how many grams of kno3are required to prepare 250ml of a .700m solution

The pH of the potassium hydrogen phthalate buffer before the addition of a base or an acid is equal to its pKa, which is 5.51.

To calculate the pH of the potassium hydrogen phthalate buffer before the addition of a base or an acid, you'll need to consider the molarity of the buffer solution and the pKa of potassium hydrogen phthalate. The pKa value for potassium hydrogen phthalate is 5.51. We can use the Henderson-Hasselbalch equation to determine the pH of the buffer:

pH = pKa + log ([A-]/[HA])
Given that the buffer is not yet titrated with any acid or base, the ratio of [A-]/[HA] is 1. Therefore, the equation becomes: pH = pKa + log (1)

Since the log (1) is 0, the equation simplifies to:
pH = pKa = 5.51

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The initial concentration (molarity) of acetic acid is 0.0462 M. The pH of the solution is 1.75,  the van't Hoff i factor for acetic acid is 2,

The dissociation reaction of acetic acid in water is:

CH₃COOH + H₂O ⇌ CH₃COO- + H₃O+

The equilibrium constant expression for this reaction is:

Ka = [CH₃COO⁻][H₃O⁺]/[CH₃COOH]

Given that acetic acid is 4% dissociated in solution, we can assume that the concentration of acetic acid remaining is 96% of the initial concentration, and the concentration of both acetate and hydronium ions formed is 4% of the initial concentration.

g) What was the initial concentration (molarity) of acetic acid?

Let's assume that the initial concentration of acetic acid is x M. Then, the concentration of acetate and hydronium ions formed is 0.04x M. The concentration of acetic acid remaining is (1-0.04)x M = 0.96x M.

Using the equilibrium constant expression, we can write:

Ka = [CH₃COO⁻][H3O⁺]/[CH₃COOH]

Ka = (0.04x)(0.04x)/(0.96x)

Ka = 0.00176

We know that the equilibrium constant expression for a weak acid can be written as Ka = [H3O⁺][A-]/[HA]. In this case, HA represents acetic acid and A- represents acetate ion. Since the concentration of acetate and hydronium ions formed is 0.04x M, and assuming that they are equal due to the 1:1 stoichiometry of the dissociation reaction, we can write:

Ka = [H3O⁺][CH₃COO⁻]/[CH₃COOH]

0.00176 = (0.04x)/(0.96x)

x = 0.0462 M

Therefore, the initial concentration (molarity) of acetic acid is 0.0462 M.

h) What is the pH?

The concentration of hydronium ions formed in the solution can be calculated using the equation:

[H3O⁺] = Ka[CH₃COOH]/[CH₃COO⁻]

[H3O⁺] = (0.00176)(0.0462)/(0.00462)

[H3O⁺] = 0.0177 M

The pH of the solution can be calculated using the equation:

pH = -log[H3O⁺]

pH = -log(0.0177)

pH = 1.75

Therefore, the pH of the solution is 1.75.

i) What is the van't Hoff i factor?

The van't Hoff i factor represents the number of particles that are formed when a solute dissolves in a solvent. In this case, acetic acid is a weak acid, which means that it partially dissociates in water to form acetate and hydronium ions. Therefore, the van't Hoff i factor for acetic acid is 2, since 2 particles are formed when it dissolves in water (one molecule of acetic acid and one hydronium ion).

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Sometimes, the best thing to do in the event of a crude oil spill is to do nothing, particularly if the spill is in a remote location with limited human or wildlife exposure.

This is because any attempt to clean up the spill could potentially do more harm than good, such as disrupting fragile ecosystems or causing further damage to the environment. In these cases, it is often recommended to simply monitor the spill and let nature take its course in breaking down and absorbing the oil. However, if the spill poses a significant threat to human health or the environment, action must be taken to contain and mitigate the spill.

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Living organisms in an ecosystem are made up of organic matter, which includes cells, proteins, lipids, carbohydrates, and nucleic acids.

Organic refers to products or items that are made from all-natural ingredients that have been grown or harvested without the use of synthetic fertilizers, pesticides, or other artificial substances. Organic products are produced in accordance with certain production standards that promote the conservation of natural resources and biodiversity. Organic farming methods are designed to create a sustainable and healthy environment, as well as provide economic benefits.

These organic molecules are the building blocks of living things and are produced by living organisms.

Dead organisms in an ecosystem are made up of inorganic matter, which includes minerals, rocks, and soil. These inorganic molecules are the remnants of dead organisms and are produced through the breakdown of living things.

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If an alkaline developer is used in dye penetrant inspections, it can cause the dye to wash out, making it difficult or impossible to detect any flaws or defects in the surface being inspected.

The alkaline developer can also react with the dye, altering its chemical properties and making it ineffective for future inspections.

This can lead to inaccurate or incomplete inspections, which can have serious consequences if the surface being inspected is critical for safety or performance.

It is important to always use the correct type of developer for the specific dye penetrant being used to ensure accurate and reliable results.

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Half-life is the time required for half of the quantity of a substance to undergo a specified reaction, decay, or transformation. After 4.5 years, there will be 7.34 × 10^-7 mol of tritium left.

The first-order rate law is given by:

rate = k [T]

where [T] is the concentration of tritium and k is the rate constant. The half-life of tritium is 12.5 years, which means that:

t1/2 = 0.693/k

Solving for k:

k = 0.693/t1/2 = 0.693/12.5 yr = 0.0554 yr⁻¹

Using the first-order integrated rate law:

㏑ ([T]/[T]₀) = -kt

where [T]₀ is the initial concentration of tritium, we can solve for [T]

㏑ ([T]/1.00 × 10⁻⁶mol) = -0.0554 yr⁻¹ x 4.5 yr

[T]/1.00 × 10⁻⁶ mol = e^-0.249 yr⁻¹

[T] = (1.00 × 10⁻⁶  mol) x e^-0.249 yr⁻¹

[T] = 7.34 × 10⁻⁷ mol

Therefore, after 4.5 years, there will be 7.34 × 10⁻⁷ mol of tritium left.

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The solubility of aluminum hydroxide at 25°C is approximately 2.8 x 10⁷ g/L (option d).

1: The solubility product constant (Ksp) equation for aluminum hydroxide (Al(OH)₃) is:

Ksp = [Al³⁺][OH⁻]₃

When Al(OH)₃ dissolves, it forms one Al³⁺ ion and three OH⁻ ions. Therefore, [Al³⁺] = s and [OH⁻] = 3s.
Ksp = (s)(3s)³
4.6 x 10⁻³³ = s(27s³)

2: Divide by 27:
s⁴ = (4.6 x 10⁻³³)/27

3: Take the fourth root:
s = (4.6 x 10⁻³³/27)^(1/4)
s = 1.8 x 10⁻⁸ mol/L

4: Now, we need to convert the solubility from mol/L to g/L:
1.8 x 10⁻⁸ mol/L * (26.98 g/mol Al + 3 * 15.999 g/mol O + 3 * 1.007 g/mol H) = 2.8 x 10⁻⁷ g/L

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The de Broglie wavelength of the beta particle is approximately  2.69 x 10^-15 m.

To find the de Broglie wavelength of the beta particle, we need to use the formula:

λ = h / p

where λ is the wavelength, h is the Planck's constant (6.626 x 10^-34 J s), and p is the momentum of the particle. We can find the momentum of the beta particle using the formula:

p = m * v

where m is the mass of the particle and v is its velocity. Plugging in the given values, we get:

p = (9.109 x 10^-28 g) * (2.70 x 10^8 m/s)
p = 2.46 x 10^-19 kg m/s

Now we can calculate the wavelength:

λ = (6.626 x 10^-34 J s) / (2.46 x 10^-19 kg m/s)
λ = 2.69 x 10^-15 m

Therefore, the de Broglie wavelength of the beta particle is 2.69 x 10^-15 m.

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Increasing the pressure on the system will shift the equilibrium to the right, favoring the formation of products D and E.

According to Le Chatelier's principle, an increase in pressure will shift the equilibrium towards the side with fewer moles of gas. Since there are fewer moles of gas on the product side, an increase in pressure will favor the forward reaction, producing more D and E. Conversely, decreasing the pressure will shift the equilibrium towards the side with more moles of gas, favoring the reverse reaction. Increasing the concentration of D will not have an effect on the equilibrium because it is a product, and decreasing the concentration of B will also not have an effect because it is a reactant.

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(a) The mean concentration of organic aerosol from the glyoxal formation pathway is 0.63 µg C m³.


1. Calculate glyoxal formation rate: (5 x 10¹¹molecules/cm²s) * (10% yield) = 5 x 10¹⁰ molecules/cm²s

2. Convert to molecules/m³s: (5 x 10¹⁰ molecules/cm²s) * (1 m²/10⁴ cm²) = 5 x 10¹⁴ molecules/m³s

3. Calculate aerosol formation rate: (5 x 10¹⁴ molecules/m³s) * (1/6 aerosol yield) = 8.33 x 10¹³ molecules/m³s

4. Convert to mass of aerosol formed in 3 days: (8.33 x 10¹³ molecules/m³s) * (3 days) * (24 hr/day) * (3600 s/hr) * (12 g/mol) * (1 mol/6.022 x 10²³ molecules) = 1.89 µg C m³

5. Divide by mixing depth: (1.89 µg C m³) / (1 km) = 0.63 µg C m³

(b) The glyoxal formation pathway is not significant as its contribution (0.63 µg C m³) is less than the typical U.S. observations of 2 µg C m³ for the concentration of organic aerosol.

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The compound immediately preceding succinate in the citric acid cycle is fumarate.

During the citric acid cycle, also known as the Krebs cycle or TCA cycle, the molecule acetyl-CoA is oxidized to produce energy in the form of ATP, as well as reducing agents such as NADH and FADH2. In the fourth step of the cycle, succinate is produced by the oxidation of succinyl-CoA, which is derived from the previous step where alpha-ketoglutarate is oxidized.

Before succinyl-CoA is formed, however, the molecule fumarate is produced by the oxidation of the previous intermediate, malate. So, the correct order of the citric acid cycle constituents leading up to succinate is malate, fumarate, succinate, and then the cycle continues with the production of oxaloacetate.

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The compound immediately preceding succinate in the citric acid cycle is fumarate.

During the citric acid cycle, also known as the Krebs cycle or TCA cycle, the molecule acetyl-CoA is oxidized to produce energy in the form of ATP, as well as reducing agents such as NADH and FADH2. In the fourth step of the cycle, succinate is produced by the oxidation of succinyl-CoA, which is derived from the previous step where alpha-ketoglutarate is oxidized.

Before succinyl-CoA is formed, however, the molecule fumarate is produced by the oxidation of the previous intermediate, malate. So, the correct order of the citric acid cycle constituents leading up to succinate is malate, fumarate, succinate, and then the cycle continues with the production of oxaloacetate.

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The decay of 83^214 Bi to 82^214 Pb occurs through the emission of a beta particle.

The decay of 83^214 Bi (Bismuth-214) to 82^214 Pb (Lead-214) occurs through the emission of a beta particle.

Step-by-step explanation:

1. Identify the initial nuclide: 83^214 Bi (Bismuth-214), where 83 is the atomic number (protons) and 214 is the mass number (protons + neutrons).

2. Identify the final nuclide: 82^214 Pb (Lead-214), where 82 is the atomic number and 214 is the mass number.

3. Observe the change in atomic number: The atomic number decreases by 1 (from 83 to 82), which indicates that a beta particle (electron) is emitted.

4. Confirm that the mass number remains the same (214) as it does not change during beta decay.

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all four elements (A, B, C, and D) have 2 unpaired electrons in their ground state

To determine how many of the following elements have 2 unpaired electrons in the ground state, let's examine the electron configurations for each element: A. C (Carbon), B. Te (Tellurium), C. Hf (Hafnium), and D. Si (Silicon).

A. Carbon (C) has an electron configuration of 1s² 2s² 2p². In the 2p subshell, there are two unpaired electrons.

B. Tellurium (Te) has an electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁴. In the 5p subshell, there are two unpaired electrons.

C. Hafnium (Hf) has an electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d². In the 5d subshell, there are two unpaired electrons.

D. Silicon (Si) has an electron configuration of 1s² 2s² 2p⁶ 3s² 3p². In the 3p subshell, there are two unpaired electrons.

So, all four elements (A, B, C, and D) have 2 unpaired electrons in their ground state.

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To find the specific heat of benzene, we can use the formula: Q = mcΔT where Q is the heat added (3450 J), m is the mass of the benzene sample (150 g).

c is the specific heat capacity we want to find, and ΔT is the temperature increase (35.8°C - 22.5°C). First, let's calculate ΔT: ΔT = 35.8°C - 22.5°C = 13.3°C Now, we can rearrange the formula to find c: c = Q / (mΔT) c = 3450 J / (150 g × 13.3°C) c ≈ 1.73 J/(g°C)

So, the specific heat capacity of benzene is approximately 1.73 J/(g°C). Plugging in the given values: 3450 J = 150 g * c * (35.8°C - 22.5°C) Solving for c: c = 1.74 J/g°C Therefore, the specific heat of benzene is 1.74 J/g°C if 3450 J of heat is added to a 150 g sample of benzene and its temperature increases from 22.5°C to 35.8°C.

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From the following formula ΔG°f = ΣnΔG°f(products) - ΣmΔG°f(reactants), the standard free energy of formation of 1.00 mol of CuO(s) is -157.3 kJ.

The standard free energy of formation of 1.00 mol of CuO(s) can be calculated using the equation: ΔG°f = ΣnΔG°f(products) - ΣmΔG°f(reactants)
;where ΔG°f is the standard free energy of formation, n and m are the stoichiometric coefficients of the products and reactants respectively.

The standard free energy of formation of CuO(s) is -157.3 kJ/mol. Therefore, the standard free energy of formation of 1.00 mol of CuO(s) can be calculated as:

ΔG°f = (1 mol) (-157.3 kJ/mol) = -157.3 kJ

So, the standard free energy of formation of 1.00 mol of CuO(s) is -157.3 kJ.

The standard free energy of formation (ΔGf°) of a compound is the change in free energy when 1.00 mol of the substance is formed from its constituent elements under standard conditions (1 atm pressure and 298 K temperature). For CuO(s), this refers to the formation of 1.00 mol of solid copper(II) oxide from its elements, copper (Cu) and oxygen (O₂).

To find the ΔGf° value for CuO(s), you can refer to a table of standard free energies of formation. According to such tables, the ΔGf° for CuO(s) is approximately -129.7 kJ/mol. This negative value indicates that the formation of CuO(s) from its elements is a spontaneous process under standard conditions.

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Entropy of C₂H₇OH(g) is anticipated to be higher than entropy of C₂H₇OH(l).

The amount of microstates in a distribution affects how likely it is that a system will exist with all of its constituent parts. The most likely distribution is the one with the highest entropy since entropy rises logarithmically with the number of microstates.

The number of various configurations of molecule location and kinetic energy at a specific thermodynamic state is referred to as microstates. a method that makes more people available

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Titrating too much NaOH into the flask would cause the normality of the solution to decrease due to the increase in volume without a corresponding increase in the number of equivalents (moles of OH-) being added.

To explain this reasoning, let's consider the following steps:

1. When titrating NaOH into the flask, you are adding more moles of hydroxide ions (OH-) to the solution.
2. As you add more NaOH, the concentration of the hydroxide ions in the solution will increase.
3. Normality is defined as the number of equivalents of solute per liter of solution (N = equivalents/L).
4. In this case, an "equivalent" refers to the number of moles of hydroxide ions (OH-).
5. As you continue to add NaOH beyond the equivalence point, the volume of the solution in the flask will also increase.
6. Because normality takes both the number of equivalents (moles of OH-) and the volume of the solution into account, increasing the volume without adding more equivalents of the solute will result in a lower normality.

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The positions on the N-phenylacetamide ring that could undergo bromination are the ortho, meta, and para positions.

Bromination is an electrophilic aromatic substitution reaction, where a bromine atom is introduced to the aromatic ring. N-phenylacetamide has an amide group attached to the phenyl ring. The amide group is a weakly electron-withdrawing group due to resonance and inductive effects, making it a meta-directing group. However, it is not strong enough to completely prevent bromination at ortho and para positions.

Therefore, N-phenylacetamide can undergo bromination at all three positions, but the meta position is more likely due to the amide group's influence. It is important to note that the presence of a catalyst that can enhance the reactivity of bromine and influence the selectivity of the reaction. The positions on the N-phenylacetamide ring that could undergo bromination are the ortho, meta, and para positions.

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For a 10-ml graduated cylinder, record the measurement to the nearest 0.1 ml while for a 50-ml graduated cylinder, record the volume to the nearest 0.01 ml.

To record the position of the meniscus to the correct precision (uncertainty) in each of the two instruments, follow these steps:

In a 10-mL graduated cylinder:
1. Observe the position of the meniscus, which is the curved surface of the water in the cylinder.
2. To determine the correct precision, note the smallest graduation on the cylinder, typically 0.1 mL.
3. Record the volume to one decimal place (0.1 mL) by estimating the position of the meniscus between the graduation marks.

In a 50-mL buret:
1. Observe the position of the meniscus, which is the curved surface of the water in the buret.
2. To determine the correct precision, note the smallest graduation on the buret, typically 0.1 mL.
3. Record the volume to one decimal place (0.1 mL) by estimating the position of the meniscus between the graduation marks.

Remember to always read the meniscus at eye level and record the measurements with the correct precision (uncertainty) as specified by the instrument.

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Consider a buffer made by adding 44.9 g of (CH[tex]_2[/tex])[tex]_2[/tex]NH[tex]_2[/tex]I to 250.0 ml of 1.42 m (CH[tex]_2[/tex])[tex]_2[/tex]NH. 10.29 is the pH of this buffer.

pH is a numerical indicator of how acidic or basic aqueous or other liquid solutions are. The phrase, which is frequently used in chemistry, biology, and agronomy, converts the hydrogen ion concentration, which typically ranges between 1 and 1014 gram-equivalents per litre, into numbers between 0 and 14. The hydrogen ion concentration in pure water, which has a pH of 7, is 107 gram-equivalents per litre, making it neutral (neither acidic nor alkaline). A solution with a pH below 7 is referred to as acidic, and one with a pH over 7 is referred to as basic, or alkaline.

moles of  (CH[tex]_2[/tex])[tex]_2[/tex]NH = 1.42 mol/L x 0.250 L

                                 = 0.355 mol

(CH[tex]_2[/tex])[tex]_2[/tex]NH[tex]_2[/tex]I →  (CH[tex]_2[/tex])[tex]_2[/tex]NH[tex]_2[/tex]⁺ + I⁻

Kb = [  (CH[tex]_2[/tex])[tex]_2[/tex]NH[tex]_2[/tex]⁺ ][OH⁻] / [ (CH[tex]_2[/tex])[tex]_2[/tex]NH  ]

[ (CH[tex]_2[/tex])[tex]_2[/tex]NH[tex]_2[/tex]⁺ ] = Kb x [ (CH₂)₂NH ]

                      = (5.4 x 10⁻⁴) x 0.355 mol

                      = 1.92 x 10⁻⁴ M

[  (CH[tex]_2[/tex])[tex]_2[/tex]NH ] =(CH[tex]_2[/tex])[tex]_2[/tex]NH +  (CH[tex]_2[/tex])[tex]_2[/tex]NH[tex]_2[/tex]⁺

                    = 0.355 mol + 1.92 x 10⁻⁴ mol

                    = 0.3552 mol

Kb = Kw / Ka

Ka = Kw / Kb

   = 1.0 x 10⁻¹⁴ / 5.4 x 10⁻⁴

   = 1.85 x 10⁻¹¹

pKa = -log(Ka)

      = -log(1.85 x 10⁻¹¹)

      = 10.73

pH = pKa + log( [ (CH₃)₂NH₂⁺ ] / [ (CH₂)₂NH ] )

pH = 10.73 + log(1.92 x 10⁻⁴ M / 0.3552 M)

    = 10.29

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To prepare a buffer of pH 4, we need to use the Henderson-Hasselbalch equation:

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

We can assume that acetic acid (HA) will be the major species in solution and the acetate ion (A^-) will be the minor species.

pH = 4

pKa of acetic acid = 4.76

Substituting these values into the Henderson-Hasselbalch equation, we get:

4 = 4.76 + log([A^-]/[HA])

log([A^-]/[HA]) = -0.76

[A^-]/[HA] = 10^(-0.76)

[A^-]/[HA] = 0.184

Since we know the concentration of acetic acid is 1.0 M, we can find the concentration of the acetate ion by multiplying the concentration of acetic acid by the ratio [A^-]/[HA]:

0.184 = [A^-]/1.0

[A^-] = 0.184 M

Now, we can use the equation for the dissociation of sodium acetate:

NaC2H3O2(aq) ↔ Na+(aq) + C2H3O2^-(aq)

The equilibrium constant for this reaction is:

K = [Na+(aq)][C2H3O2^-(aq)]/[NaC2H3O2(aq)]

Since the sodium acetate is a strong electrolyte, it will dissociate completely, so we can assume that the concentration of NaC2H3O2(aq) is equal to the concentration of sodium acetate added. Therefore, we can simplify the equilibrium constant expression to:

K = [Na+][C2H3O2^-]

We can find the concentration of sodium ion by multiplying the concentration of acetate ion by the ratio of sodium ion to acetate ion, which is 1:1 since the compound is NaC2H3O2:

[Na+] = [C2H3O2^-] = 0.184 M

We can look up the value of the equilibrium constant for this reaction (K = 1.8 x 10^-5), so we can solve for the concentration of NaC2H3O2:

1.8 x 10^-5 = (0.184 M)^2/[NaC2H3O2]

[NaC2H3O2] = 0.184^2/1.8 x 10^-5

[NaC2H3O2] = 1.89 M

Now, we can use the formula for calculating the amount (in moles) of a compound needed to make a solution:

moles = concentration x volume (in liters)

We have a volume of 50.0 mL = 0.0500 L and a concentration of 1.89 M, so:

moles of NaC2H3O2 = 1.89 M x 0.0500 L = 0.0945 moles

Finally, we can use the molar mass of NaC2H3O2·3H2O to convert moles to mass:

mass = moles x molar mass

The molar mass of NaC2H3O2·3H2O is:

Na: 1 x 22.99 g/mol = 22.99 g/mol

C: 2 x 12.01 g/mol = 24.02 g/mol

H: 6 x 1.01 g/mol = 6.06 g/mol

O: 7 x 16.00 g

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-22.0 J/K is the change in entropy when 35. g of ethanol freezes at -114.3 °C .

The amount of ethanol that freezes can be calculated using its molar mass:

Molar mass of ethanol = 46.07 g/mol

Number of moles of ethanol = mass / molar mass = 35 g / 46.07 g/mol = 0.760 mol

The heat released during the freezing of 0.760 mol of ethanol can be calculated using the heat of fusion:

ΔH = nΔHf = (0.760 mol)(4.6 kJ/mol) = 3.5 kJ

The change in entropy (ΔS) can be calculated using the following equation:

ΔS = -ΔH / T

where ΔH is the heat released during the freezing of ethanol and T is the temperature at which the freezing occurs in Kelvin.

The temperature of the freezing is -114.3 °C = 158.85 K

ΔS = -(3.5 kJ) / (158.85 K) = -22.0 J/K

Therefore, the change in entropy when 35. g of ethanol freezes at -114.3 °C is -22.0 J/K.

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Ionic attractions are weaker in media with high dielectric constants, such as water and aqueous buffers, because the dielectric constant measures a substance's ability to reduce the electrostatic forces between charged particles.

Ionic attractions refer to the electrostatic interactions between ions, which can either attract or repel one another depending on the charges involved. Dielectric constants are a measure of a solvent's ability to reduce the strength of these electrostatic interactions between ions.

In media with high dielectric constants, such as water and aqueous buffers, the solvent molecules have a greater ability to shield the charges of the ions. This means that the electrostatic attractions between ions are weaker, as the ions are less able to interact directly with one another.

This effect can be explained by considering the way in which ions interact with their surroundings. In low dielectric constant solvents, the ions are surrounded by a tightly packed layer of solvent molecules, which effectively shield their charges from other ions. This means that the ions are able to interact more strongly with one another, as there is less interference from the solvent molecules.

In contrast, in high dielectric constant solvents, the solvent molecules are more loosely packed around the ions. This means that there is more space for the solvent molecules to move around, which reduces the strength of the interactions between the ions. The net effect of this is that ionic attractions are weaker in media with high dielectric constants, such as water and aqueous buffers.

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The coefficient for O2 in the balanced equation is 2, which is a whole number as requested in your question.

To balance the chemical equation involving C2H4, O2, CO2, and H2O. Here's a step-by-step explanation:

1. Write the unbalanced chemical equation: C2H4 + O2 → CO2 + H2O

2. Balance the carbon (C) atoms: Since there are two carbon atoms in C2H4, we need two CO2 molecules to balance the carbon atoms.
  C2H4 + O2 → 2CO2 + H2O

3. Balance the hydrogen (H) atoms: There are four hydrogen atoms in C2H4, so we need two H2O molecules to balance the hydrogen atoms.
  C2H4 + O2 → 2CO2 + 2H2O

4. Balance the oxygen (O) atoms: There are now four oxygen atoms on the right side of the equation (two from each CO2 and two from the two H2O molecules). To balance the oxygen atoms, we need two O2 molecules on the left side.
  C2H4 + 2O2 → 2CO2 + 2H2O

The balanced chemical equation is: C2H4 + 2O2 → 2CO2 + 2H2O

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The numbers in the names of the ketones, such as 2-propanone, 2-butanone, 2-pentanone, and 3-pentanone, refer to the position of the carbonyl group (C=O) in the carbon chain of the molecule.

Here's a breakdown of each ketone:

1. 2-Propanone: This ketone has three carbon atoms (propane) with the carbonyl group on the second carbon atom. Its structure is CH3-C(=O)-CH3.
2. 2-Butanone: This ketone has four carbon atoms (butane) with the carbonyl group on the second carbon atom. Its structure is CH3-CH2-C(=O)-CH3.
3. 2-Pentanone: This ketone has five carbon atoms (pentane) with the carbonyl group on the second carbon atom. Its structure is CH3-CH2-CH2-C(=O)-CH3.
4. 3-Pentanone: This ketone also has five carbon atoms (pentane), but the carbonyl group is on the third carbon atom. Its structure is CH3-CH2-C(=O)-CH2-CH3.

The numbers in the names of these ketones indicate the position of the carbonyl group within the carbon chain.

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The Kb value for CN- at 25°C is approximately 1.61×10^-5.

The BKB (base dissociation constant) of CN- at 25°C can be calculated using the relationship:

Kb = Kw / Ka

where Kw is the ion product constant of water (1x10^-14) and Ka is the acid dissociation constant of HCN (6.2x10^-10).

Plugging in the values, we get:

Kb = (1x10^-14) / (6.2x10^-10)
Kb = 1.61x10^-5

Therefore, the BKB value for CN- at 25°C is 1.61x10^-5.

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The solubility product constant (Ksp) expression for cobalt(II) hydroxide (Co(OH)2) is: Ksp = [Co2+][OH-][tex]^2[/tex]

Since cobalt(II) hydroxide is a sparingly soluble compound, we can assume that it dissociates in water to a very small extent, and that the concentration of Co2+ is negligible compared to the initial concentration of OH-. Therefore, we can simplify the expression to:

Ksp ≈ [OH-][tex]^2[/tex]

Taking the square root of both sides of the equation and substituting the value of Ksp gives:

[OH-] = sqrt(Ksp) = sqrt(5.9 x 10^-15) = 7.68 x 10[tex]^-8[/tex] M

The hydroxide ion concentration in a saturated solution of cobalt(II) hydroxide is 7.68 x 10[tex]^-8[/tex] M.

To find the pH, we can use the relation between pH and [OH-]:

pH = -log [OH-] = -log (7.68 x 10[tex]^-8[/tex] = 7.11

Therefore, the pH of a saturated solution of cobalt(II) hydroxide is approximately 7.11.

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When 0.1 M HNO3 is added to a solution containing MgCO3 at 298 K with Ksp = 4.0 x 10^-5, the solubility of MgCO3 will increase.

Solubility is the capacity of a substance to dissolve when mixed with a solvent to give rise tot a solution.
HNO3 is a strong acid that will react with the MgCO3 to form soluble products. The reaction is:
MgCO3 (s) + 2HNO3 (aq) → Mg(NO3)2 (aq) + CO2 (g) + H2O (l)

Ksp is the solubility product constant. If there is an increase in solubility, the Ksp value tends to increase as well.
The addition of HNO3 will cause the MgCO3 to dissolve more to maintain the equilibrium, thus increasing its solubility in the solution.

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A total of 18 moles of electrons are transferred when 1 mole of potassium dichromate is mixed with 3 moles of nickel.

The balanced chemical equation for the given reaction is:

14H+ (aq) + Cr2O7 2- (aq) + 3Ni (s) → 2Cr3+ (aq) + 3Ni2+ (aq) + 7H2O (l)

From the equation, we can see that 6 moles of electrons are transferred per mole of Ni (s). Therefore, when 3 moles of nickel react, 18 moles of electrons are transferred.

To determine the number of moles of electrons transferred when 1 mole of potassium dichromate is mixed with 3 moles of nickel, we need to first calculate the limiting reactant. The balanced equation shows that 3 moles of nickel require 1 mole of potassium dichromate to react completely. Therefore, 1 mole of potassium dichromate will react with 3 moles of nickel.

As we know that 18 moles of electrons are transferred when 3 moles of nickel react, we can conclude that 6 moles of electrons are transferred when 1 mole of nickel reacts. Therefore, when 1 mole of potassium dichromate is mixed with 3 moles of nickel, a total of 6 x 3 = 18 moles of electrons are transferred.

In summary, 18 moles of electrons are transferred when 1 mole of potassium dichromate is mixed with 3 moles of nickel.

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