CHEMISTRY HIGH SCHOOL

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell:________ Pt(s)
|H2(g,1atm)
|H+(aq,1.0M)
|Au3+(aq,?M)
|Au(s).
What is the concentration of Au3+ in the solution if Ecell is 1.23 V ?

Answers

Answer 1
Answer:

Answer: The concentration of Au^(3+) is 1.096* 10^(-6)

Explanation:

The given chemical cell follows:

Pt(s)|H_2(g,1atm)|H^+(aq,1.0M)||Au^(3+)(aq,?M)|Au(s)

Oxidation half reaction:H_2(g,1atm)\rightarrow 2H^(+)(aq,1.0M)+2e^-;E^o_(2H^(+)/H_2)=0.0V       ( × 3)

Reduction half reaction:Au^(3+)(aq,?M)+3e^-\rightarrow Au(s);E^o_(Au^(3+)/Au)=1.50V       ( × 2)

Net cell reaction:3H_2(g,1atm)+2Au^(3+)(aq,?M)\rightarrow 6H^(+)(aq,1.0M)+2Au(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_(cell) of the reaction, we use the equation:

E^o_(cell)=E^o_(cathode)-E^o_(anode)

Putting values in above equation, we get:

E^o_(cell)=1.50-(0.0)=1.50V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_(cell)=E^o_(cell)-(0.059)/(n)\log ([H^(+)]^6)/([Au^(3+)]^2)

where,

E_(cell) = electrode potential of the cell = 1.23 V

E^o_(cell) = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

[H^(+)]=1.0M

[Au^(3+)]=?M

Putting values in above equation, we get:

1.23=1.50-(0.059)/(6)* \log(((1.0)^6)/([Au^(3+)]^2))

[Au^(3+)]=-1.0906* 10^(-6),1.096* 10^(-6)

Neglecting the negative value because concentration cannot be negative.

Hence, the concentration of Au^(3+) is 1.096* 10^(-6)

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Which refers to a mixture that contains more than one phase in which the characteristics of the particles vary throughout the mixture?

Answers

This type of mixture is a heterogeneous mixture. It is any mixture that is not uniform. Heterogeneous mixtures always have more than one phase.It is made up of substances which stays physically separate with each other. Best example is a mixture of sand and sugar. If you get two samples from different points of the mixture and analyze it, you will see that the concentrations of the components are different and that the two solid phases never combine with each other.

Answer:

C. Heterogeneous

Explanation:

As a Ca atom undergoes oxidation to Ca2+, the number of neutrons in its nucleus(1) decreases
(2) increases
(3) remains the same

Answers

Answer:

Hence the correct answer is remains the same

Explanation:

As a Ca atom undergoes oxidation to Ca2+,as follows:

Ca --- Ca 2+ + 2e-

In the process of oxidation calcium loss 2 electron only there is no change in the number of protons as well neutrons therefore there is no change in the number of neutrons in its nucleus

Hence the correct answer is remains the same

Oxidation:

Oxidation is a process in which either 1 or all following changes occurs:

1. Gaining of oxygen atoms

2. Loss of electrons

3. Loss of hydrogen atom.

4. Increasing oxidation number.

Reduction:

Reduction is a process in which either 1 or all following changes occurs:

1. Loss of oxygen atoms

2. Gaining of electrons

3. Gaining of hydrogen atom.

4. Decreasing oxidation number.
Oxidation refers to a change in electrons (In this case, loss of electrons). It has no impact on the number of neutrons, or for that matter on any sub-atomic particles in the nucleus. So the number of neutrons (3) remains the same

A medium-temperature star like the Sun would most likely emit its most intense radiation as _____. A. Ultraviolet light B. Visible light C. Microwaves D. Radio waves

Answers

Answer:

The correct answer is option B, that is, visible light.

Explanation:

The Sun is a prime source of ultraviolet rays. Although the Sun discharges all of the distinct types of electromagnetic radiation, 99 percent of its rays are in the form of ultraviolet rays, visible light, and infrared rays. The Sun emits radiation all through the electromagnetic spectrum, that is, from extremely high-energy X-rays to ultra-lengthy wavelength radio waves, and everything that comes in between. However, the peak of the emission takes place in the visible part of the spectrum.

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(3) return to lower energy states within the atoms
(4) move to higher energy states within the atoms

Answers

Answer:

  • The correct answer is (3).  
  • Return to lower energy states within the atoms

Explanation:

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Answers

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A solid can evaporate by melting into a liquid, which then evaporates; or by changing instantly into a vapor, or subliming. The rate of evaporation of a substance depends on its surface temperature, the pressure, and the humidity. Evaporation is the procedure by which a liquid or a solid changes into a vapor. A substance may evaporate by changing into a vapor at the surface, as when water evaporates from an uncovered dish.

Calculate the percentage of yield when 20 grams of sodium chloride solution reacts with an excess amount of silver nitrate solution knowing that 45 grams of silver chloride precipitated​

Answers

Percentage % yield = 91.8%

Given:

20 g NaCl

45 g AgCl

To find:

% yield=?

The balanced chemical equation will be:

NaCl + AgNO₃ ⇒ AgCl + NaNO₃

First, we need to calculate moles of NaCl.

\text{Number of Moles}=\frac{\text{Given Mass}}{\text{Molar Mass}} \n\n\text{Number of Moles}=(20)/(58.44) \n\n\text{Number of Moles}= 0.342 moles

Then, calculate moles of AgCl from equation:

\text{mol of AgCl}= 1/1 * \text{ mol NaCl}\n\n\text{mol of AgCl}= 1/1 * 0.342\n\n\text{mol of AgCl}= 0.342

Mass AgCl(theoretical) :

\text{Mass of AgCl}= mol * MW\n\n\text{Mass of AgCl}= 0.342 x 143,32 g/mol\n\n\text{Mass of AgCl}= 49.02 g

Now, calculate for percentage yield.

\%yield =(actual)/(theoretical) * 100\%\n\n\%yield = (45/49.02) * 100\%\n\n\%yield = 91.8\%

The percentage yield will be 91.8%.

Learn more:

brainly.com/question/21091465

%yield = 91.8

Further explanation

Given

20 g NaCl

45 g AgCl

Required

%yield

Solution

Reaction

NaCl + AgNO₃ ⇒ AgCl + NaNO₃

mol NaCl :

= mass : MW

= 20 g : 58,44 g/mol

= 0.342

mol AgCl from equation :

= 1/1 x mol NaCl

= 1/1 x 0.342

= 0.342

Mass AgCl(theoretical) :

= mol x MW

= 0.342  x 143,32 g/mol

= 49.02 g

%yield = (actual/theoretical) x 100%

%yield = (45/49.02) x 100%

%yield = 91.8
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