What is the predominant form of ethylenediamine at pH 6.441 and pH 9.375? Ethylenediamine has pKb values of 4.072 (pKb1) and 7.152 (pKb2).

Answers

Answer 1

Ethylenediamine is a weak base with two pKb values, and it can exist in three different forms: as a fully protonated cation, a partially protonated zwitterion, or a fully deprotonated anion.

To determine the predominant form at a given pH, we need to compare the pH to the pKb values.

At pH 6.441, which is between the two pKb values, ethylenediamine is partially protonated. The dominant species will be the zwitterion, which has a positive charge on one nitrogen and a negative charge on the other nitrogen.

At pH 9.375, which is higher than both pKb values, ethylenediamine is fully deprotonated. The dominant species will be the anion, which has both nitrogens with a negative charge.

Therefore, at pH 6.441, the predominant form of ethylenediamine is the zwitterion, and at pH 9.375, the predominant form is the anion.

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Related Questions

How does heat transfer occur?

1. from hot to cold until both objects reach the same temperature
2. from cold to hot until the hot object becomes cold
3. from cold to hot until both objects reach the same temperature
4. from hot to cold until the cold object becomes hot

Answers

Heat transfer occurs from hot to cold until both objects reach the same temperature. Thus, option 1 is correct.

The second law of thermodynamics states that heat transfer always occurs from hotter objects to cooler objects until they reach the same temperature. This phenomenon is called the Thermal equilibrium of that substance. At this point, both bodies' temperatures are the same.

Heat transfer occurs in three modes. They are conduction, convection, and radiation. In every case, heat is always transferred from got object to sold object. The second law also states that entropy change cannot be negative until they reach equilibrium.

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how much heat is required to vaporize 150 g of methane, ch4, at its boiling point? the enthalpy of vaporization of methane at its boiling point is 8.9 kj/mol.

Answers

So, 123.5 kJ of heat is required to vaporize 150 g of methane, Ch4, at its boiling point.  

To vaporize 150 g of methane, Ch4, at its boiling point of -161.6°C, we need to find the amount of heat required. The amount of heat required is the enthalpy of vaporization multiplied by the mass of the substance.

The enthalpy of vaporization of methane, Ch4, is 8.9 kJ/mol. The mass of methane, Ch4, is 150 g.

So, the amount of heat required to vaporize 150 g of methane, Ch4, at its boiling point is:

8.9 kJ/mol * 150 g = 123.5 kJ

Therefore, 123.5 kJ of heat is required to vaporize 150 g of methane, Ch4, at its boiling point.  

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Investigators isolate a crystalline metabolite from the urine of a rabbit that has been fed a straight-chain fatty acid containing a terminal phenyl group. The addition of 22.2 mL of 0.100 M NaOH completely neutralized a 302 mg sample of the metabolite in aqueous solution. What is the probable molecular weight of the metabolite? M_r: ___________ What is the name of the metabolite? 1. phenylpropionic acid 2. phenylacetic acid 3. phenylbutyric acid 4. benzoic acid Did the straight-chain fatty acid contain an even or an odd number of methylene (-CH2-) groups? 1. Even: β oxidation only breaks down straight-chain fatty acids with an even number of carbon groups. 2. Even: β oxidation removes two-carbon units and the end product contains a two-carbon unit. 3.Odd: β oxidation removes two-carbon units and the end product contains a three-carbon unit. 4. Odd: during β oxidation, the phenyl group does not count as an equivalent to a terminal methyl

Answers

Considering β oxidation, the correct answer is option 2: Even: β oxidation removes two-carbon units and the end product contains a two-carbon unit.

This is because β oxidation breaks down straight-chain fatty acids, resulting in a two-carbon unit end product, such as the metabolite phenylacetic acid.

The molecular weight (M_r) of the metabolite can be determined using the neutralization reaction and the given data.

Since 22.2 mL of 0.100 M NaOH neutralized the 302 mg sample, we can calculate the moles of NaOH used:

Moles of NaOH = (Volume of NaOH) × (Concentration of NaOH)

                          = 0.0222 L × 0.100 mol/L

                          = 0.00222 mol

Since the reaction is a 1:1 ratio,

moles of metabolite = moles of NaOH = 0.00222 mol.

Now, we can calculate the molecular weight:

Molecular weight = (Mass of metabolite) / (Moles of metabolite)

                             = 302 mg / 0.00222 mol

                             = 136 g/mol

The molecular weight suggests that the metabolite is phenylacetic acid (M_r = 136 g/mol).

The name of the metabolite is option 2: phenylacetic acid.

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a mixture of 100 g of k2cr2o7 and 200 g of water is stirred at 60 degrees c until no more of the salt dissolves. the resulting solution is poured off into a separate beaker, leaving the undissolved solid behind. the solution is now colled to 20 degrees c. what mass of k2cr2o7 crystallizes from the solution during the cooling?

Answers

The mass of K₂Cr₂O₇ that crystallizes from the solution during cooling is approximately 21.6 g.

What mass of K₂Cr₂O₇ crystallizes from the solution during the cooling?

At 60°C, the solubility of K₂Cr₂O₇ in water is approximately 121 g/L.

The amount of K₂Cr₂O₇ that dissolves in 200 g of water at 60°C is:

(121 g/L) x (0.200 L) = 24.2 g

The amount undissolved will be 100 g - 24.2 g = 75.8 g of K₂Cr₂O₇ remains undissolved.

At 20°C, the solubility of K₂Cr₂O₇ in water is approximately 13 g/L.

The amount of K₂Cr₂O₇ that can remain in the solution at 20°C is:

(13 g/L) x (0.200 L) = 2.6 g

The amount  of K₂Cr₂O₇ precipitated upon cooling will be 24.2 g - 2.6 g = 21.6 g

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A galvanic cell using Au+3 Au and Ni2+/Ni was set up at 336 K and the non-standard cell potential was determined to be 1.7030 V

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The non-standard cell potential of the galvanic cell using Au+3 Au and Ni2+/Ni at 336 K is 1.7030 V, which indicates that the concentration of one or both of the ions in the half-cells is not equal to 1 M. By using the Nernst equation, we determined that the concentration of Ni2+ in both half-cells is 1 M, and the concentration of Au+3 in the anode and cathode is 1 M and 1.68 x 10^-9 M, respectively.


First, let's start with a brief explanation of a galvanic cell. A galvanic cell, also known as a voltaic cell, is an electrochemical cell that converts chemical energy into electrical energy. It consists of two half-cells that are connected by a salt bridge or a porous barrier. Each half-cell contains a metal electrode and a solution of its respective ion. In a galvanic cell, one half-cell acts as an anode where oxidation occurs, and the other half-cell acts as a cathode where reduction occurs. The flow of electrons from the anode to the cathode generates an electric current.

Now let's apply this concept to the given galvanic cell using Au+3 Au andIn conclusion, the non-standard cell potential of the galvanic cell using Au+3 Au and Ni2+/Ni at 336 K is 1.7030 V, which indicates that the concentration of one or both of the ions in the half-cells is not equal to 1 M. By using the Nernst equation, we determined that the concentration of Ni2+ in both half-cells is 1 M, and the concentration of Au+3 in the anode and cathode is 1 M and 1.68 x 10^-9 M, respectively.. The chemical equation for the overall reaction in this cell can be written as follows:

Au+3 + Ni → Au + Ni2+

At the anode, Au is oxidized to Au+3:

Au → Au+3 + 3e-

At the cathode, Ni2+ is reduced to Ni:

Ni2+ + 2e- → Ni

The overall cell reaction is the sum of the half-reactions:

Au + Ni2+ → Au+3 + Ni

The non-standard cell potential, Ecell, can be calculated using the Nernst equation:

Ecell = E°cell - (RT/nF)ln(Q)

where E°cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred in the balanced chemical equation, F is the Faraday constant, and Q is the reaction quotient.

Since the cell potential given is non-standard, we can assume that the reaction quotient is not equal to 1. We can use the concentrations of the species in the half-cells to calculate Q. The given equation shows that the concentration of Au+3 in the anode is equal to the concentration of Ni2+ in the cathode. Therefore, we can assume that the concentrations of both ions are equal, and we can use the concentration of either ion in the calculation. Let's use the concentration of Ni2+:

Q = [Au+3]/[Ni2+] = 1

The standard reduction potentials for the half-reactions can be found in a table of standard reduction potentials. The reduction potential for Au+3 + 3e- → Au is +1.498 V, and the reduction potential for Ni2+ + 2e- → Ni is -0.257 V. Since we want the oxidation potential for Au → Au+3, we need to reverse the reduction potential for Au+3:

Au+3 → Au + 3e- E° = -1.498 V

The standard cell potential, E°cell, can be calculated as the difference between the reduction potential of the cathode and the oxidation potential of the anode:

E°cell = E°cathode - E°anode

E°cell = -0.257 V - (-1.498 V)

E°cell = 1.241 V

Now we can use the Nernst equation to calculate the non-standard cell potential:

Ecell = E°cell - (RT/nF)ln(Q)

Ecell = 1.241 V - (0.0083145 x 336 / (2 x 96485))ln(1)

Ecell = 1.241 V

However, the given non-standard cell potential is 1.7030 V, which is higher than the calculated value of 1.241 V. This indicates that the concentration of one or both of the ions in the half-cells is not equal to 1 M. To determine the actual concentrations, we need to use the given non-standard cell potential and the Nernst equation:

Ecell = E°cell - (RT/nF)ln(Q)

1.7030 V = 1.241 V - (0.0083145 x 336 / (2 x 96485))ln(Q)

ln(Q) = -20.162

Q = e^-20.162

Q = 1.68 x 10^-9

Since we know that the concentration of Au+3 in the anode is equal to the concentration of Ni2+ in the cathode, we can use the concentration of Ni2+ to calculate the concentrations of both ions:

Q = [Au+3]/[Ni2+] = 1.68 x 10^-9

Let x be the concentration of Ni2+ in both half-cells. Then the concentration of Au+3 in the anode is x, and the concentration of Au+3 in the cathode is 1.68 x 10^-9 x. Therefore, we can write the equation for the reaction quotient as:

Q = [1.68 x 10^-9 x] / x = 1.68 x 10^-9

Simplifying this equation, we get:

x = 1 M

Therefore, the concentration of Ni2+ in both half-cells is 1 M, and the concentration of Au+3 in the anode is also 1 M. The concentration of Au+3 in the cathode is 1.68 x 10^-9 M.

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what does it mean for the cell potential to be negative or zero

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A negative or zero cell potential indicates that the reaction is not spontaneous or has reached equilibrium, respectively.

Cell potential, measured in volts, represents the difference in electrical potential between the two half-cells of an electrochemical cell.

A positive cell potential indicates a spontaneous reaction, while a negative value shows that the reaction is non-spontaneous and will not occur without external energy input.

When the cell potential is zero, the reaction has reached equilibrium, meaning there is no net change in the concentrations of reactants and products.



Summary: Negative cell potential means the reaction is non-spontaneous, and a zero cell potential signifies equilibrium.

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what is the oxidation state of iron in k 3 [ f e ( c 2 o 4 2 − ) 3 ] ⋅ 3 h 2 o ?

Answers

The oxidation state of iron in [tex]K_3[Fe(C_2O_4)_3].3H_2O[/tex] is +3.

The given compound is potassium tris(oxalate)ferrate(III) trihydrate, represented as[tex]K_3[Fe(C_2O_4)_3].3H_2O[/tex]. In this complex, the oxalate ion [tex](C_2O_4)_2-[/tex] is a bidentate ligand that can coordinate with the iron(III) ion [tex](Fe_3+)[/tex] through two oxygen atoms in each ligand. The oxalate ion has a charge of -2, so each ligand contributes two electrons to the iron atom through the oxygen atoms, giving the iron atom a total of 6 electrons.

Since the oxalate ligand is negatively charged and each ligand is bidentate, the overall charge of the complex is -3. Thus, the oxidation state of iron in the complex must be +3, as it has to balance the -3 charge of the oxalate ligands.

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a 15.0- l l scuba diving tank contains a helium-oxygen (heliox) mixture made up of 25.7 g g of he h e and 4.29 g g of o2 o 2 at 298 k k . Calculate the mole fraction of each component in the mixture.Express your answers separated by a comma.

Answers

The mole fraction of helium is 0.981 and the mole fraction of oxygen is 0.019.

To calculate the mole fraction of helium and oxygen in the mixture, we need to first determine the moles of each component present:

moles of He = 25.7 g / 4.003 g/mol = 6.42 mol

moles of O2 = 4.29 g / 31.999 g/mol = 0.134 mol

The total moles of the mixture is:

total moles = 6.42 mol + 0.134 mol = 6.55 mol

The mole fraction of helium is:

mole fraction of He = moles of He / total moles = 6.42 mol / 6.55 mol = 0.981

The mole fraction of oxygen is:

mole fraction of O2 = moles of O2 / total moles = 0.134 mol / 6.55 mol = 0.019

Therefore, the mole fraction of helium is 0.981 and the mole fraction of oxygen is 0.019.

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Write and balance the following single
replacement reaction.
f) Cr + H₂PO4 -
(HINT: Use Cr³+ )

Answers

In this replacement reaction, Cr displaces H from H₂PO₄ in a chemical reaction. It is an illustration of a redox process, in which oxidation and reduction take place at the same time.

The single replacement reaction between Cr and H₂PO₄ can be represented as follows:

Cr + H₂PO₄ → CrPO₄ + H₂

In this process, H₂PO₄ is reduced to H₂ gas, Cr₃⁺ and PO₄³⁻ ions combine to create CrPO₄, and Cr is oxidized from its elemental state to a Cr³⁺ ion.

In this replacement reaction, the left side has one Cr atom and one H₂PO₄ molecule, whereas the right side has one CrPO₄ molecule and one H₂ molecule.

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Grams to moles: convert 16 feet into meters using conversion factor: 1 meter =3 feet.

6. 7. 8.

Answers

Okay, let's solve this step-by-step:

1 foot = 0.305 meters (conversion factor)

16 feet = 16 * 0.305 meters

= 4.848 meters

So, 16 feet = 4.848 meters

a procedural chanin in this experiment would be required if a student wanted to determine the acidity of tomato juice by titraing a juice sample with naoh solution. briefly explain

Answers

A procedural chain would be required in this experiment because it involves multiple steps that must be followed in a specific order to achieve accurate results. The student would need to first prepare the tomato juice sample, followed by adding a specific indicator to the sample to help visualize the endpoint of the titration.

Next, the student would need to carefully add the standardized NaOH solution to the sample while continuously stirring until the endpoint is reached. Finally, the student would need to record the volume of NaOH solution used to neutralize the sample, which can be used to calculate the acidity of the tomato juice. A procedural chain ensures that each step is carried out correctly and in the right order to obtain reliable and accurate results.


To determine the acidity of tomato juice by titrating a juice sample with NaOH solution, a procedural chain is required. First, prepare the tomato juice sample by filtering out any pulp or seeds. Next, measure a specific volume of the sample using a burette or pipette and place it into a conical flask. Add a few drops of a suitable indicator, such as phenolphthalein, which changes color at the endpoint of the titration. Then, fill a burette with the standardized NaOH solution. Slowly add the NaOH solution to the tomato juice while stirring until the endpoint is reached, indicated by a color change. Record the volume of NaOH used, and use the volume and concentration to calculate the acidity of the tomato juice.

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a buffer solution is made by adding 30.8 g of sodium acetate (NaCH3COO, molar mass= 82.03g/mol) to 250mL of 1.0M acetic acid(CH3COOH, Ka+ 1.8x10^-5). Assume no volume change. what is the pH of the buffer solution?

Answers

The pH of the buffer solution is approximately 4.94 because we need to use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution. Additionally, we need to use some stoichiometry to determine the number of moles of sodium acetate added to the solution.

First, let's use the stoichiometry to determine the number of moles of sodium acetate added to the solution. We know that the mass of sodium acetate added is 30.8 g, and the molar mass of sodium acetate is 82.03 g/mol. So, we can calculate the number of moles of sodium acetate as follows:

moles of NaCH3COO = mass of NaCH3COO / molar mass of NaCH3COO
moles of NaCH3COO = 30.8 g / 82.03 g/mol
moles of NaCH3COO = 0.3757 mol

Next, let's use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution. The Henderson-Hasselbalch equation is:

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

where pKa is the dissociation constant of the weak acid, [A-] is the concentration of the conjugate base (sodium acetate), and [HA] is the concentration of the weak acid (acetic acid).

We know that the concentration of acetic acid is 1.0 M, and the dissociation constant (Ka) is 1.8x10^-5. Therefore, the pKa of acetic acid is:

pKa = -log(Ka)
pKa = -log(1.8x10^-5)
pKa = 4.74

We also know that the concentration of sodium acetate is:

concentration of NaCH3COO = moles of NaCH3COO / volume of solution
concentration of NaCH3COO = 0.3757 mol / 0.250 L
concentration of NaCH3COO = 1.503 M

Now, we can substitute these values into the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])
pH = 4.74 + log(1.503/1.0)
pH = 4.74 + 0.176
pH = 4.92

Therefore, the pH of the buffer solution is 4.92.

but it was necessary to show all the steps involved in the calculation.
To determine the pH of the buffer solution, we will use the Henderson-Hasselbalch equation:

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

First, we need to calculate the pKa from the Ka value:
pKa = -log(Ka) = -log(1.8x10^-5) = 4.74

Next, we need to find the moles of sodium acetate (A-) and acetic acid (HA) in the solution:

Moles of sodium acetate = (30.8 g) / (82.03 g/mol) = 0.375 mol

Since the solution is 250 mL of 1.0 M acetic acid:
Moles of acetic acid = (1.0 mol/L) * (0.250 L) = 0.250 mol

Now we can plug these values into the Henderson-Hasselbalch equation:

pH = 4.74 + log([0.375]/[0.250]) = 4.74 + log(1.5) ≈ 4.94

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what complex ion can form between silver ions and thiosulfate ions?Select the correct answer below: a. [Ag(S2O3)3]^3–b. [Ag(S2O3)2]^3–c. [AgS2O3]^3–d. Ag2S2O3

Answers

When thiosulfate ions (S2O3^2-) are added to a solution of silver ions (Ag+), they can form a complex ion due to the presence of the lone pair of electrons on the sulfur atoms.

The thiosulfate ion has a polydentate ligand property, which means that it can form a complex ion by binding to the metal ion at multiple sites.

The complex ion that can form between silver ions and thiosulfate ions is [Ag(S2O3)2]^3-. This complex ion contains two thiosulfate ions bonded to a central silver ion, and it has a net charge of 3-. The thiosulfate ions in the complex are coordinated to the silver ion through their sulfur atoms, leaving their oxygen atoms free to interact with the surrounding solvent molecules.

The formation of this complex ion can be attributed to the soft-soft interaction between silver ions and sulfur atoms in thiosulfate ions. Silver ions have a soft character due to their low electronegativity, and they tend to interact favorably with soft bases such as sulfur. Thiosulfate ions, on the other hand, have a soft sulfur atom, which can form a bond with silver ions. This results in the formation of a stable complex ion between the two species.

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________ is isoelectronic with scandium.V2+Ca2+V2-VV3+

Answers

Isoelectronic species are atoms, ions or molecules that have the same number of electrons, even if their atomic numbers and charges are different.

In order to determine which of the given species is isoelectronic with scandium (Sc), we need to determine the electron configuration of Sc.

Sc has an atomic number of 21, so its electron configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 3d1

Among the given species, V2+ has 21 electrons, the same as scandium, so it is isoelectronic with scandium. V2+ has lost two electrons compared to the neutral vanadium atom (V), so its electron configuration is: 1s2 2s2 2p6 3s2 3p6 3d3

Ca2+ has 18 electrons, so it is not isoelectronic with scandium. Its electron configuration is: 1s2 2s2 2p6 3s2 3p6

V2- has 23 electrons, so it is not isoelectronic with scandium. Its electron configuration is: 1s2 2s2 2p6 3s2 3p6 3d5 4s2

VV3+ has 18 electrons, so it is not isoelectronic with scandium. Its electron configuration is: 1s2 2s2 2p6 3s2 3p6 3d2.

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Use this information to solve the following problems: 145
grams of sodium chloride are added to 900 grams of water.
Percent by mass?

Answers

The percent by mass of sodium chloride in the solution is approximately 13.88%.

To calculate the percent by mass of sodium chloride in the solution, we need to determine the mass of sodium chloride and the total mass of the solution.

Given:

Mass of sodium chloride = 145 grams

Mass of water = 900 grams

The total mass of the solution is the sum of the mass of sodium chloride and the mass of water:

Total mass of the solution = Mass of sodium chloride + Mass of water

Total mass of the solution = 145 grams + 900 grams

Total mass of the solution = 1045 grams

Now, we can calculate the percent by mass of sodium chloride using the formula:

Percent by mass = (Mass of sodium chloride / Total mass of the solution) * 100%

Plugging in the values:

Percent by mass = (145 grams / 1045 grams) * 100%

Percent by mass ≈ 13.88%

This means that for every 100 grams of the solution, approximately 13.88 grams is sodium chloride. The percent by mass is a way to express the concentration of a solute in a solution. In this case, it represents the ratio of the mass of sodium chloride to the total mass of the solution.

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write separate balanced equations for the neutralization reaction of hydrochhloric acid in part b

Answers

The balanced equation for the neutralization reaction of hydrochloric acid with sodium hydroxide is HCl + NaOH → NaCl + H2O. This equation represents the formation of salt and water from the combination of acid and base.

In part B, the neutralization reaction involves hydrochloric acid reacting with a base, sodium hydroxide, to form salt and water. The balanced chemical equation for this reaction is as follows:
HCl + NaOH → NaCl + H2O
This equation represents the reaction between one molecule of hydrochloric acid and one molecule of sodium hydroxide, resulting in one molecule of salt (sodium chloride) and one molecule of water.
It is important to note that this reaction is an example of an acid-base neutralization reaction, in which an acid and a base react to form a salt and water. In this case, hydrochloric acid is the acid and sodium hydroxide is the base. When they react, they cancel each other out and form a neutral salt and water.
In summary, the balanced equation for the neutralization reaction of hydrochloric acid with sodium hydroxide is HCl + NaOH → NaCl + H2O. This equation represents the formation of salt and water from the combination of acid and base.

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polonium-214 decays in part by a series of steps in which one alpha-particle and two beta-particles are released. which nuclide results from this series of decays?

Answers

Polonium-214 decays in part by the series of steps in which one alpha-particle as well as two beta-particles are released. The nuclide that results from this series of decays is polonium-210.

Polonium-214 undergoes alpha decay to become lead-210, which is the daughter nuclide of the first decay. Lead-210 then undergoes beta decay to become bismuth-210, which is the daughter nuclide of the second decay. Finally, bismuth-210 undergoes beta decay to become polonium-210, which is the final daughter nuclide of the series. Therefore, the nuclide that results from this series of decays is polonium-210.

Polonium-210 is a radioactive element that undergoes alpha decay. During alpha decay, the atomic nucleus of the polonium-210 atom loses two protons and two neutrons, resulting in the emission of an alpha particle (helium nucleus) from the nucleus.

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The dry solute has a mass of 0.086 g.
0.113 g of water evaporated while heating.
What is the concentration in
g solute / mL solvent at 30.1 °C?
Remember: 1 g = 1 mL for water
A. 0.88 g/mL
C. 0.0860 g/mL
B. 0.113 g/mL
D. 0.761 g/mL

Answers

Answer: D. 0.761

Explanation:

Fission of uranium-235 produces energy and _____.

A. Heavier isotopes of uranium

B. Lighter isotopes of uranium

C. Isotopes of smaller elements

D. Isotopes of larger elements

(I looked this up and didn't find an answer so any help is appreciated)

Answers

B. Lighter isotopes of uranium

an element crystallizes in the face-centered cubic unit cell. the volume of an atom of the element is 1.05x107 pm3, and the density of the element is 22.55g/cm3. calculate the molar mass of the element.

Answers

The molar mass of the element is 285.6 g/mol.

Density = mass/volume

22.55 g/cm³ = molar mass/(1.05 x 10⁷ pm³)

1 cm = 10⁸ pm (the conversion factor)

molar mass = 22.55 g/cm³ x (1.05 x 10⁷ pm³/ 10⁸ pm³) = 2.366 x 10² g/mol

The volume of the unit cell can be calculated using the edge length (a) as follows:

The volume of fcc unit cell = a³

1.05 x 10⁷ pm³ = a³

a = (1.05 x 10⁷ pm³)^(1/3) = 2.833 pm

The atomic radius (r) of the element can be calculated from the edge length (a) of the unit cell as follows:

r = a/(2√2)

r = (2.833 pm)/(2√2) = 1.994 pm

Using the atomic radius, we can estimate the atomic mass of the element using the periodic table:

Atomic mass = 4/3 x π x (1.994 pm)³ x (1.6605 x 10⁻²⁴ g/pm³)

Atomic mass = 4.754 x 10⁻²³ g

Finally, we can calculate the molar mass of the element:

Molar mass = Atomic mass x Avogadro's number

Molar mass = (4.754 x 10⁻²³ g) x (6.022 x 10²³ mol⁻¹)

Molar mass = 285.6 g/mol

Molar mass is a term used in chemistry to refer to the mass of one mole of a substance. A mole is a unit of measurement that is used to quantify a substance's amount in terms of the number of particles (atoms, molecules, ions, etc.). Molar mass is expressed in grams per mole (g/mol) and is calculated by summing the atomic masses of all the atoms present in one molecule of the substance.

Molar mass plays a crucial role in many aspects of chemistry, including stoichiometry, which involves the quantitative relationship between reactants and products in chemical reactions. By knowing the molar mass of a substance, one can calculate the number of moles of the substance present in a given amount of the substance. Additionally, molar mass is used to convert between mass and moles, which is important in many chemical calculations.

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if 3 moles of kcl was used to produce kno3. if % yield is 34%, what is the actual yield(in moles)? agno3 kcl--> agcl kno3

Answers

The actual yield in moles of [tex]KClO_3[/tex]  is 3 moles.  

The percentage yield of 34% to an actual yield in moles, you can use the following formula:

Actual yield in moles = (Percentage yield * moles of KCl) / (moles of  [tex]KClO_3[/tex]  * 100%)

Since you know that 3 moles of KCl were used to produce 3.4 moles of  [tex]KClO_3[/tex] , you can rearrange the equation to solve for the percentage yield:

Percentage yield = (moles of  [tex]KClO_3[/tex]   / 3.4 moles of KCl) * 100%

Now you can plug in the values you know to solve for the percentage yield:

Percentage yield = (3.4 / 3) * 100%

Percentage yield = 106.7%

Since the percentage yield is greater than 100%, this means the actual yield in moles of  [tex]KClO_3[/tex]  is greater than the amount of KCl used to produce it. Therefore, the actual yield in moles of  [tex]KClO_3[/tex]  is equal to the amount of KCl used to produce it, which is 3 moles.

The actual yield in moles of [tex]KClO_3[/tex]   is:

3 moles

Therefore, the actual yield in moles of [tex]KClO_3[/tex] is 3 moles.  

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identify the compound with covalent bonds. a) ch4 b) ne c) kbr d) mg e) nacl

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The compound with covalent bonds among the options provided is (a) CH4, also known as methane. Covalent bonds are formed when atoms share electrons to complete their outer electron shells, leading to the creation of a compound. Methane is an example of a covalent compound, as it consists of a carbon atom sharing electrons with four hydrogen atoms. This electron-sharing results in a stable molecule with a filled outer electron shell for each atom involved.

In contrast, the other options are not covalent compounds:
- (b) Ne: Neon is a noble gas with a stable electron configuration, so it does not form bonds with other elements.
- (c) KBr: Potassium bromide is an ionic compound formed from the transfer of an electron from potassium to bromine.
- (d) Mg: Magnesium is an element, not a compound, and as such does not have any covalent bonds within itself.
- (e) NaCl: Sodium chloride, commonly known as table salt, is an ionic compound formed from the transfer of an electron from sodium to chlorine.

In summary, among the given choices, methane (CH4) is the compound that exhibits covalent bonding.

a) CH4(Methane)..Covalent bonds are formed when two or more nonmetals share electrons. In the given options, CH4 (methane) is the compound that consists of covalent bonds.


It is composed of one carbon atom bonded to four hydrogen atoms. Carbon and hydrogen are both nonmetals and share electrons to form covalent bonds. Methane is a common example of a covalent compound and is often used as a representative molecule to explain covalent bonding.
A covalent bond is a chemical bond formed by the sharing of electrons between atoms. Unlike ionic bonds, where there is a transfer of electrons from one atom to another, covalent bonds involve the mutual sharing of electrons. This sharing allows atoms to achieve a more stable electron configuration, typically by filling their outermost electron shell. Covalent compounds tend to have lower melting and boiling points compared to ionic compounds and are often found in the gaseous or liquid state at room temperature.

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which one of the following alcohols will be soluble in water? view available hint(s) for part a 3-hexanol 2-propanol 2-butanol 2-pentanol

Answers

Answer: 2-propanol

Explanation:

Of the four alcohols listed (3-hexanol, 2-propanol, 2-butanol, 2-pentanol), only 2-propanol is expected to be soluble in water.

This is because 2-propanol is a small molecule with a hydroxyl (-OH) group that is highly polar. It can form hydrogen bonds with water molecules, which helps it to dissolve.

Therefore, 3-hexanol, 2-butanol, and 2-pentanol are not expected to be very soluble in water. They may dissolve to some extent due to their polar -OH group, but their nonpolar hydrocarbon chains will make them less soluble overall.

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how many moles of hydrogen would be need to completely hydrogenate two moles of arachidonic acid (shown below) in the presence of a ni catalyst? ch3(ch2)4(ch

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To determine how many moles of hydrogen would be needed to completely hydrogenate two moles of arachidonic acid in the presence of a Ni catalyst, we first need to determine the number of double bonds in arachidonic acid. Arachidonic acid has four double bonds, so we need to add four moles of hydrogen to completely hydrogenate it.

One mole of hydrogen gas (H2) contains two moles of hydrogen atoms, so we need eight moles of hydrogen gas to add four moles of hydrogen atoms to arachidonic acid. Therefore, to completely hydrogenate two moles of arachidonic acid in the presence of a ni catalyst, we would need 16 moles of hydrogen gas.

To completely hydrogenate two moles of arachidonic acid, you need to determine the number of double bonds present in the molecule. Arachidonic acid has the chemical formula CH₃(CH₂)₄(CH=CHCH₂)₄(CH₂)₂COOH, which contains 4 double bonds.
In order to hydrogenate one double bond, you need 1 mole of hydrogen (H₂). Therefore, for each mole of arachidonic acid, you'll need 4 moles of hydrogen.  Since you have two moles of arachidonic acid, you will need 2 x 4 = 8 moles of hydrogen to completely hydrogenate both moles of arachidonic acid in the presence of a Ni catalyst.

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what is the ph of a solution made by adding 0.40 g of naoh to 100 ml of a 0.4 m solution of ch3cooh? is it a buffering solution?

Answers

The pH of the solution is 4.55, which is made by adding 0.40 g of NaOH to 100 ml of a 0.4 m solution of CH₃COOH. This solution is not a buffering solution.

To determine the pH of the solution, we need to first calculate the moles of CH₃COOH present in 100 ml of the 0.4 M solution;

moles of CH₃COOH = M x V = 0.4 x 0.1 = 0.04 mol

Next, we need to calculate number of moles of NaOH added to the solution;

moles of NaOH = mass / molar mass = 0.40 / 40.00 = 0.01 mol

The reaction between CH₃COOH and NaOH is;

CH₃COOH + NaOH → CH₃COO⁻ Na⁺ + H₂O

The number of moles of CH₃COOH that reacted with NaOH is equal to the number of moles of NaOH added to the solution, which is 0.01 mol.

The remaining moles of CH₃COOH in solution is;

0.04 mol - 0.01 mol=0.03 mol

To find the concentration of CH₃COO⁻ in solution, we need to divide the moles by the total volume of the solution;

concentration of CH₃COO⁻ = moles / volume = 0.03 mol / 0.1 L = 0.30 M

The pH of the solution can be calculated using the Henderson-Hasselbalch equation;

pH = pKa + log([A⁻]/[HA])

where pKa is the dissociation constant of acetic acid, [A⁻] is the concentration of the acetate ion, and [HA] is the concentration of undissociated acetic acid.

The pKa of acetic acid is 4.76.

[A⁻] = 0.30 M

[HA] = 0.04 mol / 0.1 L = 0.40 M

pH = 4.76 + log(0.30/0.40) = 4.55

Therefore, the pH of the solution will be 4.55.

This solution is not a buffering solution as its pH is not within one unit of the pKa of the acid.

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if a sailor exhales 137.5 ml of co2 per minute at 23.5°c and 0.830 atm, what mass of sodium peroxide is needed per sailor in a 24-hour period?

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Approximately 779.33 grams of sodium peroxide would be needed per sailor in a 24-hour period, considering the given rate of CO2 exhaled by the sailor.

To calculate the mass of sodium peroxide needed per sailor in a 24-hour period, we need to determine the number of moles of CO2 produced by the sailor and then use the stoichiometry of the reaction between sodium peroxide (Na2O2) and CO2 to find the corresponding amount of sodium peroxide.

First, let's calculate the number of moles of CO2 produced per minute by the sailor using the ideal gas law:

PV = nRT

Where:

P = pressure (0.830 atm)

V = volume (137.5 mL or 0.1375 L)

n = number of moles of CO2 (to be determined)

R = ideal gas constant (0.0821 L·atm/(K·mol))

T = temperature (23.5°C or 296.65 K)

Rearranging the equation, we have:

n = PV / RT

n = (0.830 atm) × (0.1375 L) / (0.0821 L·atm/(K·mol)) × (296.65 K)

n ≈ 0.00697 moles CO2/min

To find the total moles of CO2 produced in a 24-hour period, we multiply this value by the number of minutes in 24 hours:

Total moles of CO2 = (0.00697 moles CO2/min) × (24 hours × 60 minutes/hour)

Total moles of CO2 ≈ 10.018 moles CO2

Now, let's determine the stoichiometry of the reaction between sodium peroxide (Na2O2) and CO2:

2 Na2O2 + 2 CO2 → 2 Na2CO3 + O2

According to the balanced equation, two moles of sodium peroxide react with two moles of CO2.

Therefore, the number of moles of sodium peroxide required is equal to the number of moles of CO2 produced:

Moles of Na2O2 = Total moles of CO2 ≈ 10.018 moles CO2

Finally, to calculate the mass of sodium peroxide, we need to multiply the moles of Na2O2 by its molar mass, which is 77.98 g/mol:

Mass of Na2O2 = Moles of Na2O2 × Molar mass of Na2O2

Mass of Na2O2 ≈ 10.018 moles × 77.98 g/mol

Mass of Na2O2 ≈ 779.33 g

Therefore, approximately 779.33 grams of sodium peroxide would be needed per sailor in a 24-hour period, considering the given rate of CO2 exhaled by the sailor.

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Which term is defined as the amount of time it takes for 50% of a sample of radioactive nuclei to decay?

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The term that is used to defined the amount of time it takes for 50% of a sample of radioactive nuclei to decay is half life.

Generally, the half-life, in radioactivity is defined as the interval of time that is required for one-half of the atomic nuclei of a radioactive sample to decay (that change spontaneously into other nuclear species by emitting particles and energy), or equivalently, the time interval which is required for the number of disintegrations per second of a radioactive.

The half-life of a radioactive isotope is also defined as the amount of time it takes for one-half of the radioactive isotope to decay. The half-life of a specific radioactive isotope stays constant and it remains unaffected by conditions and is independent of the initial amount of that isotope.

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What must be true about the heat flow in an endothermic reaction?
Responses

Heat is absorbed by the system and released by the surroundings
Heat is released by the system and absorbed by the surroundings
Cold is absorbed by the system and released by the surroundings
Cold is released by the system and absorbed by the surroundings

Answers

Heat is absorbed by the system and released by the surroundings.

What is an endothermic reaction?

An endothermic process is any thermodynamic process that causes an increase in the system's enthalpy H. A closed system often absorbs thermal energy from its surroundings, resulting in heat transfer into the system.

An exothermic reaction is a chemical reaction that produces energy in the form of light or heat. This is the inverse of an endothermic process. In chemical terms, this is: products + reactants + energy

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the value in electron volt per atom which represent the first ionisation of oxygen and nitrogen atom are:

Answers

N and 0 atoms have first ionization potentials of 14.6 and 13.6 eV, respectively.

Option A is correct.

Because of its electronic configuration, which is half-full, nitrogen has a greater ionization energy than oxygen. After losing an electron, oxygen adopts a half-full electronic configuration, resulting in a lower ionization energy than nitrogen.  Oxygen has a lower first ionization energy than nitrogen.

What is the first ionization order?

The first ionization energy fluctuates uniformly throughout the periodic table. Over time, the ionization energy increases from left to right and decreases from top to bottom in groups. Accordingly, helium has the biggest first ionization energy, while francium has quite possibly of the most minimal.

Incomplete question :

The first ionization potential in electron volts of nitrogen and oxygen atoms respectively are given by

A. 14.6, 13.6

B. 13.6, 14.6

C. 13.6, 13.6

D . 14.6, 14.6

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suggest two chemical tests that one could use to distinguish between each the following pairs of compounds. explain all expected observations. 3-bromo-3-methyl-1-pentene and bromobenzene

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One chemical test that can be used to distinguish between 3-bromo-3-methyl-1-pentene and bromobenzene is a silver nitrate test. In this test, 3-bromo-3-methyl-1-pentene will give a white precipitate with silver nitrate due to the presence of the alkene functional group, while bromobenzene will not react with silver nitrate. Another test that can be used is a sodium bicarbonate test. 3-bromo-3-methyl-1-pentene will give effervescence with sodium bicarbonate due to the presence of the acidic proton on the carbon adjacent to the bromine atom, while bromobenzene will not react with sodium bicarbonate. These tests can be used to differentiate between the two compounds based on their chemical properties.

To distinguish between 3-bromo-3-methyl-1-pentene and bromobenzene, you can perform the following two chemical tests:
1. Bromine Water Test: When 3-bromo-3-methyl-1-pentene reacts with bromine water, it undergoes an addition reaction, leading to the decolorization of the reddish-brown bromine water. On the other hand, bromobenzene does not react with bromine water due to the absence of an alkene double bond, and the color remains unchanged.

2. Baeyer's Test: In this test, 3-bromo-3-methyl-1-pentene reacts with potassium permanganate (KMnO4) solution, causing the purple color to disappear, indicating the presence of an alkene. Bromobenzene will not react with KMnO4, and the purple color will persist.
These tests help differentiate between the two compounds based on their reactions and observed changes in color.

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