27.3 eV or [tex]\( -4.37 \times 10^{-9} \) J/m³.[/tex]

When compared to oxygen gas, hydrogen gas emits approximately 2.82 times faster.

Rate of effusion refers to the speed at which a gas escapes or diffuses through a small opening or porous membrane into a vacuum or another gas. It is a measure of how quickly gas molecules can move and pass through a barrier.

The rate of effusion for a gas is determined by its molar mass. According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The molar mass of hydrogen is 2.02 g/mol, while the molar mass of oxygen is 32.00 g/mol. Therefore, the molar mass of oxygen is significantly larger than that of hydrogen.

Using Graham's law, we can calculate the ratio of the rates of effusion for hydrogen and oxygen:

Rate of effusion of hydrogen / Rate of effusion of oxygen = √ (Molar mass of oxygen / Molar mass of hydrogen)

Rate of effusion of hydrogen / Rate of effusion of oxygen = √ (32.00 g/mol / 2.02 g/mol)

= 2.82

Therefore, on comparing hydrogen gas effuses approximately 2.82 times faster than oxygen gas.

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Final answer:

Hydrogen gas effuses approximately 4 times faster than oxygen gas according to Graham's law of effusion.

Explanation:

According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Since hydrogen has a smaller molar mass than oxygen, it will effuse at a faster rate.

For example, if we consider the rate of effusion of hydrogen to be 1, then the rate of effusion of oxygen would be √((molar mass of hydrogen)/(molar mass of oxygen)).

Therefore, hydrogen gas effuses at a rate that is approximately 4 times faster than oxygen gas.

Answer: Partial pressure of [tex]N_{2}[/tex] at a depth of 132 ft below sea level is 2964 mm Hg.

Explanation:

It is known that 1 atm = 760 mm Hg.

Also,   [tex]P_{N_{2}} = x_{N_{2}}P[/tex]

where,    [tex]P_{N_{2}}[/tex] = partial pressure of [tex]N_{2}[/tex]

                 P = atmospheric pressure

            [tex]x_{N_{2}}[/tex] = mole fraction of [tex]N_{2}[/tex]

Putting the given values into the above formula as follows.

      [tex]P_{N_{2}} = x_{N_{2}}P[/tex]

    [tex]593 mm Hg = x_{N_{2}} \times 760 mm Hg[/tex]

       [tex]x_{N_{2}}[/tex] = 0.780

Now, at a depth of 132 ft below the surface of the water where pressure is 5.0 atm. So, partial pressure of [tex]N_{2}[/tex] is as follows.

         [tex]P_{N_{2}} = x_{N_{2}}P[/tex]

                  = [tex]0.78 \times 5 atm \times \frac{760 mm Hg}{1 atm}[/tex]

                  = 2964 mm Hg

Therefore, we can conclude that partial pressure of [tex]N_{2}[/tex] at a depth of 132 ft below sea level is 2964 mm Hg.

Answer:

Second reaction

NO2 + F -------> NO2F

Rate of reaction:

k1 [NO2] [F2]

Explanation:

NO2 + F2 -----> NO2F + F slow step1

NO2 + F -------> NO2F fast. Step 2

Since the first step is the slowest step, it is the rate determining step of the reaction

Hence:

rate = k1 [NO2] [F2]

Answer:

Cl(g)+NO_2Cl(g)-->NO_2(g)+Cl_2(g)

Explanation:

The main difference causing variation in mean stem length between the plants from the two dishes is the presence or absence of light, which affects photosynthesis and plant growth.

The most probable cause for the difference in mean stem length between plants in dish A (no light) and dish B (light cycle) is the effect of light on plant growth. Plants in dish A, with no exposure to light, likely did not undergo photosynthesis effectively, resulting in shorter stems. On the other hand, the plants in dish B received a 14-hour light cycle, allowing them to photosynthesize and grow taller.

In terms of experimental variables, the key difference between the groups is the presence of light, a critical factor for photosynthesis, and consequently, plant growth. This is supported by the fact that other conditions were kept consistent for both dishes.

Answer:

[tex]\Delta _fS=0.2724\frac{J}{K}[/tex]

Explanation:

Hello,

In this case, we define the entropy change for such freezing process as:

[tex]\Delta _fS=\frac{n_{Hg}\Delta _fH}{T_f}[/tex]

Thus, we compute the moles that are in 5.590 g of liquid mercury:

[tex]n_{Hg}=5.590 gHg*\frac{1molHg}{200.59gHg} =0.02787molHg[/tex]

Hence, we compute the required entropy change, considering the temperature to be in kelvins:

[tex]\Delta _fS=\frac{0.02787mol*2.29\frac{kJ}{mol} }{(-38.9+273.15)K}\\\\\Delta _fS=2.724x10^{-4}\frac{kJ}{K} *\frac{1000J}{kJ} \\\\\Delta _fS=0.2724\frac{J}{K}[/tex]

Best regards.

Answer:

ΔS = -0.272 J/K

Explanation:

Step 1: Data given

Its normal freezing point is –38.9 °C

molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol

Mass of Hg = 5.590 grams

Step 2:

ΔG = ΔH - TΔS

At the normal freezing point, or any phase change in general,  

ΔG =0

0  =  Δ

Hfus

−

Tfus  Δ

S

fus

Δ

S

fus = Δ

Hfus

/Tfus

Δ

S

fus = 2290 J/mol / 234.25 K

Δ

S

fus = 9.776 J/mol*K

Since fusion is  from solid to liquid. Freezing is the opposite process, so the entropy change of freezing is -9.776 J/mol*K

Step 3: Calculate moles Hg

Moles Hg = 5.590 grams / 200.59 g/mol

Moles Hg = 0.02787 moles

Step 4: Calculate the entropy change of the system:

Δ

S = -9.776 J/mol*K * 0.02787 moles

ΔS = -0.272 J/K

Final answer:

The balanced chemical equation for the overall reaction is 2NO(g) + O2(g) → 2NO2(g). The experimentally-observable rate law, assuming the second step is rate-determining, is rate = k[NO]^2[O2], where k is the overall rate constant.

Explanation:

The balanced chemical equation for the overall reaction combining the two steps 2NO(g) → N2O2(g) and NO2(g) + O2(g) → 2NO2(g) is:

2NO(g) + O2(g) → 2NO2(g)

To write the experimentally-observable rate law for the overall reaction, we must identify the rate-determining step. Assuming the second step is the rate-determining step, and given that the first step is a fast equilibrium, the overall rate can be expressed as:

rate = k2[N2O2][O2]

Since [N2O2] is the intermediate formed in the first step and we know that the rate of formation of N2O2 is proportional to the square of [NO] concentration, we can express [N2O2] in terms of [NO]. Substituting into the rate law, we get:

rate = k2k1[NO]2[O2]

Here, k = k1k2 represents the overall rate constant for the reaction. Therefore, the rate law for the overall chemical reaction is:

rate = k[NO]2[O2]

Answer:

The energy released in the decay process = 18.63 keV

Explanation:

To solve this question, we have to calculate the binding energy of each isotope and then take the difference.

The mass of Tritium = 3.016049 amu.

So,the binding energy of Tritium =  3.016049 *931.494 MeV

= 2809.43155 MeV.

The mass of Helium 3 = 3.016029 amu.

So, the binding energy of Helium 3 = 3.016029 * 931.494 MeV

= 2809.41292 MeV.

The difference between the binding energy of Tritium and the binding energy of Helium is: 32809.43155 - 2809.412 = 0.01863 MeV

1 MeV = 1000keV.

Thus, 0.01863 MeV = 0.01863*1000keV = 18.63 keV.

So, the energy released in the decay process = 18.63 keV.

Answer:

Explanation:

find the solution below

Answer:  The number of Ni2+ ions is 1.46 × 10^18 ions

The number of Ag+  ions is 2.92 × 10^18 ions

Explanation: Please see the attachments below