Which of the following explains the relationship between surface area of a solid solute and its dissolution in a solvent? a.) As surface area increases, the time necessary for complete dissolution decreases. b.) As surface area decreases, the time necessary for complete dissolution decreases. c.) Surface area does not affect the time necessary for complete dissolution.

The Ka for the weak acid is 2.45 x 10^-6 of concentration 0.0102m .

To calculate Ka, first, we need to calculate the concentration of H+ and the initial concentration of acid. The weak acid is monoprotic, meaning it can donate only one hydrogen ion (H+) to water.Therefore, it will dissociate as follows: HA + H2O ⇔ A- + H3O+where HA is the acid molecule, and A- is its corresponding conjugate base.

The H3O+ is also known as a hydronium ion. The first step is to calculate the concentration of H3O+.The pH of the solution is 2.68.Hence, pH = -log[H3O+]2.68 = -log[H3O+][H3O+] = 1.58 x 10^-3The concentration of H3O+ is 1.58 x 10^-3 M. Since the weak acid is monoprotic, the initial concentration of acid is equal to the concentration of the conjugate base of the weak acid, which we get from the dissociation equilibrium.

The equilibrium expression for the dissociation of a weak acid is given as follows: Ka = [A-][H3O+]/[HA]We need to find the value of Ka. We have already calculated the value of [H3O+].So, Ka = [A-][1.58 x 10^-3 M]/0.0102 MWe need to calculate the value of [A-].

From the equilibrium equation for weak acid: HA + H2O ⇔ A- + H3O+0.0102 M x1.58 x 10^-3 M Here, x is the concentration of A-.So, 1.58 x 10^-3 M = x, which is also the concentration of the conjugate base of the weak acid. So, Ka = [A-][H3O+]/[HA] = (1.58 x 10^-3 M)^2/0.0102 M= 2.45 x 10^-6Therefore, Ka for the weak acid is 2.45 x 10^-6.

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The colors of red litmus paper in acidic, neutral, and basic solutions are: C. red, colorless, and blue respectively.

Red litmus paper is used to test whether or not a solution is acidic in chemistry. It's utilized to detect the acidity or alkalinity of a substance. In acidic or neutral solutions, red litmus paper remains red. It will turn blue when it comes into contact with basic solutions. Red litmus paper is a pH indicator. It alters color based on the pH of the substance in which it is dissolved.

Litmus paper is a pH paper that is produced using lichen dyes. It's a paper that has been treated with litmus, which is a water-soluble mixture of different dyes obtained from lichens. Litmus paper's two colors, blue and red, are produced from litmus. The blue litmus paper turns red in acidic solutions and turns blue in basic solutions. Conversely, red litmus paper turns blue in basic solutions and remains red in acidic or neutral solutions. Therefore, Option C is Correct.

The Question was Incomplete, Find the full content below :

The colours of red litmus paper in acidic, neutral, and basic solutions are:

A. red, orange and blue respectively

B. blue, violet and red respectively

C. red, colourless and blue respectively

D. red, red and blue respectively

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The passage data regarding the thermal stability and the enzyme activity of the mkr681h is the most consistent with the conclusion regarding the role of Arg681 in the cct is the Arg681 is the engaged in catalytic function of enzyme.

The R681H is denotes that the amino acid 681 that is the arginine, R is the changed to the histidine (H). The Enzyme activity will depends on the principally on the enzyme’s intrinsic catalytic efficiency, and its concentration, and the initial substrate concentration, in the presence of the inhibitors or the allosteric activators, the temperature, and the pH.

This will most strongly suggests that the Arg681 will be involved in the catalytic function of  normal enzyme.

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The percent yield of a reaction is calculated by taking the amount of product produced and dividing it by the amount of reactant used, and then multiplying by 100 to get a percentage.

In this case, 55.0 g of ethene was produced from 100.0 g of ethanol, so the percent yield is 55.0 g divided by 100.0 g, multiplied by 100, which gives a percent yield of 55%.

This percent yield indicates the efficiency of the reaction, as the higher the percent yield, the more efficient the reaction is. A percent yield of 55% means that the reaction was relatively efficient, as the large majority of the reactant was converted into product. If the percent yield was much lower than this, it could indicate that there were some issues with the reaction and that it was not as efficient as it could be.

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The reaction rate will decrease in the second experiment when the concentration of both compounds is lowered. This is because the rate of a chemical reaction is directly proportional to the concentration of the reactants.

The rate of a chemical reaction is directly proportional to the concentration of the reactants.

This is because increasing the concentration of the reactants increases the number of reactant molecules available to collide with each other, which results in a faster reaction rate.

Other factors that can affect the rate of a chemical reaction include temperature, pressure, the presence of a catalyst, and the surface area of solid reactants. Increasing temperature, pressure, or the surface area of solid reactants generally leads to a faster reaction rate, while the presence of a catalyst can increase the rate of a reaction without being consumed in the process.

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The element which would have no unpaired electrons in its ground state is beryllium.

The ground state refers to the lowest energy level in which an electron can exist in an atom. An electron in the ground state has no excess energy, therefore it is stable and unlikely to be disrupted by an external force. The energy of the ground state is defined as zero, and all other states have greater energy than it.

The electronic configuration of Be is [tex]1s^2 2s^2[/tex]. In the ground state, both of the 2s electrons in beryllium are paired, so there are no unpaired electrons.

The number of unpaired electrons in the other elements listed is:

Li: one unpaired electron in the 2s orbital

C: two unpaired electrons in the 2p orbitals

O: two unpaired electrons in the 2p orbitals

B: one unpaired electron in the 2p orbital

The element that would have no unpaired electrons in its ground state is Be (beryllium).

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Answer: A

Explanation: An exothermic reaction generates heat. Unless the reaction is cooled in some way, its temperature increases. If you increase the temperature with an external heater, it slows the reaction down or reverse its direction.

67 L of N2 gas at STP is equivalent to 2.67 moles of N2 ga

To calculate the number of moles of N2 gas, we can use the ideal gas law:

PV = nRT

where:

P is the pressure of the gas

V is the volume of the gas

n is the number of moles of gas

R is the universal gas constant

T is the temperature of the gas in kelvin

Assuming standard temperature and pressure (STP), the pressure is 1 atmosphere and the temperature is 273 K. The volume of gas is 67 L.

So, we have:

PV = nRT

n = (PV) / (RT)

n = (1 atm) x (67 L) / ((0.0821 L·atm/mol·K) x (273 K))

n = 2.67 mol

Therefore, 67 L of N2 gas at STP is equivalent to 2.67 moles of N2 gas.

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In this case, the correct reaction showing how FeCO3 has increased solubility when forming the complex ion Fe(CN)64- is B) FeCO3 (s) + 6 CN- (aq) <-> Fe(CN)64- (aq) + CO32- (aq). So, the correct option is B.

This reaction shows that when FeCO3 is combined with six CN- ions, it forms the complex ion Fe(CN)64-, which is soluble in water. This increases the solubility of FeCO3. The reaction also produces CO32-, which is also soluble in water. There is no increased solubility in FeCO3 when forming the complex ion Fe(CN)64- in all other options. Therefore, the correct answer is option B.

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The relative pH at the equivalence point of the titration of a weak acid with a strong base is pH > 7.

The equivalence point in titration is the point at which the amount of added titrant is just enough to completely neutralize the analyte solution. The equivalence point can be found from an acid-base titration curve by the inflection point of the curve. In titration, the equivalence point occurs when the number of moles of titrant is equal to the number of moles of the analyte. For the titration of a weak acid with a strong base, the equivalence point will have a pH greater than 7. This is because the strong base will completely neutralize the weak acid and any excess base will increase the pH of the solution beyond neutrality.

In other words, the solution has become basic because of the excess hydroxide ions added from the titrant, despite the fact that the original substance being analyzed (acetic acid) was acidic. Therefore, the relative pH at the equivalence point of the titration of a weak acid with a strong base is greater than 7.

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The type of soap affects its cleansing ability and dishwasher soap because of the specialized formulation of dishwasher soap, which is specifically designed to remove grease and dirt from dishes.

Here is an experiment that explores the effect of the type of soap on the cleanliness of dishes:

Title: The Effect of Soap Type on Dish Cleaning Performance

Introduction: Soap is a commonly used cleaning agent for removing dirt and grease from various surfaces, including dishes. There are different types of soap available in the market, each with its unique composition and cleaning properties. In this experiment, we will investigate the effect of soap type (shampoo, hand soap, dishwasher) on dish cleaning performance.

Hypothesis: We hypothesize that the type of soap used will have a significant effect on dish cleaning performance, with dishwasher soap being the most effective due to its specialized formulation.

Materials:

Three different types of soap (shampoo, hand soap, and dishwasher soap)

Measuring cup

Three identical dirty dishes

Sink with running water

Stopwatch

Paper towel

Digital scale

Procedure:

Measure out 30 mL of each soap type into separate containers.

Weigh each of the three dirty dishes using a digital scale and record the weights.

Wet one dish in the sink and apply 10 mL of the first soap type to the dish.

Rub the dish with a paper towel for 30 seconds.

Rinse the dish with running water for 10 seconds.

Dry the dish with a paper towel and weigh it. Record the weight and note the cleanliness of the dish.

Repeat steps 3-6 with the remaining two types of soap on the other two dishes.

Repeat steps 3-7 for each soap type two more times with new dirty dishes, for a total of three trials per soap type.

Data Analysis:

Calculate the difference in weight between the dirty and cleaned dishes for each trial.

Calculate the average weight difference for each soap type.

Plot the average weight difference for each soap type on a bar graph.

Analyze the graph to determine the effect of soap type on dish cleaning performance.

Conclusion:

Based on the results of this experiment, we can conclude that the type of soap used has a significant effect on dish cleaning performance. The dishwasher soap was found to be the most effective in cleaning dishes, followed by hand soap and then shampoo.

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The mass of [tex]SO_3[/tex] produced by reacting 6.3g of [tex]SO_2[/tex] with oxygen in the synthesis reaction is 7.875g.

Given the mass of [tex]SO_2[/tex] reacted = 6.3g

[tex]2SO_2(g) + O_2(g) -- > 2SO_3(g)[/tex]

We can see that 2 moles of [tex]SO_2[/tex] produce 2 moles of [tex]SO_3[/tex].

The mole ratio of [tex]SO_2[/tex] : [tex]SO_3[/tex] = 1 : 1

The molar mass of Sulfur dioxide = 64g/mol.

The number of moles of Sulfur dioxide reacted = 6.3/64 = 0.098mol

Since the mole ratio is 1 the moles of [tex]SO_3[/tex] produced = 0.098

The molar mass of Sulfur trioxide = 80g/mol

The mass of [tex]SO_3[/tex] produced = 0.098 * 80 = 7.875g

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You need 163.8 mL of concentrated sulfuric acid (18M) to prepare 9.0 L of a 1.00M solution.

The given problem can be solved using the molarity equation which is:
Molarity (M) = moles of solute (n) / liters of solution (L)
The balanced chemical equation for sulfuric acid (H2SO4) is:H2SO4 → 2H+ + SO42
The molecular weight of H2SO4 is 98g/mol.
Hence, 18M H2SO4 solution contains 98g/Liter.98g / 18M = 5.44 g/mL5.44 g/mL x 1000 mL/L = 5440 g/L5440 g / 98 g/mol = 55.102M
Concentrated sulfuric acid is 18M.

So, the volume of concentrated sulfuric acid (18M H2SO4) that is needed to prepare 9.0 L of 1.00M solution can be determined as follows:
Number of moles of H2SO4 in 9.0 L of 1.00M H2SO4 solution = 1.0 mol/L × 9.0 L = 9.0 mol
Total number of moles of H2SO4 in the final solution (1.00M) can be calculated as:9.0 mol / 55.102 mol/L = 0.1638 L = 163.8 mL (rounded to 2 decimal places)Therefore,

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The volume of a gas will increase if the number of moles of gas is quadrupled at constant pressure and temperature: the volume of the gas will also increase four times its original volume.

This can be explained using the Ideal Gas Law, which states that the volume of a gas is proportional to the number of moles of gas when pressure and temperature remain constant. Therefore, if the number of moles of gas is increased by a factor of four, the volume of the gas will also increase by a factor of four.

To understand this concept better, let us consider the following example. Let us assume that there is a certain amount of gas, A, which contains one mole of gas at a constant pressure and temperature. This gas will occupy a certain volume, V1.

If the number of moles of gas is quadrupled to four moles, the volume of the gas will become four times the original volume, V2. Therefore, the volume of the gas, V2, is four times the original volume, V1.

This example demonstrates that if the number of moles of gas is increased at constant pressure and temperature, the volume of the gas will also increase proportionately. Therefore, if the number of moles of gas is quadrupled at constant pressure and temperature, the volume of the gas will also increase four times its original volume.

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According to Avogadro's Law, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas. Therefore, if the number of moles of gas is quadrupled while keeping the temperature and pressure constant, the volume of the gas will also quadruple.

Mathematically, we can express this relationship as:

V ∝ n

where V is the volume of the gas, n is the number of moles of the gas, and the symbol ∝ means "is proportional to".

If we quadruple the number of moles of gas, then we have:

n' = 4n

where n' is the new number of moles of gas, and n is the original number of moles of gas.

Using the relationship between volume and number of moles, we can write:

V' ∝ n'

Substituting n' = 4n, we get:

V' ∝ 4n

Simplifying, we get:

V' = 4V

Therefore, if the number of moles of gas is quadrupled at constant pressure and temperature, the volume of the gas will also quadruple.

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Answer:

Since iodine is a non-polar ion, it will melt in a non-polar solvent like hexane. While water is a polar solvent, lodine does not react in it.

The concentration of Ni2+ ions at equilibrium is approximately 4.4×10⁻⁹ M.

A more detailed explanation of the answer.

To find the concentration of Ni2+ ions at equilibrium, we need to use the formation constant (Kf) of Ni(NH3)6 2+ and perform an ICE table (Initial, Change, Equilibrium) calculation. The given Kf value is 5.5×10⁸.

Student question: What is the concentration of Ni2+ ions at equilibrium when 0.160 moles of NiCl2 are added to a liter of 1.20 M NH3 solution?

1. Balanced equation for the reaction:
Ni2+ + 6 NH3 <=> Ni(NH3)6 2+

2. Create the ICE table:

|     | Ni2+ | NH3   | Ni(NH3)6 2+ |
|-----|------|-------|------------|
| I   | 0.16 | 1.20  | 0          |
| C   | -x   | -6x   | +x         |
| E   | 0.16-x | 1.20-6x | x      |

3. Write the expression for Kf:
Kf = [Ni(NH3)6 2+]/([Ni2+][NH3]⁶) = 5.5×10⁸

4. Substitute the equilibrium values from the ICE table:
5.5×10^8 = (x)/((0.16-x)(1.20-6x)⁶)

5. Since Kf is very large, we can assume that x (change in concentration) is very small compared to the initial concentrations. Thus, we can approximate 0.16-x ≈ 0.16 and 1.20-6x ≈ 1.20.

6. Simplify and solve for x:
5.5×10⁸ = (x)/((0.16)(1.20)⁶)
x ≈ 4.4×10⁻⁹ M

The concentration of Ni2+ ions at equilibrium is approximately 4.4×10⁻⁹ M.

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First, we need to calculate the number of moles of aluminum that reacted:

Molar mass of aluminum = 27 g/mol

Number of moles of aluminum = 0.84 g / 27 g/mol = 0.031 mol

According to the balanced chemical equation, 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride. So, 0.031 moles of aluminum will react with:

0.031 mol Al x (3 mol Cl2 / 2 mol Al) = 0.0465 mol Cl2

Now, we can use the molar gas volume to calculate the volume of chlorine gas used:

Volume of Cl2 = (0.0465 mol Cl2) x (24 dm³/mol) = 1.116 dm³ or 1116 mL (rounded to 3 significant figures)

Therefore, the volume of chlorine gas used in the reaction is 1.116 dm³ or 1116 mL.

The solubility of strontium sulfate, SrSO₄ in 0.36 M sodium sulfate, Na₂SO₄ solution is approximately 1.06 x 10⁻⁶ M.

Solubility is the maximum amount of solute that can dissolve in a solvent at a given temperature and pressure to form a saturated solution.

The solubility of strontium sulfate (SrSO₄) in a 0.36 M sodium sulfate (Na₂SO₄) solution can be calculated using the solubility product constant (Ksp) and the common ion effect. The Ksp of SrSO₄ is 3.80 x 10⁻⁷. To find the solubility of SrSO₄, we can set up an expression using the Ksp value and the concentration of Na₂SO₄:

Ksp = [Sr²⁺][SO₄²⁻]

Since Na₂SO₄ dissociates into 2 Na⁺ ions and 1 SO₄²⁻ ion, the initial concentration of SO₄²⁻ ions from Na₂SO₄ is 0.36 M. Let x be the solubility of SrSO₄, then:

3.80 x 10⁻⁷ = [x][(0.36 + x)]

As x is significantly smaller than 0.36, we can assume x is negligible in the equation:

3.80 x 10⁻⁷ ≈ [x][0.36]

Solving for x:

x ≈ 1.06 x 10⁻⁶ M

Thus, the solubility of SrSO₄ in a 0.36 M Na₂SO₄ solution is approximately 1.06 x 10⁻⁶ M.

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The statement "the cage size of the zeolites is in the centimeters scale" is False.

Zeolite is a crystalline and porous alumino-silicate mineral consisting of hydrated alkaline metals and alkali earth metals. These minerals have microporous structures that make them useful in industrial and medical applications, among other things. They have a diverse array of applications, including as catalysts, adsorbents, and molecular sieves. Zeolites are minerals that have a unique framework that is capable of trapping and holding a variety of molecules within their microporous structure.

Zeolites are small in size, typically between 0.3 and 2 microns. The cavities or pores within these crystals, known as cages, are in the range of 4 to 12 Angstroms in size (1 Angstrom = 0.1 nm). These cavities are small, which allows the zeolite to selectively filter molecules based on their size, shape, and chemical properties.}

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The molar mass of an unknown gas that effuses through an opening at a rate of 3.16 times slower than neon gas is 199.6 g/mol.

The unknown gas effuses through an opening at a rate 3.16 times slower than neon gas. We need to estimate the molar mass of this unknown gas.

According to Graham's Law, the rate of effusion is inversely proportional to the square root of the molecular mass of the gas. The effusion rate is given for both neon and the unknown gas, thus we can say that the ratio of the effusion rates is equal to the ratio of the square roots of their molecular masses.

Let M₁ and M₂ be molar masses of the unknown gas and Neon gas respectively. Then the ratio of the effusion rates would be,

R₂/R₁ = √M₁/√M₂

Where R₁ and R₂ are the rates of effusion of the unknown gas and Neon gas respectively.

It is given that the effusion rate of the unknown gas is 3.16 times slower than that of Neon gas.

So, R₁ = 1/3.16 * R₂ or R₂ = 3.16 R₁

Putting these values in the above equation and squaring both sides, we get:

M₁/M₂ = (3.16R1/R1)²

M₁/M₂ = 9.98

M₁ = 9.98 × M₂

Now, the molar mass of Neon is 20g/mol (Neon is monoatomic and its atomic mass is 20).

We can substitute this value to find out the molar mass of the unknown gas.

M₁ = 9.98 × 20

M₁ = 199.6 g/mol

Hence, the estimated molar mass of the unknown gas is 199.6 g/mol.

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The calcium sample, which contains 2.71 x 1020 particles, thus, has a mass of 0.0181 grammes.

4.52 moles of caco3 are present. For c, a c o 3, the molar mass is 100 grammes per mole. As a result, the mass of c c 3 is equal to moles times molar mass, or 4.52 moles times 100 grammes per mole, which is 452 grammes.

We may use the techniques below to determine the mass of a sample of calcium that contains 2.71 x 1020 particles:

Calculate the number of moles of calcium:

Number of moles = Number of particles / Avogadro's number

= 2.71 x 10²⁰ / 6.022 x 10²³

= 0.000450 mol

Calculate the mass of calcium in grams:

Mass (g) = Number of moles x Atomic mass (g/mol)

= 0.000450 mol x 40.08 g/mol (atomic mass of calcium)

= 0.0181 g

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In the experiment, sodium bisulfite (NaHSO3) in water is used to destroy any unreacted bromine (Br2) or to trap the Br2 and not allow it to escape from the reaction setup. The reaction is shown below but cannot be described using conventional organic curved-arrow pushing.

The resulting mixture is placed in acid waste for the following reasons:

Sodium bisulfite's addition to the reaction mix in the procedure is done to destroy any unreacted bromine (Br2) or to trap the Br2 and prevent it from escaping the reaction setup. Following the reaction, it is necessary to neutralize the mixture with sodium carbonate or another base. After that, the neutralized mixture should be properly disposed of in an acid waste container. Thus, the resulting mixture is placed in acid waste.

Sodium bisulfite is used in excess to the amount of bromine to ensure that all of the bromine is captured or reacted. The resulting mixture is extremely acidic as a result of the reaction. As a result, the mixture must be neutralized before being disposed of.

The most straightforward approach to neutralizing it is to add a basic substance like sodium carbonate, which reacts with the acidic mixture to create water and sodium sulfate (Na2SO4).

As a result, when sodium bisulfite is added in the procedure, the resulting mixture is put into acid waste.

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The long will the reaction will take to reduce the concentration of to the 50% of its original value is 1386 sec. The option A is correct.

The expression is as :

d[N₂O₅] / dt = -0.0005 [N₂O₅]

[N₂O₅] (t) = [N₂O₅] (0) . e^-0.0005t

[N₂O₅](t) = Ce^-0.0005t

The reaction concentration of to the 50% of its original value is 0.5 C

0.5 C = Ce^-0.0005t

By dividing both side by C

0.5 = e^-0.0005t

Now taking the natural logarithm on the both side, we get

ln0.5 = lne^-0.0005t

ln 0.5 = - 0.0005t

t = - 0.693 / - 0.0005

t = 1386 sec

The time taken is 1386 sec. The correct option is A.

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Thus question is incomplete, the complete question is :

Experiments show that if the chemical reaction N2O5→2NO2+1/2O2, takes place at 45°C, the rate of reaction of dinitrogen pentoxide is proportional to its concentration as follows: -d[N2O5]/dt = 0.0005[N2O5] How long will the reaction take to reduce the concentration of N2O5 to 50% of its original value? select the correct answer. question 4 options:

A) t = 1386 sec

B) t = 211 sec

C) t = 2345 sec

D) t = 111 sec

The order of increasing reactivity of the following alkyl halides in an E2 reaction is (CH3)2C(Br)CH2CH2CH3 < Intermediate reactivity, (CH3)2CHCH2CH(Br)CH3 < Highest reactivity, (CH3)2CHCH2CH2CH2Br < Lowest reactivity.

In an E2 reaction, the rate of reaction is affected by the size and the polarizability of the leaving group, the bulkiness of the alkyl groups, and the steric hindrance. In this case, the size and polarizability of the leaving group increases from (CH3)2C(Br)CH2CH2CH3 < (CH3)2CHCH2CH(Br)CH3 < (CH3)2CHCH2CH2CH2Br, making the reactivity increase in the same order.

The bulkiness of the alkyl groups has the opposite effect; the bulkier the alkyl groups, the lower the reactivity of the alkyl halide. The alkyl groups in the compounds are in the order (CH3)2C(Br)CH2CH2CH3 < (CH3)2CHCH2CH2CH2Br < (CH3)2CHCH2CH(Br)CH3, making the reactivity increase in the reverse order.

Lastly, steric hindrance affects the rate of reaction as well. The steric hindrance decreases from (CH3)2C(Br)CH2CH2CH3 < (CH3)2CHCH2CH(Br)CH3 < (CH3)2CHCH2CH2CH2Br, leading to the highest reactivity of (CH3)2CHCH2CH(Br)CH3.

Overall, this leads to the order of reactivity (CH3)2C(Br)CH2CH2CH3 < Intermediate reactivity, (CH3)2CHCH2CH(Br)CH3 < Highest reactivity, (CH3)2CHCH2CH2CH2Br < Lowest reactivity.

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Complete Question:

rank the following alkyl halides in order of increasing reactivity in an E2 reaction. Be sure to answer all parts

1. (CH3)2C(Br)CH2CH2CH3

2. (CH3)2CHCH2CH(Br)CH3

3. (CH3)2CHCH2CH2CH2Br

lowest reactivity: ?

Intermediate reactivity: ?

Highest reactivity: ?

The term "civic engagement" describes people actively participating in their communities and society as a whole. The lack of water scarcity in many areas of the world is a serious problem brought on by causes including population increase, climate change, and pollution.

In response to changing supply and demand throughout time, water availability varies. As demand rises and/or the amount or quality of the water supply declines, there is an increase in water scarcity.

When there is a water shortage, the primary issue is that people cannot get fresh, clean drinking water. The human body can only exist for a very short time without water, and not drinking other issues with water are covered in the section below.

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Question:

What is civic engagement in 150 words?

What are your tips on water scarcity?

To calculate the moles of anhydrous salt, divide the molar mass by the molar mass of the anhydrous salt.

The molar mass of an anhydrous salt is the sum of the molar mass of each of its components.

To determine the smallest whole number ratio of moles of water to moles of anhydrous salt, divide the number of moles of water by the number of moles of anhydrous salt.

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The solution now has a concentration of 0.113 M.

When a solution is diluted, the amount of solute remains the same, but the volume of the solution increases. Therefore, the concentration of the solution decreases.

In this case, 100.0 mL of a 0.565 M solution of KBr is diluted to a total volume of 500.0 mL. The amount of KBr in the original solution can be calculated as follows:

amount of KBr = concentration x volume = 0.565 mol/L x 0.1000 L = 0.0565 moles

When this solution is diluted to a volume of 500.0 mL, the amount of KBr remains the same:

amount of KBr = 0.0565 moles

The new concentration of the solution can be calculated using the following equation:

new concentration = amount of solute / new volume

new volume = 500.0 mL = 0.5000 L

new concentration = 0.0565 moles / 0.5000 L = 0.113 M

Therefore, the new concentration of the solution is 0.113 M.

To learn more about concentration refer to:

brainly.com/question/10725862

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