What is the IUPAC name for the following compound (problem C)

The rate determining step for the reactivity for the solvolysis of 2-chloro-norbornane depends only on the decomposition of a single molecular species which is the 2-chloro-norbornane itself. For unimolecular reactions, the mechanism pathway being followed is that of an SN1 mechanism.

Answer:

SN1 mechanism

M₂ = (M₁V₁) / V₂

Dilution represents the addition of a solvent (water) without adding solutes. In dilution, the mole of the solute remains, so the concentration of the solution will drop.

When calculating dilution factors, the units of volume and concentration must remain consistent.

Dilution calculations can be typically performed following the formula:

[tex]\boxed{ \ M_1V_1 = M_2V_2 \ }[/tex].

with,

The equation which represents the correct way to find the concentration of the dilute solution (M₂) is

[tex]\boxed{ \ M_2 = \frac{M_1V_1}{V_2} \ }[/tex].

_ _ _ _ _ _ _ _ _ _

Example:

How much must be dissolved to carry out 5 liters of 0.4 M methanol solution? How much water do you require adding?

[tex]\boxed{ \ V_1 = \frac{M_2V_2}{M_1} \ }[/tex]

[tex]\boxed{ \ V_1 = \frac{0.4 \times 5}{25} \ }[/tex]

[tex]\boxed{ \ V_1 = \frac{0.4}{5} \ }[/tex]

[tex]\boxed{ \ V_1 = \frac{0.8}{10} \ }[/tex]

[tex]\boxed{ \ V_1 = \frac{8}{100} \ }[/tex]

_ _ _ _ _ _ _ _ _ _

Notes:

[tex]\boxed{ \ molarity = \frac{moles \ of \ solute}{liters \ of \ solution} \ }[/tex]

M₂ = (M₁V₁) / V₂ is the equation which represents the correct way to find the concentration of the dilute solution.

In chemistry, concentration is the ratio of solute to solvent or solution in a given volume. The amount of material in various solutions is quantified and compared using this crucial characteristic. Chemistry requires concentration in many different areas, such as chemical reactions, analyte detection, and solution preparation. Concentration can be expressed in a number of different ways, including molarity (M), mass/volume percent (%), parts per million (ppm), and mole fraction. Depending on the needs of the experiment or application, each approach offers a distinctive viewpoint on the solute concentration in a solution and is applied in various situations.

M₂ V₂= (M₁V₁)

M₂ = (M₁V₁) / V₂

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Final answer:

To find the number of moles of nitrogen in 0.215 g of N2O, calculate the molar mass of N2O (44 g/mol), then use the mass to calculate the moles of N2O. Finally, multiply by 2 since there are two nitrogen atoms per molecule of N2O, resulting in approximately 0.00978 moles of nitrogen.

Explanation:

To calculate the number of moles of nitrogen in 0.215 g of N2O, we first need the molar mass of N2O. The formula N2O contains two nitrogen atoms (N) and one oxygen atom (O). Since N has a molar mass of 14 g/mol, and O has a molar mass of 16 g/mol, the molar mass of N2O is calculated as:

[tex](2 mol N) times (14 g/mol) + (1 mol O) times (16 g/mol) = 28 g/mol + 16 g/mol = 44 g/mol.[/tex]

Next, we use the formula:

Number of moles (n) = mass (m) / molar mass (M)

n = 0.215 g / 44 g/mol

Therefore, the moles of N2O in 0.215 g is calculated as approximately:

n = 0.00489 moles of N2O

To find the number of moles of nitrogen (N) in N2O, we need to remember that each molecule of N2O contains two atoms of nitrogen, so we multiply the moles of N2O by 2:

n(N) = 0.00489 moles of N2O

n(N) = 0.00978 moles of nitrogen

A compound in chemistry is a substance that is formed when two or more elements are chemically bound together, with fixed ratios of the elements. Examples include water (H2O) and carbon dioxide (CO2).

In chemistry, a compound is a substance formed when two or more elements are chemically bound together. A given compound will always contain the same elements in fixed ratios. For instance, water (H2O) is a compound because it is made of two hydrogen atoms and one oxygen atom. Similarly, carbon dioxide (CO2) is a compound composed of one carbon atom and two oxygen atoms. Therefore, the identification of a substance as a compound depends on its elemental composition.

A compound is a substance made up of two or more different elements chemically bonded together. From the given options, the substance that is a compound is sodium chloride (NaCl). This is because sodium chloride is formed by the chemical bond between sodium (Na) and chlorine (Cl). It has a fixed ratio of sodium to chlorine atoms, and its properties are different from those of the individual elements.

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The net ionic equation for overall reaction is 2 KOH(aq) + H₃PO₄(aq) → K₂HPO₄(aq) + 2 H₂O(l)

Kalium hydroxide (KOH) and phosphoric acid (H₃PO₄) in water react to form potassium hydrogen phosphate (K₂HPO₄) and water. The balanced chemical equation for this reaction is:

2 KOH(aq) + H₃PO₄(aq) → K₂HPO₄(aq) + 2 H₂O(l).

In this process, potassium hydroxide contributes OH⁻ to phosphoric acid, which donates H⁺. The hydroxide and hydrogen ions produce water molecules. In the meantime, the remaining ions, K⁺ and HPO₄²⁻, create potassium hydrogen phosphate.

Since potassium hydroxide is a strong base and phosphoric acid is weak, the reaction completes, yielding the products. This net ionic equation simplifies the process by concentrating on chemical change species.

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The net ionic equation for the reaction between potassium hydroxide and phosphoric acid, with excess base, is H3PO4(aq) + 3OH-(aq) → 3H2O(l) + PO4^3-(aq).

When aqueous solutions of potassium hydroxide (KOH) and phosphoric acid (H3PO4) are combined, the hydroxide ions (OH-) from the strong base KOH will react with the hydrogen ions (H+) from the weak acid H3PO4 to form water (H2O) and the phosphate ion (PO43-). Given the excess base, we will see the complete neutralization of H3PO4. The net ionic equation, considering H3PO4 does not fully dissociate, is as follows:

H3PO4(aq) + 3OH-(aq) → 3H2O(l) + PO43-(aq)

The key to writing the net ionic equation is recognizing that the potassium ions (K+) and excess hydroxide ions (OH-) remain in solution and thus are spectator ions, not part of the net ionic equation.

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Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The given balanced ionic equation is,

[tex]Ni(NO_3)_2(aq)+Na_2S(aq)\rightarrow NiS(s)+2NaNO_3(aq)[/tex]

The ionic equation in separated aqueous solution will be,

[tex]Ni^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+S^{2-}(aq)\rightarrow NiS(s)+2Na^+(aq)+2NO_3^-(aq)[/tex]

In this equation, [tex]Na^+\text{ and }NO_3^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]Ni^{2+}(aq)+S^{2-}(aq)\rightarrow NiS(s)[/tex]

(b) The given balanced ionic equation is,

[tex]NaNO_3(aq)+KBr(aq)\rightarrow KNO_3(s)+NaBr(aq)[/tex]

The ionic equation in separated aqueous solution will be,

[tex]Na^+(aq)+NO_3^-(aq)+K^+(aq)+Br^-(aq)\rightarrow KNO_3(s)+Na^+(aq)+Br^-(aq)[/tex]

In this equation, [tex]Na^+\text{ and }Br^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]NO_3^-(aq)+K^+(aq)\rightarrow KNO_3(s)[/tex]

(c) The given balanced ionic equation is,

[tex]Li_2SO_4(aq)+BaCl_2(aq)\rightarrow BaSO_4(s)+2LiCl(aq)[/tex]

The ionic equation in separated aqueous solution will be,

[tex]2Li^+(aq)+SO_4^{2-}(aq)+Ba^{2+}(aq)+2Cl^-(aq)\rightarrow BaSO_4(s)+2Li^+(aq)+2Cl^-(aq)[/tex]

In this equation, [tex]Li^+\text{ and }Cl^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]SO_4^{2-}(aq)+Ba^{2+}(aq)\rightarrow BaSO_4(s)[/tex]

Answer:

Fluoride is a stronger oxidizing agent than lithium.

Explanation:

The mass ratio of oxygen to nitrogen in the first compound is 1.1413, and in the second compound, it is 4.5810. The ratio of these mass ratios is approximately 4.015.

The question involves calculating the mass ratio of oxygen to nitrogen for two nitrogen-oxygen compounds and then finding the ratio of these mass ratios.

For the first compound:
Mass of O: 53.3 g
Mass of N: 46.7 g
Mass ratio of O to N: 53.3 g / 46.7 g = 1.1413

For the second compound:
Mass of O: 82.0 g
Mass of N: 17.9 g
Mass ratio of O to N: 82.0 g / 17.9 g = 4.5810

Now, we calculate the ratio of the mass ratio of the second compound to that of the first compound:
4.5810 / 1.1413 = 4.015

Therefore, the ratio of the mass ratio of oxygen to nitrogen in the second compound to the first one is approximately 4.015.

The compound that produces 4-ethyl-3-hexanol in the presence of nickel and hydrogen is [tex]\boxed{{\text{4 - ethylhexan - 3 - one}}}[/tex]. (Refer to the attached image)

Further Explanation:

The reduction reaction is a process in organic reactions where hydrogen atoms with or without the presence of catalyst like (Ni, Pt, and Pd.) are added to organic molecules like alcohols, phenols, aldehyde, and alkenes. It is also defined as a reaction where a less electronegative species is added to a more electronegative species.

Reduction of aldehyde or ketone compounds in the presence of nickel catalysts and hydrogen gives alcohol as the product.

A general reduction reaction is as follows:  

[tex]{\text{RCHO}}\;{\text{ + }}\;{{\text{H}}_{\text{2}}}{\text{/Ni}} \to {\text{RC}}{{\text{H}}_{\text{2}}}{\text{OH}}[/tex]

In the hydrogenation reactions of aldehyde, the reduction of the carbonyl group takes place. The reaction of an aldehyde with hydrogen gas in presence of nickel catalyst leads to the formation of a primary alcohol, whereas the reaction of ketone with hydrogen gas in presence of nickel catalyst leads to the formation of a secondary alcohol.

[tex]{\text{4 - ethylhexan - 3 - one}}[/tex] reacts with hydrogen gas in the presence of nickel catalyst to undergo reduction reaction to form 4-ethyl-3-hexanol. (Refer to the attached image)

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Answer details:

Grade: High School

Subject: Chemistry

Chapter: Reduction Reaction

Keywords: Reduction reaction, hydrogen, addition, 4-ethyl-3-hexanol, aldehyde, ketone, double, single, nickel, catalyst, 4-ethylhexan-3-one, hydrogenation and 4-ethyl-3-hexanol.

Answer: The element having given valence electronic configuration is beryllium.

Explanation:

Electronic configuration is defined as the representation of electrons around the nucleus of an atom. Number of electrons in an atom are determined by the atomic number of that atom.

Valence electrons are defined as the electrons present in the outermost shell of an atom.

We are given:

Valence electronic configuration of an atom = [tex]2s^2[/tex]

So, the actual electronic configuration of atom will be [tex]1s^22s^2[/tex]

Total number of electrons = 2 + 2 = 4

So, the atomic number of given element is 4 and the element is beryllium

Hence, the element having given valence electronic configuration is beryllium.

To find the pH of the buffer solution, the Henderson-Hasselbalch equation is used with the provided concentrations of weak acid and conjugate base, yielding a pH value of approximately 5.26.

To calculate the pH of the buffer solution containing a weak acid and its conjugate base, we can use the Henderson-Hasselbalch equation:

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

where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

Given that the weak acid concentration is 0.120 M its conjugate base concentration is 0.440 M, and the Ka for the weak acid is 2.0 × 10^-5, we first calculate pKa:

pKa = -log(Ka) = -log(2.0 × 10^-5) = 4.70

Next, we use the Henderson-Hasselbalch equation:

pH = 4.70 + log(0.440/0.120) = 4.70 + log(3.667) ≈ 4.70 + 0.564 = 5.264

The pH of the buffer solution is therefore approximately 5.26.

Final answer:

To calculate the pH of the buffer solution, first determine the pKa from the given Ka and then apply the Henderson-Hasselbalch equation using the concentrations of the weak acid and its conjugate base. The pH of the given buffer solution is approximately 5.27.

Explanation:

The pH of a buffer solution can be determined using the Henderson-Hasselbalch equation, which is pH = pKa + log(·[A−]/[HA]·), where [A−] is the concentration of the conjugate base and [HA] is the concentration of the acid. In this case, the weak acid has a given Ka value of 2.0 × 10⁻µ, which allows us to calculate its pKa as the negative logarithm of Ka (pKa = -log(Ka)).

First, calculate the pKa:

pKa = -log(Ka)
  = -log(2.0 × 10⁻µ)
  = 4.70

Now, apply the Henderson-Hasselbalch equation:

pH = pKa + log([A−]/[HA])
  = 4.70 + log(0.440 M/0.120 M)
  = 4.70 + log(3.67)
  = 4.70 + 0.565
  = 5.265

Therefore, the pH of the buffer solution is approximately 5.27.

Chemists add buffers to solutions to stabilize pH levels, critical for maintaining the proper environment for chemical and biological processes. Buffers resist pH changes by balancing the effects of added acids or bases through weak acids or bases and their salts.

A chemist might add a buffer to a solution to maintain a stable pH when acids or bases are added. This is critical because many chemical reactions and biological processes require a consistent pH to function properly. Buffers work by utilizing a weak acid or base and their corresponding salts to neutralize added acids or bases without significantly changing the solution's pH. For instance, a buffer system can consist of acetic acid and sodium acetate; the acetic acid can react with any added base, while the sodium acetate reacts with any added acid, both acting to dampen fluctuations in pH levels.

In biological systems, for example, buffers help maintain conditions that are necessary for enzyme function and cellular processes. There are practical limits to buffer concentrations due to the potential formation of unwanted precipitates and the tolerance of the system to dissolved salts.

Another important role of buffers includes calibration in analytical chemistry. Adding substances like Na2SO4 to a buffer can adjust ions' concentrations that are crucial for the accuracy of a calibration curve, which is especially useful in detecting low concentrations of certain compounds.