A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
Br(g)
Cl2(g)
I2(g)
F2(g)
B. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2S(g)
H2O(g)
H2O2(g)
C. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous)
C(s, diamond)
C(s, graphite)

Answers

Answer 1

Answer:

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Explanation:

Hello,

In this case, we can apply the following principles to explain the order:

- The greater the molar mass, the larger the standard molar entropy.

- The greater the molar mass and the structural complexity, the larger the standard molar entropy.

- The greater the structural complexity, the larger the standard molar entropy.

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

This is due to the fact that the greater the molar mass, the larger the standard molar entropy.

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

This is due to the fact that the greater the molar mass and the structural complexity, the larger the standard molar entropy as the hydrogen peroxide has four bonds and weights 34 g/mol as well as hydrogen sulfide that has two bonds only.

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Since the molecular complexity is greater in the amorphous carbon (messy arrangement), mid in the graphite and lower in the diamond (well organized).

Regards.


Related Questions

Why don't siblings look exactly alike

Answers

Answer:

Your genes play a big role in making you who you are. ... But brothers and sisters don't look exactly alike because everyone (including parents) actually has two copies of most of their genes. And these copies can be different. Parents pass one of their two copies of each of their genes to their kids.

Given the following reaction and data, A + B → Products

Experiment A (M) B (M) Rate (M/s)
1 1.50 1.50 0.320
2 1.50 2.50 0.320
3 3.00 1.50 0.640

Required:
a. What is the rate law of the reaction?
b. What is the rate constant?

Answers

Answer:

a. Rate = k×[A]

b. k = 0.213s⁻¹

Explanation:

a. When you are studying the kinetics of a reaction such as:

A + B → Products.

General rate law must be like:

Rate = k×[A]ᵃ[B]ᵇ

You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.

If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1

Rate = k×[A]¹[B]ᵇ

In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]

Rate = k×[A][B]⁰

Rate = k×[A]

b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:

Rate = k×[A]

0.320M/s = k×[1.50M]

k = 0.213s⁻¹

Which of the following elements can't have an expanded octet? answers A. oxygen B. phosphorous C. chlorine d. sulfer

Answers

answer is oxygen .

oxygen is an exception in octet rule

Among the following given elements,oxygen is an element which cannot have an expanded octet.

What is an expanded octet?

Expanded octet is a condition where an octet has more than 8 electrons and which is called as hyper-valency state. This concept is related to hybrid orbital theory and Lewis theory. Hyper-valent compounds  are not less common  and are of equal stability as the compounds which obey octet rule.

Expansion of octet is possible for elements from third period on wards only as they have low-lying empty d - orbitals which can accommodate more than eight electrons.

Expanded octet is not applicable to oxygen as it is second period of periodic table and has less than ten electrons and even does  not have the 2d -orbitals   due to which it  does not fulfill the criteria of an element to have an expanded octet.

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A substance, W has a concentration of 0.02mol when its molar mass was found to be

74.0 gmol-1 . Another substance V contains 1.00x1023 atoms and has molar mass of

40.0gmol-1. Which of the two substances has the greater mass (in grams)? [L =

6.02x1023]

ii) A 250 cm3

solution contains 14.63g of sodium chloride (NaCl). Calculate the

concentration of the solution in moldm-3

[Na= 23, Cl = 35.5]​

Answers

Answer:

Explanation:

mass of W in gram  = mole x molecular weight

= .02 x 74 = 1.48 gm

mass of V in gram

first of all we shall calculate the no of moles of V

1 mole = 6.0 x  10²³ atoms

1 x 10²³ atoms = 1 / 6 moles

mass of V in grams

= 40 / 6

= 6.67 grams .

So V has greater mass .

ii )

molecular weight of sodium chloride

= 58.5 gm

14.63 gram of sodium chloride

= 14.63 / 58.5 = .25 moles

250 cm³ = 250 x 10⁻³ dm³

So 250 x 10⁻³ dm³ of solution contains .25 moles of salt

1 dm³ of solution will contain .25 / 250 x 10⁻³ mole

= 1 mole

so concentration of solution is 1 mole per dm³

Calculate the amount of heat needed to melt of ice () and bring it to a temperature of . Round your answer to significant digits. Also, be sure your answer contains a unit symbol.

Answers

Answer:

your answer is given below

Explanation:

Take it stepwise, and you are going to have to look up the various heat values.

You have 36.5grams of ice, presumably at 0C.  You will need to add heat to take the ice at 0 C to water at 0 C. (Latent heat of melting)  Then you add in a different heat value to take the water at 0 C to water at 82.3 C. (Specific heat of water)

Add the two heat amounts together.

According to specific heat capacity, to calculate the amount of heat needed to melt ice given mass of ices is multiplied by specific heat of ice and the temperature change which it undergoes.

What is specific heat capacity?

Specific heat capacity is defined as the amount of energy required to raise the temperature of one gram of substance by one degree Celsius. It has units of calories or joules per gram per degree Celsius.

It varies with temperature and is different for each state of matter. Water in the liquid form has the highest specific heat capacity among all common substances .Specific heat capacity of a substance is infinite as it undergoes phase transition ,it is highest for gases and can rise if the gas is allowed to expand.

It is given by the formula ,

Q=mcΔT

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A star is observed from two positions of Earth in its orbit, in summer and winter. Which of these is the best method to calculate the approximate distance of the star from Earth? measure the parallax and use it in calculations measure the red shift of emitted light and use it in calculations use doppler effect to calculate the shift in light traveling from star to Earth in winter use doppler effect to calculate the shift in light traveling from star to Earth in summer

Answers

Answer:

measure the parallax and use it in calculations

Explanation:

got it right on test

There are many more stars at different distances from the earth. The distance to the stars calculated in light years and it is measured using parallax method.Thus option a is correct.

What is parallax method?

Parallax method is used to measure the approximate distance of stars from earth. It uses the position of nearby star from two points opposite to earth and the small angular displacement observed from the remote stars are noted.

The orbit radius of earth and distance to the stars can be calculated from the parallactic angle p, that is one second of arc. Thus the distance is described in the units parsec.

The distance to the stars are usually calculated in light years. One parsec equals 3.26 light years. The nearest star to earth is named as proxima century having the distance parallax 0.76813'' which equals 4.24 light years. Thus, parallax is inversely proportional to the distance.

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Which describes the molecule below

Answers

Answer:

Option D. A lipid with three unsaturated fatty acid.

Explanation:

The molecule in the diagram above contains three fatty acid.

A careful observation of the molecule reveals that each of the three fatty acids contains a double bond.

The presence of a double bond in a compound shows that the compound is unsaturated.

Thus, we can say that the molecule is a lipid with three unsaturated fatty acid.

Consider these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)|Cu(s) Ag+(aq)|Ag(s) -0.40 V -0.76 V ‑0.25 V +0.34 V +0.80 V Based on the data above, which species is the best reducing agent?

Answers

Answer:

The best reducing agent is Zn(s)

Explanation:

A reducing agent must to be able to reduce another compound, by oxidizing itself. Consequently, the oxidation potential must be high. The oxidation potential of a compound is the reduction potential of the same compound  with the opposite charge. Given the reduction potentials, the best reducing agent will be the compound with the most negative reduction potential. Among the following reduction potentials:

Cd₂⁺(aq)|Cd(s) ⇒ -0.40 V

Zn²⁺(aq)|Zn(s) ⇒ -0.76 V

Ni²⁺(aq)|Ni(s) ⇒‑0.25 V

Cu²⁺(aq)|Cu(s) ⇒ +0.34 V

Ag⁺(aq)|Ag(s) ⇒ +0.80 V

The most negative is Zn²⁺(aq)|Zn(s) ⇒ -0.76 V

From this, the most reducing agent is Zn. Zn(s) is oxidized to Zn²⁺ ions with the highest oxidation potential (0.76 V).

Suppose a small amount of a solid is added to water and, after a short time, all the solid has dissolved. Which of the following statements is most likely to be true?
A) The solution is supersaturated with solute.
B) The solution is saturated with solute.
C) The solution is unsaturated with solute.
D) The solution is either unsaturated or supersaturated with solute.
E) The solution is either saturated or supersaturated with solute.

Answers

Answer:

the option e is correct I think

The solution is either saturated or supersaturated with solute

Answer:

the option e is correct I think

The solution is either saturated or supersaturated with solute

Explanation:

If a reaction is first-order with respect to a particular reactant, when the concentration of that reactant is increased by a factor of 2, the reaction rate will _____.

Answers

Answer:

It would increase by a factor of 2

Explanation:

The rate law for a fist order reaction is given as;

A --> B

rate = k [A]

upon doubling the concentration, we have;

rate = k [A]

rate 2 = k 2 [A]

Dividing both equations;

rate 2 / rate 1 = k 2 [A] / k [A]

rate 2 / rate 1  = 2 / 1

The ratio between rate 2 and rate 1 is 2 : 1. This means that the reaction rate would also increase by a factor of 2.

Which of the following is a half-reaction? A. Zn+CuSO4−> B. 2Cl−−>Cl2+2e− C. H2+1/2O2−>H2O D. −>Cu+ZnSO4

Answers

Answer:

2Cl——>Cl2+2e-

Explanation:

It shows an electron loss or gain

To find the pH of a solution of NH4Br directly, one would need to use:__________
Select the correct answer below:
a) the Kb of NH3 to find the hydroxide concentration
b) the Ka of NH+4 to find the hydronium concentration
c) the Kb of NH3 to find the hydronium concentration
d) the Ka of NH+4 to find the hydroxide concentration

Answers

Answer:

b) the Ka of NH₄⁺ to find the hydronium concentration

Explanation:

The equilbrium of NH₄⁺ (The conjugate acid of NH₃, a weak base), is:

NH₄⁺ ⇄ NH₃ + H⁺

Where Ka of the conjugate acid is:

Ka = [NH₃] [H⁺] / [NH₄⁺]

Thus, if you know Ka of NH₄⁺ and its molar concentration you can calculate  [H⁺], the hydronium concentration, to find pH (Because pH =  -log [H⁺])

Thus, right option is:

b) the Ka of NH₄⁺ to find the hydronium concentration

A rock has a mass of 15.8 g and causes the water level in a graduated cylinder to raise from 22.3 mL to 32.5 mL. What is the density of the rock in Kg/mL?​

Answers

Answer:

Explanation:

mass -  15.8 g = 0.0158 kg

volume = 32.5 - 22.5 = 10.2 ml

density = mass / volume

= 0.0158 / 10.2

= 0.00154 kg/ml

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The K sp for silver(I) phosphate is 1.8 × 10 –18. Determine the silver ion concentration in a saturated solution of silver(I) phosphate.

Answers

Answer:

[tex][Ag^+]=4.82x10^{-5}M[/tex]

Explanation:

Hello,

In this case, the dissociation reaction for silver phosphate is:

[tex]Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4^{3-}(aq)[/tex]

Therefore, the equilibrium expression is:

[tex]Ksp=[Ag^+]^3[PO_4^{3-}][/tex]

And in terms of the reaction extent [tex]x[/tex] is:

[tex]Ksp=1.8x10^{-18}=(3x)^3(x)[/tex]

Thus, [tex]x[/tex] turns out:

[tex]1.8x10^{-18}=27x^4\\\\x=\sqrt[4]{\frac{1.8x10^{-18}}{27} } \\\\x=1.61x10^{-5}M[/tex]

In such a way, the concentration of the silver ion is:

[tex][Ag^+]=3x=3*1.61x10^{-5}M=4.82x10^{-5}M[/tex]

Best regards.

Calculate the solubility of Mg(OH)2 in water at 25 C. You'll find Ksp data in the ALEKS Data tab. Round your answer to significant digits.

Answers

Answer:

1.12 × 10⁻⁴ M

Explanation:

Step 1: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 2: Make an ICE chart

We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.

       Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

I                                0                    0

C                              +S                +2S

E                                S                  2S

The solubility product constant is:

Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³

S = 1.12 × 10⁻⁴ M

Which of the following ionic lattices would have the highest melting point?
A. Potassium oxide
w
B. Boron nitride
C. Beryllium oxide
D. Lithium chloride

Answers

Answer:

I think, berryllium oxide, is answer.

Explanation:

Hope it helps you....

The ionic lattices would have the highest melting point Potassium oxide. option A is correct.

what is ionic lattice?

An ionic compound is a giant structure of ions. The ions have a regular, repeating arrangement called an ionic lattice. The lattice is formed because the ions attract each other and form a regular pattern with oppositely charged ions next to each other.

Ionic compounds are held together by electrostatic forces between oppositely charged ions.

These forces are usually referred to as the ionic lattice contains such a large number of ions, that a lot of energy is needed to overcome this ionic bonding so ionic compounds have high melting and boiling points.

therefore, sodium oxide has the highest melting point. option A is correct

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A complexometric titration can also be used to determine the amount of calcium in milk. The calcium concentration in milk is typically 1,200 mg/L. How would you alter the procedure used in this experiment to determine milk calcium content

Answers

Answer:

d

Explanation:answer is d on edg 2020

Name four types of salts​

Answers

Answer:

Any ionic molecule formed of a base and an acid, which dissolves in water to produce ions is known as a salt. The four common types of salts are:  

1. NaCl or sodium chloride is the most common kind of salt known. It is also known as table salt.  

2. K2Cr2O7 or potassium dichromate refers to an orange-colored salt formed of chromium, potassium, and oxygen. It is toxic to humans and is also an oxidizer, which is a fire hazard.  

3. CaCl2 or calcium chloride looks like table salt due to its white color. It is broadly used to withdraw ice from roads. It is hygroscopic.  

4. NaHSO4 or sodium bisulfate produces from hydrogen, sodium, oxygen, and sulfur. It is also known as dry acid. It has commercial applications like reducing the pH of swimming pools and spas and others.  

What is the change in entropy (AS) when a solid substance decomposes and

produces gaseous products?

Answers

Explanation:

Entropy refers to the degree of disorderliness in a system. Generally solids have  a greater degree of order when compared to gas in which the molecules/particles move randomly in all directions.

The change in entropy from solid to gaseous phase is a positive change. Because there is increase disorderliness of the system.

Will a precipitate (ppt) form when 20.0 mL of 1.1 × 10 –3 M Ba(NO 3) 2 are added to 80.0 mL of 8.4 × 10 –4 M Na 2CO 3?

Answers

Answer:

A precipitate will form, BaCO₃

Explanation:

When Ba²⁺ and CO₃²⁻ ions are in an aqueous media, BaCO₃(s), a precipitate, is produced following its Ksp expression:

Ksp = 5.1x10⁻⁹ = [Ba²⁺] [CO₃²⁻]

Where the concentrations of the ions are the concentrations in equilibrium

For actual concentrations of a solution, you can define Q, reaction quotient, as:

Q = [Ba²⁺] [CO₃²⁻]

If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form.

Actual concentrations of Ba²⁺ and CO₃²⁻ are:

[Ba²⁺] = [Ba(NO₃)₂] = 1.1x10⁻³ × (20.0mL / 100.0mL) = 2.2x10⁻⁴M

[CO₃²⁻] = [Na₂CO₃] = 8.4x10⁻⁴ × (80.0mL / 100.0mL) = 6.72x10⁻⁴M

100.0mL is the volume of the mixture of the solutions

Replacing in Q expression:

Q = [Ba²⁺] [CO₃²⁻]

Q = [2.2x10⁻⁴M] [6.72x10⁻⁴M]

Q = 1.5x10⁻⁷

As Q > Ksp

A precipitate will form, BaCO₃

When HCl is added to water, the [H3O+] = 0.6 M. What is the [OH-]?

What is the pH of the solution?

Answers

Answer:

pH=0.22.

[tex][OH^-]=1.66x10^{-14}M[/tex]

Explanation:

Hello,

In this case, since the pH is directly computed from the given concentration of hydronium ions:

[tex]pH=-log([H_3O^+])=-log(0.6M)=0.22[/tex]

It is widely known that the pH and POH are directly related via:

[tex]pH+pOH=14[/tex]

Therefore, the pOH is:

[tex]pOH=14-pH\\\\pOH=14-0.22=13.78[/tex]

Thus, the concentration of hydroxyl ions are computed from the pOH:

[tex]pOH=-log([OH^-]}\\\\[/tex]

[tex][OH^-]=10^{-pOH]=10^{-13.78}[/tex]

[tex][OH^-]=1.66x10^{-14}M[/tex]

Regards.

What can you learn about the pH of a substance with the conductivity test? hint: gives you no info on concentration.

Answers

Answer:

See explanation

Explanation:

So, I'm gonna take a shot at this one and say this:

With a strongly acidic/basic solution, you'll get a high conductivity when preforming a conductivity test.

The more acidic or basic a substance is, the higher the electrical conductivity.

Based on how high or low the conductivity is, it will give you an idea of the substance's pH.

Hope that made since or gave you an idea of what you're looking for. Good luck :)

26.0 g of a liquid that has a density of 1.44 g/mL needs to be measured out in a graduated cylinder . What volume should be used

Answers

Answer: The volume of liquid used will be 18.055 mL.

what are the monomers of bakelite​

Answers

Answer:

Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.

Answer: The monomers of bakelite are formaldehyde and phenol

Explanation:

Given the following equivalents, make the following conversion: 1.00 knop = ? knips

4 clips = 5 blips
1 knop = 6 bippy
3 blip = 18 pringle
1 clip = 10 knip
10 bippy = 8 pringle

Answers

Answer:

[tex]6.4knips[/tex]

Explanation:

Hello,

In this case, given the stated equivalences, we can use the following proportional factor in order to compute the required knips:

[tex]knips=1.00knop*\frac{6bippy}{1knop} *\frac{8pringle}{10bippy}* \frac{3blip}{18pringle} *\frac{4clips}{5blips} *\frac{10knip}{1clip} \\\\=6.4knips[/tex]

Regards.

1 knop=6.4 knips

First convert knop to bippy:-

[tex]1\ knop\times\frac{6\ bippy}{1\ knop} =6\ bippy[/tex]

Now, Convert 6 bippy to pringle:-

[tex]6\ bippy\times\frac{8\ pringle}{10\ bippy} =4.8\ pringle[/tex]

Now, convert 4.8 pringle to blip:-

[tex]4.8\ pringle\times\frac{3\ blip}{18\ priangle} =0.8\ blip[/tex]

Now, convert 0.8 blip to clips as follows:-

[tex]0.8\ blip\times\frac{4\ clips}{5\ blip} =0.64\ clip[/tex]

Now, convert 0.64 clip to knips:-

[tex]0.64\ clip\times\frac{10\ knip}{1\ clip} =6.4\ knip[/tex]

Hence, the following conversion is as follows:-

1.00 knop=6.4 knips

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A compound is found to contain 27.29 % carbon and 72.71 % oxygen by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound

Answers

Answer:

CO2

Explanation:

Mass of carbon = 27.29

Mass of oxygen = 72.21

Step 1:

We have to first convert these masses to moles

Carbon = 2.29/2.01 = 2.274

Oxygen = 72.71/16 = 4.544

Step 2:

We have to divide the mols by the smallest mol to get simplest ratio of whole number.

The smallest mol is 2.274. we have to divide the mols by this.

2.274/2.274 = 1

4.544/2.274 = 2

Our empirical formula is therefore CO2

Calculate the molar hydrogen ion concentration of each of the following biological solutions given the pH:
(a) gastric juice, pH= 1.80
(b) urine, pH 4.75 56.

Answers

Answer: A

Explanation: Calculate the molar hydrogen ion concentration of each of the following biological solutions given the pH, Urine pH= 4.90

An oxidation-reduction reaction in which 3 electrons are Transferred has ∆G° = +18.55 kJ at 25°C? What is the value of E°?

Answers

Answer:

The correct answer is: 0.064 V

Explanation:

In a oxidation-reduction reaction, the relation between Gibbs free energy (ΔGº) and cell potential (Eºcell) is given by:

[tex]\Delta G^{0} = -nFE^{0} _{cell}[/tex]

where n is the number of electrons that are transferred in the reaction and F is the Faraday constant (96,500 C/mol e-). Given: ∆G° = +18.55 kJ and n= 3 mol e-, we calculate Eºcell as follows:

+18.55 kJ = (-3 mol e-) x (96500 C/mol e-) x Eºcell

Eºcell= (+18.55 kJ)/(-3 mol e-) x (96500 C/mol e-)

Eºcell= (18550 J)/ (289500 C)

Eºcell= 0.064 J/C

Since 1 Volt= 1 Joule/1 Coulomb, thus:

Eºcell= 0.064 J/C = 0.064 V

A laboratory technician combines 35.9 mL of 0.258 M chromium(II) chloride with 35.8 mL 0.338 M potassium hydroxide. How many grams of chromium(II) hydroxide can precipitate

Answers

Answer:

0.52 g of chromium(II) hydroxide, Cr(OH)2.

Explanation:

We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.

This can be obtained as follow:

Molarity of CrCl2 = 0.258 M

Volume = 35.9 mL = 35.9/1000 = 0.0359 L

Mole of CrCl2 =?

Molarity = mole /Volume

0.258 = mole of CrCl2 /0.0359

Cross multiply

Mole of CrCl2 = 0.258 x 0.0359

Mole of CrCl2 = 0.0093 mole

Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.

This can be obtained as follow:

Molarity of KOH = 0.338 M

Volume = 35.8 mL = 35.8/1000 = 0.0358 L

Mole of KOH =.?

Molarity = mole /Volume

0.338 = mole of KOH /0.0358

Cross multiply

Mole of KOH = 0.338 x 0.0358

Mole of KOH = 0.0121 mole.

Next, we shall write the balanced equation for the reaction. This is given below:

2KOH + CrCl2 → Cr(OH)2 + 2KCl

From the balanced equation above,

2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

2 mole of KOH reacted with 1 mole of CrCl2.

Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.

From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.

Therefore, KOH is the limiting reactant.

Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.

In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.

The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:

From the balanced equation above,

2 mole of KOH reacted to produce 1 mole of Cr(OH)2.

Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.

Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.

This is illustrated below:

Mole of Cr(OH)2 = 0.00605 mole

Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol

Mass of Cr(OH)2 =..?

Mole = mass /Molar mass

0.00605 = mass of Cr(OH)2/86

Cross multiply

Mass of Cr(OH)2 = 0.00605 x 86

Mass of Cr(OH)2 = 0.52 g

Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.

alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decimal places.

Answers

A 1.0-L buffer solution contains 0.100 mol  HC2H3O2 and 0.100 mol  NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

Answer:

The pH of this solution = 5.06  

Explanation:

Given that:

number of moles of CH3COOH = 0.100 mol

volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex]  = 0.10 M

The chemical equation for this reaction is :

[tex]\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}[/tex]

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is  = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex]  

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added  to the original buffer, the concentration of NaOH added will be = [tex]\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}[/tex] = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

                  [tex]\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}[/tex]

Initial             0.10               0.035         0.10                  -

Change        -0.035          -0.035       + 0.035              -

Equilibrium    0.065              0              0.135               -

By using  Henderson-Hasselbalch equation:

The pH of this solution = pKa + log [tex]\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}[/tex]

The pH of this solution = 4.74 + log [tex]\mathtt{\dfrac{0.135}{0.065}}[/tex]

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06    to two decimal places

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