A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74 atm. When the experiment is repeated using 0.020 mol of ethanoic acid instead of methyl methanoate, the measured pressure is lower than 0.74 atm. The lower pressure for ethanoic acid is due to the following reversible reaction.CH3COOH(g)+CH3COOH(g) ⇋ (CH3COOH)2(g)+Assume that when equilibrium has been reached, 50 percent of the ethanoic acid molecules have reacted.i. Calculate the total pressure in the vessel at equilibrium at 450 K.ii. Calculate the value of the equilibrium constant, Kp, for the reaction at 450 K


Answer 1


Starting moles of ethanol acid = 0.020 mol

At the equilibrium 50 % of the ethanol acid molecules reacted

∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %

                                                                   = 0.010 mol

Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol

Moles of the product (CH3COOH)^(2) gas formed are calculated as

0.010 mol CH3COOH * 1 mol (CH3COOH)^(2) / 2 mol CH3COOH

= 0.005 mol (CH3COOH)^(2)

Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol (CH3COOH)^(2)

That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Now Calculate the pressure  :

0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas

P1/n1 = P2/n2

P2 = P1*n2 / n1

      = 0.74 atm * 0.015 mol / 0.020 mol

     = 0.555 atm

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it's Fr. which is francium.


An atomic radius is defined to be one-half the distance between the nuclei of two atoms, assuming a spherical atom since, according to the quantum mechanical model of the atom, electrons are located within a probability cloud surrounding the nucleus which has no sharp boundary.

Notice that, in general, there are two main trends of atomic radii in the Periodic Table of Elements.

The first trend illustrates that atomic radii increase when going down a group in the periodic table. This is because when moving downwards in a group, every subsequent atom gains an additional principal energy level, which leads to electron shielding. Electron shielding refers to the decreased attraction between the electrons that occupy the higher principal energy level and the nucleus of the atom due to the shielding of electrons in the lower principal energy level.

The second trend outlines that atomic radii decrease when going across the period from left to right. For elements within a period, individual electrons occupy the same principal energy level. Likewise, when an electron is added, a new proton is also added to the nucleus, providing the nucleus with a stronger positive charge and hence leading to a higher effective nuclear charge. This increase in nuclear attraction pulls the electrons closer towards the nucleus, leading to a decrease in atomic radius.

Therefore, given the option between beryllium, calcium, barium, and strontium, the element with the largest atomic radius is barium since all the elements given are in Group II, however, barium is the element furthest down the group and therefore have electrons occupying the highest principal energy level compared to other elements.

What type of compound is disulfur dichloride?a. organic compound
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c. covalent compound
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It is formed by covalent bonds, in which the atoms share a pair of valence electrons

Lead(II) sulfide was once used in glazing earthenware. It will also react with hydrogen peroxide to form lead(II) sulfate and water. How many grams of hydrogen peroxide are needed to react completely with 265 g of lead(II) sulfide?


As per the balanced equation the amount of hydrogen peroxide required completely reacts with 265 g of lead sulphide is 150.6 g.

What is hydrogen  peroxide ?

Hydrogen peroxide is covalent compound formed by two hydrogen and two oxygens. It is used as an oxidising agent. Hydrogen peroxide reacts with lead sulphide to give lead sulphate and water and the balanced reaction is given below:

\rm  PbS+ 4 H_(2)O_(2) \rightarrow PbSO_(4) + 4 H_(2)O

As per the balanced equation 4 moles of hydrogen peroxide is required to react with one mole of lead sulphide. One mole of lead sulphide is 239.30 g and one mole of hydrogen peroxide is 34 g/mol

4 moles of hydrogen peroxides weighs 4 ×34 = 136 g. Thus, 136 g of hydrogen peroxide is needed for 239.3 g of PbS. Therefore, the mass of hydrogen peroxide needed to react with 265 g of PbS is calculated as follows:

mass = (136 ×265 g ) / 239.3

         = 150.6 g.

Hence, amount of hydrogen peroxide required completely reacts with 265 g of lead sulphide is 150.6 g.

To find more about hydrogen peroxide, refer the link below;



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C is the answer: chemical waste container

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This question is incomplete


This question is incomplete but some general explanation provides a clear answer to what is been asked in the question.

An ionic/electrovalent compound is a compound whose constituent atoms are joined together by ionic bond. Ionic bond is a bond involving the transfer of valence electron(s) from an atom (to form a positively charged cation) to another  atom (to form a  negatively charged anion). The atom transferring is usually a metal while the atom receiving is usually a non-metal.

For example (as shown in the attachment), in the formation of NaCl salt, the sodium (Na) transfers the single electron (valence) on it's outermost shell to chlorine (Cl) which ordinarily has 7 electrons on it's outermost shell but becomes 8 after receiving the valence electron from sodium. It should also be noted that Na is a metal while Cl is a non-metal.

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;