what is true about a polar covalent bond? select the correct answer below: they are characterized by a partial positive charge on one atom and a partial negative charge on the other. the electrons are shared equally between the atoms in the bond. the electrons are transferred from one atom to the other in the bond. the electrons are absorbed into the nucleus of one atom in the bond.

Answer:

Radioactive isotopes have many useful applications. In medicine, for example, cobalt-60 is extensively employed as a radiation source to arrest the development of cancer. Other radioactive isotopes are used as tracers for diagnostic purposes as well as in research on metabolic processes.

Explanation:

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The solid acetic acid at its freezing point 24 g C₂HO₂ x (1 mol C₂HO₂/60.05 g C₂HO₂) x (11.7 kJ/mol C₂HO₂) = 278.08 kJ.

Acetic acid is a colorless, corrosive liquid with a distinct vinegar-like odor. It is also known as ethanoic acid and is an important component of vinegar. It is a weak organic acid and its molecular formula is CH3COOH. Acetic acid is produced both synthetically and naturally. Commercially, it is produced via a two-step process involving the oxidation of methanol. Acetic acid is a versatile chemical used in various industries, such as pharmaceuticals, food and beverage, coatings and inks, and chemical manufacturing. It is also used as an acidulant, a preservative, and a flavoring agent.

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Answer:

Cu + H₂SO4 → CuSO4 + SO2 + H₂

To balance this equation, we need to ensure that the number of atoms of each element is equal on both sides of the equation. Here's how to balance it step by step:

Balance the Copper atoms: There is 1 Copper atom on each side, so no changes are needed.

Cu + H₂SO4 → CuSO4 + SO2 + H₂

Balance the Sulfur atoms: There is 1 Sulfur atom on each side, so no changes are needed.

Cu + H₂SO4 → CuSO4 + SO2 + H₂

Balance the Hydrogen atoms: There are 2 Hydrogen atoms on the right side, but only 1 on the left. We can balance this by adding a coefficient of 2 in front of H₂.

Cu + H₂SO4 → CuSO4 + SO2 + 2H₂

Balance the Oxygen atoms: There are 4 Oxygen atoms on the right side, but only 2 on the left. We can balance this by adding a coefficient of 2 in front of CuSO4 and a coefficient of 1/2 in front of SO2.

Cu + H₂SO4 → 2CuSO4 + SO2 + 2H₂

Now the equation is balanced, with 1 Copper atom, 1 Sulfur atom, 4 Hydrogen atoms, and 6 Oxygen atoms on each side. I hope this helps, and good luck on your final exam tomorrow!

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The pH of a solution that is 0.10 M HF and 0.10 M NaF (the conjugate base) is given as follows:

pH is calculated as follows: [H+] = √(Ka × [acid])/[conjugate base][H+] = √(3.5 × 10⁻⁴ × 0.10)/0.10[H+] = 0.0187 M.

The pH is calculated using the following formula: pH = -log[H+]pH = -log(0.0187) pH = 1.73.

The pH of the given solution is 1.73.

In conclusion, the pH of a solution that is 0.10 M HF and 0.10 M NaF (the conjugate base) is 1.73.

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The name and chemical symbol of the given element whose electronic configurations are shown is:

a. Germanium and its symbol is Ge.

b. Niobium and its symbol is Nb.

The chemical symbol of an element is the symbol that is used to represent the atom of the element usually based on the name o the element.

The name and chemical symbol of the given element whose electronic configurations are shown is determined as follows:

a.) [Ar]4s¹3d¹⁰4p²5p¹

The atomic number of the element is 32

The element whose atomic number is 32 is Germanium and its symbol is Ge.

b.) [Kr]5s²4d²5p¹

The atomic number of the element is 41

The element whose atomic number is 32 is Niobium and its symbol is Nb.

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The compound in which the octet is expanded to include 12 electrons is SF6 (sulphur hexafluoride). In SF6, the sulphur atom has six fluorine atoms surrounding it, and in order to bond with all six fluorine atoms, the sulphur atom must have an expanded octet, meaning it has 12 electrons in its outermost energy level.

The octet is expanded to include 12 electrons in compounds where the central atom can accommodate more than eight electrons. Such compounds typically involve elements from the 3rd period or below. A common example is sulfur hexafluoride (SF6), where sulfur has an expanded octet of 12 electrons.

The inorganic compound sulphur hexafluoride is a colourless, odourless, nonflammable, and nontoxic gas. With six fluorine atoms joined to a central sulphur atom, SF6 has an octahedral structure. As might be expected for a non-polar gas, SF6 dissolves poorly in water but readily in non-polar organic solvents. At sea level, it has a density of 6.12 g/L, which is significantly higher than the density of air (1.225 g/L). It is often carried as a compressed gas that has been liquefied.

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C/ If Fe is greater than Fg, the negatively charged oil droplet will move freely toward the negatively charged plate.

In the Millikan oil droplet experiment, the negatively charged oil droplets are subjected to an electric field created by the charged plates. The electric force (Fe) acts on the oil droplet in a direction opposite to the gravitational force (Fg). When Fe is greater than Fg, the electric force overcomes the gravitational force, causing the negatively charged oil droplet to experience an upward force. As a result, the oil droplet moves freely upward toward the negatively charged plate.

Option B is incorrect because if Fe is equal to Fg, the forces balance each other, resulting in a stationary droplet. However, the question states that Fe is increased until it is greater than Fg, implying that the droplet is no longer stationary but moves in response to the electric force.

Therefore, option C is the correct answer, as it describes the effect of an electric field on the motion of a negatively charged oil droplet in the Millikan oil droplet experiment.

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Metal melting transitions and glass melting transitions are two types of phase transitions that occur when a solid material is heated to a certain temperature.

Some primary differences between metal melting transitions and glass melting transitions include:
1. Order of structure: Metals have a crystalline structure, whereas glasses have an amorphous (non-crystalline) structure. In a metal, atoms are arranged in an ordered pattern, while in a glass, atoms are arranged randomly.
2. Melting point: Metals generally have a well-defined melting point at which the solid-to-liquid phase transition occurs. Glass, on the other hand, does not have a specific melting point but instead undergoes a gradual softening over a temperature range, known as the glass transition temperature.
3. Phase transition: During metal melting transitions, the crystalline structure breaks down, and the metal changes from a solid to a liquid state. In glass melting transitions, the amorphous structure becomes more fluid and less viscous, but there is no distinct phase change as in metals.
4. Heat capacity: Metals typically have a lower heat capacity compared to glasses. This means metals require less energy to change temperature, making them easier to melt or heat up than glasses.
5. Thermal expansion: Metals generally have a lower coefficient of thermal expansion compared to glasses. This means metals expand less when heated, which affects their melting transitions and thermal properties.
6. Conductivity: Metals are good conductors of heat and electricity, while glasses are poor conductors. This difference affects the melting transitions, as metals can more easily transfer heat throughout their structure, leading to a more uniform melting process. In contrast, glasses have a more localized heating process, which can cause a more gradual softening behavior.

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Fine hair typically has fewer layers of cuticle scales compared to other hair types.

The cuticle is the outermost layer of the hair shaft that serves as a protective barrier. It consists of multiple layers of scale-like structures called cuticle scales. The number of cuticle scale layers can vary depending on individual hair characteristics, including hair type and texture.

In the case of fine hair, which is characterized by a smaller diameter and a smoother texture, it generally has fewer layers of cuticle scales compared to thicker or coarser hair types. Fine hair tends to have a more delicate structure with thinner cuticle scales. This can make fine hair more prone to damage and breakage, as it offers less protection to the inner layers of the hair shaft.

While the exact number of cuticle scale layers in fine hair can vary from person to person, it is typically lower compared to other hair types. The reduced number of cuticle scales in fine hair contributes to its specific characteristics, such as its softness and smoothness.

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The heat of solution (δH) for sodium hydroxide is -44.5 kJ/mol. We need to calculate the amount of energy involved when 5.0 g of sodium hydroxide is dissolved in water. Equation 3 is given as:NaOH(s) → Na+(aq) + OH-(aq)The molar mass of NaOH is 40.0 g/mol.

We need to find out the number of moles of NaOH in 5.0 g of NaOH. Number of moles = Mass of the substance/Molar mass of the substance= 5.0 g/40.0 g/mol= 0.125 mol Now, we need to calculate the amount of energy involved when 0.125 mol of NaOH is dissolved in water. Energy involved = δH × Number of moles of NaOH= -44.5 kJ/mol × 0.125 mol= -5.56 kJ Thus, the amount of energy involved when 5.0 g of NaOH is dissolved in water is -5.56 kJ. The negative sign indicates that the reaction is exothermic.

In chemical thermodynamics, the heat of solution is the heat released or absorbed when a substance dissolves in a solvent at a constant pressure. If the value of heat of solution is negative, it indicates that the reaction is exothermic, and if it is positive, it indicates that the reaction is endothermic. The heat of solution for NaOH is -44.5 kJ/mol, which means that the dissolution of NaOH in water is an exothermic process.

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We have that for the Question it can be said that the NaOH combines with CH_3COOH to produce CH_3COONa (Salt)

From the question we are told

how should the ph of a 0.1 m solution of nac2h3o2 compare with that of a 0.1 m solution of kc2h3o2?

Generally

with  the ph of a 0.1 m solution of nac2h3o2 compared with that of a 0.1 m solution of kc2h3o2 ,we see that the salt produce  is a weak acid and strong akali salt

We see that the salt produced in water gives a base from the derived weak acid

The salt produce is CH_3COONa

Therefore

the NaOH combines with CH_3COOH to produce CH_3COONa (Salt)

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3C⁻⁻ on reaction with HI gives I⁻⁻⁻ as the main products and not H⁻ and C₂H₅I.

When 3C⁻⁻ is reacted with HI, the reaction product obtained is I⁻⁻⁻ as the main product. The C₂H₅I and H⁻ are not produced in significant quantities and cannot be considered the main product.The 3C⁻⁻ compound reacts with HI in the presence of a solvent to produce hydrogen gas, H⁻, C₂H₅I, and I⁻⁻⁻. The primary product obtained is I⁻⁻⁻ because it is stable and has a higher energy than C₂H₅I and H⁻.However, the reaction can be controlled to obtain C₂H₅I and H⁻ as the primary products by changing the reaction conditions. The reaction must be carried out in anhydrous conditions and at a low temperature so that the reaction proceeds in the desired direction.

3C⁻⁻ on reaction with HI gives I⁻⁻⁻ as the main products and not H⁻ and C₂H₅I. However, the reaction can be controlled to obtain C₂H₅I and H⁻ as the primary products by changing the reaction conditions.

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Answer: I think it is ‘Genetically diverse’

Explanation:

The option that shows a physical change would be a cube of aluminum metal being flattened to create aluminum foil. Option C.

In chemistry, physical changes refer to reactions that only alter the physical properties of substances and not their chemical properties.

It is opposed to chemical changes which are changes that alter the chemical properties of substances, often along with their physical properties.

The changes to the color of a liquid when another liquid is added is a chemical change because the original color may not be recoverable.

The combination of two clear liquids resulting in a cloudy solution is a chemical change. So also is the combination of two clear liquids resulting in the beaker becoming hot.

The only physical change is the flattening of an aluminum cube to create aluminum foil. The chemical properties of the aluminum remain intact.

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Tempering Valve should be installed to cool the water.

A tempering valve is also called an anti-scald valve.

The thermostat controls the temperature inside the tank between 120F to 140F.

A thermostat is just a switch that is actuated by temperature. The thermostat will activates one of the elements in response to a "request for heat" when it detects a water temperature below its predetermined point.

At 140 degrees F temperature water can cause a burn to the skin in five seconds.

The Tempering valve is the mixing chamber, where cold water is mixed with hot water.  

It functions as a mixing device, mixing hot and cold water to generate the proper water temperature that is emitted from a certain outlet.

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Explanation:

Being made of charged particles can also do some things that gases can't do, such as conduct electricity. For instance, if you have gas in a liquid state, all of its particles will behave the same way once it's cool to room temperature.

the chemical equation will be XY2

Answer:

hydrogen bonding among water molecules

Explanation:

The number of moles of calcium chloride ( CaCl₂) are contained in the given sample is 0.045 moles.

Given that :

The mass of the calcium chloride, CaCl₂ = 5 g

The molar mass of the calcium chloride, CaCl₂ = 110.98 g /mol

The number of moles can be calculated by the formula given below :

The number of moles = mass / molar mass

where,

Mass = 5 g

Molar mass = 119.98 g/mol

The number of moles = 5 g / 110.98 g /mol

The number of moles = 0.045 moles.

The moles of the calcium chloride ,  CaCl₂ is 0.045 mol.

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1. The volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2.The appropriate initial condition for the ODE is S(0) = 0, as there is no salt initially in the tank.

3. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. The solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^{2/2} + 30t)} dt) / e^{(t^{2/2} + 30t)})\)

5.  the salt concentration in the tank as t→infinity is zero.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being added and drained. The tank is being filled at a rate of 2 liters per minute and drained at a rate of 1 liter per minute.

Since the tank starts with an initial volume of 30 liters, the volume of water in the tank at time t can be calculated using the equation:
Volume(t) = Initial volume + (Rate of filling - Rate of draining) * t

Volume(t) = 30 + (2 - 1) * t

So, the volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2. Let S(t) denote the amount of salt in the fish tank at time t in grams.

To show that S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t),

we need to take the derivative of S(t) with respect to t and substitute it into the given ODE.

Taking the derivative of S(t), we have:

S'(t) = 0 - (1+0)S(t) + 0

S'(t) = -S(t)

Substituting this into the given ODE, we get:

-S(t) = 70 - (t+30)S(t)

Simplifying the equation, we have:

S'(t) = 70 - (t+30)S(t)

Therefore, S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t).

The appropriate initial condition for the ODE is S(0) = 0,

as there is no salt initially in the tank.

3. This ODE is a first-order linear ordinary differential equation. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. To solve the initial value problem for S(t) using the method of integrating factors, we first rewrite the ODE in standard form:

S'(t) + (t+30)S(t) = 70

The integrating factor is given by:
\(\mu(t) = e^{(\int (t+30) dt)} = e^{(t^2/2 + 30t)\)

Multiplying both sides of the equation by μ(t), we have:
\(e^{(t^2/2 + 30t)} * S'(t) + e^{(t^2/2 + 30t)} * (t+30)S(t) = 70 * e^{(t^2/2 + 30t)\)

Applying the product rule to the left side of the equation, we get:
\((e^{(t^{2/2} + 30t) * S(t))' = 70 * e^{(t^{2/2} + 30t)})\)

Integrating both sides of the equation with respect to t, we have:
\(\int (e^{(t^2/2 + 30t)} * S(t))' dt = \int (70 * e^{(t^2/2 + 30t))} dt\)

Using the fundamental theorem of calculus, the left side becomes:
\(e^{(t^2/2 + 30t)} * S(t) = \int (70 * e^{(t^2/2 + 30t))} dt\)

Simplifying the right side by integrating, we get:
\(e^{(t^2/2 + 30t)} * S(t) = 70 * \int e^{(t^2/2 + 30t)} dt\)

At this point, the integration of \(e^{(t^2/2 + 30t)\) becomes difficult to express in terms of elementary functions.

Hence, the solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^2/2 + 30t)} dt) / e^{(t^2/2 + 30t)\)

5. As t approaches infinity, the exponential term \(e^{(t^2/2 + 30t)\) becomes very large, causing the salt concentration S(t) to approach zero. Therefore, the salt concentration in the tank as t→infinity is zero.

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The salt concentration in the tank as t approaches infinity is 70/3.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being poured into the tank and the rate at which water is being drained from the bottom.

At a rate of 2 litres per minute, water is being poured into the tank. So after t minutes, the amount of water poured into the tank is 2t litres.

At a rate of 1 litre per minute, water is being drained from the tank. So after t minutes, the amount of water drained from the tank is t litres.

Since the tank was initially filled with 30 litres of fresh water, the volume of water in the tank at time t is given by:
Volume(t) = 30 + 2t - t
Volume(t) = 30 + t

2. Let S(t) denote the amount of salt in the fish tank at time t. To determine the ODE for S(t), we need to consider the salt being poured into the tank and the salt being drained from the tank.

The salt concentration in the water being poured into the tank is 35 grams per litre. So the amount of salt being poured into the tank per minute is 35 * 2 = 70 grams.

The amount of salt being drained from the tank per minute is S(t)/Volume(t) * 1.

Therefore, the ODE for S(t) is:
S'(t) = 70 - S(t)/Volume(t)

The initial condition for this ODE is S(0) = 0, since there was no salt in the tank initially.

3. The ODE S'(t) = 70 - S(t)/Volume(t) is a first-order linear ODE. It is not separable since the variables S(t) and Volume(t) are mixed together.

4. To solve the initial value problem for S(t), we can rewrite the ODE as:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)

This is a linear ODE of the form y'(t) + p(t)y(t) = g(t), where p(t) = 1/Volume(t) and g(t) = 70 * Volume(t).

To solve this type of ODE, we can multiply both sides by an integrating factor, which is the exponential of the integral of p(t).

The integrating factor is exp(integral of 1/Volume(t) dt) = exp(ln(Volume(t))) = Volume(t).

Multiplying both sides of the ODE by the integrating factor, we get:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)
Volume(t) * S'(t) + Volume(t) * S(t) = 70 * Volume(t)^2
( Volume(t) * S(t) )' = 70 * Volume(t)^2

Integrating both sides with respect to t, we get:
Volume(t) * S(t) = 70/3 * Volume(t)^3 + C

S(t) = 70/3 * Volume(t)^2 + C/Volume(t)

Using the initial condition S(0) = 0, we can solve for C:
0 = 70/3 * 30^2 + C/30
C = -70000

Therefore, the solution for S(t) is:
S(t) = 70/3 * Volume(t)^2 - 70000/Volume(t)

5. As t approaches infinity, the volume of water in the tank becomes very large. In this case, we can approximate the volume of the tank as t, since the rate at which water is being poured in is 2 litres per minute. So the salt concentration in the tank as t approaches infinity is given by:
S(t)/Volume(t) = (70/3 * t^2 - 70000/t) / t
As t approaches infinity, the second term (-70000/t) approaches 0, so the salt concentration in the tank as t approaches infinity is:
S(t)/Volume(t) = 70/3 * t^2 / t = 70/3 * t

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Answer: An atom can be considered unstable in one of two ways. If it picks up or loses an electron, it becomes electrically charged and highly reactive. Such electrically charged atoms are known as ions. Instability can also occur in the nucleus when the number of protons and neutrons is unbalanced.

Explanation:

Answer:

An atom is the smallest particles of an element that takes part in a chemical reaction.

Explanation:

The three tiny particles are usually referred to as subatomic particles. They are Protons, Neutrons and Electrons.The protons and neutrons in the nucleus are collectively referred to as nucleus

A. The pH of the buffer is 9.25.

B. The pH of the buffer is 4.09.

C. The pH of the buffer is 4.75.

the pH of the buffer is determined by the pKb value of NH₃ and the concentrations of NH₃ and NH₄Cl. By using the Henderson-Hasselbalch equation, the pH is calculated to be 9.25. This indicates that the buffer is basic in nature.

The pH of the buffer is calculated by considering the dissociation of CH₃COOH and the concentration of CH₃COO⁻. Using the Henderson-Hasselbalch equation, the pH is found to be 4.09. This suggests that the buffer is acidic.

The pH of the buffer is determined by the dissociation of HC₇H₅O₂ and the concentration of C₇H₅O₂⁻. Applying the Henderson-Hasselbalch equation, the pH is calculated to be 4.75. This indicates that the buffer is slightly acidic.

Overall, the pH values of the buffers are influenced by the equilibrium between the weak acid and its conjugate base. These calculations demonstrate the ability of buffers to resist drastic changes in pH when small amounts of acid or base are added.

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Answer:

Ver explanación e imagen adjunta.

Explanation:

El compuesto se conoce generalmente como N, N-dietilbenzamida.

Es una amida porque tiene la fórmula general RCONR.

La molécula se llama N, N-dietil benzamida porque los dos grupos etilo están unidos al átomo de nitrógeno como se muestra en la molécula.

Answer:

The theoretical yield of urea = 120.35kg

The percent yield for the reaction = 72.70%

Explanation:

Lets calculate -

The given reaction is -

\(2NH_3(aq)+CO_2\) →\(CH_4N_2O(aq)+H_2O (l)\)

Molar mass of urea \(CH_4N_2O\)= 60g/mole

Moles of \(NH_3\) = \(\frac{62.8kg/mole}{17g/mole}\) (since \(Moles=\frac{mass of substance}{mass of one mole}\))

                     = 4011.76 moles

Moles of \(CO_2\) = \(\frac{105kg}{44g/mole}\)

                = \(\frac{105000g}{44g/mole}\)

                = 2386.36 moles

Theoritically , moles of \(NH_3\) required = double the moles of \(CO_2\)

    but , \(4011.76<2\times 2386.36\) , the limiting reagent is \(NH_3\)

Theoritical moles of urea obtained = \(\frac{1 mole CH_4N_2O}{2mole NH_3}\times4011.76 mole NH_3\)

                                                      = \(2005.88mole CH_4N_2O\)

Mass of 2005.88 mole of \(CH_4N_2O\) =\(2005.88 mole \times\frac{60g CH_4N_2O}{1mole CH_4N_2O}\)

                                                     = 120352.8g

                                                     \(120352.8g\times \frac{1kg}{1000g}\)

                                                     = 120.35kg

Therefore , theroritical yeild of urea = 120.35kg

Now , Percent yeild = \(\frac{87.5kg}{120.35kg}\times100\)

                                 72.70%

Thus , the percent yeild for the reaction is 72.70%

Answer:

B. Fuel cell cars reduce use of non-renewable fossil fuels

Explanation:

Ksp = 2.74*10^-11

Step 1: First balance the equation:

CaC2O4 (s) -> Ca2+ (aq) + C2O4 2- (aq)

Step 2: Write Ksp From above Equation

Ksp=[Ca2+][C2O4]  ( Ignore Calcium oxalate as it is a solid)

Step 3: convert mg to g and then g/L into Molarity (Mol/L) to find the concentrations used in finding Ksp:

0.67mg/L x (1 g/ 1000 mg) x (1 mol/ 128.1 g) = 5.23e-6 M

Now,

Therefore

Ksp=(Ca2+)(C2O4)

Ksp=(5.23e-6)(5.23*10^-6)

Ksp = 2.74*10^-11

What is its Ksp?

It denotes Constant of  products Solubility, also this is used to refer to the eqillibrium constant that exist between a solid substance and its solution.

Therefore, the Ksp of cac2o4 in this case will be 2.74*10^-11

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An example which doesn't depict physical weathering is acid rain falling on limestone.

Physical weathering involves the breakdown of rock into smaller particles through physical processes such as friction between rocks as a result of wind blowing sand onto rocks, freeze-thaw in colder climates etc and it is usually reversible.

Chemical weathering on the other hand involves the  breakdown of rock into smaller particles through chemical processes or reaction  such as acid rain falling on limestone and it is usually irreversible.

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The correct order of reactivity towards electrophilic aromatic substitution is:

b.) Phenol > benzene > chlorobenzene > benzoic acid

Phenol is more reactive than benzene towards electrophilic substitution reactions due to the presence of the -OH group (hydroxyl group) attached to the benzene ring. The lone pair of electrons on the oxygen atom in the -OH group can donate electron density to the ring, making it more nucleophilic and facilitating electrophilic attack.

Benzene, although less reactive than phenol, can still undergo electrophilic substitution reactions due to its aromaticity and delocalized electron system.

Chlorobenzene is less reactive than both phenol and benzene because the chlorine atom is an electron-withdrawing group. It withdraws electron density from the benzene ring, making it less nucleophilic and less prone to electrophilic substitution.

Benzoic acid is the least reactive among the options given. The carboxylic acid group (-COOH) is an even stronger electron-withdrawing group than the chlorine atom in chlorobenzene. It further reduces the electron density on the benzene ring, decreasing its reactivity towards electrophilic substitution.

Therefore, the correct order of reactivity towards electrophilic substitution is option b.

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the complete question is:

The correct order of reactivity towards electrophilic aromatic substitution is?

a.) Benzene > phenol > benzoic acid > chlorobenzene

b.) Phenol > benzene > chlorobenzene > benzoic acid

c.) Chlorobenzene > benzoic acid > phenol > benzene

d.) Benzoic acid > chlorobenzene > benzene > phenol

Answer:

Many transition metals will have electron configurations slightly different than those predicted by using the Aufbau principle, Pauli exclusion principle and Hund's rule.

For example, using these rules would produce an electron configuration for copper of [Ar] 4s2 3d9. The actual electron configuration for copper will move one electron from the 4s sublevel into the 3d sublevel resulting in [Ar] 4s1 3d10. The copper 1+ ion will have an electron configuration of [Ar] 4s0 3d10 and the copper 2+ ion will have an electron configuration of [Ar] 4s0 3d9. Many other transition elements will show movement of electrons from the highest energy s sublevel into the d sublevel one energy below.

Explanation: here cutie hope it helps ;)