The atomic and mass numbers for four different atoms are given below. Which two are isotopes?

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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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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Answer:

7,9,11

Explanation:

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

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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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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Answer:

Explanation:

to put this plainly after putting strong electrolytes in water they dissolve to nothing but ions they give you an example and definition of a strong electrolyte I suggest you read those and see if you come up with what I gave you they also tell you that not all strong electrolytes dissolve as strongly as others that are in the example.

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?

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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Hello its me Sanya yadav

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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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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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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Answer:

32(molecular mass has no unit )

Explanation:

(16)(o2)

16×2

=32

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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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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ANSWER: From this experiment, Jeff can conclude that the heat capacity of oil is lower than that of water which means that oil requires less energy per gram of liquid in order to change its temperature.

Answer:

Atom

Explanation:

That describes an ATOM

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

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 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 symmetrical acid anhydride will undergo a reaction with an alcohol to give an ester and a carboxylic acid. An asymmetrical acid anhydride will undergo a reaction with an alcohol to give two esters and two carboxylic acids.

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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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1. The first step is to convert your given masses into moles. This way, you can see which reactant is in excess and which one is limiting. A good flow of thought would be: kilograms of SiO2/Carbon→ grams SiO2/Carbon→moles SiO2/Carbon.

2. Repeat this step for the carbon reactant.

3. By comparison, we can tell that the carbon reactant is in excess, which means that the SiO2 reactant is limiting. Because of this, we’ll use the amount of SiO2 to find our percent yield. There’s only so much SiO2 that can react with the carbon.

4. You can now calculate the actual and theoretical yields. The theoretical yield is larger than the actual yield because it does not have physical interactions like loss of energy to the environment and friction to interfere with the reaction.

5. Now compare your two values of actual and theoretical. The formula for percent mass is [(Actual yield)/(Theoretical yield)]*100%. Our final answer for this question is the percent yield of the reaction is 90.8% We can interpret this as 90.8% of the SiO2 reacting will produce solid silicon.

Hope this helps!!

[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 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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Answer:

subscript

Explanation:

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

               the ' ₂ '  is a subscript

Answer:

synergy

Explanation:

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

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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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