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3 years ago

Suppose you were dissolving a metal such as zinc with hydrochloric acid

1 answer:

8 0

So are you asking how would the particle size of the zinc affect the rate of its dissolution?
you must now that the bigger the particle the lesser the dissolution.
The reaction rate will go as the total surface area, therefore, the smaller the particles the faster the reaction. 
This relationship is NOT linear.

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The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

Answer:

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Cell having 1st and 2nd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

Cell having 1st and 3rd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

Cell having 1st and 4th half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

Cell having 2nd and 3rd half reactions:

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

Cell having 3rd and 4th half reactions:

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$

Hence,

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0

3 years ago

A student has two compounds in two separate bottles but with no labels on either one. One is an unbranched alkane, octane, C8H18

Aleks [24]

Answer:

a) both substances are insoluble in water

b) both substances are soluble in ligroin

c) both substances suffer combustion, octane produces more CO₂ than hexene.

d) both substances are less dense than waterl, with hexene having the lowest density.

e) only hexene would react with bromine

f) only hexene would react with permanganate

Explanation:

a) both substances are non-polar and water is polar

b) both substances are non-polar and lingroin is non-polar

c) C₈H₁₈ + 17.5O₂ → 8CO₂ + 9H₂O

C₆H₁₂ + 9O₂ → 6CO₂ + 6H₂O

d) water = 997 kg/m³

ocatne = 703 kg/m³

hexene = 673 kg/m³

e) bromine test is used to detect unsaturations

f) permanganate test is used to detect unsaturations

7 0

3 years ago

The quinine in tonic water shines a strange blue-white colour when lit with UV light. Explain what is happening here.

Alecsey [184]

Answer:

Here's what I get.

Explanation:

Quinine contains phosphors, substances that glow when they are hit with certain wavelengths of light.

The phosphors in quinine absorb UV light, which is invisible to our eyes.

Electrons in the phosphors absorb the UV energy and are excited to higher energy levels.

When the electrons drop back to lower energy levels, they emit some of this energy as a glowing blue visible light.

5 0

3 years ago

A rock weighing 26.0g placed graduated cylinder displacing the volume from 13.12ml to 25.3ml what is the density of the rock in

tia_tia [17]

density = mass /volume

1)mass of rock = 26.0 gram

2)volume of rock = volume of water it has displaced= 25.3-13.12 = 12.18

put the value of mass and volume in first equation and get the density value

3 0

3 years ago

What is 5 divided by 2.7

Strike441 [17]

5 divided by 2.7 is 0.54

3 0

3 years ago