A student sets up the following equation to solve a problem in solution stoichiometry.
(The ? stands for a number the student is going to calculate.)
Enter the units of the student's answer.

The age of rock that is containing 0.628 g Uranium-238 is 2032631864.35 years.

Half-life can be given as the time required by the object to reduced to half of its initial concentration. The concentration remained can be given as:

Final concentration = Initial concentration * 1/2 ^ (t/t1/2)

The initial concentration of rock has been the remaining uranium-238 and lead-206 cumulative concentration. Thus, the initial concentration is given as:

Initial concentration = Uranium-238 + Lead-206

Initial concentration = 0.068 g + 0.025 g

Initial concentration = 0.093g

The final concentration of Uranium-238 remained = 0.68 g

The half life given = 4.5 × 10⁹ years

The age (t) of the rock can be given as:

[tex]0.068=0.093\left(\frac{1}{2}\right)^{\frac{t}{4.5\times \:10^9}}\\\\\frac{t\times \:10^{-9}}{4.5}\ln \left(\frac{1}{2}\right)=\ln \left(\frac{68}{93}\right)\\\\t=2032631864.35961[/tex]

Thus, the age of rock that has remaining 0.068 g Uranium-238 is 2032631864.35 years.

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Styrene is an organic chemical used to make rubber, latex, resins, etc. It can be best converted using the anti-Markovnikov rule by (1) BH₃ - THF and (2) H₂O₂, NaOH. Thus, option B is correct.

The anti-Markovnikov rule defines the attachment of the substituent group to the least substituted carbon than the more substituted groups. This is due to the induction and hyperconjugation effect.

To convert styrene, the alcohol group from sodium hydroxide must be added to the least substituted carbon of the compound using other reagents like BH₃ - THF and hydrogen peroxide.

Therefore, option B. (1) BH₃ - THF and (2) H₂O₂, NaOH can be used to convert styrene.

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Your question is incomplete, but most probably your full question was, What is the best method to convert styrene, C₆H₅CH=CH₂, to 2-phenyl ethanol with minimal side product formation?

Answer:

Atom

Explanation:

That describes an ATOM

Answer:

3.97 x 10²² molecules

Explanation:

To find the moles, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

Before you can plug the values into the equation, you need to (1) convert the pressure from torr to atm (by dividing by 760), then (2) convert the volume from mL to L (by dividing by 1000), and then (3) convert the temperature from Celsius to Kelvin (by adding 273).

P = 800.0 torr / 760 = 1.05 atm               R = 0.08206 atm*L/mol*K

V = 1500 mL / 1000 = 1.5 L                       T = 19.0 °C + 273 = 292 K

n = ? moles

PV = nRT

(1.05 atm)(1.5 L) = n(0.08206 atm*L/mol*K)(292 K)

1.579 = n(23.96)

0.0659 = moles

Now, you need to convert moles to molecules using Avogadro's Number.

Avogadro's Number:

6.022 x 10²³ molecules = 1 mole

 0.0659 moles          6.022 x 10²³ molecules
------------------------  x  -------------------------------------  =  3.97 x 10²² molecules
                                               1 mole

A chemistry laboratory device is a clay triangle to support the crucible when heated to high temperatures. It consists of a three-sided triangle frame made of clay or ceramic material.

The heat can pass through the open sides and is evenly distributed throughout the crucible. Although the soil triangle is a common and commonly used piece of laboratory equipment, it is difficult to ascertain who actually invented it.

The design of the clay triangle probably changed over time as a result of diverse contributions from scientists and chemists working on building laboratory equipment. The soil triangle may not have had a clear beginning, as is the case with many scientific instruments, but it is considered an evolution in laboratory technology overall.

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The equation described by the kb value is 5.21 x [tex]10^-^3[/tex].

The potential of Hydrogen is what pH is formally known as. The negative logarithm of the concentration of H+ ions is known as pH. Thus, the definition of pH as the amount of hydrogen is provided. The hydrogen ion concentration in a solution is described by the pH scale, which also serves as a gauge for the solution's acidity or basicity.

Assuming  PO₄³⁻ (the phosphate anion).

PO₄³- + H₂O ==> HPO₄²⁻ + OH⁻

Kb = [HPO₄²⁻][OH⁻] / [PO₄³⁻]

We can find the [OH⁻] from the pH of 12.70.

pH + pOH = 14

14.0 - 12.7 = 1.3 = pOH

[OH] = 1x[tex]10^-^1^.^3[/tex]

[OH-] = 5.0x[tex]10^-^2[/tex] M

[HPO₄²⁻] = 5.0x[tex]10^-^2[/tex] M

Kb = (5.0x[tex]10^-^2[/tex])2 / 0.48

     = 2.5x[tex]10^-^3[/tex] / 0.48

Kb = 5.21 x [tex]10^-^3[/tex]

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The prediction is that B. The electrons will flow to the zinc anode where a negative charge will build up and eventually halt the reaction.

This is known as a chemical element, of the periodic table, that is essential to life and is one of the most widely used metals. Zinc is of considerable commercial importance.

Without the salt bridge, positive and negative charges will build up around the electrodes causing the reaction to stop.

Hence, we know that the purpose of the salt bridge is to keep the solutions electrically neutral and allow the free flow of ions from one cell to another.

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According to Beer-Lambert's law, the solution allows for the passage of 3/4 of the incoming light intensity at that wavelength.

Calculation:

According to Beer-Lambert's law,

A = ∈ * C * l

A = log(I₀/I)

T = I₀/I

log (I₀/I) = ∈ * C * l

Here,

A= absorbance of the solution

l= path length= 2.67 mm = 0.267 cm (∵ 1mm = 0.1 cm)

T= transmittance

C= concentration of the given solution = 3.17mmol/L = 3.17* 10⁻³ mol/L

I= transmitted light

I₀= incident light

∈= molar absorption coefficient = 712 L mol⁻¹cm⁻¹

A = ∈ * C * l= 712 * 3.17 * 0.267 *10⁻³ = 602.63* 10⁻³ =0.602

T = [tex]10^{-A}[/tex] = [tex]10^{-0.602}[/tex] = 0.25

so, the percentage of transmittance = 0.25 * 100 = 25 %

Therefore, the intensity reduction will be equal to the original radiation intensity less the transmittance.

Give the initial intensity = 100

So, the reduction = 100 - 25 = 75.

Hence the fraction of incident light intencity= 75/100 = 3/4

Therefore the answer is 3/4.

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Density of ZnS given that the zn-s distance and bond angle are 0. 234 nm and 109. 5o, respectively if the atomic weights of zinc and sulfur are 65. 41 g/mol and 32. 06 g/mol. is 4.1109 g/cm³

For density In the crystal structure, we determine the angle's value for one set of ZnS bonds. θ θ + ∅ + 90° = 180° θ = 90° - ∅ θ = 90° - (109.5° / 2) After determining that = 35.25°, we derive the value of x from the geometry: distance angle d = 0.234 x = dsin = 0.234 sin35.25°) = 0.135 nm = 0.135 107 cm; following this, we get the length of the unit cell: a = 4x a = 4(0.135) a = 0.54 nm = 0.54 107 cm; and The formula for n' is (number of corner atoms in unit ell contribution of each corner atom in unit cell) + (number of face center atoms in unit cell contribution of each face center atom in unit cell). n' = 8 × 1/8) + (6 × 1/2) = 1 + 3 = 4

Density = P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)] = 389.88 / 94.84 = 4.1109 g/cm³

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Intermolecular hydrogen force must be overcome to convert liquid NH3 to NH3 vapour.

Sublimation is the phase transition of a substance directly from the solid to the gaseous phase without passing through the liquid phase.

There are the 3 forces that exist between the 2 molecules of NH3

1. Dispersion force: these forces exist between any 2 molecules regardless of anything else.

2. Hydrogen bonding: H is directly bonded to the extremely electronegative atom N, O and F, hence is nearly stripped off electron due to the large electronegativity difference. Hydrogen being small atom further reduces in size even allowing it to approach other molecules very closely. Here, N, O, and F gathers more partial negative charge.

Due to the closure approach between the molecule and strong partial charge, hydrogen bonding is generally more important than dipole bond.

3. dipole dipole interaction: NH3 is a polar molecule with a permanent separation of charge. Each molecules behaves like a tiny magnet and has positive and negative poles so 2 NH3 molecules attract each other. Dipole dipole interactions are stronger than dispersion force in the small molecules.

Thus we concluded that the hydrogen bond should overcome to convert NH3 to vapours.

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Ksp(the solubility product constant) = [Cu⁺] [I⁻]

So, the Ksp for Cui would be:

Ksp = (2.26 × 10⁻⁶) (2.26 × 10⁻⁶) = 5.11 x 10⁻¹²

Formula used:

K = [tex]K_{sp} = [A^+]^a [B^-]^b[/tex], where

Ksp = solubility product constant

A⁺    = cation in an aquious solution

B⁻ = anion in an aqueous solution

a, b = relative concentrations of a and b

The equilibrium constant for a solid material dissolving in an aqueous solution is the SOLUBILITY PRODUCT CONSTANT, Ksp. It stands for the degree of solute dissolution in solution. A substance's Ksp value increases with how soluble it is.

Take into account the general dissolving response (in aqueous solutions) below:

                           aA(s)⇔cC(aq)+dD(aq)

The molarities or concentrations of the products (cC and dD) must be multiplied in order to find the Ksp. Any product that has a coefficient in front of it must be raised to the power of that coefficient (and also multiply the concentration by that coefficient).

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436 atoms of Pb 207 are left in the rock.

Alpha and beta decays are the mechanism through which uranium transforms into the lead. While 235U and its daughter nuclides only experience seven alpha and four beta decays, 238U and its daughter nuclides suffer eight alpha and six beta decays overall.

When there are too few neutrons in relation to protons, alpha decay occurs. There is the discharge of an alpha particle, which consists of two protons and two neutrons.

When there are more neutrons than protons, beta decay can happen. A high-speed electron is ejected from a proton that transforms spontaneously from a neutron.

Since gamma radiation is emitted as a wave rather than a particle, the atomic structure of an atom remains unaltered. Atomic fission is the primary cause of neutron emission. At one time, many neutrons can be released.

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[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Let's solve ~

Initial concentration of weak acid HA = 0.200 M

and dissociation constant ([tex]{ \alpha} [/tex]) is :

[tex]\qquad \sf  \dashrightarrow \: \alpha = \frac{dissociation \: \: percentage}{100} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \alpha = \frac{9.4}{100} = 0.094[/tex]

Now, at initial stage :

At equilibrium :

Now, we know :

[tex]\qquad \sf  \dashrightarrow \: { K_a = \dfrac{[H+] [A-]}{[HA]}} [/tex]

( big brackets represents concentration )

[tex]\qquad \sf  \dashrightarrow \: { K_a = \dfrac{0.0188×0.0188}{0.1812}} [/tex]

[tex]\qquad \sf  \dashrightarrow \: { K_a = \dfrac{0.00035344}{0.1812}} [/tex]

[tex]\qquad \sf  \dashrightarrow \: { K_a \approx 0.00195 } [/tex]

[tex]\qquad \sf  \dashrightarrow \: {K_a \approx 1.9 × {10}^{-3} } [/tex]

According to the problem, we have a 0.200 M solution of a weak acid HA that is 9.4% ionized. To find the Ka (acid dissociation constant) for HA, we can use the percent ionization formula:

From this formula, we know that [H+] (the concentration of hydrogen ions) is equal to 9.4% of [HA]initial, and [HA]initial is equal to the initial concentration of the weak acid, which is 0.200 M. Solving for [H+], we get:

Now we can use the equation for Ka:

We don't know the concentration of the conjugate base (A-) at this point, but we can assume that it is equal to [H+] because the weak acid is only slightly ionized. Therefore, we can substitute [A-] = [H+] = 0.0188 M into the equation and solve for Ka:

[tex]\sf Ka = \dfrac{(0.0188)^2}{(0.200 - 0.0188)} = \bold{1.8 \times 10^{-3}}[/tex]

So the answer is [tex]\bold{ B.\: 1.8 \times 10^{-3}}[/tex].

I hope this helps!

The pressure of the mixture can be found out adding the Partial Pressure of the gas.

Partial pressure of gas is the pressure if it alone occupied the entire volume of original mixture.

Moles = given mass / molar mass

Given,

Mass of Nitrogen = 32g

Molar Mass of Nitrogen = 28g

Mass of Neon = 16g

Molar Mass of Neon = 20g

Volume of gas = 4L

Temperature of vessel = 0°C = 273K

Partial Pressure of nitrogen = P₁

Partial Pressure of neon = P₂

Moles of nitrogen, n₁= 32/28 = 1.14

Moles of neon, n₂= 16/20 = 0.8

Using Ideal gas equation for Nitrogen,

∴ P₁V = n₁RT

⇒ P₁ (1) = 1.14 × 0.082 × 273

⇒ P₁ = 25.52atm

Using Ideal gas equation for Neon,

∴ P₂V = n₂RT

⇒ P₂ (1) = 0.8 × 0.082 × 273

⇒ P₂ = 17.9 atm

Total pressure of the mixture can be found out by adding the partial pressures of both the gases.

Total pressure P = P₁  + P₂  

P = 25.52 + 17.9 atm

P = 43.42 atm

Hence, total pressure of the mixture is 43.42atm

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The initial ph of the 0. 10M ch3cooh  (aq) solution is  -0.64.

The salt sodium acetate is present at the equivalency point. It is a salt of a strong base and a weak acid. The expression "its pH" provides the information.

pH = 7 + 0.5p + 0.5logC

where C is the salt concentration. Monobasic acid is Acetic acid here.

25 ml of 0.10 M NaOH solution are needed for the titration of 25 ml of 0.10 M acetic acid.

Total volume = 25 + 25 = 50ml

The salt concentration = C = 0.10×  = 0.05

The pH for weak acid is given by,

pH = [tex]\frac{1}{2}[/tex] (p[tex]K_{a}[/tex] + logC)

pH =  [tex]\frac{1}{2}[/tex] (1.8× + log0.05)

pH =  [tex]\frac{1}{2}[/tex] [1.8× + (-1.30)]

pH =  [tex]\frac{1}{2}[/tex] (1.8× -1.30)

pH = -0.64

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Divide the number of moles of NaOH by the number of liters of NaOH solution needed to reach the titration's endpoint to determine the molarity of the NaOH solution.

"The process of calculating the quantity of a material A by adding measured increments of substance B, the titrant, with which it reacts until exact chemical equivalency is obtained (the equivalence point)" is the definition of titration.

Titration is a method of chemical analysis where the amount of a sample's ingredient is determined by adding an exact known amount of a different substance to the measured sample in which the desired constituent interacts in a specific, known proportion.

The equivalence point, or the point at which chemically equivalent amounts of the reactants have been combined, is to be detected by the titration. The stoichiometry of the reaction determines how many reactants have been combined at the equivalence point.

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Answer:

7,9,11

Explanation:

this is because water includes 0H, which would mean that it is more than 6

Answer:

subscript

Explanation:

H₂ O   means there are two hydrogen atoms and one oxygen atom

               the ' ₂ '  is a subscript

The mass of AgCl precipitated is 0.215g.

Precipitate is to form an insoluble compound either by reacting two salts or by changing the temperature to affect the solubility of the compound.

To calculate mass of AgCl precipitate -

The first step is to make a balanced chemical equation.

2AgNO3 + CaCl2 ---> 2AgCl + Ca(NO3)2

Molecular Weights:

CaCl2 = 110.98 g/mol

AgNO3 =170.01g/mol

AgCl= 143.45 g/mol

Volume:

CaCl2: 30.0mL=0.03L

AgNO3: 15.0mL=0.015 L

mole  CaCl2 = 30/1000 x 0.2 =0.006

mole  AgNO3 =15/1000 x 0.1 = 0.0015

CaCl2 is in excess so 0.0015 mole AgCl  will form

0.0015 x 143.5 = 0.215g

The 0.215 mass in grams of AgCl will precipitate.

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The equilibrium potential and the standard free energy change can give the standard potential of the electrodes. The standard cell potential is 0.32 V.

A reaction's standard free energy change (∆Gº') is the energy produced when the reactants undergo changes to form the product. It is given as,

ΔG° = -nFE° cell

Given,

Standard free energy change (ΔG°) = − 61. 0 kJ

Number of moles of electrons (n) = 2

Faraday's constant (F) = 96500 C

The standard cell potential (E° cell) is calculated as:

E° cell = ΔG° ÷ -nF

E° cell = − 61000 ÷ -(2 × 96500)

E° cell = 61000 ÷ 193000

= 0.32 V

Therefore, 0.32 V is the standard cell potential of the cell.

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Answer:

Density = 0.27 decagrams/cm³

Explanation:

From conversion tables, we know that;

1 g/cm³ = 0.1 decagrams/cm³

We are given;

Density of aluminium = 2.7 g/cm³

Thus;

Density = 2.7 * 0.1

Density = 0.27 decagrams/cm³

Answer:

270000 decagrams/m³

Explanation:

1.

Density=mass/Volume

=2.7g/1cm3

=(2.7/1000)/(1/1000000)

=2.7x1000

Density=2700kg/ m3

= 270000 decagrams/m³

2.

1000g=1kg

1g=1/1000kg

1cm3= ? m3

100cm=1m

1cm=1/100 m

1cm3=1/1000000 m3

The group of unsaturated hydrocarbons which 2 carbons are double bonded together, with H bonded to the left, and C H 2 bonded below left, above right, and below right is derived from Alkenes

Organic compounds are compounds which contains carbon and hydrogen

Some few classes or organic compounds or hydrocarbons are as follows:

So therefore, the group of unsaturated hydrocarbons which 2 carbons are double bonded together, with H bonded to the left, and C H 2 bonded below left, above right, and below right is derived from Alkenes

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Answer:

The answer is A.) Alkene

All the answers in that section are:

Alkene

Aromatic Hydrocarbon

Alkene

Alkyne

Explanation:

Here's the picture as proof. Hope it helps!

A combustion reaction will always be associated with a change in entropy that will be positive.

The entropy of a system is the measurement of the disorder of the system which happens in the thermodynamic reaction.

During association the value of entropy  will be positive and during dissociation the value will be negative.

During association the number of molecule increases due to which the entropy of the combustion increases and the value will be positive .

Therefore , combustion reaction will always be associated with a change in entropy that will be positive.

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The iron ion's charge is +2 in iron (II) sulfate, which is denoted by the symbol (II).

Information about iron (II) sulfate:

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The pH of the solution at 25 degree celsius of 1.3 × 10⁻⁶ moles of a sample of Sr(OH)₂ is 10.02.

The pH of any solution gives an idea about the acidic and basic nature of the solution and the equation of pH will be represented as:

pH + pOH = 14

Given that,

Moles of Sr(OH)₂ = 1.3 × 10⁻⁶ mol

Volume of solution = 25mL = 0.025L

The concentration of Sr(OH)₂ in terms of molarity = 1.3×10⁻⁶/0.025

                                                                                    = 5.2×10¯⁵M

Dissociation of Sr(OH)₂ takes place as:

                               Sr(OH)₂ → Sr²⁺ + 2OH⁻

From the stoichiometry of the reaction 1 mole of Sr(OH)₂ produces 2 moles of OH⁻.

Given that the base is a strong base and that it entirely dissociates into its ions, the hydroxide ion concentration is 5.2×10¯⁵×2 = 1.04×10¯⁴ M.

pOH = -log[OH⁻]

pOH = -log(1.04×10¯⁴)

pOH = 3.98

Now we put this value on the first equation we get,

pH = 14 - 3.98 = 10.02

Therefore, the value of pOH is 10.02.

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Considering the definition of pH, the hydronium ion concentration [H₃O⁺] of the solution is  1×10⁻³ M.

pH is the Hydrogen Potential. It is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.

Mathematically, pH is calculated as the negative base 10 logarithm of the activity of hydrogen ions:

pH= - log [H⁺]= -log [H₃O⁺]

The numerical scale that measures the pH of substances includes the numbers from 0 to 14. The pH value 7 corresponds to neutral substances. Acidic substances are those with a pH lower than 7, while basic substances have a pH higher than 7.

In this case, being pH= 3, you can replace this value in the definition of pH:

3= -log [H₃O⁺]

Solving:

[H₃O⁺]= 10⁻³ M

[H₃O⁺]= 1×10⁻³ M

Finally, the hydronium ion concentration [H₃O⁺] of the solution is  1×10⁻³ M.

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Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.

Commonly, sacrificial electrodes are employed to stop another metal from corroding or oxidising. A metal that is more reactive than the metal being shielded must serve as the sacrificial electrode. Magnesium, aluminium, and zinc are the three metals most frequently used in sacrificial anodes.

Manganese-Magnesium (Mn-Mg) electrode is more suited for on-shore pipelines where the electrolyte (soil or water) resistivity is higher since it has the highest negative electropotential of the three. In order to replenish any electrons that could have been lost during the oxidation of the shielded metal, the highly active metal offers its electrons.

Therefore, Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.

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Answer:

32(molecular mass has no unit )

Explanation:

(16)(o2)

16×2

=32

Answer:

synergy

Explanation:

when two substances combine effects to be greater than just their sum of their individual effects