A 1.000 L vessel is filled with 1.000 mole of N2,
2.000 moles of H2, and 3.000 moles of NH3.
When the reaction
N2(g) + 3 H2(g) ⇀↽ 2 NH3(g)
comes to equilibrium, it is observed that the
concentration of NH3 is 2.07 moles/L. What
is the numerical value of the equilibrium constant Kc?

Answers

Answer 1

[tex]NH_{3}[/tex]The equilibrium constant (Kc) for the given reaction is 0.025.

What is Equilibrium?

Chemical equilibrium is described by the equilibrium constant, which is a numerical value that quantitatively expresses the ratio of concentrations (or partial pressures) of reactants and products at equilibrium. The equilibrium constant is denoted by the symbol K, and its value depends on the specific chemical reaction and the temperature at which the reaction occurs.

Let x be the change in the concentration of [tex]NH_{3}[/tex]at equilibrium, then the equilibrium concentrations of [tex]N_{2}[/tex], [tex]H_{2}[/tex]and [tex]NH_{3}[/tex] are:

[[tex]N_{2}[/tex]]eq = (1.000 - x) M

[[tex]H_{2}[/tex]]eq = (2.000 - 3x) M

[[tex]NH_{3}[/tex]eq = (3.000 + 2x) M

Substituting these equilibrium concentrations into the equilibrium constant expression, we get:

At equilibrium, the concentration of [tex]NH_{3}[/tex] is 2.07 M, so we have:

[[tex]NH_{3}[/tex]]eq = 2.07 M

Substituting this value into the equilibrium concentrations, we get:

(3.000 + 2x) = 2.07

Solving for x, we get:

x = -0.465

Substituting this value of x into the equilibrium concentrations and into the equilibrium constant expression, we get:

[[tex]N_{2}[/tex]eq = (1.000 - x) M = 1.465 M

[[tex]H_{2}[/tex]]eq = (2.000 - 3x) M = 3.395 M

[[tex]NH_{3}[/tex]]eq = (3.000 + 2x) M = 1.170 M

Kc = 2.07 / (1 * [tex]3^{3}[/tex])

Kc = 0.025

So, the equilibrium constant (Kc) for the given reaction is 0.025.

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Related Questions

a solution was made by a 1/8 dilution of the stock solution followed by a 1/4 dilution of the resultant solution. what is the dilution of the final solution? question 9 options: 1/32 dilution 1/2 dilution 1/10 dilution 1/12 dilution

Answers

The dilution of the final solution is 1/32 dilution. Option (a) is the correct answer.

To work out the general weakening of the last arrangement, we really want to duplicate the weakening elements of each step.

The primary weakening is a 1/8 weakening, and that implies that the centralization of the arrangement is diminished by a component of 1/8. Hence, the resultant arrangement is 1/8 of the first focus.

The subsequent weakening is a 1/4 weakening, and that implies that the convergence of the resultant arrangement is decreased by an element of 1/4.

To find the general weakening, we increase the weakening variables of each step:

1/8 x 1/4 = 1/32

Subsequently, the weakening of the last arrangement is 1/32 weakening. Choice (a) is the right response.

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the gain or loss of electrons from an atom results in the formation of a (an)

Answers

The formation of ions is an essential process in chemistry and is involved in many chemical reactions and compounds.

Atoms are composed of protons, neutrons, and electrons. The number of protons in an atom determines its atomic number and the element it represents. The electrons in an atom occupy different energy levels or shells, and these electrons participate in chemical reactions. The outermost shell of electrons, called the valence shell, is particularly important in chemical reactions because it determines the chemical properties of the atom.

When an atom gains or loses electrons, it becomes charged and is called an ion. The process of gaining or losing electrons is called ionization. When an atom loses one or more electrons, it becomes a positively charged ion called a cation. Cations have a smaller number of electrons than protons and have a net positive charge. For example, when the element sodium (Na) loses one electron, it becomes a sodium ion (Na+).

On the other hand, when an atom gains one or more electrons, it becomes a negatively charged ion called an anion. Anions have a larger number of electrons than protons and have a net negative charge. For example, when the element chlorine (Cl) gains one electron, it becomes a chloride ion (Cl-).

The formation of ions is a fundamental process in many chemical reactions. Ions can combine with each other to form ionic compounds, which are compounds composed of ions held together by electrostatic forces. For example, sodium ions (Na+) and chloride ions (Cl-) can combine to form sodium chloride (NaCl), which is common table salt.

Overall, the formation of ions is an essential process in chemistry and is involved in many chemical reactions and compounds.

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how many moles of atoms are there in 1.00 lb (454g) of lead

Answers

we need to use the atomic weight of lead to convert the given weight in grams to moles. The atomic weight of lead is 207.2 g/mol.



First, let's convert the given weight in pounds to grams: 1.00 lb = 454 g
Next, let's calculate the number of moles of lead atoms in 454 g of lead: moles of lead atoms = (454 g) / (207.2 g/mol) = 2.19 mol.


Therefore, there are 2.19 moles of lead atoms in 1.00 lb (454g) of lead. To calculate the number of moles of atoms in 1.00 lb (454g) of lead, you need to use the formula: moles = mass (g) / molar mass (g/mol)

The molar mass of lead (Pb) is 207.2 g/mol. Using the given mass of 454g, the calculation is as follows:
moles = 454g / 207.2 g/mol = 2.19 moles
So, there are 2.19 moles of atoms in 1.00 lb (454g) of lead.

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To determine how many moles of atoms are there in 1.00 lb (454g) of lead, you'll need to follow these steps:

Step 1: Convert weight to grams.
1.00 lb of lead is already given as 454g.

Step 2: Find the molar mass of lead.
Lead (Pb) has a molar mass of approximately 207.2 g/mol.

Step 3: Calculate the number of moles.
To find the moles, divide the mass of lead (454g) by its molar mass (207.2 g/mol).

Moles = 454g / 207.2 g/mol

Your answer: There are approximately 2.19 moles of atoms in 1.00 lb (454g) of lead.

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if groundwater contaminant is not visible does that mean it is safe to drink? Explain

Answers

It depends on what you meant by saying not visible. Of it is not visible by using accurate measuring equipment then I think so, but if you mean that all transparent water is drinkable, then no. Think about this. When you put salt in water, you can't see it but it is still there: if you taste the water you can tell that there's salt in there. Let's say that instead of salt there are some bacteria, or some other type of salt which is not appropriate to drink at high levels, such as nitrates. I personally wouldn't recommend drinking from any type.of water unless you are not sure about its purity

which types of lipids would not have their fatty acids completely hydrolyzed by treatment with acid or alkali?

Answers

Answer: Sphingolipids

Explanation: Sphingolipids are a type of lipid that would not have their fatty acids completely hydrolyzed by the treatment with acid or alkali treatment. This is because sphingolipids contain a unique type of fatty acid called a "long chain base" that is attached to the rest of the molecule through an amide bond, rather than an ester bond.

The amide bond is resistant to acid or alkali hydrolysis, so the fatty acid portion of the sphingolipid molecule would remain intact even after treatment with acid or alkali.

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How many moles of h2 can be produced from x grams of mg in magnesium-aluminum alloy? the molar mass of mg is 24. 31 g/mol?

Answers

The number of moles of H₂ that can be produced from x grams of Mg is (x / 24.31)

The balanced chemical equation for the reaction between Mg and HCl is,

Mg + 2HCl → MgCl₂ + H₂

This equation shows that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H₂. Therefore, the number of moles of H₂ that can be produced from x grams of Mg can be calculated as follows:

Calculate the number of moles of Mg in x grams:

Number of moles of Mg = mass of Mg / molar mass of Mg

Number of moles of Mg = x / 24.31

Use the mole ratio between Mg and H₂ to calculate the number of moles of H₂ produced:

Number of moles of H₂ = Number of moles of Mg × (1 mole of H₂ / 1 mole of Mg)

Number of moles of H₂ = (x / 24.31) × (1/1)

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select the best possible answer. does the equilibrium favor the reactants or products in this substitution reaction?

Answers

The correct answer is option B. Equilibrium Favors the Products. Equilibrium is a state of balance in which the concentrations of reactants and products remain constant over time.

The concentrations of the reactants and products do not change in a substitution reaction once it has reached equilibrium.

This indicates that the forward response rate and the reverse reaction rate are equal. In this situation, the reaction favours the production of the products over the reactants since the equilibrium favours the products.

This indicates that the forward reaction is occurring at a faster rate than the reverse reaction.

As a result, the equilibrium will favour the products, and their concentrations will be higher than those of the reactants.

Complete Question:

Select  the best possible answer to this question:

Which of the following best describes the equilibrium of this substitution reaction?

A. Equilibrium favors the reactants

B. Equilibrium favors the products

C. Equilibrium is unaffected

D. Equilibrium is reversed

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you make a stock solution of 2.4831 grams of your unknown acid using a 100.00 ml volumetric flask. then you use 20.00 ml of that stock for a titration, which requires 23.85 ml of 0.108 m naoh to reach the first equivalence point. how many moles of naoh were used to reach the first equivalence point?

Answers

0.2582 moles of NaOH were required to arrive at the initial equivalence point.

The number of moles of NaOH used to reach the first equivalence point can be calculated using the molarity of the base and the volume of it used in the titration.

To do this, the formula M1V1=M2V2 is used, where M1 is the molarity of the base (0.108 M for NaOH), V1 is the volume of the base used (23.85 ml), M2 is the molarity of the acid (unknown), and V2 is the volume of acid used (20 ml).

Therefore, the number of moles of NaOH used to reach the first equivalence point is 0.2582 moles.

In summary, by measuring the amount of NaOH required to reach the first equivalence point and applying the molarity and volume of the acid and base, respectively, the number of moles of NaOH used can be calculated as 0.2582 moles.

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. the two main sources for the increase of carbon dioxide in the atmosphere are . select one:

Answers

Answer:

combustion

respiration by humans

Explanation:

burning of wood leaves release carbon dioxide which is a green house gas and detrimental to the climate

the nadh cofactor has a midpoint potential of -320 mv vs nhe. what fraction of a population of these cofactors would be in the nad form in a ph 7.0 solution with a potential of -300 mv vs nhe? -350 mv vs nhe?

Answers

The fraction of NADH cofactors in the NAD form in a pH 7.0 solution with a potential of -300 mV vs NHE is 0.015. The fraction of NADH cofactors in the NAD form in a pH 7.0 solution with a potential of -350 mV vs NHE is 0.065.

The NADH/NAD couple has a midpoint potential of -320 mV vs NHE. At pH 7.0, the NADH/NAD couple has an Nernst potential of -320 mV. To calculate the fraction of NADH cofactors in the NAD form at a given potential, we use the Nernst equation:

E = E0 - (RT/nF) ln ([NAD]/[NADH])

where E0 is the standard potential (-320 mV), R is the gas constant, T is the temperature, n is the number of electrons transferred (2 for NADH/NAD), F is the Faraday constant, and [NAD]/[NADH] is the ratio of the oxidized to reduced forms of the cofactor.

Solving for [NAD]/[NADH], we get:

[NAD]/[NADH] = e^((E-E0) nF/RT)

Plugging in the values for E and T, and assuming a 1:1 ratio of NADH to NAD, we get:

[NAD]/[NADH] = e^((E-E0) nF/RT) = e^((E-E0)/59.16)

At -300 mV vs NHE, we get:

[NAD]/[NADH] = e^((-300+320)/59.16) = e^(-0.533) = 0.59

So the fraction of NADH cofactors in the NAD form is

0.59/(1+0.59) = 0.015.

At -350 mV vs NHE, we get:

[NAD]/[NADH] = e^((-350+320)/59.16) = e^(-0.495) = 0.61

So the fraction of NADH cofactors in the NAD form is

0.61/(1+0.61) = 0.065.

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how many amperes are required to deposit 0.218 grams of cobalt metal in 626 seconds, from a solution that contains ions.

Answers

To calculate the amperes required to deposit 0.218 grams of cobalt metal in 626 seconds, we need to use Faraday's law of electrolysis, which states that the amount of substance deposited at an electrode is proportional to the quantity of electric charge passed through the electrode.


The formula to calculate the amount of substance deposited is:
mass (in grams) = (current (in amperes) x time (in seconds) x atomic weight) / (number of electrons transferred x Faraday's constant)
For cobalt, the atomic weight is 58.93 g/mol, and the number of electrons transferred during the deposition process is 2 (since cobalt has a +2 oxidation state). Faraday's constant is 96,485 coulombs/mol.
Substituting the values given in the question, we get:
0.218 g = (current x 626 x 58.93) / (2 x 96,485)
Solving for current, we get:
current = (0.218 x 2 x 96,485) / (626 x 58.93)
current = 0.353 amperes
Therefore, 0.353 amperes are required to deposit 0.218 grams of cobalt metal in 626 seconds from a solution that contains cobalt ions.

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a sample of air at 7.50 atm is cooled from 448k to 224k if the volume reamins constant what is the final pressure

Answers

the final pressure of the air sample when cooled from 448 K to 224 K, with constant volume, is approximately 3.75 atm.

The final pressure of the air sample can be determined using the combined gas law, which states that P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature, respectively. Since the volume remains constant, we can use this formula to solve for P2:
P1/T1 = P2/T2
Plugging in the given values, we get:
7.50 atm / 448 K = P2 / 224 K
Simplifying and solving for P2, we get:
P2 = (7.50 atm / 448 K) * 224 K
P2 = 3.75 atm
Therefore, the final pressure of the air sample is 3.75 atm.
We'll be using the Combined Gas Law formula to solve this, but since the volume remains constant, we can simplify it to Gay-Lussac's Law.
Gay-Lussac's Law: P1/T1 = P2/T2
Where:
P1 = initial pressure = 7.50 atm
T1 = initial temperature = 448 K
P2 = final pressure (what we're solving for)
T2 = final temperature = 224 K
Step 1: Rearrange the equation to isolate P2:
P2 = (P1/T1) * T2
Step 2: Plug in the given values:
P2 = (7.50 atm / 448 K) * 224 K
Step 3: Calculate the final pressure:
P2 = (0.0167410714286) * 224 K
P2 ≈ 3.75 atm
So, the final pressure of the air sample when cooled from 448 K to 224 K, with constant volume, is approximately 3.75 atm.

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iodine-123 is a radioactive isotope used to study thyroid gland functions. it decays in a first-order process with a half life of 13.1 h. you receive a 10.00 g sample for some of your experiments, but you have to work fast before it is all gone. calculate the number of hours it will take for 8.40 g of your sample to decay.

Answers

It will take 11.1 hours for 8.40 g of the iodine-123 sample to decay.

The decay of iodine-123 is a first-order process, meaning that the rate of decay is proportional to the amount of iodine-123 present. The half-life of iodine-123 is 13.1 hours, which means that half of the original sample will decay in that time.

To calculate the time it will take for 8.40 g of the sample to decay, we can use the following formula for first-order decay:

ln(Nt/N0) = -kt

where Nt is the amount remaining at time t, N0 is the initial amount, k is the rate constant, and t is the time.

We can rearrange this formula to solve for t:

t = ln(Nt/N0) / (-k)

We know that Nt/N0 = 8.40 g / 10.00 g = 0.84, and we can calculate k from the half-life:

t1/2 = ln(2) / kk = ln(2) / t1/2k = ln(2) / 13.1 hk = 0.0528 h⁻¹

Plugging these values into the formula for t, we get:

t = ln(0.84) / (-0.0528 h⁻¹)t = 11.1 hours

Therefore, it will take 11.1 hours for 8.40 g of the iodine-123 sample to decay.

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suppose of zinc chloride is dissolved in of a aqueous solution of potassium carbonate. calculate the final molarity of zinc cation in the solution. you can assume the volume of the solution doesn't change when the zinc chloride is dissolved in it.

Answers

The final molarity of zinc cation in the solution is 0.0122 M.

Assuming complete dissociation of zinc chloride, we can write the balanced chemical equation as:

[tex]ZnCl_2 (aq) + K_2CO_3 (aq) - > Zn_2+ (aq) + 2K+ (aq) + 2Cl- (aq) + (CO_3) ^{2-} (aq)[/tex]

First, we need to calculate the moles of zinc chloride present in the solution:

moles of ZnCl2 = (0.25 g / 136.30 g/mol) = 0.001833 mol

Since 1 mole of ZnCl2 produces 1 mole of Zn2+, the final molarity of zinc cation in the solution will be:

Molarity of [tex]Zn_{2+[/tex]= moles of [tex]Zn_{2+[/tex]

volume of solution in liters moles of Zn2+ = 0.001833 mol

volume of solution = 0.150 L

Molarity of Zn2+ = 0.001833 mol / 0.150 L = 0.0122 M.

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A buffer contains 0.17 mol of propionic acid (C2H5COOH) and 0.21 mol of sodium propionate (C2H5COONa) in 1.20 L.Part AWhat is the pH of this buffer?Part BWhat is the pH of the buffer after the addition of 0.02 mol of NaOH?Part CWhat is the pH of the buffer after the addition of 0.02 mol of HI?

Answers

The pH of the buffer is 4.992, pH of the buffer after the addition of 0.02 mol of NaOH is 5.086 and pH of the buffer after the addition of 0.02 mol of HI 4.9.

A weak base and its salt are combined with a strong acid to create a basic buffer, which has a basic pH. Aqueous solutions of ammonium hydroxide and ammonium chloride at equal concentrations have a pH of 9.25. These solutions have a pH greater than seven.

Given 0.17 mol propionic acid , 0.21 mol sodium propionate in 1.20 L

We recognize the conjugate acid-base pair  propionic acid (Ka = 1.3 x 10-5  and  pKa = 4.9)

Part A:

[acid] = 0.17 / 1.2 = 0.1417 M

[base] = 0.21/1.2 = 0.175 M

pH of the buffer = 4.9 + log (0.175/0.1417) = 4.9 + 0.092 = 4.992

Part B:

moles acid = 0.17 - 0.02 = 0.15

[acid] = 0.15/1.2 = 0.125 M

moles salt = 0.21 + 0.02 = 0.23

[base] = 0.23/1.2 = 0.1917 M

pH of the buffer = 4.9 + log (0.1917/0.125) = 4.9 + 0.186 = 5.086

Part C:

moles of acid = 0.17 + 0.02 = 0.19

[acid] = 0.19/1.2 = 0.1583

moles of base = 0.21 - 0.02 = 0.19

[base] = 0.19/1.2 = 0.1583

pH of the buffer = 4.9 + log (0.1583/0.1583) = 4.9.

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which term best describes a solution in a typical kitchen that has as much dissolved solute as it can hold? responses

Answers

The term that best describes a solution in a typical kitchen that has as much dissolved solute as it can hold is "saturated solution."

A saturated solution is a solution that contains the maximum amount of solute that can dissolve in a given solvent at a particular temperature and pressure. If more solute is added to a saturated solution, it will not dissolve and will form a separate phase or precipitate.

In a kitchen setting, a common example of a saturated solution is a solution of table salt (sodium chloride) in water. At room temperature, water can dissolve a certain amount of salt, and once this limit is reached, the solution becomes saturated. If more salt is added to the solution, it will not dissolve and will settle at the bottom of the container.

It is important to note that the solubility of a substance can vary depending on factors such as temperature and pressure. Therefore, a solution that is saturated at one temperature or pressure may not be saturated under different conditions.

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if each orange sphere represents 0.010 mol of sulfate ion, how many moles of acid and of base reacted?

Answers

The number of moles of acid and base that react depends on the stoichiometry of the chemical reaction and the amounts of reactants used

Without additional information about the chemical reaction or system being referred to, we cannot determine the number of moles of acid and base that reacted.

If we assume that the orange spheres represent sulfate ions in a specific reaction, then we would need to know the stoichiometry of the reaction to determine the number of moles of acid and base that reacted.

For example, if the reaction involved sulfuric acid ([tex]H_2SO_4[/tex]) and sodium hydroxide (NaOH) and the orange spheres represent sulfate ions ([tex](SO_4)^{2-[/tex]), then the balanced chemical equation would be:

[tex]H_2SO_4 + 2NaOH - > Na_2SO_4 + 2H_2O[/tex]

In this case, we would need to know the amount of sodium hydroxide used to determine the number of moles of acid and base that reacted. If we know the number of orange spheres representing sulfate ions and the amount of sodium hydroxide used, we can determine the moles of acid and base that reacted.

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Nitrates and phosphates are examples of
a. disease-causing agents
b. oxygen-demanding wastes
c. sediment
d. organic chemicals
e. inorganic plant nutrients

Answers

Nitrates and phosphates are examples of inorganic plant nutrients. They are essential elements for plant growth and are commonly found in fertilizers. Option (e)

Nitrates are made up of nitrogen and oxygen and are converted into nitrites by bacteria. They are essential for the production of proteins and nucleic acids in plants. Phosphates, on the other hand, are made up of phosphorous and oxygen and are involved in energy transfer and storage in plants. However, excess nitrates and phosphates can also contribute to environmental problems such as eutrophication, which can lead to oxygen depletion in bodies of water and harm aquatic life. Proper management of fertilizers and waste disposal is important to prevent negative impacts on the environment.

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Nitrates and phosphates are examples of inorganic plant nutrients. They are essential elements for plant growth and are commonly found in fertilizers.

Nitrates are made up of nitrogen and oxygen and are converted into nitrites by bacteria. They are essential for the production of proteins and nucleic acids in plants. Phosphates, on the other hand, are made up of phosphorous and oxygen and are involved in energy transfer and storage in plants. However, excess nitrates and phosphates can also contribute to environmental problems such as eutrophication, which can lead to oxygen depletion in bodies of water and harm aquatic life. Proper management of fertilizers and waste disposal is important to prevent negative impacts on the environment.

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when cuso4.5h2o is dissolved in water what are the major species present in the solution besides the solvent molecules?

Answers

When CuSO4·5H2O (copper(II) sulfate pentahydrate) is dissolved in water, the major species present in the solution, besides the solvent molecules (water), are Cu²⁺ (copper(II) ions) and SO₄²⁻ (sulfate ions).

The dissolution process involves the dissociation of CuSO4·5H2O into its constituent ions:
CuSO4·5H2O → Cu²⁺ + SO₄²⁻ + 5H2O
The water molecules serve as the solvent, and the Cu²⁺ and SO₄²⁻ ions are the solute, forming the solution.

The compound dissociates in water, releasing the Cu2+ and SO42- ions, which become hydrated by water molecules. The five water molecules in the formula unit of the compound (CuSO4·5H2O) become part of the solvent and do not exist as distinct species in the solution.

So, in summary, the major species present in a solution of CuSO4·5H2O in water are Cu2+ cations and SO42- anions, along with water molecules as the solvent.

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When CuSO4.5H2O (copper(II) sulfate pentahydrate) is dissolved in water, the major species present in the solution besides the solvent molecules are Cu²⁺ (copper(II) ions) and SO₄²⁻ (sulfate ions).

When CuSO4·5H2O is dissolved in water, the major species present in the solution besides the solvent molecules are Cu2+ ions and SO42- ions. The Cu2+ ions and SO42- ions come from the dissociation of the CuSO4 compound in water, while the H2O molecules are present as the solvent. The Cu2+ ions and SO42- ions interact with the water molecules through hydration and solvation, respectively, which affects the physical and chemical properties of the solution. The dissolution process can be represented by the following equation:

CuSO4.5H2O (s) → Cu²⁺ (aq) + SO₄²⁻ (aq) + 5H2O (l)

In this equation, CuSO4.5H2O dissociates into its constituent ions, Cu²⁺ and SO₄²⁻, while the water molecules from the pentahydrate become part of the solvent.

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the same gas makes up most of the atmosphere of mars and venus. this gas is: a. water vapor b. carbon dioxide c. ozone d. nitrogen e. ammonia gas

Answers

The gas that makes up most of the atmosphere of Mars and Venus is carbon dioxide (b).

What is the atmosphere of Mars and Venus composed of?


Both planets have atmospheres predominantly composed of carbon dioxide, with Venus having around 96.5% [tex]CO_{2}[/tex] and Mars having about 95% [tex]CO_{2}[/tex]. The combination of geological history, lack of liquid water, and limited biological activity are the main factors that have resulted in carbon dioxide being abundant in the atmospheres of Mars and Venus.

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The gas that makes up most of the atmosphere of Mars and Venus is carbon dioxide. In fact, the atmosphere of Venus is nearly 97% carbon dioxide, while Mars has an atmosphere that is about 95% carbon dioxide.

This is in contrast to Earth, which has an atmosphere that is mostly nitrogen and oxygen, with only a small percentage of carbon dioxide. The high levels of carbon dioxide in the atmospheres of Mars and Venus contribute to their extremely hot temperatures, as the gas traps heat from the sun and creates a greenhouse effect. Additionally, the lack of a strong magnetic field on these planets means that they are more vulnerable to the stripping away of their atmospheres by solar winds. Understanding the composition of the atmospheres of other planets is important for astrobiology and the search for life beyond Earth.

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a buffer solution has 0.750 m h2co3 and 0.650 m hco3−. if 0.020 moles of hcl is added to 275 ml of the buffer solution, what is the ph after the addition? the pka of carbonic acid is 6.37.

Answers

The pH after the addition of the 0.020 moles of HCl is added to 275 ml of the buffer solution is 6.40.

A buffer solution is an acidic or basic aqueous solution made up of a combination of a weak acid and its conjugate base, or vice versa (more specifically, a pH buffer or hydrogen ion buffer). When a modest amount of a strong acid or base is applied to it, the pH hardly changes at all.

A multitude of chemical applications employ buffer solutions to maintain pH at a practically constant value. Numerous biological systems employ buffering to control pH in the natural world.

275mL buffer 1L/1,1000 mL 0.75 mol H2CO3/ 1L Solution = 0.206 mol H2CO3

275 mL buffer 1L/ 1,000 mL 0.65 mol HCO3- / 1L Solution= 0.179 mol HCO3-

pH = 6.37 + log(0.179 mol + 0.020 mol / 0.206 mol + 0.020 mol)

pH = 6.37 + 0.0293

pH =  6.40.

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Hydrogen peroxide solution consists of what two chemicals?
A. Hydrogen peroxide and water
B. Hydrogen peroxide and gasoline
C. Water and gasoline
D. Water and CO2

Answers

Hydrogen peroxide solution consists of two chemicals: hydrogen peroxide and water.

Hydrogen peroxide, chemical formula H2O2, is a clear, colorless liquid that is commonly used as a disinfectant, bleaching agent, and oxidizer. It is a powerful oxidizing agent and can decompose spontaneously, releasing oxygen gas. When it is dissolved in water, it forms a solution known as hydrogen peroxide solution, which is used in a variety of applications.

The solution typically contains about 3-10% hydrogen peroxide, with the remaining percentage being water. The concentration of hydrogen peroxide in the solution can vary depending on its intended use.

In summary, the two chemicals that make up hydrogen peroxide solution are hydrogen peroxide and water.

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Hydrogen peroxide solution consists of two chemicals, i.e. A. Hydrogen peroxide and water



What is Hydrogen Peroxide?

Hydrogen peroxide ([tex]H_{2}O_{2}[/tex]) is a chemical compound that consists of two hydrogen atoms (H) and two oxygen atoms (O), hence the chemical formula [tex]H_{2}O_{2}[/tex]. It is a pale blue liquid that is a powerful oxidizer and has various uses as a disinfectant, bleaching agent, and antiseptic. Hydrogen peroxide is often used as a solution in water, where it can readily decompose into water ([tex]H_{2}O[/tex]) and oxygen ([tex]O_{2}[/tex]) through a spontaneous reaction, releasing oxygen gas as bubbles. The decomposition of hydrogen peroxide in water is an exothermic reaction, meaning it releases heat as it occurs.

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an experimental plot of ln(k) vs. 1/t is obtained in lab for a reaction. the slope of the best-fit line for the graph is -2905 k. what is the value of the activation energy for the reaction in kj/mol?

Answers

Multiplying the slope (-2905 k) by the gas constant (0.008314 kJ/mol K) gives the activation energy: 24.1 kJ/mol.

The slant of the best-fit line for the diagram of ln(k) versus 1/T is equivalent to - Ea/R, where Ea is the actuation energy for the response, R is the gas consistent, and T is the temperature in Kelvin. To decide the actuation energy, we really want to improve this condition to address for Ea.

Ea = - slant x R

We realize that the slant of the best-fit line is - 2905 K, and R is 8.314 J/(mol·K). In any case, the slant should be changed over completely to units of J/(mol·K) by duplicating by 1000, since we need the actuation energy in units of kJ/mol. Accordingly:

Ea = - (- 2905 K x 8.314 J/(mol·K)) x (1/1000 kJ/J)

Ea = 24.1 kJ/mol

The initiation energy for the response is 24.1 kJ/mol. This worth addresses the base energy expected for the reactants to defeat the energy hindrance and structure items. The higher the initiation energy, the more slow the response rate, as well as the other way around.

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154.42g of oxygen gas (O2) react with an excess of ethane (C2H6) produces how many moles of water vapor (H20)?

Answers

For every 60 grammes of ethane, 108 grammes of water are produced. We therefore obtain 10.8 g of water from the combustion of 6 g of ethane. As a result, is created in 0.6 moles.

How are moles determined when vapour pressure is involved?

The mole fraction of the solvent must be multiplied by the partial pressure of the solvent in order to determine an ideal solution's vapour pressure. The vapour pressure would be 2.7 mmHg, for example, if the mole fraction is 0.3 and the partial pressure is 9 mmHg.

One mol of the solute is contained in one thousand grammes of the solvent (water) in a one molal solution. It follows that the solution's vapour pressure is 12.08 kPa.

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when a 2.5 liter vessel is filled with an unknown gas at stp, it weighs 2.75 g more than when it is evacuated. determine the molar mass of the unknown gas

Answers

The molar mass of the unknown gas is 27.0 g/mol.

According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. At STP, the pressure is 1 atm, the volume is 2.5 L, and the temperature is 273.15 K.

To find the number of moles of gas present, we can rearrange the ideal gas law equation to solve for n:

n = PV/RT

Substituting the values at STP, we get:

n = (1 atm) x (2.5 L) / [(0.08206 L atm/mol K) x (273.15 K)]

n = 0.1018 moles

The difference in weight between the gas-filled vessel and the evacuated vessel is 2.75 g, which is the weight of 0.1018 moles of the unknown gas.

So the molar mass of the gas can be calculated as:

molar mass = mass / moles

molar mass = 2.75 g / 0.1018 mole

molar mass = 27.0 g/mol

Therefore, the molar mass of the unknown gas is 27.0 g/mol.

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The molar mass of the unknown gas is 27.0 g/mol.

According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. At STP, the pressure is 1 atm, the volume is 2.5 L, and the temperature is 273.15 K.

To find the number of moles of gas present, we can rearrange the ideal gas law equation to solve for n:

n = PV/RT

Substituting the values at STP, we get:

n = (1 atm) x (2.5 L) / [(0.08206 L atm/mol K) x (273.15 K)]

n = 0.1018 moles

The difference in weight between the gas-filled vessel and the evacuated vessel is 2.75 g, which is the weight of 0.1018 moles of the unknown gas.

So the molar mass of the gas can be calculated as:

molar mass = mass / moles

molar mass = 2.75 g / 0.1018 mole

molar mass = 27.0 g/mol

Therefore, the molar mass of the unknown gas is 27.0 g/mol.

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when the reaction had finished, the solution was acidid. 25.0 ml of .5 mol l-1 na2co3 solution was required to neutralize the excess acid. what mass of magnesium carboante was orignally used

Answers

The mass of magnesium carbonate originally used was 2.108 g.

To solve this problem, we need to use stoichiometry and the concept of molarity. We know that the excess acid was neutralized by 25.0 ml of 0.5 mol L-1 Na2CO3 solution. This means that the amount of acid that reacted with the magnesium carbonate is equal to the amount of Na2CO3 in the solution.

First, let's calculate the amount of Na2CO3 in the solution:
0.5 mol L-1 x 0.025 L = 0.0125 mol Na2CO3

Since magnesium carbonate reacts with two moles of acid per mole of MgCO3, the amount of acid that reacted with the MgCO3 is twice the amount of Na2CO3:
0.0125 mol Na2CO3 x 2 = 0.025 mol H+

Now we can use the molarity of the acid to calculate the volume of acid that reacted with the MgCO3:
0.025 mol H+ / 0.1 mol L-1 = 0.25 L

Finally, we can use the volume of acid and the molarity of the acid to calculate the amount of MgCO3 that was originally used:
0.25 L x 0.1 mol L-1 = 0.025 mol MgCO3

To convert moles to mass, we need to use the molar mass of MgCO3:
0.025 mol x 84.31 g mol-1 = 2.108 g

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which category of amino acid contains r groups that are hydrophobic? which category of amino acid contains r groups that are hydrophobic? polar acidic basic non-polar basic and acidic

Answers

The amino acid that contains the R groups that are hydrophobic are the non - polar.

The Amino acids are the building blocks of the molecules of the  proteins. These contains the one hydrogen atom and the one amine group, the one carboxylic acid group and the one side chain that is the R group will be attached to the central carbon atom.

The side chains of the non polar amino acids includes the long carbon chains or the carbon rings, which makes them bulky. These are the hydrophobic, that means they repel the water. Therefore the non-polar amino acids are the hydrophobic.

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a 20.00-ml sample of 0.3000 m hbr is titrated with 0.15 m naoh. what is the ph of the solution after 40.3 ml of naoh have been added to the acid? group of answer choices 10.87 1.00 3.13 13.14 11.05

Answers

The pH of the solution is 4.72. So the answer is not one of the provided choices.

To solve this problem, we can use the equation for the reaction between HBr and NaOH:

[tex]HBr + NaOH - > NaBr + H_2O[/tex]

We know the initial concentration of HBr is 0.3000 M, and the volume is 20.00 mL, so the initial moles of HBr is:

0.3000 M × 0.02000 L = 0.00600 moles HBr

When 40.3 mL of 0.15 M NaOH is added, we can calculate the moles of NaOH added:

0.15 M × 0.0403 L = 0.006045 moles NaOH

Since the reaction between HBr and NaOH is 1:1, the moles of HBr remaining is:

0.006045 moles NaOH - 0.00600 moles

HBr = 4.5 × 10^-5 moles HBr

We can calculate the new volume of the solution:

20.00 mL + 40.3 mL = 60.3 mL = 0.0603 L

Now, we can calculate the new concentration of HBr:

(4.5 × 10^-5 moles HBr) / (0.0603 L) = 0.000746 M HBr

Finally, we can calculate the pH using the equation for the dissociation of HBr in water:

[tex]HBr + H_2O - > H_3O+ Br^-[/tex]

The equilibrium expression is:

Ka = [H3O+][Br-] / [HBr]

Since the concentration of HBr is very small, we can assume that it dissociates completely, so:

[H3O+] = [Br-] = xKa = x^2 / 0.000746

Solving for x, we get:

x = √(Ka × 0.000746) = √(8.7 × 10^-10 × 0.000746) = 1.89 × 10^-5 M.

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ssurr is negative for a phase transition that still happens. which of the following systems is the only one for which this is possible? group of answer choices a condensation process. a deposition process. a freezing process. a vaporization process.

Answers

The only system for which ssurr can be negative but the phase transition still occurs is a freezing process. Option 3 is correct.

During a freezing process, the temperature of the system decreases, which results in a decrease in entropy. The surroundings gain heat and have a positive change in entropy, causing the ssurr to be negative. However, the phase transition still occurs because the system's change in enthalpy, ΔH, is negative and more than compensates for the decrease in entropy. This results in a negative ΔG and a spontaneous process.

In other phase transitions, such as vaporization, condensation, and deposition, the ssurr must be positive or zero for the process to be spontaneous, as the increase in entropy of the surroundings compensates for the decrease in entropy of the system. Hence Option 3 is correct.

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If 51 grams of NH4CI is dissolved at 50°C, how many additional grams NH4CI would be
needed to make the solution saturated at 80°C

Answers

We would need an additional 636.75 grams of NH4CI to make solution saturated at 80°C.

What is meant by solubility?

Maximum amount of a solute that can dissolve in any given amount of solvent at a particular temperature is called as solubility.

According to solubility curve for NH4CI, solubility of NH4CI in water increases with temperature. At 50°C, solubility of NH4CI is approximately 40 g/100 mL, which means that 51 grams of NH4CI would dissolve in 127.5 mL of water (51 g/40 g/100 mL x 1000 mL = 127.5 mL).

To make solution saturated at 80°C, we need to find new solubility of NH4CI at 80°C. According to the solubility curve, solubility of NH4CI in water at 80°C is approximately 90 g/100 ml.

mass of solute = (solubility at 80°C - solubility at 50°C) x volume of solvent

mass of solute = (90 g/100 mL - 40 g/100 mL) x 127.5 mL = 636.75 g

Therefore, we would need an additional 636.75 grams of NH4CI to make the solution saturated at 80°C.

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