A 74.28-g sample of Ba(OH) is dissolved in enough water fo make 2.450 liters of solution. How
627871
many mL of this solution must be diluted with water in order to make 1.000 L of 0.100 M Ba(OH)?

Answer:

CH3-0-

Explanation:

An anion is always a better nucleophile than a neutral molecule.

Nucleophilicity is parallel to basicity. Acidity: HI > HCN > H₂O > EtOH. So, among the options provided, the best nucleophile is A) [tex]EtO}^-[/tex].

A chemical species known as a nucleophile is one that tends to give an electron pair to an electrophile during a chemical reaction, often an atom or an ion. The word "nucleophile," which refers to someone who is drawn to positively charged regions in a chemical reaction, is derived from the Latin terms "nucleus," which means "nucleus or core," and "philein," which means to love.

Many chemical processes, including addition, elimination, and nucleophilic substitution reactions, depend heavily on nucleophiles. In these reactions, molecules that are electron-deficient and have a propensity to attract electrons—known as electrophiles—are attacked by nucleophiles.

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From the following options, which would you expect to be the best nucleophile?

A) [tex]EtO}^-[/tex]

B) [tex]{OH}^-[/tex]

C) [tex]{CN}^-[/tex]

D) [tex]{I}^-[/tex]

b) It diffuses into the alveoli is the correct option. The gas goes from a higher partial pressure to a lower partial pressure along its own gradient.

The partial pressure of CO2 at the blood is thus 45 mm Hg and at the alveoli is 40 mm Hg. After then, CO2 travels from the blood to the alveoli.  In this mechanism, CO2 cannot be broken down into carbon and oxygen. Compared to O2, CO2 is more soluble in blood. For CO2 diffusion, a very small partial pressure gradient is required. It just takes a 5-mmHg pressure gradient for CO2 gas to diffuse. Hence, CO2 will transfer from blood into alveoli.

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Complete Question:

The partial pressure of carbon dioxide is 45 mm Hg in the blood and 40 mm Hg in the alveoli. What happens to the carbon dioxide?

a) It diffuses into the blood.

b) It diffuses into the alveoli.

c) The gradient is too small for carbon dioxide to diffuse.

d) It decomposes into carbon and oxygen

Answer:

where is the photo

Explanation:

where is it???

Total two electrons are transferred between atoms of magnesium and flourine. Ionic bond is formed between atoms by donating two electrons of magnesium atom to the two fluorine atoms.

Every element has atomic orbitals that contain a specific number of electrons . Out of these electrons, only outermost shell electrons of the atom participating in bond formation with another atom. We have a ratio of magnesium to fluoride ions, mole ratio = 1 : 2 , that one magnesium and two flourine atoms are there. Electronic configuration of Mg (12)

= 1s²2s²2p⁶3s²

Similarly, Electronic configuration of F (17)

= 1s²2s²2p⁶3s²3p⁵

Thus, the valency of Magnesium is two and Fluorine is one. Ionic compound is formed by magnesium cation and fluoride anion. Thus, magnesium'valence electrons (2 electrons) will be transferred completely to form an ionic bond with two fluorine atoms having 7 valence electrons each and form a MgF₂ molecule. The Lewis structure of MgF₂ compound is present in above figure.

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Withdrawing heat from a solution of two gases will lower the temperature, affecting the probability of a successful reaction on a per-collision basis.

When temperature decreases, the average kinetic energy of the gas molecules also decreases. Consequently, the molecules move slower, resulting in fewer collisions between them. This reduces the chances of successful reactions occurring during each collision.

Moreover, a lower temperature means that the energy barrier for a reaction, known as activation energy, becomes relatively higher compared to the average kinetic energy of the molecules. Activation energy is the minimum amount of energy required for a reaction to occur.

The reaction rate, which is the speed at which a reaction proceeds, is influenced by temperature. The Arrhenius equation, which quantifies the relationship between temperature and reaction rate, shows that a decrease in temperature results in a slower reaction rate.

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Answer:

The molar mass of ammonium sulfate ((NH4)2SO4) is:

2 x (14.01 g/mol) (for the nitrogen atoms in the ammonium ions)

8 x (1.01 g/mol) (for the hydrogen atoms in the ammonium ions)

1 x (32.06 g/mol) (for the sulfur atom)

4 x (16.00 g/mol) (for the oxygen atoms in the sulfate ion)

= 132.14 g/mol

To convert the mass of 4392.3 g of ammonium sulfate to moles, we need to divide this mass by the molar mass:

moles = 4392.3 g / 132.14 g/mol

moles = 33.20 mol (rounded to two decimal places)

Therefore, there are 33.20 moles in 4392.3 g of ammonium sulfate.

33.20 mol are in 4392.3 g of ammonium sulfate. The mole designates  6.022×10²³  units, which is a very large number.

In chemistry, a mole, usually spelt mol, is a common scientific measurement unit for significant amounts of very small objects like atoms, molecules, or even other predetermined particles. The mole designates 6.022×10²³ units, which is a very large number. Under the International Units of Measure (SI), the mole is defined as this number as of May 20, 2019, according the General Meeting on Weights as well as Measurements.

The amount of atoms discovered through experimentation to be present in 12 grammes of carbon-12 was originally used to define the mole. In honour of a Italian scientist Amedeo Avogadro, the amount of cells inside a mole is also known as Avogadro's number or Avogadro's constant (1776–1856).

The molar mass of ammonium sulfate=2 x (14.01 g/mol)+8 x (1.01 g/mol)+1 x (32.06 g/mol)+4 x (16.00 g/mol) = 132.14 g/mol

moles = 4392.3 g / 132.14 g/mol

moles = 33.20 mol

Therefore, 33.20 mol are in 4392.3 g of ammonium sulfate. The mole designates  6.022×10²³  units, which is a very large number.

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In general, the concentrations of each species at equilibrium will depend on the stoichiometry of the reaction, the equilibrium constant (K), and the initial concentrations of the reactants and products.

At equilibrium, the forward and reverse reaction rates are equal, and the concentrations of the reactants and products no longer change with time. The concentrations of each species at equilibrium will depend on the initial concentrations and the equilibrium constant. If the equilibrium constant is large, then the reaction will favor the products at equilibrium. If the equilibrium constant is small, then the reaction will favor the reactants at equilibrium.

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Answer:

A daughter element is the element formed when a radioactive element undergoes radioactive decay

Explanation:

may I get brainliest thanks

Once you have the correct values for pressure, volume, and temperature, follow these following steps to calculate the mass and number of moles of dinitrogen monoxide gas collected.

To calculate the mass and number of moles of dinitrogen monoxide gas collected, we will use the ideal gas law equation:
PV = nRT
where:
P = pressure in the flask
V = measured volume of the flask
n = number of moles of gas
R = ideal gas constant (0.0821 L atm / K mol)
T =temperature in Kelvin
However, the student question did not provide values for the pressure, volume, or temperature. Assuming that you have these values, you can proceed as follows:
1. Convert the given temperature to Kelvin if it's in Celsius (K = °C + 273.15).
2. Plug in the given values for P, V, and T into the ideal gas law equation
3. Solve for n (number of moles):
n = PV / RT
4. Calculate the mass of dinitrogen monoxide (N₂O) gas using the molar mass (44.013 g/mol):
mass = n × molar mass
5. Round your answers to the appropriate significant digits.

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If the calibration is to 2 ml, calibration lines are marked on the cylinder every 2 ml. Therefore, we can calculate the liquid volume in the cylinder to the nearest 2 ml.

Calibration in chemistry is the process of ensuring that any scientific method or instrument produces accurate results. If the calibration is 2 ml, then there are calibration lines on the cylinder spaced 2 ml apart.

Formally, calibration is the process of comparing a measurement device that needs to be calibrated to a device or reference standard that can be tracked. The term "calibrator" may also be used to describe the reference standard. According to logic, the calibration of the instrument should be more precise than the reference standard.

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the wavelength of this gamma ray is approximately 3.50 x [tex]10^-12[/tex] meters, which is in the range of typical gamma ray wavelengths.

The relationship between the frequency (f) and the wavelength (λ) of a wave is given by the formula:

c = fλ

where c is the speed of light in vacuum, which is approximately 3.00 x [tex]10^8[/tex] m/s.

To find the wavelength of a gamma ray with a frequency of [tex]8.56 x 10^19[/tex]Hz, we can use this formula and solve for λ:

c = fλ

λ = c / f

Substituting the given values:

λ = (3.00 x [tex]10^8[/tex] m/s) / (8.56 x  Hz)

λ ≈ 3.50 x[tex]10^-12\\[/tex] m

Frequency is a measure of how many cycles or waves of a periodic signal occur per unit of time. It is usually measured in Hertz (Hz), which represents one cycle per second. For example, a sound wave with a frequency of 440 Hz corresponds to the musical note A above middle C.

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According to the BOD values given, the rate constant (k) is approximately 0.0291 per day at 20°C.

BOD or biochemical oxygen demand is the amount of dissolved oxygen absorbed by aerobic bacteria growing on the organic material present in a water sample at a certain temperature during a specific period measured analytically.

To calculate the rate constant (k), we can use the following formula:

k = ln(BOD1/BOD2)/t

Where,

BOD1 = Initial BOD concentration

BOD2 = BOD concentration after time t

t = time elapsed in days

Given that,

BOD concentration at the end of 7 days (BOD2) = 60.0 ml/l

Ultimate BOD concentration (BOD1) = 85.0 ml/l.

The time elapsed (t) = 7 days.

Therefore,

k = (ln(85.0/60.0))/7

k = 0.0291 per day

Thus, the rate constant (k) is approximately 0.0291 per day at 20°C.

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The iupac name for the following compound is 3,6-dichloroaniline. Option d is correct choice.

The IUPAC name of a compound is a systematic name that follows a set of rules defined by the International Union of Pure and Applied Chemistry (IUPAC). In the case of the given compound with two chlorine atoms and an amino group attached to a benzene ring, the correct IUPAC name is based on the positions of the chlorine atoms relative to the amino group.

To name the compound, we need to number the carbons in the benzene ring starting from one of the carbons adjacent to the amino group. The positions of the two chlorine atoms are then indicated by their numbers, separated by a comma, before the name of the parent compound, which is "aniline" in this case.

By following this procedure, we can see that the correct IUPAC name for the compound is 3,6-dichloroaniline, as the chlorine atoms are located at positions 3 and 6 relative to the amino group. Hence option d is correct choice.

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Solubility of [tex]CaF$_2$[/tex] at 25°C with Ksp of 4.0 x [tex]10$^{-11}$[/tex] is approximately 6.34 x [tex]10$^{-4}$[/tex] mol/L.

The dissolvability of a substance is characterized as the most extreme measure of the substance that can break up in a given dissolvable at a specific temperature and strain. The dissolvability of calcium fluoride [tex](CaF$_2$)[/tex] at 25°C can be determined utilizing its Ksp esteem, which is 4.0 x [tex]10$^{-11}$[/tex].The Ksp articulation for [tex]CaF$_2$[/tex] is:

Ksp = [tex][Ca$^{2+}$][F$^{-}$]$^2$[/tex]

Let the dissolvability of [tex]CaF$_2$[/tex] be addressed by the variable x, so the convergences of [tex]Ca$^{2+}$[/tex] and [tex]F$^{-}$[/tex] in arrangement will be equivalent to x.Subbing these qualities into the Ksp articulation gives:

Ksp = [tex]x$\times$x$^2$ = x$^3$[/tex]

Reworking the condition gives:

x = [tex](Ksp)$^{1/3}$[/tex]

Subbing the worth of Ksp = 4.0 x [tex]10$^{-11}$[/tex] into the above condition gives:x = [tex](4.0 x 10$^{-11}$)$^{1/3}$ = 6.34 x 10$^{-4}$[/tex] mol/L

In this manner, the dissolvability of calcium fluoride at 25°C is around 6.34 x [tex]10$^{-4}$[/tex] mol/L, adjusted to two decimal spots for comfort. This really intends that at harmony, 6.34 x [tex]10$^{-4}$[/tex] moles of calcium fluoride can break up in one liter of water at 25°C.

On the off chance that how much [tex]CaF$_2$[/tex] in arrangement is higher than this worth, it will encourage out of the arrangement, and assuming it is lower, more [tex]CaF$_2$[/tex] will break down until the balance focus is reached.

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Answer: [1 ]      3-ethyl- 1,1-dimethyl-cyclohexane

              [2]       1-methyl-ethoxy-propanol

               [3]      4-oxo-pentan-2-one

               [4]      2-ethyl-4-hydroxy-3-methyl-pentanoic acid

Explanation:

The initial concentration of 'A' was approximately 1.371 M.

To determine the initial concentration of 'A' given the reaction rate constant, time, and the remaining amount of 'A', you can use the integrated rate law equation for a first-order reaction:

[A]t = [A]₀ * e^(-kt)

where:
[A]t = concentration of 'A' at time t (0.048 M)
[A]₀ = initial concentration of 'A'
k = reaction rate constant (0.012 M⁻¹ s⁻¹)
t = time in seconds (27 minutes = 27 * 60 = 1620 seconds)

Now, let's solve for [A]₀:

0.048 M = [A]₀ * e^(-0.012 M⁻¹ s⁻¹ * 1620 s)

To find [A]₀, divide both sides by e^(-0.012 * 1620):

[A]₀ = 0.048 M / e^(-0.012 * 1620)

Now, calculate the value:

[A]₀ ≈ 0.048 M / e^(-19.44)

[A]₀ ≈ 0.048 M / 0.000035

[A]₀ ≈ 1.371 M

So, the initial concentration of A was approximately 1.371 M.

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The average rate of reaction for the given time interval of 0 s to 205 s is calculated out to be -0.00088 M/s.

To calculate the average rate of reaction for each successive time interval, we need to use the formula:

Average rate of reaction = (Change in concentration) / (Change in time)

For the time interval from 0 s to 50 s, the change in concentration is (0.42 M - 0.50 M) = -0.08 M, and the change in time is (50 s - 0 s) = 50 s. Therefore, the average rate of reaction is:

Average rate of reaction = (-0.08 M) / (50 s) = -0.0016 M/s

Similarly, we can calculate the average rate of reaction for each successive time interval as follows:

----------------------------------------------------------------------------------------------------------

Time interval  Change in concentration Change in time Average    

                                                                                                 rate of reaction

----------------------------------------------------------------------------------------------------------

0 s to 50 s        -0.08 M                        50 s                  -0.0016 M/s

50 s to 100 s        -0.06 M                        50 s              -0.0012 M/s

100 s to 150 s               -0.05 M                       50 s                    -0.0010 M/s

150 s to 200 s       -0.05 M                       50 s                   -0.0010 M/s

200 s to 250 s       -0.04 M                       50 s                   -0.0008 M/s

----------------------------------------------------------------------------------------------------------

To find the average rate of reaction for the time interval from 0 s to 205 s, we need to add up the changes in concentration and divide by the total change in time:

Average rate of reaction = (-0.08 M - 0.06 M - 0.05 M - 0.05 M - 0.04 M) / (205 s - 0 s) = -0.00088 M/s

Therefore, the average rate of reaction for the time interval from 0 s to 205 s is -0.00088 M/s.

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The complete question is :

The following table shows the concentration of a reactant at various times during a chemical reaction. Calculate the average rate of reaction for each successive time interval. What is the average rate of reaction for the time interval from 0 s to 205 s?

Time (s)         Concentration (M)

0                     0.50

50                     0.42

100                     0.36

150                     0.31

200                     0.26

250                     0.22

Answer: 24.31g/mol

Explanation:

The correct answer is To determine the mass of Mg used in the reaction, we need to know the number of moles of Mg that reacted with the 2.26 moles of HCl. To do this, we can use the mole ratio from the balanced chemical equation:

Mg + 2 HCl → MgCl2 + H2 According to the stoichiometry of this equation, 1 mole of Mg reacts. With 2 moles of HCl to produce 1 mole of MgCl2 and 1 mole of H2. Therefore, to determine the amount of Mg used in the reaction, we need to first calculate the number of moles of HCl that reacted, and then use the mole ratio between Mg and HCl to find the number of moles of Mg. Given that 2.26 moles of HCl were used in the reaction, we can use the mole ratio of the balanced equation to calculate the number of moles of Mg moles of Mg = 1/2 x moles of HCl = 1/2 x 2.26 mol = 1.13 mol Now that we know the number of moles of Mg used, we can calculate the mass of Mg using its molar mass: mass of Mg = moles of Mg x molar mass of Mg = 1.13 mol x 24.31 g/mol = 27.5 g Therefore, the mass of Mg used in the reaction was 27.5 grams.

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According to the Beer lamberts law the absorbance is directly proportional to the concentration of the sample.

Beer-lambert Law states that when the extinction coefficient and the path length are constant, the absorbance is proportional to the concentration of the sample. According to this law the absorbance is directly proportional to the length of the light path.  It is equals to the inner width of the cuvette. It  is evident that the path length affects absorbance. We know that with a longer optical path length, the light has to travel through more solution and can hit more molecules or atoms and be absorbed more of the solution. The absorbance is directly proportional to the concentration of the given sample.

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The  complete question is,

Does stopper affect absorbance?

In cellular respiration, one molecule of glucose (C6H12O6) is oxidized to produce energy in the form of ATP through a series of metabolic reactions, including the citric acid cycle.

During the citric acid cycle, acetyl-CoA, which is produced from the breakdown of glucose in earlier steps of cellular respiration, enters the cycle and reacts with oxaloacetate to form citrate. Through a series of enzymatic reactions, citrate is eventually converted back to oxaloacetate, producing CO2, NADH, FADH2, and ATP in the process.

For each glucose molecule oxidized by cellular respiration, two molecules of pyruvate are produced during glycolysis. Each pyruvate molecule then enters the mitochondria and is converted to one molecule of acetyl-CoA, which enters the citric acid cycle. Therefore, for each glucose molecule oxidized, two molecules of acetyl-CoA enter the citric acid cycle.

During the citric acid cycle, two molecules of CO2 are produced for each molecule of acetyl-CoA that enters the cycle. Therefore, for each glucose molecule oxidized by cellular respiration, a total of four molecules of CO2 are produced in the citric acid cycle (two for each acetyl-CoA molecule).

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There are 1.69 x 10²² molecules of NaClO4 in 4.446 g of NaClO4.

Avogadro's number (6.022 x 10²³) is the basis of calculations that convert grams to atoms or molecules.

The molar mass of NaClO4 is 22.99 + 35.45 + 4(16.00) = 146.99 g/mol, which is found by adding the atomic masses of each component. This is then converted to moles by dividing the mass of the sample by the molar mass.

The number of molecules can be found by multiplying the number of moles by Avogadro's number, which gives the number of molecules. The following is the calculation:

mass of NaClO4 = 4.446 g
molar mass of NaClO4 = 146.99 g/mol

moles of NaClO4 = mass / molar mass = 4.446 g / 146.99 g/mol = 0.03022 mol

Number of molecules = Avogadro's number × moles = (6.022 x 10²³) x 0.03022 = 1.69 x 10²² molecules

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Theoretical yield of iron (III) oxide is 4.92 g ; percent yield for this reaction is approximately 73.3%.

Quantity of a product obtained from reaction is expressed in terms of yield of the reaction. Amount of product predicted by stoichiometry is called theoretical yield and whereas amount obtained actually is called  actual yield.

Balanced chemical equation for the reaction is: 4Fe + 3O₂→ 2Fe₂O₃

Molar mass of Fe = 55.85 g/mol

So, Number of moles of Fe = mass/molar mass = 3.44 g/55.85 g/mol = 0.0615 mol

and Number of moles of Fe₂O₃ = 0.0615 mol Fe × (2 mol Fe₂O₃/4 mol Fe) = 0.0308 mol Fe₂O₃

Molar mass of Fe2O3 = 159.69 g/mol

Theoretical yield of Fe₂O₃ = number of moles × molar mass = 0.0308 mol × 159.69 g/mol = 4.92 g

Therefore, theoretical yield of iron (III) oxide is 4.92 g.

As percent yield = (actual yield / theoretical yield) x 100%

So, percent yield = (4.32 g / 5.89 g) x 100% ≈ 73.3%

Therefore,  percent yield for this reaction is approximately 73.3%.

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No, the freezing point depression does not equal the freezing point of the pure solvent minus the freezing point of the solution.

The freezing point depression is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution. In other words, it is the amount by which the freezing point of the solvent is lowered when a solute is added to it.

The freezing point depression is related to the concentration of the solute in the solution, as well as the properties of the solvent and solute. The greater the concentration of the solute, the greater the freezing point depression.

The relationship between freezing point depression, solute concentration, and solvent properties is described by the equation ΔTf = Kf·m, where ΔTf is the freezing point depression, Kf is the freezing point depression constant of the solvent, and m is the molality of the solute in the solution.

In summary, the freezing point depression is not equal to the freezing point of the pure solvent minus the freezing point of the solution, but rather the difference between the freezing point of the pure solvent and the freezing point of the solution, and is related to the concentration of the solute in the solution.

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Answer:

A

Explanation:

I believe it's A sorry if any part of my answer is wrong i tried my best to help you improve

For the compound Ba(OH)2, the formulas of all the species that you can expect to be present in aqueous solution are:Ba2+ and 2OH-Ba2+ represents the cationic part of the compound, while 2OH- represents the anionic part of the compound.

In an aqueous solution of Ba(OH)2, the ions would dissolve and exist as free ions. The Ba2+ ion would carry a +2 charge, while the OH- ions would each carry a -1 charge. The balanced chemical equation for the compound dissociation in water is:Ba(OH)2 + 2H2O → Ba2+ + 2OH- + 2H2O

Thus, the formulas of all the species that you can expect to be present in an aqueous solution of Ba(OH)2 are Ba2+ and 2OH-.

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The white solid formed was likely a different compound due to the use of incorrect reagents and conditions during the reaction.

When an ester is hydrolyzed with HCl, the reaction typically proceeds via an acid-catalyzed mechanism, in which the ester is protonated to form a tetrahedral intermediate that quickly breaks down into a carboxylic acid and an alcohol. On the other hand, when an ester is hydrolyzed with NaOH, the reaction proceeds via a base-catalyzed mechanism, in which the ester is deprotonated to form a negatively charged intermediate that quickly breaks down into a carboxylate ion and an alcohol.

In this case, the student hydrolyzed the intermediate ester with 6 M HCl instead of 3 M NaOH. This likely led to the protonation of the intermediate and the formation of a carboxylic acid and an alcohol. The student then boiled the reaction mixture with 50% NaOH instead of 6 M HCl. This would have led to the deprotonation of the carboxylic acid and the formation of a carboxylate ion and an alcohol. However, since the NaOH concentration was much higher than what was needed for the reaction, it's possible that the reaction was not selective and other side reactions occurred as well.

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The equilibrium partial pressure of CH4(g) is 0.417 atm.

For the following reaction, Kp = 0.262 at 1000°C: C(s) + 2H2(g) ⇌ CH4(g) At equilibrium, the partial pressure of H2 is 1.26 atm.

The equilibrium constant (Kp) is defined as the ratio of the partial pressures of products to the partial pressures of reactants with each concentration term raised to a power equivalent to its coefficient in the balanced chemical equation.

At equilibrium, the partial pressure of H2 is 1.26 atm. CH4 partial pressure can be calculated by applying the equilibrium constant to this value. Here are the steps for calculating the equilibrium partial pressure of CH4(g): Write the equilibrium equation and the corresponding Kp expression.

Calculate the value of Kp.Substitute the known partial pressure of H2 into the equilibrium expression and solve for the unknown equilibrium partial pressure of CH4(g).At equilibrium, C(s) + 2H2(g) ⇌ CH4(g)The Kp expression is: Kp = PCH4/PH2²Kp = 0.262PCH4 = (Kp)(PH2²)PCH4 = (0.262)(1.26²)PCH4 = 0.417 atm

Therefore, the equilibrium partial pressure of CH4(g) is 0.417 atm.

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The pH of the solution is 11.13.

First, let's write the balanced chemical equation for the reaction:

NH₂NH₂ + H₂O ↔ NH₂NH₃⁺ + OH⁻

The Kb expression for this reaction is,

Kb = [NH₂NH₃⁺][OH⁻] / [NH₂NH₂]

Since we're given the concentration of NH₂NH₂ and NH₂NH₃⁺, we can use the Kb expression to find [OH⁻]:

Kb = [NH₂NH₃⁺][OH⁻] / [NH₂NH₂]

1.7 × 10^-6 = (0.387-x)x / (0.258+x)

where x is the concentration of  ion in mol/L.

Solving for x, we get:

x = 7.43 × 10^-4 M

Therefore, the concentration of OH⁻ ion in the solution is 7.43 × 10^-4 M. To find the pH of the solution, we can use the equation:

pH = 14 - pOH = 14 - (-log[OH⁻]) = 11.13.

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The empirical formula of the compound is CH2O.

To find the empirical formula of the compound, we need to determine the ratios of the atoms in the compound. To do this, we need to find the number of moles of each element present in the compound.

The molar mass of carbon is 12.01 g/mol, the molar mass of hydrogen is 1.008 g/mol, and the molar mass of oxygen is 16.00 g/mol. The number of moles of carbon in the sample is:0.1565 g × 1 mol/12.01 g ≈ 0.0130 mol. The number of moles of hydrogen in the sample is:0.02627 g × 1 mol/1.008 g ≈ 0.0261 molThe number of moles of oxygen in the sample is:

0.4170 g × 1 mol/16.00 g ≈ 0.0261 mol

Now that we know the number of moles of each element, we can determine the simplest whole number ratio of the elements in the compound. We can do this by dividing each number of moles by the smallest number of moles. In this case, the smallest number of moles is 0.0130 mol.

We get:

Carbon: 0.0130 mol ÷ 0.0130 mol = 1

Hydrogen: 0.0261 mol ÷ 0.0130 mol = 2

Oxygen: 0.0261 mol ÷ 0.0130 mol = 2

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