by what factor will the rate of the reaction change if the ph decreases from 5.00 to 2

The mass of hydrogen present in the 300 mL sample is approximately 18.14 grams. To determine the mass of hydrogen present in the sample, we need to account for the partial pressure of hydrogen and correct for the presence of water vapor.

The total pressure in the sample is the sum of the partial pressure of hydrogen and the vapor pressure of water:

Total pressure = Partial pressure of hydrogen + Vapor pressure of water

The partial pressure of hydrogen can be calculated using Dalton's law of partial pressures:

Partial pressure of hydrogen = Total pressure - Vapor pressure of water

Now, we can use the ideal gas law equation to calculate the number of moles of hydrogen:

PV = nRT

where:

P = Partial pressure of hydrogen (in atm)

V = Volume of hydrogen (in L)

n = Number of moles of hydrogen

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

T = Temperature (in Kelvin)

Let's convert the volume from milliliters to liters:

Volume of hydrogen = 300 mL = 300/1000 L = 0.3 L

Now, we can rearrange the ideal gas law equation to solve for the number of moles:

n = PV / RT

n = (729 torr * 0.3 L) / (0.0821 L·atm/(mol·K) * 294.15 K) [21°C converted to Kelvin]

Performing the calculation:

n = (218.7 torr·L) / (24.11 L·atm/(mol·K))

n ≈ 9.07 mol

Finally, we can calculate the mass of hydrogen using the molar mass of hydrogen (H₂):

Mass of hydrogen = Number of moles * Molar mass of hydrogen

Molar mass of hydrogen = 2 g/mol

Mass of hydrogen = 9.07 mol * 2 g/mol

Mass of hydrogen ≈ 18.14 g

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The pH of a 0.17 M solution of this weak acid is approximately 2.99, To determine the pH of a 0.17 M solution of a monoprotic weak acid, we can use the acid dissociation constant (Ka) and the equilibrium expression: Ka = [H+][A-] / [HA].

where [H+] is the concentration of hydrogen ions, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

Assuming that the acid dissociates to a negligible extent, we can make the approximation [HA] ≈ initial concentration of the acid, and [A-] ≈ 0. Then, the equilibrium expression becomes:

Ka = [H+]² / [HA]

Solving for [H+], we get:

[H+] = sqrt(Ka x [HA])

[H+] = sqrt(4.1 x 10⁻⁶ x 0.17)

[H+] = 1.01 x 10⁻³ M

To find the pH, we can use the equation:

pH = -log[H+]

pH = -log(1.01 x 10⁻³)

pH = 2.99

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0.0095 moles of H2C2O4 must be dissolved in water to create 764 ml of a solution with a pH of 2.31.

The pH of a solution is related to the concentration of hydrogen ions (H+) in the solution. The pH scale ranges from 0 to 14, with lower values indicating higher concentrations of H+. The pH of 2.31 indicates a H+ concentration of 7.14 × 10^-3 mol/L.

H2C2O4 is a weak acid that undergoes the following reaction in water: [tex]H2C2O4 + H2O ⇌ H3O+ + HC2O4-[/tex]. The Ka of H2C2O4 is 5.9 × 10^-2. Using the pH and Ka values, we can set up an equation to find the concentration of H2C2O4:

[tex]Ka = [H3O+][HC2O4-]/[H2C2O4][/tex]

[tex][H2C2O4] = [HC2O4-] = x[/tex]

[tex][H3O+] = 7.14 × 10^-3 mol/L[/tex]

[tex]5.9 × 10^-2 = (7.14 × 10^-3)^2 / x[/tex]

[tex]x = 0.0095 mol/L[/tex]

The volume of the solution is 764 mL = 0.764 L. Therefore, the number of moles of H2C2O4 required is:

moles = concentration × volume = 0.0095 mol/L × 0.764 L = 0.0073 mol

Therefore, 0.0095 moles of H2C2O4 must be dissolved in water to create 764 ml of a solution with a pH of 2.31.

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The [H3O+] of the 0.10 M NH4Cl solution in H2O at 25°C is approximately 7.5 x 10^-6.

To calculate the [H3O+] of a 0.10 M solution of NH4Cl in H2O at 25°C, we first need to determine the Kb for NH3 and the Ka for NH4+. Since Kb for NH3 is given as 1.8 x 10^-5, we can use the relationship between Ka, Kb, and Kw (the ion product of water) to find the Ka for NH4+:
Kw = Ka × Kb
Kw = 1.0 x 10^-14 (at 25°C)
So, Ka for NH4+ = Kw / Kb = (1.0 x 10^-14) / (1.8 x 10^-5) = 5.56 x 10^-10.
Now, we can use the Ka expression for the dissociation of NH4+ to solve for [H3O+]:
NH4+ (aq) ↔ NH3 (aq) + H3O+ (aq)
Ka = [NH3][H3O+] / [NH4+]
Let x be the concentration of [H3O+]. Then:
5.56 x 10^-10 = (x)(x) / (0.10 - x)
Assuming x << 0.10, we can simplify the equation to:
5.56 x 10^-10 ≈ x^2 / 0.10
Now, solve for x (concentration of [H3O+]):
x^2 ≈ 5.56 x 10^-11
x ≈ 7.46 x 10^-6
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To draw the structure and give the systematic name of compounds with the molecular formula C5H12, we need to understand the different types of hydrogens present in the compound. Hydrogens can be classified as primary, secondary, or tertiary depending on the carbon they are attached to.

a) A compound with only primary and secondary hydrogens will have five carbons with three primary and two secondary hydrogens attached to them. The structure of this compound is a straight chain of five carbons with a methyl group attached to the second carbon. The systematic name of this compound is 2-methyl pentane.

b) A compound with only primary hydrogens will have five carbons with three primary hydrogens attached to them. The structure of this compound is also a straight chain of five carbons. The systematic name of this compound is pentane.

c) A compound with one tertiary hydrogen will have five carbons with one tertiary hydrogen attached to them. The structure of this compound is a branched chain with a methyl group attached to the first carbon and a tert-butyl group attached to the fourth carbon. The systematic name of this compound is 2,2-dimethylbutane.

d) A compound with two secondary hydrogens will have five carbons with two secondary hydrogens attached to them. The structure of this compound is also a branched-chain with a methyl group attached to the first carbon and an isopropyl group attached to the third carbon. The systematic name of this compound is 2-methyl-2-isopropylpentane.

In conclusion, the structure and systematic names of compounds with the molecular formula C5H12 can be determined by identifying the types of hydrogens present in the compound and using this information to draw the structure and assign the systematic name.

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Nitrobenzoic acid can have a total of 6 possible isomers. One of these isomers is o-nitrobenzoic acid.Tribromobenzoic acid can have a total of 4 possible isomers. One of these isomers is 2,4,6-tribromobenzoic acid.

Isomers are different compounds with the same molecular formula but different arrangements or orientations of atoms. In the case of nitrobenzoic acid, the isomers differ in the position of the nitro (-NO2) group on the benzene ring. The "o-" in o-nitrobenzoic acid indicates that the nitro group is located in the ortho position, which is adjacent to the carboxyl group (-COOH) on the benzene ring.

Similarly, in tribromobenzoic acid, the isomers differ in the position of the bromine (-Br) substituents on the benzene ring. The numbering in 2,4,6-tribromobenzoic acid indicates that the bromine atoms are located in the 2nd, 4th, and 6th positions on the benzene ring.

Overall, these compounds demonstrate the concept of isomerism, where different arrangements of atoms lead to distinct chemical structures and properties.

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True. The addition of Br2 to cyclopentene follows an electrophilic addition mechanism where the double bond of cyclopentene acts as the nucleophile attacking one of the Br2 molecules.

This results in the formation of a cyclic intermediate with a bridging bromine atom. The intermediate then breaks down to form the trans-1,2-dibromocyclopentane product. The "trans" in the name refers to the relative positions of the two bromine atoms on the cyclopentane ring. This reaction is stereospecific and yields only the trans isomer. The addition of Br2 to cyclopentene is an important reaction in organic chemistry and is commonly used for the synthesis of other compounds. In conclusion, the statement is true and can be explained by the electrophilic addition mechanism that occurs during the reaction.

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The answer using two significant figures is ΔSuniv = -130.08 J/K.

To find ΔSuniv, we need to first find ΔG∘rxn, which is the change in Gibbs free energy. We can do this using the equation:

ΔG∘rxn = ΔH∘rxn - TΔS∘rxn

We are given the values of ΔH∘rxn, ΔS∘rxn, and T:

ΔH∘rxn = 84 kJ = 84000 J (convert kJ to J)
ΔS∘rxn = 144 J/K
T = 303 K

Now we can plug these values into the equation:

ΔG∘rxn = 84000 J - (303 K)(144 J/K)

ΔG∘rxn = 84000 J - 43632 J

ΔG∘rxn = 40368 J

Now that we have the value of ΔG∘rxn, we can find ΔSuniv using the equation:

ΔSuniv = (-ΔG∘rxn) / T

Plugging in the values:

ΔSuniv = (-40368 J) / (303 K)

ΔSuniv = -133.08 J/K

Since we need to express the answer using two significant figures, the final value of ΔSuniv will be:

ΔSuniv = -130 J/K

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The calculation of ΔSuniv requires the use of the equation ΔSuniv = ΔSsys + ΔSsurr, where ΔSsys is the change in entropy of the system, and ΔSsurr is the change in entropy of the surroundings.

To determine ΔSuniv, we need to convert ΔH∘rxn from kJ to J, which gives ΔH∘rxn = 84000 J. Then, we can plug in the values for ΔH∘rxn, ΔSrxn, and T into the equation:

ΔSuniv = ΔSsys + ΔSsurr = ΔSrxn - ΔH∘rxn/T

ΔSuniv = (144 J/K) - (84000 J)/(303 K) = -87 J/K

The negative value for ΔSuniv indicates that the process is not spontaneous under the given conditions. This means that the reaction is not favorable at the given temperature and that the system requires an external input of energy to occur.

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Acid will increase the solubility of salicylic acid in water and base will decrease the solubility of salicylic acid in water.

Salicylic acid, an organic acid, breaks down to lose a proton to the carboxylic acid functional group in an aqueous solution. An intramolecular in hydrogen bond is created when the resultant carboxylate ion () interacts intramolecularly with the hydrogen atom within the hydroxyl group (-OH). Acid will increase the solubility of salicylic acid in water and base will decrease the solubility of salicylic acid in water.

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All acid-base reactions in water involve the transfer of a proton (H+) from the acid to the base, forming a conjugate base and a conjugate acid. This is because water can act as both an acid and a base, and the proton transfer reaction follows the same general mechanism regardless of the specific acid or base involved. The reaction involved can be represented as:

acid + base → conjugate base + conjugate acid

For example, in the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), the H+ ion from HCl is transferred to the OH- ion from NaOH, forming water (H2O) and the conjugate base of HCl (Cl-) and the conjugate acid of NaOH (Na+). This reaction can be generalized to any acid-base reaction in water.

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The solubility of [tex]CaF_{2}[/tex] in a solution of [tex]Ca(NO_{3})_{2}[/tex] will be represented by the concentration term of 2F- (option b).

When[tex]CaF_{2}[/tex] dissolves in water, it dissociates into [tex]Ca_{2}[/tex]+ and F- ions. However, in the presence of[tex]Ca(NO_{3})_{2}[/tex], the common ion effect will occur, which will shift the equilibrium of [tex]CaF_{2}[/tex] dissociation to the left, decreasing its solubility.

The common ion effect occurs because [tex]Ca(NO_{3})_{2}[/tex] provides additional [tex]Ca_{2}[/tex]+ ions to the solution, which, in turn, react with F- ions, forming [tex]CaF_{2}[/tex]and decreasing the concentration of free F- ions.

Thus, the concentration of F- ions will determine the solubility of [tex]CaF_{2}[/tex] in a solution of [tex]Ca(NO_{3})_{2}[/tex]. Therefore, the concentration term for the solubility product expression of [tex]CaF_{2}[/tex] in this solution will be [F-]2. Hence, option (b) 2F- is the correct answer.

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The correct order of increasing entropy for the compounds CBr4, C2Br6, and C3Br8 is:

**c. CBr4, C3Br8, C2Br6**.

Entropy is a measure of the degree of disorder or randomness in a system. In general, larger and more complex molecules tend to have higher entropy due to increased molecular motion and conformational possibilities. Among the given compounds, CBr4 has the fewest number of bromine atoms and the simplest molecular structure, resulting in lower entropy. C3Br8, on the other hand, has the most bromine atoms and the most complex structure, leading to higher entropy. C2Br6 falls in between these two compounds in terms of complexity and, thus, has intermediate entropy.

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If we want to prepare 1 L of 0.150 M CuSO4, take 375 mL of the 0.4 M CuSO4 solution and dilute it with water until the final volume reaches 1000 mL.

To prepare 0.150 M CuSO4 from 0.4 M CuSO4, you need to perform a dilution.

The formula for dilution is C1V1 = C2V2, where C1 and C2 are the initial and final concentrations, and V1 and V2 are the initial and final volumes. In this case, C1 = 0.4 M and C2 = 0.150 M.

Let's assume you want to prepare 1 L (1000 mL) of the 0.150 M solution.

This makes V2 = 1000 mL.

Using the formula: (0.4 M) * V1 = (0.150 M) * (1000 mL).

Solving for V1, we get V1 = (0.150 M * 1000 mL) / 0.4 M = 375 mL.

So, to prepare 1 L of 0.150 M CuSO4, take 375 mL of the 0.4 M CuSO4 solution and dilute it with water until the final volume reaches 1000 mL. Make sure to mix the solution thoroughly to ensure uniform concentration.

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1. The molar hydroxide ion concentration in an aqueous solution in which the [H+] = 2.5 x 10⁻⁷ M is 4.0 x 10⁻⁸ M.

2. There are three peptide bonds present in the polypeptide shown below.

To find the molar hydroxide ion concentration (OH⁻), you can use the ion product of water, which is a constant at a given temperature.

1. At 25°C, this constant (Kw) is 1.0 x 10⁻¹⁴ the equation wil be

Kw = (H⁺) × (OH⁻)

OH⁻ = Kw / H⁺

Substituting the given [H+] = 2.5 x 10⁻⁷ M into the equation

[OH-] = (1.0 x 10⁻¹⁴) / (2.5 x 10⁻⁷)

[OH-] = 4.0 x 10⁻⁸ M

2. In a polypeptide, every amino acid is connected to the next one through a peptide bond. The peptide bonds are formed between the carboxyl group of the first amino acid (N2N) and the amino group of the second amino acid (CH2C), the carboxyl group of the secnd amino acid (CH2C) and the amino grup of the third amino acid (NHCHC)

             O          O           O

             ||            ||            ||

N₂NCH₂CNHCHCNHCHCOH

                       |            |

                     CH₃     CH₂OH

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We need to add approximately 1.74 kg of nickel to 2.43 kg of copper to yield a solidus temperature of 1300°C. When copper and nickel are mixed together, they form an alloy.

The solidus temperature of the alloy depends on the proportions of copper and nickel in the mixture. To calculate the amount of nickel that must be added to 2.43 kg of copper to yield a solidus temperature of 1300°C, we need to use the lever rule equation. The lever rule equation relates the weight of each component in the alloy to the solidus temperature of the alloy. The equation is:

((Wn - Wc) / (Ws - Wc)) = ((Ts - Tc) / (Tn - Ts))

where:

Wn = weight of nickel to be added

Wc = weight of copper

Ws = weight of the resulting alloy

Ts = solidus temperature of the resulting alloy

Tc = solidus temperature of copper

Tn = solidus temperature of nickel

We are given the weight of copper (2.43 kg) and the solidus temperature of copper (1084°C). We are also given the desired solidus temperature of the alloy (1300°C) and the solidus temperature of nickel (1455°C).

We can use the lever rule equation to solve for the weight of nickel that must be added to the copper to yield the desired solidus temperature of 1300°C.

First, we rearrange the equation to solve for the weight of nickel:

Wn = ((Ts - Tc) / (Tn - Ts)) * (Ws - Wc)

Then, we substitute the known values:

Wn = ((1300°C - 1084°C) / (1455°C - 1300°C)) * (Wn + 2.43 kg - 2.43 kg)

We simplify this equation to get:

Wn = (216°C / 155°C) * Wn

Wn = 1.3935 * Wn

Finally, we divide both sides by 1.3935 to get:

Wn ≈ 1.74 kg

Therefore, we need to add approximately 1.74 kg of nickel to 2.43 kg of copper to yield a solidus temperature of 1300°C.

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The decomposition of hydrogen peroxide to form water and oxygen is an example of a disproportionation reaction due to the change in oxidation states of oxygen.

The decomposition of hydrogen peroxide to form water and oxygen is indeed an example of a disproportionation reaction. In this reaction, hydrogen peroxide (H₂O₂) undergoes a self-oxidation and reduction process simultaneously.

In hydrogen peroxide, the oxygen atom is in the -1 oxidation state. During the disproportionation reaction, one oxygen atom in H₂O₂ is reduced to a -2 oxidation state, forming water (H₂O), while the other oxygen atom is oxidized to a 0 oxidation state, resulting in the formation of oxygen gas (O₂).

This simultaneous oxidation and reduction of the same element (oxygen in this case) within a single compound is characteristic of a disproportionation reaction.

The oxidation state of the oxygen changes from -1 to -2 and 0, demonstrating the disproportionation process.

Therefore, the statement that the decomposition of hydrogen peroxide to form water and oxygen is an example of a disproportionation reaction is valid, and the given reason explaining the change in oxidation states supports this assertion.

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When, a buffer will be prepared by mixing 86.4 ml of 1.05 m hbr and 274 ml of 0.833 M ethylamine. Then, the pH of the buffer after 0.068 mol NaOH is added is 5.72.

To solve this problem, we use the Henderson-Hasselbalch equation;

pH = pKa + log([base]/[acid])

First, we need to find the concentrations of the acid and base in the buffer solution;

[acid] = 1.05 M (HBr)

[base] = 0.833 M (ethylamine)

The pKa of HBr is -9, so we can assume that the concentration of H⁺ions is equal to the concentration of HBr. Therefore, the pH of the buffer before adding NaOH is;

pH = -log[H⁺] = -log(1.05) = 0.978

To calculate pH after adding 0.068 mol NaOH, we need to determine the new concentrations of the acid and base. We know that 0.068 mol NaOH will react with some of the HBr in the buffer, so we calculate how much HBr will be left.

1 mol HBr reacts with 1 mol NaOH, so 0.068 mol NaOH will react with 0.068 mol HBr. The amount of HBr remaining in the buffer is;

0.068 mol HBr - 0.068 mol NaOH = 0.054 mol HBr

The concentration of HBr is now;

[acid] = 0.054 mol / 0.3604 L = 0.1499 M

To calculate the concentration of the conjugate base, we need to determine how much of the ethylamine will react with the remaining H⁺ ions. Since ethylamine is a weak base, we need to use the [tex]K_{b}[/tex] equation;

[tex]K_{b}[/tex] = [BH⁺][OH⁻] / [B]

We can assume that all of the remaining H⁺ ions will react with the ethylamine to form the conjugate acid. The amount of ethylamine that reacts can be calculated using the stoichiometry of the reaction;

C₂H₅NH₂ + H⁺ → C₂H₅NH₃⁺

1 mol C₂H₅NH₂reacts with 1 mol H⁺, so 0.054 mol H⁺ will react with 0.054 molC₂H₅NH₂. The amount of C₂H₅NH₂ remaining in the buffer is;

.833 mol - 0.054 mol = 0.779 mol

The concentration of the conjugate base is;

[base] = 0.779 mol / 0.3604 L = 2.160 M

Now we use the Henderson-Hasselbalch equation to calculate the pH;

pH = pKa + log([base]/[acid])

pH = 9 - log(2.160/0.1499)

pH = 5.72

Therefore, the pH of the buffer after 0.068 mol NaOH is added is 5.72.

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The energy of a photon, E, is related to its wavelength, λ, by the equation:  the wavelength of the light is approximately 293 nm.

Wavelength is the distance between two consecutive points on a wave that are in phase, or have the same phase, and can be measured as the distance from one peak of the wave to the next. Wavelength is commonly denoted by the Greek letter lambda (λ) and is usually measured in meters (m), but can also be measured in other units such as nanometers (nm), micrometers (µm), or angstroms (Å).

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1 mole is equal to 6. 022 x 10 23 particles, which is also known as the Avogadro's constant. To compute the amount of moles of any material in the sample, just divide the substance's stated weight by its molar mass.

To calculate the number of moles of Cu2+ in the 15.5 g sample of soluble plant fertilizer, we need to first convert the weight percentage of copper(II) sulfate to its molar mass.

The molar mass of CuSO4 is 159.609 g/mol (63.546 g/mol for Cu and 2 x 32.066 g/mol for SO4).

0.0700% by weight means that there are 0.0700 g of CuSO4 in every 100 g of fertilizer.

Therefore, in the 15.5 g sample of fertilizer, there are:

0.0700 g CuSO4/100 g fertilizer x 15.5 g fertilizer = 0.01085 g CuSO4

To convert grams to moles, we divide by the molar mass:

0.01085 g CuSO4 / 159.609 g/mol = 6.81 x 10^-5 moles CuSO4

Since CuSO4 dissociates in water to form one Cu2+ ion and one SO4 2- ion, the number of moles of Cu2+ in the sample is the same as the number of moles of CuSO4:

6.81 x 10^-5 mol Cu2+

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The product of the reaction between CH₃CH₂COOH and CH₃CH₂OH will produce ethyl ethanoate (CH₃COOCH₂CH₃) and water (H₂O).

This is an esterification reaction, which is a type of condensation reaction that occurs between a carboxylic acid and an alcohol in the presence of an acid catalyst, typically sulfuric acid (H₂SO₄).

The reaction involves the removal of a water molecule from the carboxylic acid and alcohol to form the ester and water. Ethyl acetate is a colorless liquid with a fruity odor and is commonly used as a solvent in various applications, such as in the manufacture of coatings, adhesives, and pharmaceuticals.

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The reaction is first order in A. This means that if the concentration of reactant A is doubled, the reaction rate will also double.

The order of a reaction is the exponent to which the concentration of the reactant is raised in the rate equation. From the information given in the question, we know that if [A] doubles and [B] stay the same, and as a result, the rate is cut in half. This suggests that the reaction rate is directly proportional to the concentration of reactant A raised to the power of 1 (first order) and inversely proportional to the concentration of the other reactant B.  

It is important to note that the order of a reaction cannot be determined solely by looking at the balanced chemical equation, but rather requires experimental data to determine the rate equation.

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Approximately 51.37 liters of nitrogen gas at STP would react with 37.2 grams of magnesium, considering the stoichiometry of the balanced chemical equation for the reaction.

To calculate the volume of nitrogen gas at STP that would react with 37.2 grams of magnesium, we first need to determine the number of moles of magnesium. The molar mass of magnesium (Mg) is 24.31 g/mol, so we can calculate the number of moles by dividing the given mass by the molar mass:

moles of Mg = 37.2 g / 24.31 g/mol = 1.528 mol.

From the balanced chemical equation for the reaction between magnesium and nitrogen gas, we know that 3 moles of nitrogen gas react with 2 moles of magnesium:

3N2 + 2Mg -> 2Mg3N2.

Therefore, we can conclude that 2 moles of magnesium would react with 3 moles of nitrogen gas. Using this ratio, we can calculate the number of moles of nitrogen gas:

moles of N2 = (3/2) * moles of Mg = (3/2) * 1.528 mol = 2.292 mol.

At STP (standard temperature and pressure), 1 mole of any ideal gas occupies 22.4 liters. Therefore, the volume of nitrogen gas would be:

volume of N2 = 2.292 mol * 22.4 L/mol = 51.37 L.

Thus, approximately 51.37 liters of nitrogen gas at STP would react with 37.2 grams of magnesium.

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The pKa of lactic acid is 3.08, which means that at this pH, half of the acid molecules are dissociated and half are not.

So, the correct answer is D.

To find the pH of a 0.10M solution of lactic acid, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA]) where [A-] is the concentration of the conjugate base (lactate) and [HA] is the concentration of the acid (lactic acid).

At the start, both concentrations are equal to 0.10M.

Plugging in the values, we get:

pH = 3.08 + log([0.10]/[0.10])

pH = 3.08 + log(1) pH = 3.08

Therefore, the pH of a 0.10M solution of lactic acid is 3.08.

Hence the answer of the question is D.

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The bag contains 56 marbles. (35 yellow marbles can be expressed in the ratio as 5 yellow marbles for every 8 orange marbles.)

If a bag contains 35 yellow marbles, we can determine the total number of marbles in the bag using the given ratio. According to the ratio provided, for every 5 yellow marbles, there are 8 orange marbles. We can set up a proportion to find the total number of marbles in the bag.

Let x be the total number of marbles in the bag. The proportion can be written as: 5 yellow marbles / 8 orange marbles = 35 yellow marbles / x

Cross-multiplying, we get: 5x = 35 * 8

5x = 280

Dividing both sides by 5, we find: x = 56

Therefore, the bag contains 56 marbles.

According to the given ratio of 5 yellow marbles for every 8 orange marbles, we can set up a proportion to find the total number of marbles in the bag. By cross-multiplying, we find that 5 times the total number of marbles is equal to 35 times 8. Simplifying the equation, we get 5x = 280. Dividing both sides of the equation by 5, we find that the total number of marbles in the bag, represented by x, is equal to 56. Therefore, the bag contains 56 marbles in total. The given information of having 35 yellow marbles helps us determine the overall quantity of marbles in the bag using the provided ratio.

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The pH of the solution is 7, which indicates a neutral solution.

Given that the solution has a hydroxide-ion (OH⁻) concentration of 1.0 x 10⁻⁷ mol/L, we need to determine the hydrogen-ion (H⁺) concentration first to calculate the pH of the solution.

Step 1: Use the ion product of water (Kw) to find the H⁺ concentration.
Kw = [H⁺][OH⁻]
Kw (at 25°C) = 1.0 x 10⁻¹⁴

Step 2: Plug in the given OH⁻ concentration and solve for H⁺ concentration.
1.0 x 10⁻¹⁴ = [H⁺](1.0 x 10⁻⁷)
[H⁺] = (1.0 x 10⁻¹⁴) / (1.0 x 10⁻⁷)
[H⁺] = 1.0 x 10⁻⁷ mol/L

Step 3: Calculate the pH using the pH formula.
pH = -log10[H⁺]

Step 4: Plug in the H⁺ concentration and solve for pH.
pH = -log10(1.0 x 10⁻⁷)
pH = 7

The pH of the solution is 7, which indicates a neutral solution.

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The pH of the solution with a hydroxide-ion concentration of 1.0 x 10⁻⁷ mol per liter is 7.

The pH of a solution is a measure of its acidity or alkalinity and is determined by the concentration of hydronium ions (H₃O⁺). However, in this case, we are given the hydroxide-ion concentration (OH⁻), which is related to the concentration of hydronium ions through the self-ionization of water:

H₂O ⇌ H⁺ + OH⁻

In pure water, the concentration of H⁺ ions is equal to the concentration of OH⁻ ions, which is 1.0 x 10⁻⁷ mol per liter. This corresponds to a neutral solution.

The pH scale is logarithmic and is defined as the negative logarithm (base 10) of the H⁺ concentration:

pH = -log[H⁺]

Since the solution is neutral, the H⁺ concentration is also 1.0 x 10⁻⁷ mol per liter. Substituting this value into the pH equation:

pH = -log(1.0 x 10⁻⁷)

pH = 7

Therefore, the pH of the solution with a hydroxide-ion concentration of 1.0 x 10⁻⁷ mol per liter is 7, indicating a neutral solution.

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The pair that is not a conjugate acid-base pair according to Brønsted-Lowry is H₃O+/OH-.

A conjugate acid-base pair consists of two species that differ by the transfer of a proton (H+). In this context, we can analyze each pair:

1. H₃O+/OH-: This is not a conjugate pair, as OH- needs to gain a proton to become H₂O, not  H₃O+ .
2. CH₃OH₂⁺/CH₃OH: This is a conjugate pair, as CH₃OH can accept a proton to become CH₃OH₂⁺.
3. HI/I-: This is a conjugate pair, as I- can accept a proton to become HI.
4. HSO₄⁻/SO₄²⁻: This is a conjugate pair, as SO₄⁻² can accept a proton to become HSO₄⁻.
5. H₂/H-: This is a conjugate pair, as H- can accept a proton to become H₂.
Therefore, the pair H₃O⁺/OH⁻is not a conjugate acid-base pair according to Brønsted and Lowry's theory.

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Out of the four elements given, only two elements, namely Cl (chlorine) and Xe (xenon), can form compounds with an expanded octet.

This is because they have vacant d-orbitals in their valence shell, which allows them to accommodate more than eight electrons and form compounds with an expanded octet. On the other hand, I (iodine) and O (oxygen) already have a complete octet in their valence shell and cannot form compounds with an expanded octet.
Out of the elements you mentioned (I, O, Cl, Xe), two elements can form compounds with an expanded octet. These are Iodine (I) and Xenon (Xe). They can accommodate more than 8 electrons in their valence shell due to the availability of empty d-orbitals in their respective electron configurations.

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The balanced equation in basic solution is:

Fe2+ + MnO4- + H2O → Fe3+ + Mn2+

To balance the given equation using the half-reaction method in basic solution, we first need to split the equation into two half-reactions:

Oxidation half-reaction: Fe2+ → Fe3+

Reduction half-reaction: MnO4- → Mn2+

Step 1: Balancing the Oxidation Half-Reaction

Fe2+ → Fe3+

We can balance the oxidation half-reaction by adding one electron to the left-hand side of the equation:

Fe2+ + e- → Fe3+

Step 2: Balancing the Reduction Half-Reaction

MnO4- → Mn2+

We start by identifying the oxidation state of each element in the reaction.

MnO4-: Mn has an oxidation state of +7, and each oxygen atom has an oxidation state of -2. The overall charge of the ion is -1, so the oxidation state of Mn + the sum of the oxidation states of the oxygens must equal -1. Therefore, we have:

MnO4-: Mn(+7) + 4(-2) = -1

Mn2+: Mn has an oxidation state of +2.

To balance the reduction half-reaction, we first balance the oxygen atoms by adding 4 OH- ions to the right-hand side of the equation:

MnO4- + 4OH- → MnO2 + 2H2O + 4e-

Next, we balance the hydrogen atoms by adding 2 H2O molecules to the left-hand side of the equation:

MnO4- + 4OH- + 3H2O → MnO2 + 8OH- + 4e-

Step 3: Balancing the Overall Equation

Now that we have balanced the oxidation and reduction half-reactions, we can combine them to get the overall balanced equation:

Fe2+ + MnO4- + 4OH- + 3H2O → Fe3+ + Mn2+ + 8OH-

Finally, we simplify the equation by canceling out the OH- ions on both sides of the equation:

Fe2+ + MnO4- + H2O → Fe3+ + Mn2+

Therefore, the balanced equation in basic solution is:

Fe2+ + MnO4- + H2O → Fe3+ + Mn2+

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The two double bonds are drawn between the carbon at the 1-position and the adjacent carbons, which both have a negative charge. This structure shows that the positive charge is delocalized throughout the ring, making the carbocation more stable.

Resonance structures are important in determining the stability of carbocations. To complete the resonance structure drawn, we need to add one curved arrow to show the movement of an electron pair that results in the positive charge moving to the 1-position of the ring. This movement of electrons creates a new bond between the carbon at the 1-position and the adjacent carbon, which now has a positive charge.
To complete the resonance structure, we need to draw two double bonds that have a positive charge at the 1-position of the ring.
Overall, resonance structures are important in stabilizing carbocations by spreading out the positive charge throughout the molecule. By completing the resonance structure with two double bonds that have a positive charge at the 1-position of the ring, we can see the importance of delocalization of charge in creating a more stable carbocation.

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