Calculate the volume of 0.500 M C2H3OH and 0.500 M CH3O-Na required to prepare 0.100 L of pH = 5.00 buffer with a buffer strength of 0.100 M. The pKa of C2H302H is 4.75. C2H302H: Number C2H3O2Na: Number

We know the formula to calculate the volume of the solution is :V= n/CWhere,V is the volume of the solution n is the number of moles of the solute.C is the concentration of the solution In this question, the number of moles of the solute is 2.5 and the concentration of the solution is 2.0M.The correct option is (b) 4.5 L.

Therefore, we have, V = n/CV= 2.5 / 2.0V= 1.25 LSo, 1.25 L solution is produced by dissolving 2.5 moles of solute in a 2.0 M solution.Now we have to calculate how many liters of solution is produced from 2.5 moles of solute when a 2.0 M solution is required. Concentration of the solution is given by the formula :C= n/V Where, C is the concentration of the solution.n is the number of moles of the solute. V is the volume of the solution Let's plug in the given values,2.0 M = 2.5/ VV = 2.5 / 2.0 MV = 1.25 LSo, 1.25 L solution is produced from 2.5 moles of solute when a 2.0 M solution is required. Answer: b.4.5 L

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The concentration of iron(II) chloride contaminant in the original groundwater sample is 109.5 mg/L or 109.5 ppm.

To calculate the concentration of iron (II) chloride contaminant in the original groundwater sample, follow the steps below:

Step 1: Write the balanced chemical equation for the reaction between iron(II) chloride and silver nitrate.feCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Fe(NO3)2(aq)

Step 2: Calculate the moles of silver nitrate used.

The molarity of silver nitrate = 48.0 mM or 0.0480 M

The volume of silver nitrate = 200. mL or 0.200 L

Number of moles of silver nitrate = Molarity × Volume= 0.0480 M × 0.200 L= 0.00960 mol

Step 3: Determine the number of moles of silver chloride formed. The balanced equation shows that 1 mole of iron(II) chloride reacts with 2 moles of silver nitrate to form 2 moles of silver chloride.

Moles of AgCl = (moles of AgNO3 used ÷ 2) = 0.00960 mol ÷ 2= 0.00480 mol

Step 4: Convert moles of silver chloride to mass.

The molar mass of AgCl = 143.32 g/molMass of AgCl = Moles of AgCl × Molar mass= 0.00480 mol × 143.32 g/mol= 0.689 g or 689 mgStep 5: Calculate the concentration of iron(II) chloride in the original groundwater sample.Mass of iron(II) chloride = Mass of AgCl × (1 mol FeCl2 ÷ 2 mol AgCl)× (126.75 g FeCl2 ÷ 1 mol FeCl2)= 689 mg × (1 mol FeCl2 ÷ 2 mol AgCl) × (126.75 g FeCl2 ÷ 1 mol FeCl2)= 21943.625 mg or 21.9 gThe original volume of groundwater sample = 200. mL or 0.200 L

Concentration of iron(II) chloride in the groundwater sample = (Mass of iron(II) chloride ÷ Volume of sample)× (1 L ÷ 1000 mL)= (21.9 g ÷ 0.200 L) × (1 L ÷ 1000 mL)= 109.5 mg/L or 109.5 ppmT

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The compound methylamine (CH3NH2) contains a covalent bond between the carbon and nitrogen atom, and in the bond, the carbon atom is slightly positive (+δ), So the correct option is C. slightly positive.

The carbon atom has an electronegativity value of 2.55 while the nitrogen atom has an electronegativity value of 3.04. Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The electronegativity difference between the carbon and nitrogen atom creates a polar bond, with nitrogen pulling electrons towards itself and becoming slightly negative, while carbon loses some electron density and becomes slightly positive in the C-N bond.

Methylamine (CH3NH2) is an organic compound that belongs to the primary amines. It is formed by replacing one hydrogen atom in ammonia with a methyl group (-CH3). The molecule is polar due to the presence of the C-N bond that makes the nitrogen slightly negative and carbon slightly positive

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The pentose phosphate pathway (PPP) is a metabolic pathway that generates NADPH and ribose 5-phosphate (R5P) in mammalian cells. The pathway provides cells with the products they need for biosynthesis, such as nucleic acids, amino acids, and fatty acids.

This pathway is essential for the cell's anabolic processes and is involved in redox homeostasis. It is primarily regulated by the cell's energy requirements. If there is a greater need for NADPH, the PPP flux will increase, and if there is a greater need for R5P, the flux will decrease. When the need for R5P is greater than the need for NADPH, most of the ribulose 5-phosphate is converted into fructose 6-phosphate.

This reaction is catalyzed by the enzyme phosphopentose isomerase, which converts ribulose 5-phosphate to ribose 5-phosphate and then to fructose 6-phosphate. This conversion is irreversible, and the process is known as the oxidative phase of the PPP.

Overall, the pentose phosphate pathway is a crucial metabolic pathway for maintaining redox balance and providing cells with the biosynthetic products they require.

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When a mutation prevents dephosphorylation of glycogen synthase, glycogen levels could remain high. This is because glycogen synthase is the enzyme that forms the glycosidic bonds required for glycogen formation and glycogen is the storage form of glucose.

Glycogen is a polysaccharide that serves as the storage form of glucose in animals. Glycogen is stored in the liver and muscles. Glycogen synthase is the enzyme responsible for glycogen synthesis. Glycogen synthase is found in the liver and muscle tissue and is regulated by various hormones. Glycogen synthase converts glucose into glycogen via a condensation reaction.In glycogenesis, glycogen synthase produces α(1→4) glycosidic bonds between glucose molecules to form linear α(1→4)-linked glucose chains.

These linear chains are then branched via the action of branching enzyme, which produces α(1→6) glycosidic bonds. The result is a highly branched, complex glycogen molecule.How does glycogen levels remain high when a mutation prevents dephosphorylation of glycogen synthase?When a mutation prevents dephosphorylation of glycogen synthase, the enzyme remains in its active form, and glucose is continually converted to glycogen, resulting in high levels of glycogen. Glycogen synthase is typically activated by dephosphorylation and inactivated by phosphorylation. In the presence of a mutation that prevents dephosphorylation, the enzyme would remain in its active form, continually forming glycogen. As a result, the glycogen level would remain high. Therefore, glycogen levels can remain high when a mutation prevents dephosphorylation of glycogen synthase.

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We can substitute the chemical formula for 4-bromoaniline in this equation and obtain the final answer as:

C6H4BrNH2 + H2O ⇌ C6H4BrNH3+ + OH-Kb = [C6H4BrNH3+][OH-]/[C6H4BrNH2]

Thus, the left-hand side of the given equilibrium constant equation is

C6H4BrNH2

and the complete equation is

:C6H4BrNH2 = Kb

The equilibrium constant (Kb) is used to define the basicity of a compound. When we talk about basicity, it refers to the ability of a compound to take a proton (H+) from another molecule. Here, we need to complete the equation for the equilibrium constant of 4-bromoaniline, a weak base, with water. We know that the reaction of 4-bromoaniline with water takes the following form:

C6H4BrNH2 + H2O ⇌ C6H4BrNH3+ + OH-

We can now write the expression for the Kb of 4-bromoaniline as follows:

Kb = [C6H4BrNH3+][OH-]/[C6H4BrNH2]

We can substitute the chemical formula for 4-bromoaniline in this equation and obtain the final answer as:

C6H4BrNH2 + H2O ⇌ C6H4BrNH3+ + OH-Kb = [C6H4BrNH3+][OH-]/[C6H4BrNH2]

Thus, the left-hand side of the given equilibrium constant equation is

C6H4BrNH2

and the complete equation is:

C6H4BrNH2 = Kb

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The most likely fate of a protein with an N-terminal hydrophobic sorting signal and an additional internal hydrophobic domain of 22 amino acids correct option is B. the protein is transported to mitochondria.

A protein is a macromolecule composed of amino acid chains joined together by peptide bonds. They can perform various functions, including catalyzing metabolic reactions, replicating DNA, responding to stimuli, and transporting molecules from one location to another within cells. The N-terminal sorting signal is a short sequence of amino acids that is present at the start of a protein. The sorting signal is responsible for directing the protein to its appropriate location within the cell. A protein with an N-terminal hydrophobic sorting signal and an additional internal hydrophobic domain of 22 amino acids is transported to mitochondria.

The presence of both an N-terminal hydrophobic sorting signal and an internal hydrophobic domain suggests that the protein is destined for transport to the mitochondria. Mitochondria are the primary organelles responsible for generating cellular energy. They are surrounded by a double membrane, the innermost of which is highly selective and aids in the transport of molecules and proteins necessary for energy production.

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The reaction involves a series of reactions that produce 5-dodecanone. The following is the synthetic pathway, which includes all reactions and mechanisms.

The synthetic route for the given transformation is shown below:

The starting compound is the phenylpropionic acid, and the reaction begins with the formation of the alkene through HBO and HBr in the presence of HOOH. The alkene produced can undergo bromination to give the corresponding alkyl bromide using Br2. The intermediate formed by the reaction then reacts with H2SO4 to form an alkyl oxide ion which is then subjected to hydrolysis using H2SO4 and HgSO4 to form the corresponding alcohol. The alcohol is then subjected to a series of reactions to form the final product.The alcohol is first reacted with CH3CH2CH2CH2CH2CI to form a new alkyl iodide. The alkyl iodide is then reacted with CH3CH2CH2CH2CH2CH2CI to form another alkyl iodide. The process is repeated with CH3CH2CH2CH2CH2CH2CH2CI.

The alkyl iodide produced is then treated with NaNH2/NH3 to form the corresponding alkyne. The alkyne is then hydrogenated using H/Pt to form the corresponding alkene. The alkene is then subjected to hydrogenation again, this time using Wilkinson's Catalyst, to form the corresponding alkane. The alkane is then reacted with Lindlar's Catalyst to form the corresponding alkene. The alkene is then reacted with Na/NH3 to form the corresponding alkyne. Finally, the alkyne is subjected to ozonolysis using O3 and then subjected to reduction using DMS (dimethyl sulfide) to form the final product. The final product is 5-dodecanone, which is produced through the reactions outlined above.

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The degree of unsaturation for the given formulas is as follows:

(a) C₂₄H₂₄: 36

(b) C₇H₆BrCl: 12

(c) C₉H₁₁N: 12.5

To determine the degree of unsaturation (index of hydrogen deficiency) in a formula, we can use the formula:

Degree of unsaturation = [tex]\[(2n + 2) - \frac{h + x}{2}\][/tex]

where n is the number of carbon atoms, h is the number of hydrogen atoms, and x is the number of halogen atoms (if present).

(a) C₂₄H₂₄:

Degree of unsaturation = [tex]\[(2 \times 24 + 2) - \frac{24 + 0}{2}\][/tex]

                     = 48 - 12

                     = 36

The degree of unsaturation for C₂₄H₂₄ is 36.

(b) C₇H₆BrCl:

Degree of unsaturation = [tex]\[(2 \times 7 + 2) - \frac{6 + 1 + 1}{2}\][/tex]

                     = 14 - 2

                     = 12

The degree of unsaturation for C₇H₆BrCl is 12.

(c) C₉H₁₁N:

Degree of unsaturation = [tex]\[(2 * 9 + 2) - \frac{11 + 0}{2}\][/tex]

                     = 18 - 5.5

                     = 12.5

The degree of unsaturation for C₉H₁₁N is 12.5.

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The change in entropy is given by ΔS = ΔQ/T, where ΔQ is the heat absorbed, T is the temperature, and ΔS is the change in entropy. In this case, ΔS = 69.1 J/K.

The change in entropy is given by:

[tex]\begin{equation}\Delta S = \frac{\Delta Q}{T}[/tex]

where ΔQ is the heat absorbed, T is the temperature, and ΔS is the change in entropy.

The heat absorbed is the latent heat of fusion, which is 445 J/g. The mass of the substance is 4.3 kg, so the heat absorbed is:

ΔQ = 445 J/g * 4.3 kg = 19185 J

The temperature is 4.4°C, which is 277.6 K. Therefore, the change in entropy is:

[tex]\begin{equation}\Delta S = \frac{19185 \si{\joule}}{277.6 \si{\kelvin}} = 69.1 \si{\joule\per\kelvin}[/tex]

Therefore, the change in entropy is 69.1 J/K.

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The amount of the radioactive material reaction remaining after a certain period of time can be determined using the formula:Nt = N0(1/2)t/t₁/₂where:Nt = remaining amount of the radioactive material after the elapsed time, t.

N0 = the initial amount of the radioactive material, t₁/₂ = half-life period of the material. Therefore, the answer is 12.5 g (to three significant figures).

Given,Initial amount, N0 = 100.0 gHalf-life, t₁/₂ = 2.4 minutes Elapsed time, t = 7.2 minutesThe formula to calculate the remaining amount is:Nt = N0(1/2)t/t₁/₂Substituting the values:Nt = 100.0 g (1/2)^(7.2/2.4)Nt = 100.0 g (1/2)³Nt = 100.0 g (0.125)Nt = 12.5 gThe amount of Zn-71 remaining after 7.2 minutes has elapsed is 12.5 g.

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When moderately compressed, gas molecules have very little attraction for one another with an below. A gas is a state of matter that is highly compressible, which means that its volume can be reduced by compressing and that it expands to fill any available space.

The kinetic energy of the gas molecules is the driving force behind this behavior. The gas molecules are in constant motion, colliding with one another and with the walls of the container in which they are contained. The intermolecular forces of attraction between gas molecules are negligible when the gas is moderately compressed. In other words, when the pressure of

the gas is not too high, the attractive forces between the molecules are negligible. This is because the distance between the molecules is too great for the attractive forces to have any significant effect. The ideal gas law, PV=nRT, assumes that the molecules of a gas have zero volume and do not interact with one another. While real gases do have volume and do interact with one another, the ideal gas law is a good approximation of the behavior of gases under most conditions.

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The compounds in increasing order of solubility in water are I < II < IV < III.

The increasing order of solubility in water from the given compounds can be determined as follows:

CH3–CH2–CH2–CH3 (I) is a hydrocarbon, which is a nonpolar substance and will not dissolve in water.

Thus, it is the least soluble in water.

CH3–CH2–O–CH2–CH3 (II) is an ether compound with a polar oxygen atom in the center.

It is more soluble in water than hydrocarbons but less soluble than alcohols.

CH3–CH2–OH (III) is an alcohol compound that contains a polar -OH group.

This polar group is capable of forming hydrogen bonds with water molecules, making it the most soluble in water.

CH3–OH (IV) is another alcohol compound that is similar to compound III.

Thus, it will be more soluble in water than hydrocarbons and ether compounds but less soluble than compound III.

Therefore, the compounds in increasing order of solubility in water are I < II < IV < III.

Option A, I < III < IV < II, is the exact opposite order, and hence it is incorrect.

Option B, I < II < IV < III, is the correct order and is the answer to the question.

Option C, III < IV < II < I, is in reverse order, and therefore, it is incorrect.

Option D, I < II < III < IV, is incorrect as it places alcohol CH3–OH (IV) before CH3–CH2–OH (III) which is not possible as the former is less soluble than the latter.

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The correct order of increasing first ionization energy of the atoms is a) Be > Na > Rb, b) Se > Te, c)  Br > Ni > K, and d) Ne > Sr > Se.

Ionization is defined as the energy required to remove an electron from a neutral atom in its ground state. As the ionization energy increases, the task of removing an electron becomes more challenging. As a result, in general, the first ionization energy increases across a period and decreases down a group because the atomic radius increases.

a) Be, Na, Rb

Be has the smallest atomic radius, Na has the second smallest atomic radius, and Rb has the largest atomic radius of the three elements. Therefore, Rb has the smallest first ionization energy, Na has the second smallest first ionization energy, and Be has the largest first ionization energy. The correct order, then, is Be > Na > Rb.

b) Se, Se, Te

This group of atoms contains duplicate elements. So, Te has a larger atomic radius than Se, and the first ionization energy decreases as the atomic radius increases. The correct order is, therefore, Se > Te.

c) Br, K, Ni

Among these atoms, K has the lowest first ionization energy. Br and Ni have comparable radii, but Ni has a larger atomic radius than Br, making it easier to remove an electron from Br than from Ni. So, the correct order is Br > Ni > K.

d) Ne, Sr, Se

Neon is a noble gas, which means it has a high first ionization energy and is highly stable. The atomic radius of Sr is larger than that of Se, making it easier to remove an electron from Se. So, the correct order is Ne > Sr > Se.

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In a SiO4-4 tetrahedron, the angle between covalent bonds is 109.5 degrees.What is SiO4-4 tetrahedron?The tetrahedron with SiO4-4 as the base is known as the SiO4-4 tetrahedron. The SiO4-4 tetrahedron is an orthosilicate (anions with tetrahedral coordination). \

SiO4-4 is a tetrahedral anion that forms the basic component of most silicates. Silicates are the most abundant and important minerals on the planet, and they include quartz, feldspar, mica, zeolites, and asbestos, among others.The four oxygen atoms in the SiO4-4 tetrahedron are located at the vertices of the tetrahedron and are bound to a central silicon atom, which is also at the tetrahedron's centre.

To stabilise the structure, the Si-O bonds in the tetrahedron are covalent and directional.In SiO4-4 tetrahedron, the angle between covalent bonds is 109.5 degrees. The tetrahedron has four sides, and each side has a 109.5-degree angle. It's a three-dimensional shape with four triangular faces and a tetrahedral geometry that has the SiO4-4 tetrahedron, with a total of 8 electrons in the valence shell of the silicon atom.

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A Pythagorean triple, the sum of the squares of the two smallest numbers must be equal to the square of the largest number. That is, if a, b, and c are three numbers that form a Pythagorean triple, then a^2 + b^2 = c^2.

Now, we compare the value of a^2 + b^2 (610) to the value of c^2. The largest number is 24, so c^2 = 24^2 = 576.Since a^2 + b^2 ≠ c^2 (610 ≠ 576), the set of numbers 13, 21, and 24 do not form a Pythagorean triple. Therefore, the statement "the set of numbers 13, 21, and 24 form a Pythagorean triple" is false.

The two smallest numbers are 13 and 21.So, we have a^2 + b^2 = 13^2 + 21^2 = 169 + 441 = 610.Now, we compare the value of a^2 + b^2 (610) to the value of c^2. The largest number is 24, so c^2 = 24^2 = 576.Since a^2 + b^2 ≠ c^2 (610 ≠ 576), the set of numbers 13, 21, and 24 do not form a Pythagorean triple. Therefore, the statement "the set of numbers 13, 21, and 24 form a Pythagorean triple" is false.

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A net ionic equation is a chemical equation that shows the reaction that occurred between ions in aqueous solutions. It focuses on the ions that were changed during the reaction.

The first step of writing a net ionic equation involves writing the balanced molecular equation for the reaction. Sr(NO3)2 and K2SO4 are soluble salts that will dissociate in water to give their constituent ions. The balanced molecular equation for this reaction can be written as: Sr(NO3)2 (aq) + K2SO4 (aq) → 2KNO3 (aq) + SrSO4 (s)The next step is to determine the ions that were involved in the reaction. Only the ions that changed during the reaction are included in the net ionic equation.

The potassium and nitrate ions are not involved in the reaction. Therefore, they are excluded from the net ionic equation. The net ionic equation is:2Sr²⁺ (aq) + SO4²⁻ (aq) → SrSO4 (s)Hence, the net ionic equation for the reaction between aqueous solutions of Sr(NO3)2 and K2SO4 is 2Sr²⁺ (aq) + SO4²⁻ (aq) → SrSO4 (s).

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The oxidation state of the highlighted atoms in the chemical species is as follows:

An atom's oxidation number or oxidation state in a chemical species reveals how many electrons it has lost or gained in a compound or ion.

In OH⁻ (aq), the highlighted atom is oxygen (O), and its oxidation state is -2.

In NH₄ (aq), the highlighted atom is nitrogen (N), and its oxidation state is -3.

In I⁻ (aq), the highlighted atom is iodine (I), and its oxidation state is -1.

In Br₂(g), the highlighted atom is bromine (Br), and since it is in its elemental form, its oxidation state is 0.

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The compound that is expected to have the shortest retention time in gas chromatography is D. 3-methyl cyclohexene.

In gas chromatography, the retention time is the time taken for a compound to travel through the column and reach the detector. The retention time depends on various factors such as the volatility, polarity, and interaction with the stationary phase.

In general, less polar and more volatile compounds tend to have shorter retention times in gas chromatography. Among the given options, 3-methyl cyclohexene is the most volatile and least polar compound. It is an alkene, which is generally less polar than alcohols or cyclohexanols.

Therefore, D. 3-methyl cyclohexene is expected to have the shortest retention time in the gas chromatograph compared to the other compounds listed.

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Increasing the concentration of acetic acid in a reaction can lead to a higher percentage of acetic acid reacting and producing [tex]CH_3CO_2[/tex].

In a chemical reaction, the concentration of reactants plays a crucial role in determining the extent of the reaction. By increasing the acetic acid concentration, more acetic acid molecules will be present in a given volume. This higher concentration leads to a more significant number of collisions between acetic acid molecules, increasing the chances of successful collisions that result in the formation of [tex]CH_3CO_2[/tex].

Additionally, an increased concentration of acetic acid can shift the equilibrium of the reaction towards the formation of [tex]CH_3CO_2[/tex]. Le Chatelier's principle states that if the concentration of a reactant is increased, the equilibrium will shift in the direction that consumes that reactant. Thus, by increasing the concentration of acetic acid, the equilibrium will favour the forward reaction, resulting in a higher percentage of acetic acid reacting and producing [tex]CH_3CO_2[/tex].

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The reaction between potassium t-butoxide with (1S,2S)-1-bromo-2-methyl-1-phenylbutane leads to the formation of (1S,2S)-1-methyl-2-phenylbut-2-ene. This is the E2 reaction involving a strong base and a primary substrate.

The mechanism of the reaction between potassium t-butoxide and (1S,2S)-1-bromo-2-methyl-1-phenylbutane:Explanation: A primary substrate is involved in the reaction which undergoes E2 elimination, leading to the formation of an alkene. Alkene formation is a two-step reaction.

The stereochemistry of the product is illustrated below: Thus, the product formed by the reaction of potassium t-butoxide with (1S,2S)-1-bromo-2-methyl-1-phenylbutane is (1S,2S)-1-methyl-2-phenylbut-2-ene and the stereochemistry of the product is trans.

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The answer to how much of an 800-gram sample of potassium-40 will remain after 3.9 × 109 years of radioactive decay is option (3) "200 grams."

The amount of an 800-gram sample of potassium-40 that will remain after 3.9 × 109 years of radioactive decay can be calculated by using the radioactive decay law. The radioactive decay law states that the number of radioactive nuclei N of a sample decreases as a function of time t. This can be given by the equation N = N₀ e^(-λt)

Where N₀ is the initial number of radioactive nuclei, λ is the decay constant, and t is the time.

The decay constant is related to the half-life T½ of the radioactive isotope by the equation

T½ = ln2 / λ Given that the half-life of potassium-40 is 1.28 × 10^9 years,

we can find the decay constant as follows

λ = ln2 / T½

= ln2 / (1.28 × 10^9)

= 5.43 × 10^-10 year^-1

Substituting the given values into the radioactive decay law, we get

N = 800 e^(-5.43 × 10^-10 × 3.9 × 10^9)N ≈ 200 grams

Therefore, the answer is option (3) 200 grams.

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The conversion factor needed to convert between mass and moles of the atom fluorine is the molar mass of fluorine (F₂).

The molar mass of fluorine is 38.00 g/mol which means that one mole of fluorine weighs 38.00 grams.

When given the mass of fluorine, dividing the given mass by the molar mass of fluorine (38.00 g/mol) will give the number of moles of fluorine present. On the other hand, when given the number of moles of fluorine, multiplying the given number of moles by the molar mass of fluorine (38.00 g/mol) will give the mass of fluorine present. The formula that can be used for this conversion is:n = m / MM

where n is the number of moles, m is the mass, and MM is the molar mass. It is important to keep in mind that the molar mass of any element or compound can be found by summing the atomic masses of all the atoms in the molecule.

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The pH of a solution after adding 40 ml of 0.10 M NaOH to 20 ml of 0.20 M HClO is 1.56.

Firstly, let us write down the balanced chemical equation for the reaction of HClO and NaOH. NaOH is a strong base, and HClO is a weak acid.NaOH + HClO → NaClO + H2OThe reaction is an acid-base reaction in which the products are NaClO and H2O.The equation tells us that one mole of NaOH reacts with one mole of HClO.

The concentration of H3O+ is calculated as follows:Ka = [H3O+] [ClO-] / [HClO]3.0 × 10-8 = [H3O+] [0.04] / [0.004] [0.02]H3O+ = 0.000173 MNow we can use the definition of pH to calculate it:pH = -log[H3O+]pH = -log[0.000173]pH = 1.56.

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The concentration of OH- in the aqueous solution is 1.0 x 10-3 M.

In an aqueous solution, the concentration of hydroxide ions (OH-) can be determined based on the concentration of hydronium ions (H3O+).

The relationship between the two can be expressed using the concept of the pH scale, where pH is defined as the negative logarithm of the H3O+ concentration.

Given that the H3O+ concentration is 1.0 x 10-11 M, we can determine the concentration of OH- using the relationship Kw = [H3O+][OH-]. Kw represents the ion product of water and is equal to 1.0 x 10-14 at 25°C.

Rearranging the equation, we find [OH-] = Kw / [H3O+].

Substituting the values, we get [OH-] = (1.0 x 10-14) / (1.0 x 10-11) = 1.0 x 10-3 M.

Therefore, the concentration of OH- in the aqueous solution is 1.0 x 10-3 M.

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The concentration of the original Ba(OH)₂ solution if 15.0 ml solution of Ba(OH)₂ is neutralized with 22.7 ml of 0.200 m HCl is 151.3 mol/dm³

To determine concentration of the original Ba(OH)₂ solution, we must know he balanced chemical equation for the neutralization reaction is:

Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O

From the equation above, the stoichiometric ratio of Ba(OH)₂ and HCl is 1:2. That means one mole of Ba(OH)₂ reacts with 2 moles of HCl. The balanced chemical equation also shows that the number of moles of HCl used is the same as the number of moles of Ba(OH)₂. Hence:

moles of HCl = 0.200 mol/dm³ × 22.7 dm³ = 4.54 mol

Using the stoichiometric ratio, the moles of Ba(OH)₂ in the solution can be calculated to be:

moles of Ba(OH)₂ = 4.54 mol ÷ 2 = 2.27 mol

The volume of the Ba(OH)₂ solution is 15.0 mL, which is 0.015 dm³. Therefore, the concentration of the original Ba(OH)₂ solution can be calculated as:

concentration = moles/volume= 2.27 mol ÷ 0.015 dm³= 151.3 mol/dm³

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The concentration of the original Ba(OH)₂ solution is 0.302 M.

Given data

Volume of Ba(OH)₂ solution used = 15.0 ml

Volume of HCl used = 22.7 ml

Molarity of HCl solution used = 0.200 M

We need to calculate the concentration of Ba(OH)₂ solution, which is not known.Molar ratio of HCl and Ba(OH)₂ in a balanced chemical equation of their neutralization is;

HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

The balanced chemical equation tells us that 1 mole of HCl is required to neutralize 1 mole of Ba(OH)₂.

So, the moles of HCl used in the reaction is;

moles of HCl = molarity × volume (in liters)

moles of HCl = 0.200 M × 0.0227 L = 0.00454 mole

Since one mole of HCl reacts with 1 mole of Ba(OH)₂,

so the number of moles of Ba(OH)₂ used is also equal to 0.00454 mole. Since we know the volume of the Ba(OH)₂ solution used, we can calculate the molarity of the solution as;

molarity = moles of solute / volume of solution in liters

Molarity = 0.00454 / (15.0 / 1000) = 0.302 M

Therefore, the concentration of the original Ba(OH)₂ solution is 0.302 M.

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When you move from left to right across a row and up a column on the periodic table, the statement which is true is "It becomes more difficult to add an electron to the atom. This is due to the fact that the electrons are added to the same energy level as the valence electrons.

When you move from left to right across a row and up a column on the periodic table, the statement which is true is "It becomes more difficult to add an electron to the atom. "This is due to the fact that the electrons are added to the same energy level as the valence electrons. As a result, there are more protons in the nucleus, resulting in a stronger electrostatic pull on the valence electrons, making it more difficult to add electrons. M The periodic table is a graphical representation of the elements arranged in rows and columns based on their atomic structure. It is designed in a way to reflect the chemical and physical properties of the elements. The periodic table has eight groups and seven rows. The groups contain elements with similar properties, while the rows contain elements with the same number of electron shells.

The electron configuration of the elements determines their position in the periodic table. The valence electrons, which are found in the outermost shell, determine the element's chemical properties. Electrons are negatively charged particles that revolve around the nucleus in shells. The energy of the electrons increases with the distance from the nucleus, and it takes more energy to add an electron to a higher energy shell.When moving from left to right across a row, the number of protons in the nucleus increases, making the electrostatic attraction between the nucleus and the valence electrons stronger. This results in the electrons being held more tightly, making it more challenging to add an electron. As a result, the atom becomes smaller and more electronegative as you move across a row. When moving up a column, the number of electrons in the outermost shell increases, making the size of the atom larger. In addition, the strength of the nucleus' attraction decreases, making it easier to add an electron to the outermost shell. As a result, the atoms become more reactive as you move down the column.

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The pH of a 0.0000001 Molar HCl solution is 7.

Since HCl is a strong acid, it dissociates completely in water to form H+ and Cl- ions.

The concentration of H+ ions in the solution will be equal to the concentration of the HCl, which is 0.0000001 Molar.

Using the pH scale, we can calculate the pH of this solution as follows:pH = -log [H+]pH = -log 0.0000001pH = 7

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

The pH of a 0.0450 Molar Ba(OH)2 solution is 12.

Since Ba(OH)2 is a strong base, it dissociates completely in water to form Ba2+ and OH- ions.

The concentration of OH- ions in the solution will be twice the concentration of Ba(OH)2, which is 0.0450 Molar.

Using the pH scale, we can calculate the pH of this solution as follows:pOH = -log [OH-]pOH = -log (2 x 0.0450)pOH = 1.34pH + pOH = 14pH = 14 - 1.34pH = 12.66

The pH of the solution is 12.66, which is basic.

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30.0 mL of 0.050 M Ca(CN)2 are needed to make 25.0 mL of 0.010 M solution. Hence, Volume of 0.050 M solution containing 0.00025 mol of Ca(CN)2= 0.00025 / 0.00125 = 0.2 L or 200 mL.

Molarity of Ca(CN)2 solution = 0.050 M Molarity of solution to be made = 0.010 MVolume of solution to be made = 25.0 mLNumber of moles of Ca(CN)2 in 25.0 mL of 0.010 M solution =0.010 * 25.0 / 1000 = 0.00025 molMolar mass of Ca(CN)2 = 80.11 g/mol

Mass of Ca(CN)2 in 0.00025 mol of Ca(CN)2 = 0.00025 * 80.11 = 0.020 m gNumber of moles of Ca(CN)2 in 0.050 M solution = 0.050 * 25.0 / 1000 = 0.00125 mol Therefore, Volume of 0.050 M solution containing 0.020 mg of Ca(CN)2 = (200/1000) * 0.020 = 0.004 mL or 4.0 mL Therefore, Volume of 0.050 M solution containing 20.0 mg of Ca(CN)2 = (4.0/0.020) * 20.0 = 400.0 mL or 0.400 L.

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Phenol has the chemical formula C6H5OH. It is a weak acid and when dissolved in water it undergoes an ionization reaction as shown below C6H5OH(aq) + H2O(l) ⇌ H3O+(aq) + C6H5O-(aq).

K a = \[\frac{[H_3O^+][C_6H_5O^-]}{[C_6H_5OH]}\]The Ka for phenol is given as 1.3 × 10−10.Let x be the degree of dissociation of phenol.The initial concentration of phenol is 0.56 M.The concentration of the undissociated phenol is (0.56 - x) M.The concentrations of the H3O+ and C6H5O− ions are each x M. Applying the weak acid equilibrium reaction and Ka expression, we have;Ka = \[\frac{[H_3O^+][C_6H_5O^-]}{[C_6H_5OH]}\]1.3 × 10−10 = \[\frac{x^2}{0.56 - x}\]Since x is very small compared to 0.56,

We can safely assume that 0.56 - x ≈ 0.56.So, 1.3 × 10−10 = x2/0.56x = √(1.3 × 10−10 × 0.56)x = 1.129 × 10−6The percent ionization of phenol is given by;Percent ionization = \[\frac{x}{[C_6H_5OH]}\]Percent ionization = \[\frac{1.129 \times 10^{-6}}{0.56} \times 100\% = 0.000202 \times 100\% = 0.0202\%\]Therefore, the percent ionization of phenol in a 0.56 m aqueous solution is 0.0202%.

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