In compliance with conservation of energy, Einstein explained that in the photoelectric effect, the energy of a photon (hv) absorbed by a metal is the sum of the work function (Φ), the minimum energy needed to dislodge an electron from the metal’s surface, and the kinetic energy (Ek) of the electron: hv = Φ + Ek. When light of wavelength 358.1 nm falls on the surface of potassium metal, the speed (u) of the dislodged electron is 6.40 x 10⁵ m7s. (a) What is Ek (½mu²) of the dislodged electron? (b) What is Φ (in J) of potassium?


Answer 1


 a) 1.866 × 10 ⁻¹⁹ J      b)   3.685 × 10⁻¹⁹ J


the constants involved are

h ( Planck constant) = 6.626 × 10⁻³⁴ m² kg/s

Me of electron = 9.109 × 10 ⁻³¹ kg

speed of light = 3.0 × 10 ⁸ m/s

a) the Ek ( kinetic energy of the dislodged electron) = 0.5 mu²

Ek = 0.5 × 9.109 × 10⁻³¹ × ( 6.40 × 10⁵ )² = 1.866 × 10 ⁻¹⁹ J

b) Φ ( minimum energy needed to dislodge the electron ) can be calculated by this formula

hv =   Φ + Ek

where Ek = 1.866 × 10 ⁻¹⁹ J

v ( threshold frequency ) = c / λ where c is the speed of light and λ is the wavelength of light = 358.1 nm = 3.581 × 10⁻⁷ m

v = ( 3.0 × 10 ⁸ m/s ) / (3.581 × 10⁻⁷ m ) = 8.378 × 10¹⁴ s⁻¹

hv = 6.626 × 10⁻³⁴ m² kg/s ×  8.378 × 10¹⁴ s⁻¹ = 5.551 × 10⁻¹⁹ J

5.551 × 10⁻¹⁹ J = 1.866 × 10 ⁻¹⁹ J + Φ

Φ = 5.551 × 10⁻¹⁹ J - 1.866 × 10 ⁻¹⁹ J = 3.685 × 10⁻¹⁹ J

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Give an example of a substance that is MORE dense in its solid state when compared to its liquid state.


Answer: Wax


Density is defined as the mass contained per unit volume.


Usually solids are more denser than liquid, as molecules in solid are more strongly packed and thus have more mass per unit volume.

Liquids on the other hand contain molecules which are less tightly bound and thus thus contain less mass per unit volume as compared to solid.

Example: Solid wax is more denser than liquid wax.

If a sample contains 84.0 % of the R enantiomer and 16.0 % of the S enantiomer, what is the enantiomeric excess of the mixture



enantiomeric excess = 68%


Enantiomeric excess is a value used to determine the purity of chiral molecules. It is possible to determine enantiomeric excess (ee) using:

ee = R - S / R + S * 100

Where R is the mass (In this case percentage) of the R enantiomer and S of the S enantiomer.

Replacing with values of the problem:

ee = 84% - 16% / 84% + 16% * 100

ee = 68%

Final answer:

The enantiomeric excess of the mixture, defined as the difference between the concentrations of the R and S enantiomers, is 68.0%.


The enantiomeric excess (ee) is defined as the absolute difference between the mole percentage of the major enantiomer and the minor enantiomer in a mixture. In a sample that contains 84.0 % of the R enantiomer and 16.0 % of the S enantiomer, the enantiomeric excess is calculated as follows:

  1. Calculation: The enantiomeric excess is 84.0% (R) - 16.0% (S) = 68.0%

Therefore, the enantiomeric excess of the mixture is 68.0%.

Learn more about Enantiomeric Excess here:


Which term describes this molecular shape?A. Trigonal pyramidal
B. Trigonal planar
C. Tetrahedral
D. Linear


Answer: Trigonal Pyrimidal

Explanation: Just took test

Calculate the pH of a solution prepared by adding 20.0 mL of 0.100 M HCl to 80.0 mL of a buffer that is comprised of 0.25 M C2H5NH2 and 0.25 M C2H5NH3Cl. Kb of C2H5NH2 = 9.5 x 10-4.


Answer: The pH of resulting solution is 10.893


To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in mL)}}

  • For ethylamine:

Molarity of ethylamine solution = 0.25 M

Volume of solution = 80 mL

Putting values in above equation, we get:

0.25M=\frac{\text{Moles of ethylamine}* 1000}{80mL}\n\n\text{Moles of ethylamine}=(0.25* 80)/(1000)=0.02mol

  • For HCl:

Molarity of HCl = 0.100 M

Volume of solution = 20.0 mL

Putting values in above equation, we get:

0.100M=\frac{\text{Moles of HCl}* 1000}{20.0mL}\n\n\text{Moles of HCl}=(0.100* 20)/(1000)=0.002mol

  • For C_2H_5NH_3Cl:

Molarity of C_2H_5NH_3Cl solution = 0.25 M

Volume of solution = 80 mL

Putting values in above equation, we get:

0.25M=\frac{\text{Moles of }C_2H_5NH_3Cl* 1000}{80mL}\n\n\text{Moles of }=(0.25* 80)/(1000)=0.02mol

The chemical reaction for ethylamine and HCl follows the equation:

                  C_2H_5NH_2+HCl\rightarrow C_2H_5NH_3Cl

Initial:           0.02          0.002         0.02

Final:            0.018          -                0.022

Volume of solution = 20.0 + 80.0 = 100 mL = 0.100 L    (Conversion factor:  1 L = 1000 mL)

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:



We are given:

pK_b = negative logarithm of base dissociation constant of ethylamine = -\log(9.5* 10^(-4))=3.02



pOH = ?

Putting values in above equation, we get:


To calculate pH of the solution, we use the equation:


Hence, the pH of the solution is 10.893

The pH of the solution is 10.9


  • Volume of buffer = 80mL
  • Volume of HCL = 20.0mL
  • conc. of C2H5NH2 = 0.25M
  • conc. of C2H5NH3Cl = 0.25
  • Kb of C2H5NH2 = 9.5*10^-4

pH of a Solution

The pH of buffer can be calculated by using Henderson-Hasselbalch's equation

pOH = _pKb+ log ([salt])/([base])

The initial moles of salt present is calculated as

0.25 * 80*10^-^3 = 0.02mmoles

The initial moles of base present is calculated as

0.25*80*10^-^3 = 20mmoles

On adding HCl the following reaction will occurs

C_2H_5NH_2 + HCl \to C_2H_5NH_3Cl

This will lead to formation of extra moles of salt that is  equal to moles of acid added and eventually lead to decrease in number of moles of base by equal measure.

Moles of HCl added is

moles of HCL= 0.1 * 20 * 10^-^3 = 2mmoles

Adding the value

Moles of salt present = 20 + 2 = 22mmoles

Subtracting the value

Moles of base left = 20-2 = 18mmoles

Now using Henderson-Hasselbalch's equation we can calculate the pOH of solution

pKb = -logKb = -log (9.5*10^-^4) = 3.02

The pOH of the base can be calculated as

pOH = 3.02 + log ((22)/(18))  = 3.107

Using the above, we can solve for the pH of the solution.

pH = 14 - pOH = 10.893

The pH of the solution is 10.9

Learn more on pH of a solution using Henderson-Hasselbalch equation here;

Which particles affect the stability in of the atom


The stability of an atom is affected by the balance between the electrons, protons, and neutrons in an atom.

What are sub-atomic particles?

A particle less than an atom is referred to as a subatomic particle. A subatomic particle can either be an elementary particle, which is not made of other particles, or a composite particle, which is composed of other particles, according to the Standard Model of particle physics.

Particles smaller than an atom are referred to as subatomic particles. The three primary subatomic particles present in an atom are protons, neutrons, and electrons.

Learn more about sub-atomic particles at:


balance protons and neutrons

For the reaction, calculate how many moles of the product form when 2.73 mol of h2 completely reacts. assume that there is more than enough of the other reactant. h2(g)+cl2(g)→2hcl(g)


For the calculation of number of moles of HClthat is produced by 2.73 moles of H₂.

Considering the reaction shown below:

H₂ + Cl₂------->2HCl

This can be seen from the reaction that 1 mole of H₂ produce=2 mole of HCl

So , 2.73 mole of H_{2} will produce= 2.73\times 2 mole of HCl

That is 2.73 mole of H_{2} will produce= 5.46 mole of HCl

So 5.46 mole of HCl will be produced by 2.73 mole of H_{2}





In this case, as both of the reactants are completely consumed, one infers that the following stoichiometric relationship leads to the produced moles of the product; hydrochloric acid:


It is important to notice that based on the undergoing chemical reaction, 1 mole of hydrogen is related with 2 moles of hydrochloric acid, that is why he product's moles doubles the hydrogen's moles.

Best regards.