Consider two acids: CH3CO2H (acetic acid, pKa = 4.8) and C6H5CO2H (benzoic acid, pKa = 4.2). Which acid is the stronger acid? Select the single best answer.
a. benzoic acid
b. acetic acid

Answer: The concentration of chloride ions in the solution obtained is 1.27 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]     .....(1)

Molarity of calcium chloride solution = 0.439 M

Volume of solution = 194.4 mL

Putting values in equation 1, we get:

[tex]0.439=\frac{\text{Moles of calcium chloride}\times 1000}{194.4}\\\\\text{Moles of calcium chloride}=\frac{(0.439\times 194.4)}{1000}=0.085mol[/tex]

1 mole of calcium chloride produces 2 moles of chloride ions and 1 mole of calcium ion

Moles of chloride ions in calcium chloride = [tex](2\times 0.085)=0.170mol[/tex]

Molarity of aluminum chloride solution = 0.497 M

Volume of solution = 363 mL

Putting values in equation 1, we get:

[tex]0.497=\frac{\text{Moles of }AlCl_3\times 1000}{363}\\\\\text{Moles of }AlCl_3=\frac{(0.497\times 363)}{1000}=0.180mol[/tex]

1 mole of aluminum chloride produces 3 moles of chloride ions and 1 mole of aluminum ion

Moles of chloride ions in aluminum chloride = [tex](3\times 0.180)=0.540mol[/tex]

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.170 + 0.540) moles = 0.710 moles

Total volume of the solution = (194.4 + 363) mL = 557.4 mL

Putting values in equation 1, we get:

[tex]\text{Concentration of chloride ions}=\frac{0.710mol\times 1000}{557.4}\\\\\text{Concentration of chloride ions}=1.27M[/tex]

Hence, the concentration of chloride ions in the solution obtained is 1.27 M

In an acid-base reaction, the Brønsted–Lowry acid donates a proton and becomes a conjugate base while the Brønsted–Lowry base accepts a proton and becomes a conjugate acid. The reactions given as examples illustrate these changes with H2CO3 and H2O acting as acids in the two reactions while H2O and NH3 act as bases respectively.

In the reaction, Brønsted–Lowry acid is the species that donates a proton (H+), and the Brønsted–Lowry base is the species that accepts a proton. After the acid has lost a proton, it becomes the conjugate base. Conversely, after the base gains a proton, it becomes the conjugate acid.

For the reaction H2CO3(aq) + H2O(l) ∆ H3O+(aq) + HCO3 -(aq), H2CO3 is the Brønsted-Lowry acid and H2O is the Brønsted-Lowry base. Upon losing a proton, H2CO3 becomes HCO3- which is the conjugate base and upon gaining a proton, H2O becomes H3O+ which is the conjugate acid.

For the reaction NH3(aq) + H2O(l) ∆ NH4 +(aq) + OH-(aq), NH3 is the Brønsted-Lowry base and H2O is the Brønsted-Lowry acid. Upon gaining a proton, NH3 becomes NH4+ which is the conjugate acid and upon losing a proton, H2O becomes OH- which is the conjugate base.

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In a Brønsted-Lowry reaction, the acid donates a proton and the base accepts the proton. The species formed when an acid donates a proton becomes a conjugate base, and the species formed when a base accepts a proton becomes a conjugate acid. In the given reactions, H2CO3 and H2O function as acids, and H2O and NH3 function as bases.

In a Brønsted-Lowry acid-base reaction, an acid is a substance that donates a proton (H+) and a base is a substance that accepts a proton. A conjugate acid is the species formed when a Brønsted-Lowry base gains a proton, and a conjugate base is the species remaining after a Brønsted-Lowry acid has lost a proton.

For the reaction H2CO3(aq) + H2O(l) ∆ H3O+(aq) + HCO3 -(aq), the Brønsted-Lowry acid is H2CO3, which donates a proton to become the conjugate base, HCO3-. The Brønsted-Lowry base is H2O, which accepts a proton to become the conjugate acid, H3O+.

For the reaction NH3(aq) + H2O(l) ∆ NH4 +(aq) + OH-(aq), the Brønsted-Lowry base is NH3, which accepts a proton to become the conjugate acid, NH4+. The Brønsted-Lowry acid is H2O, which donates a proton to become the conjugate base, OH-.

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The element with the electron configuration 1s²2s²2p⁴ is **oxygen (O)**.

Here's its orbital diagram:

     ↑      ↑

1s:    |      |

     ↓      ↓

     ↑      ↑

2s:    |      |

     ↓      ↓

     ↑      ↓      ↑      ↓

2p:  -----|-----|-----|-----

     ↓      ↑      ↓      ↑

**Quantum numbers of the sixth electron:**

* Principal quantum number (n): 2 (second energy level)

* Azimuthal quantum number (l): 1 (p orbital)

* Magnetic quantum number (m_l): 0 (p_z orbital)

* Spin quantum number (m_s): +½ or -½ (two possible spin states)

**Explanation:**

- The electron configuration follows the Aufbau principle, filling orbitals from lowest energy to highest energy.

- The first two electrons fill the 1s orbital, the next two fill the 2s orbital, and the sixth electron goes into one of the three 2p orbitals (2p_x, 2p_y, or 2p_z).

- The magnetic quantum number (m_l) specifies the specific 2p orbital, and in this case, it's 0 for the 2p_z orbital.

- The spin quantum number (m_s) represents the electron's spin, which can be either +½ or -½.

Answer:

copper(I) bromide: CuBr

copper(I) oxide: Cu₂O

copper(II) bromide: CuBr₂

copper(II) oxide: CuO

iron(III) bromide: FeBr₃

iron(III) oxide: Fe₂O₃

lead(IV) bromide: PbBr₄

lead(IV) oxide: PbO₂

I hope this helped you! Brainliest would be greatly appreciated.

The formulas of the ionic compounds are as follows;

copper(I) bromide ----- CuBr

copper(I) oxide ------ Cu2O

copper(II) bromide ----- CuBr2

copper(II) oxide   ------- CuO

iron(III) bromide ------ FeBr3

iron(III) oxide ------ Fe2O3

lead(IV) bromide ---- PbBr4

lead(IV) oxide  ------ PbO2

The formula of an ionic compound is written with knowledge of the oxidation state of the metal in the ionic compound.

The oxidation state of each atom determines the subscript it carries in the formula as shown in the chemical formulas written above.

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Calcium salts in the bone provide it with compressional strength, which is the ability to withstand loads that might reduce size. This is crucial for bones to perform tasks such as standing, sitting, or lifting heavy objects.

Calcium salts play a vital role in the structure and mechanical properties of bones. Particularly, they give bone its d. compressional strength. Compressional strength refers to the capacity of a material or structure to withstand loads tending to reduce size. This property is crucial to bones as they need to resist compression from the body's weight and movements. For instance, when we stand up, sit down or lift heavy objects, our leg bones or vertebrae need to bear the compressional force.

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Calcium salts provide bones with their compressional strength by crystallizing on a matrix of collagen fibers, providing hardness and rigidity. The collagen fibers add flexibility to prevent the bones from becoming too brittle.

Calcium salts in the bone provide its d. compressional strength. This occurs when calcium phosphate and calcium carbonate combine to create hydroxyapatite, a mineral that provides bones with hardness and strength. These salts crystallize on a matrix of collagen fibers within the bone. The role of the collagen fibers is to provide bones with flexibility, preventing them from becoming too brittle.

However, it's important to note that both calcium salts and collagen fibers play crucial roles in the overall health and functionality of bones. The calcium salts allow the bones to withstand a great deal of pressure without deforming (compressional strength), while the collagen fibers give bones their resilience and flexibility. The combination of these two components allows bones to handle a variety of physical stresses.

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High ionization energies are common in elements found on the right side of the periodic table, notably the Noble Gases due to their full electron shells. In contrast, low ionization energies are associated with elements on the left side of the periodic table, particularly metals like those of the Alkali Metal group with just one or two electrons in their outermost shell.

In the periodic table, elements with relatively high IEs (Ionization Energies) are generally located on the right side, specifically in the group of Noble Gases. Noble gases like Helium or Neon have high IEs because they have full electron shells, and thus it requires a large amount of energy to remove an electron. On the other hand, the elements with relatively low IEs are located on the left side of the periodic table. This is because metals, such as those in the Alkali Metal group, have only one or two electrons in their outermost shell. These electrons are relatively easy to remove, resulting in a low ionization energy.

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Answer: 288.32g/mol

Explanation:Please see attachment for explanation

Answer:

The pH of the solution is 4.28

Explanation:

The dissolution reaction as below

CH₃COOH ⇔  CH₃COO⁻ + H⁺

[tex]K_{a} = \frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH} = 10^{-4.76}[/tex]

Assume the concentration of the ion, [H⁺] = a,

          so [CH₃COO⁻] = a and [CH₃COOH] = 3a

Then use the formula of Ka, we get

Ka = a * a / 3a = 10^-4.76 ⇔ a = 3 x 10^-4.76 = 5.21 x 10^-5

Hence pH = -log(a) = - log(5.21 x 10^-5) = 4.28

Answer:

Part A:

For electron:

[tex]\lambda_e=2.1392*10^{-10} m[/tex]

For Proton:

[tex]\lambda_p=1.16696*10^{-13} m[/tex]

Part B:

For electron:

[tex]\lambda_e=9.44703*10^{-12} m[/tex]

For Proton:

[tex]\lambda_p=2.20646*10^{-13} m[/tex]

Explanation:

Formula for wave length λ is:

[tex]\lambda=\frac{h}{mv}[/tex]

where:

h is Planck's constant=[tex]6.626*10^{-34}[/tex]

m is the mass

v is the velocity

Part A:

For electron:

[tex]\lambda_e=\frac{6,626*10^{-34}}{(9.11*10^{-31})*(3.4*10^6)} \\\lambda_e=2.1392*10^{-10} m[/tex]

For Proton:

[tex]\lambda_p=\frac{6,626*10^{-34}}{(1.67*10^{-27})*(3.4*10^6)} \\\lambda_p=1.16696*10^{-13} m[/tex]

Wavelength of proton is smaller than that of electron

Part B:

Formula for K.E:

[tex]K.E=\frac{1}{2}mv^2\\v=\sqrt{2 K.E/m}[/tex]

For Electron:

[tex]v_e=\sqrt{\frac{2*2.7*10^{-15}}{9.11*10^{-31}}} \\v_e=76990597.74\ m/s[/tex]

Wavelength for electron:

[tex]\lambda_e=\frac{6,626*10^{-34}}{(9.11*10^{-31})*(76990597.74)} \\\lambda_e=9.44703*10^{-12} m[/tex]

For Proton:

[tex]v_p=\sqrt{\frac{2*2.7*10^{-15}}{1.67*10^{-27}}} \\v_p=1798202.696\ m/s[/tex]

Wavelength for proton:

[tex]\lambda_p=\frac{6,626*10^{-34}}{(1.67*10^{-27})*(1798202.696)} \\\lambda_p=2.20646*10^{-13} m[/tex]

Wavelength of electron is greater than that of proton.

Answer:

P(total) = 1329 torr

Explanation:

Given data:

Pressure of nitrogen = 984 torr

Pressure of carbon dioxide = 345 torr

Total pressure of system = ?

Solution:

The given problem will be solve through the Dalton law of partial pressure of gases.

According to the this law,

The total pressure exerted by the mixture of gases is equal to the sum of partial pressure of individual gas.

Mathematical expression,

P(total) = P₁ + P₂ +.......+ Pₙ

Here we will put the values in formula,

P₁ = partial pressure of nitrogen

P₂ = partial pressure of carbon dioxide

P(total) = 984 torr + 345 torr

P(total) = 1329 torr

Final answer:

The total pressure of the system is calculated by adding the partial pressures of the individual gases, resulting in a total pressure of 843.42 torr.

Explanation:

To calculate the total pressure of a mixture of gases, you use Dalton's Law of Partial Pressures which implies that the overall pressure exerted by a mixture of non-reacting gases is equal to the sum of their partial pressures.

In this case,

The partial pressure of nitrogen (N₂) is given as 78% of 984 torr, which is 767.52 torr.

The partial pressure of carbon dioxide (CO₂) is given as 22% of 345 torr, which is 75.9 torr.

The total pressure is the sum of these partial pressures:

767.52 torr (N₂) + 75.9 torr (CO₂) = 843.42 torr

The total pressure of the system is 843.42 torr.

Answer:

The drawing of the mechanism and the organic product is shown below.

Explanation:

Alkynes react with bromine to form a dihaloalkene (with an equivalent of the halogen) or a tetrahaloalkane derivative (with two equivalents of the halogen), which is this case. The triple bond adds two halogen molecules as shown in the drawing, an anti addition occurring.

The quantum numbers serve for experimenting with electron positions within an atom. For each given atom or ion - Li, Br-, Cs, and B - the quantum numbers for the outermost electron exhibit the electron's location in different shells, subshells and spins.

The quantum numbers provide a unique address for each electron in an atom, listed as (n, l, ml, ms), where n is the principal quantum number, l is the azimuthal quantum number, ml is the magnetic quantum number, and ms is the spin quantum number.

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For these cases, the quantum numbers have been determined based on electron configurations and orbital positions. The solutions cover Li, Br⁻, Cs, and B electrons specifically.

Quantum numbers are used to describe the position and energy of an electron in an atom. Let's break down the quantum numbers for each of the given scenarios:

Answer:

5.48 g

Explanation:

The concentration in M represents the mol/L, or the number of moles in each L of the solution, and it can be calculated by the number of moles (n) divided by the volume of the solution (V) in L, so:

M = n/V

1.71 = n/0.100

n = 0.171 mol

The molar mass of methanol is 32.04 g/mol, and it is the mass (m) divided by the number of moles:

FM = m/n

32.04 = m/0.171

m = 5.48 g

Answer:

The set of planes responsible for this diffraction peak is the [111] set.

Explanation:

We need to calculate the interplanar spacing, dₕₖₗ for nickel.

dₕₖₗ = nλ/2 sin θ

where θ = half of the diffraction angle = 44.53°/2 = 22.265°, n is the order of reflection = 1 and λ is the wavelength of the monochromatic radiation = 0.1542 nm.

dₕₖₗ = 1×0.1542/2sin 22.265°

dₕₖₗ = 0.2035 nm

But, interplanar spacing, dₕₖₗ is related to the plane (hkl), by the relation

√(h² + k² + l²) = a/dₕₖₗ

a is the lattice parameter.

Since Nickel has an FCC structure, a = 2R√2, R is given as 0.1242 nm

√(h² + k² + l²) = 2R√2/dₕₖₗ = (2×0.1246√2)/0.2035 = 1.732

(h² + k² + l²) = 1.732² = 3

The only 3 integers the values of h, k and l which fit properly into the equation is [111]

Therefore, the answer is [111].

Hope this Helps!!

                                 √(h² + k² + l²) = a/dₕₖₗ  

              =   √(h² + k² + l²) = 2R√2/dₕₖₗ = (2×0.1246√2)/0.2035 = 1.732  

                                   =    (h² + k² + l²) = 1.732² = 3

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Answer:2

Explanation:

Ba(OH)2 contains two oxygen atoms

BaSO4 contains four oxygen atoms.

This means that barium sulphate contains two more oxygen atoms than barium hydroxide in its formula. This is clearly seen from the two formulae shown above.

The number of oxygen atom represented in the formula for barium sulfate than in the formula for barium hydroxide is 2 atoms

The molecular formula of barium sulfate = BaSO₄

The molecular formula of barium hydroxide = Ba(OH)₂

The number of oxygen atoms in BaSO₄ = 4 atoms

The number of oxygen atoms in Ba(OH)₂ = 2 atoms

Differences in atoms of oxygen = 4 – 2 = 2 atoms

From the above illustration, we can see that there are 2 more atoms of oxygen in barium sulfate, BaSO₄ than in barium hydroxide, Ba(OH)₂

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Answer:

The pH of the resultant solution is 2.4

Explanation:

From the data

          V_total=V_w+V_a

          V_total=400+200 ml

Now the concentration of solution is given as

[tex]M_1V_1=M_2V_2\\M_2=\frac{M_1V_1}{V_2}\\M_2=\frac{0.30 \times 0.2}{0.6}\\M_2=0.1 M\\[/tex]

The reaction equation is given as

Equation                            [tex]HCOOH + H_2O \rightarrow HCOO^- +H_3O^+\\[/tex]

Initial Concentration is        0.1                             0               0

Change is                            -x                                x               x

_______________________________________________

At Equilibrium                  0.1-x                              x                x

Now the equation for K_a is given as

[tex]K_a=\frac{[HCOO^-][H_3O^+]}{[HCOOH]}[/tex]

Here Ka, for pKa-=3.75,  is given as

                   [tex]K_a=10^{-pKa}\\K_a=10^{-3.75}\\K_a=1.78 \times 10^{-4}[/tex]

Substituting this and concentrations at equilibrium in equation of Ka gives

[tex]K_a=\frac{[HCOO^-][H_3O^+]}{[HCOOH]}\\1.78 \times 10^{-4}=\frac{xx}{0.1-x}\\[/tex]

Solving for x

[tex]1.78 \times 10^{-4} (0.1-x)=x^2\\[/tex]

Here 0.1-x≈0.1 so

[tex]1.78 \times 10^{-5} =x^2\\x=\sqrt{1.78 \times 10^{-5}}\\x=4.219 \times 10^{-3}[/tex]

As [tex][H_3O^+]=[H^+][/tex]=x so its value is 4.219 x10^(-3) so pH is given as

[tex]pH=-log[H^+]\\pH=-log (4.219 \times 10^{-3})\\pH=2.37 \approx 2.4[/tex]

So the pH of the resultant solution is 2.4

Answer:

2.4

Explanation:

Given that:

Initial Concentration (M₁) of our formic acid from the question= 0.30 M

Initial volume (V₁) of the formic acid = 200 mL

Volume of water added = 400 mL

pKa= 3.75

∴ we can determine the total volume of the final solution by the addition of (Initial volume (V₁) of the formic acid) + (Volume of water added)

= 200 mL + 400 mL

= 600 mL

We can find the final concentration using the formula;

M₁V₁ = M₂V₂

M₂= [tex]\frac{0.30*200}{600}[/tex]

M₂= 0.1M

∴ The concentration of the diluted formic acid(since water is added to the initial volume) = 0.1M

From our pKa which is = 3.75

Ka of formic acid (HCOOH) = [tex]10^{-3.75}[/tex]

                                             =  1.78 × 10⁻⁴

If water is added to the formic acid; the equation for the reaction can be represented as;

                                  HCOOH + H₂O    ⇒    HCOO⁻    + H₃O

Initial                            0.1                                   0              0

Change                        -x                                    +x             +x

Equilibrium                  0.1-x                                 x              x

Ka = [tex]\frac{x^2}{0.1-x}[/tex] = 1.78 × 10⁻⁴

x² = (0.1 × 1.78 × 10⁻⁴)      since x = 0

x² =  1.78 × 10⁻⁵

x = [tex]\sqrt{1.78*10^{-5}}[/tex]

x = 0.004217

x = [tex][H_3O^+]=[H^+][/tex] =  0.004217 M

pH = [tex]-log[H^+][/tex]

= -log (0.004217)

= 2.375

≅ 2.4 (to  one decimal place)

We can thereby conclude that the final pH of the formic acid solution = 2.4

Answer:

what is the amount of energy ( Δ G ) required = 36.65KJ

Explanation:

The concept of VANT HOFF ISOTHERM EQUATION was used, as well as the relationship between the gibb's free energy and the electromotive force, all the steps and appropriate calculations is as shown in the attached file in order to get the amount of energy required for the transport of the hydrogen ions.

The energy required for the transport of hydrogen ions from a cell to the stomach lumen can be calculated using the Nernst equation and the equation for Gibbs free energy.

The amount of energy required for the transport of hydrogen ions from a cell into the stomach lumen can be calculated using the Nernst equation and the equation for Gibbs free energy. The Nernst equation relates the equilibrium potential for an ion to the concentrations of the ion on either side of the membrane, while the equation for Gibbs free energy relates the change in free energy to the change in potential across the membrane and the Faraday constant. By substituting the given values into these equations, the amount of energy required can be calculated.

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Explanation:

(a)    The given data is as follows.

     Temperature (T) = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K

      [tex]\Delta H_{vap}[/tex] = 43000 J/mol

Since, both the liquid and vapors are at equilibrium. Therefore, change in free energy will be calculated as follows.

      [tex]\Delta G_{vap} = \Delta H_{vap} - T \Delta S_{vap}[/tex] = 0        

            43000 - [tex](298 \times \Delta S_{vap})[/tex] = 0

                      [tex]\Delta S_{vap}[/tex] = -144 J/mol K

Negative sign indicates an increase in entropy of the system.

Now, for 1 mole of [tex]CCl_{4}[/tex] is as follows.

     = 144 J/K

So,     [tex]S_{vapor}[/tex] - 214 = 144 J/k

                         = 358 J/K

Therefore, we can conclude that entropy of [tex]CCl_{4}[/tex] vapor is 358 J/K.

(b)  As we know that intensive variable are the variables which do not depend on the amount of a substance.

So, in the given situation only temperature will act as an intensive variable that will be required to completely specify the vapor-liquid mixture of [tex]CCl_{4}[/tex].

(a) The entropy of 1 mole of the vapor at 25°C is 358.3 J/K. (b) One intensive variables are required to completely specify the vapor-liquid mixture of CCl4.

Let's address each part of your question step by step.

(a) Entropy of 1 mole of CCl₄ vapor at 25°C (298 K)

Given data:

- The heat of vaporization [tex](\(\Delta H_{\text{vap}}\)) of CCl_4 \ is\ 43000 J/mol\ at\ 25\°C.[/tex]

- The entropy of 1 mol of liquid CCl₄ [tex](\(S_{\text{liquid}}\))[/tex] is 214 J/K at 25°C.

We need to find the entropy of 1 mole of the vapor [tex](\(S_{\text{vapor}}\))[/tex] at 25°C (298 K).

The change in entropy during vaporization [tex](\(\Delta S_{\text{vap}}\))[/tex] can be calculated using the formula:

[tex]\[ \Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T} \][/tex]

Where:

- [tex]\(\Delta H_{\text{vap}}\)[/tex] = 43000 J/mol

- T = 298 K

So,

[tex]\[ \Delta S_{\text{vap}} = \frac{43000 \, \text{J/mol}}{298 \, \text{K}} \]\[ \Delta S_{\text{vap}} = 144.3 \, \text{J/(mol} \cdot \text{K)} \][/tex]

Now, using the relation:

[tex]\[ S_{\text{vapor}} = S_{\text{liquid}} + \Delta S_{\text{vap}} \][/tex]

Substitute the values:

[tex]\[ S_{\text{vapor}} = 214 \, \text{J/(mol} \cdot \text{K)} + 144.3 \, \text{J/(mol} \cdot \text{K)} \]\[ S_{\text{vapor}} = 358.3 \, \text{J/(mol} \cdot \text{K)} \][/tex]

So, the entropy of 1 mole of CCl₄ vapor at 25°C is:

[tex]\[ S_{\text{vapor}} = 358.3 \, \text{J/(mol} \cdot \text{K)} \][/tex]

(b) Number of intensive variables to specify the vapor-liquid mixture of CCl₄

For a system in equilibrium, the number of intensive variables required to completely specify the state of the system is given by the Gibbs phase rule:

F = C - P + 2

Where:

- F  is the number of degrees of freedom (intensive variables needed).

- C is the number of components.

- P is the number of phases.

In this case:

- C  (number of components) = 1 (CCl₄).

- P (number of phases) = 2 (liquid and vapor).

Using the Gibbs phase rule:

F = 1 - 2 + 2

F = 1

Thus, one intensive variable (e.g., temperature or pressure) is required to completely specify the vapor-liquid mixture of CCl₄.

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The activation energy is 124.776[tex]\frac{kJ}{mole}[/tex]

b) The frequency factor is 1.77 ×[tex]10^{18}[/tex]

c) The rate constant is 0.00033 [tex](\frac{dm^{3} }{mole} )^{2}[/tex][tex]\frac{1}{s}[/tex]

Explanation:

From the question the elementary reaction for A and B is given as

      2A + B → 4C

The rate equation the elementary reaction is

      -[tex]r_{A}[/tex] = [tex]k[/tex][tex][A]^{2}[/tex][tex][B][/tex]

            =  [tex]k[2]^{2}[1.5][/tex]

            = 6k

     [tex]k = \frac{-r_{A} }{6}[/tex]

      When temperature changes, the rate constant change an this causes the rate of reaction to change as shown on the second uploaded image.

The relationship between temperature and rate constant can be deduced from these equation

                    [tex]k = Aexp(-\frac{E_{a} }{RT} )[/tex]

             taking ln of both sides we have

                   [tex]lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}[/tex]

        Considering the graph for the rate constant [tex]ln k[/tex] and [tex](\frac{1}{T} )[/tex] the slope from the equation is [tex]-(\frac{E_{a} }{R})[/tex] and the intercept is [tex]ln A[/tex]

From the given table we can generate another table using the equation above as shown on the third uploaded image

The graph of [tex]ln k[/tex]  vs [tex](\frac{1}{T} )[/tex]  is shown on the fourth uploaded image

  From the graph we can see that the slope is [tex]-(\frac{E_{a} }{R} ) = - 15008[/tex]

Now we can obtain the activation energy [tex]E_{a}[/tex] by making it the subject in the equation also generally R which is the gas constant is [tex]8.145 \frac{J}{kmole}[/tex]

                [tex]E_{a} = 15008 × 8,3145\frac{J}{molK}[/tex]  

                     [tex]= 124\frac{KJ}{mole}[/tex]

    Hence the activation energy is [tex]= 124\frac{KJ}{mole}[/tex]

b) From the graph its intercept is [tex]ln A = 42.019[/tex]

                                                          [tex]A = exp(42.019)[/tex]

                                                             [tex]=1.77 × 10^{18}[/tex]

Hence the frquency factor A is  [tex]=1.77 × 10^{18}[/tex]

c) From the equation of rate constant

                                          [tex]lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}[/tex]

We have

                [tex]ln k = 42.019 - 15008 * (\frac{1}{300} )[/tex]

                      [tex]k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}[/tex]

Hence the rate constant is [tex]k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}[/tex]    

Answer:

1.98x10⁻¹² kg

Explanation:

The energy of a photon is given by:

h is Planck's constant, 6.626x10⁻³⁴ J·s

c is the speed of light, 3x10⁸ m/s

and λ is the wavelenght, 671 nm (or 6.71x10⁻⁷m)

Now we multiply that value by Avogadro's number, to calculate the energy of 1 mol of such protons:

Finally we calculate the mass equivalence using the equation:

The mass equivalence of 1 mole of photons of this wavelength is equal to [tex]1.78 \times 10^{-11} \;kg[/tex]

Given the following data:

We know that the speed of light is [tex]3 \times 10^8[/tex] m/s.

Planck constant = [tex]6.626 \times 10^{-34}\;J.s[/tex]

Avogadro constant = [tex]6.02 \times 10^{23}[/tex]

To determine the mass equivalence of 1 mole of photons of this wavelength:

First of all, we would calculate the energy associated with the photons of this wavelength by using the Planck-Einstein relation:

Mathematically, Planck-Einstein relation for photon energy is given by the formula:

[tex]E = hf = h\frac{v}{\lambda}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]E = \frac{6.626 \times 10^{-34} \; \times \;3 \times 10^8 }{6.71 \times 10^{-7}} \\\\E = \frac{1.99 \times 10^{-25}}{6.71 \times 10^{-7}} \\\\E=2.96 \times 10^{-19} \;Joules[/tex]

For the energy of 1 mole of photons:

[tex]E = 2.96 \times 10^{-19} \times 6.02 \times 10^{23}\\\\E = 1.78 \times 10^5 \; Joules[/tex]

Now, we can find the mass equivalence of 1 mole of photons by using the formula:

[tex]E = mc^2\\\\m = \frac{E}{c^2} \\\\m = \frac{1.78 \times 10^5}{(3.0 \times 10^8)^2} \\\\m = \frac{1.78 \times 10^5}{(9.0 \times 10^{16})}\\\\m = 1.78 \times 10^{-11} \;kg[/tex]

Read more: https://brainly.com/question/16901506

Answer:

entropy change of the system = +71 J/K

Explanation:

for a reversible process ;

ΔSsurrounding + ΔSsystem = 0 ............(1)

but ΔSsurrounding = - 71 J/K

substitute in equation (1)

ΔSsystem = -ΔSsurrounding

ΔSsystem = -(- 71 J/K)

= +71 J/K

For a reversible process where the surroundings undergo an entropy change of -71 J/K, the entropy change of the system is 71 J/K.

Since the process is reversible, the total entropy change for the universe is zero. This implies that the sum of the entropy changes for the system and the surroundings must be zero.

ΔSuniverse = ΔSsys + ΔSsurr = 0

Given that the entropy change of the surroundings, ΔSsurr, is -71 J/K, we can solve for the entropy change of the system, ΔSsys:

ΔSsys = - ΔSsurr

ΔSsys = - (-71 J/K)

ΔSsys = 71 J/K

Therefore, the entropy change of the system for this process is 71 J/K.

Answer: The molar mass of the unknown protein is 6387.9 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

[tex]\pi=iMRT[/tex]

or,

[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure of the solution = 0.0766 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of protein = 100. mg = 0.100 g   (Conversion factor:  1 g = 1000 mg)

Molar mass of protein = ?

Volume of solution = 5.00 mL

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the solution = [tex]25^oC=[25+273]K=298K[/tex]

Putting values in above equation, we get:

[tex]0.0766atm=1\times \frac{0.100\times 1000}{\text{Molar mass of protein}\times 5}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\\pi=\frac{1\times 0.100\times 1000\times 0.0821\times 298}{0.0766\times 5}=6387.9g/mol[/tex]

Hence, the molar mass of the unknown protein is 6387.9 g/mol

Answer: The volume of sodium bicarbonate needed is 196.6 mL

Explanation:

To calculate the volume of base, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]NaHCO_3[/tex]

We are given:

[tex]n_1=2\\M_1=1.9M\\V_1=98.3mL\\n_2=1\\M_2=1.9M\\V_2=?mL[/tex]

Putting values in above equation, we get:

[tex]2\times 1.9\times 98.3=1\times 0.19\times V_2\\\\V_2=\frac{2\times 1.9\times 98.3}{1\times 1.9}=196.6mL[/tex]

Hence, the volume of sodium bicarbonate needed is 196.6 mL

Answer: 6.26atm

Explanation:Please see attachment for explanation

Answer:

The osmotic pressure is 6.26 atm

Explanation:

Step 1: Data given

Mass of testosterone = 12.4 grams

Volume of benzene = 168 mL

Temperature = 298 Kelvin

Step 2: Calculate moles of testosterone

Moles testosterone = mass / molar mass

Moles testosterone = 12.4 grams / 288.42 g/mol

Moles testosterone = 0.0430 moles

Step 3: Calculate molarity

Molarity = moles / volume

Molarity = 0.0430 moles / 0.168 L

Molarity = 0.256 M

Step 4 : Calculate the osmotic pressure

π = iMRT

⇒ with i = The Van't hoff factor = 1 (since testosterone is nonelectrolyte)

⇒ with M = the molair concentration = 0.256 M

⇒ with R = the gas constant = 0.08206 L*atm/k*mol

⇒ with T = the temperature = 298 K

π = 1*0.256*0.08206*298

π = 6.26 atm

The osmotic pressure is 6.26 atm

Answer:

d. Decomposition

Explanation:

Reversible changes -

As the name suggests the the changes which can be reversed back to the original compound , is referred to as reversible changes .

All the physical changes are reversible in nature .

For example ,

Melting of ice , as the ice is melted from solid to liquid form , the liquid can again be converted back to solid , by reducing the temperature .

Hence , the changes are reversible in nature .

The species can not be further merged back to get back the original compound , hence is not a reversible change .

Explanation:

It is known that charge on xenon nucleus is [tex]q_{1}[/tex] equal to +54e. And, charge on the proton is [tex]q_{2}[/tex] equal to +e. So, radius of the nucleus is as follows.

            r = [tex]\frac{6.0}{2}[/tex]

              = 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

              d = r + 2.5

                 = (3.0 + 2.5) fm

                 = 5.5 fm

                 = [tex]5.5 \times 10^{-15} m[/tex]     (as 1 fm = [tex]10^{-15}[/tex])

Therefore, electrostatic repulsive force on proton is calculated as follows.

              F = [tex]\frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}[/tex]

Putting the given values into the above formula as follows.

           F = [tex]\frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}[/tex]

              = [tex](9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}[/tex]

              = [tex](9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}[/tex]

              = 411.2 N

or,           = [tex]4.1 \times 10^{2}[/tex] N

Thus, we ca conclude that [tex]4.1 \times 10^{2}[/tex] N is the electric force on a proton 2.5 fm from the surface of the nucleus.

Answer:

Explanation:

Graham's law of diffusion states that the rate at which effusion occurs in two gases is equal to the square inverse of each of their molar masses.

Final answer:

Using Graham's law of effusion, we can calculate the molecular weight of the gas that diffuses 1.86 times faster than XeXe. The molecular weight of the unknown gas is approximately 37.98 g/mol.

Explanation:

To find the molecular weight of the gas that diffuses through a porous membrane 1.86 times faster than Xe, we need to use Graham's law of effusion. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

Let the molecular weight of Xe be 'a' and the molecular weight of the unknown gas be 'b'.

According to Graham's law, (Rate of effusion of unknown gas)/(Rate of effusion of Xe) = sqrt((molar mass of Xe)/(molar mass of unknown gas)).

Squaring both sides of the equation, we get, (Rate of effusion of unknown gas)^2 = (Rate of effusion of Xe)^2 * (molar mass of Xe)/(molar mass of unknown gas).

Since the rate of diffusion of the unknown gas is 1.86 times faster than Xe, we can substitute this value into the equation to get (1.86)^2 = 1.86^2 * (molar mass of Xe)/(molar mass of unknown gas).

Simplifying the equation, (molar mass of unknown gas) = (molar mass of Xe) / (1.86^2).

Substituting the molar mass of Xe, which is 131.293 g/mol, into the equation, we get (molar mass of unknown gas) = 131.293 / (1.86^2).

Calculating the value, (molar mass of unknown gas) = 131.293 / 3.4596 = 37.98 g/mol.

To prepare a 0.00200 M NaCl solution from a 0.500 M stock, dilute 0.4 mL of the stock to 100 mL with water using volumetric pipets and a flask.

To prepare 100 mL of a 0.00200 M NaCl solution from a 0.500 M stock solution, you need to perform a dilution. First, use the dilution equation C1V1 = C2V2, where C1 is the concentration of the stock solution, V1 is the volume of the stock solution needed, C2 is the concentration of the diluted solution, and V2 is the volume of the diluted solution. Plugging in the values, you get (0.500 M) V1 = (0.00200 M)(100 mL), which simplifies to V1 = (0.00200 M)(100 mL) / (0.500 M) = 0.4 mL. This is the volume of the stock solution you need to pipet into the 100-mL volumetric flask. As we only have 1-mL, 5-mL, and 10-mL volumetric pipets available, you can use the 1-mL pipet to add 0.4 mL of the stock solution to the volumetric flask, then fill the flask with water up to the 100-mL mark to achieve the desired concentration.

Answer:

Molarity = 0.5 M

Osmolarity = 0.5 x 2 = 1 Osmpl.

Molecules of Cl2 = 6.02 x [tex]10^{23}[/tex] / 4= 1.505 x [tex]10^{23}[/tex] no. of molecules

Explanation:

If we add half mole in 1L volume than molarity will obviously be 0.5 M.

The osmolarity is molarity multiplies by number of dissociates of solute that for CaCl2 are 2. So, 2 x 0.5 = 1

Half will be molecules of Ca and half will be of Cl2 for 0.5M.

Final answer:

The molarity of a 1-liter solution with half a mole of calcium chloride is 0.5 M. The osmolarity, considering the dissociation of calcium chloride into three ions, is 1.5 osmol/L. The solution would contain 6.022 x 10²³chloride ions.

Explanation:

The molarity of a solution is defined as the number of moles of a solute per liter of solution. In this case, to calculate the molarity of calcium chloride, CaCl₂, in a 1-liter solution containing half a mole, we simply divide the moles of solute by the volume of the solution in liters:

Molarity (M) = moles of solute / liters of solution

Molarity (M) = 0.5 mol / 1 L = 0.5 M

The osmolarity of a solution is the total concentration of solute particles dissolved in a solution. Since calcium chloride dissociates into three ions (one Ca²⁺ ion and two Cl
- ions), the osmolarity is as follows:

Osmolarity = molarity  x number of particles per formula unit

Osmolarity = 0.5 M  x 3 = 1.5 osmol/L

For the number of chloride ions in the solution, we must look at the stoichiometry of calcium chloride: each formula unit of CaCl₂ yields two Cl⁻ions. So, we have:

Number of Cl⁻ ions = 0.5 mol CaCl₂ x (2 mol Cl⁻ / 1 mol CaCl₂) x (6.022 x 10²³ ions/mol)

Number of Cl⁻ ions = 1 mol Cl⁻ x 6.022 x 10²³ ions/mol = 6.022 x 10²³ ions

Answer : The molality of solution is, 5.69 mole/L

Explanation :

The relation between the molarity, molality and the density of the solution is,

[tex]d=M[\frac{1}{m}+\frac{M_b}{1000}][/tex]

where,

d = density of solution  = 1.25 g/mL

m = molality of solution  = ?

M = molarity of solution  = 4.57 M

[tex]M_b =\text{molar mass of solute }(H_2SO_4)[/tex] = 98 g/mole

Now put all the given values in the above formula, we get

[tex]1.25g/ml=4.57M[\frac{1}{m}+\frac{98g/mole}{1000}][/tex]

[tex]m=5.69mol/kg[/tex]

Therefore, the molality of solution is, 5.69 mole/L

Answer:

The molality is 5.7 molal

Explanation:

Step 1: Data given

Molarity of H2SO4 = 4.57 M ( = 4.57 mol/L)

Density of the solution = 1.25 g/mL

Molar mass of H2SO4 = 98.09 g/mol

Step 2: Calculate mass of solution

Suppose we have 1L (= 1000 mL) of solution

Mass of the solution = 1.25 g/mL * 1000 mL

Mass of the solution = 1250 grams

Step 3: Calculate mass H2SO4

In 1L of a 4.57 M solution we have 4.57

Mass H2SO4 = moles H2SO4 * molar mass H2SO4

Mass H2SO4 = 4.57 moles * 98.09 g/mol

Mass H2SO4 = 448.3 grams

Step 4: Calculate mass solvent

Mass solvent = mass solution - mass H2SO4

Mass solvent = 1250 grams - 448.3 grams

Mass solvent = 801.7 grams

Step 5: Calculate molality

Molality = moles H2SO4 / mass solvent

Molality = 4.57 moles / 0.8017 kg

Molality = 5.7 molal

The molality is 5.7 molal