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Law Incorporation [45]
3 years ago
14

WILL MARK YOU BRAINLIEST!! how are a elements abundance in nature and percent composition related

Chemistry
1 answer:
Drupady [299]3 years ago
8 0
Because they both have to do with and chemistry science
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The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:
german

<u>Answer:</u>

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

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> 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

<u>Explanation:</u>

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}

  • <u>Cell having 1st and 2nd half reactions:</u>

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

  • <u>Cell having 1st and 3rd half reactions:</u>

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

  • <u>Cell having 1st and 4th half reactions:</u>

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

  • <u>Cell having 2nd and 3rd half reactions:</u>

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

  • <u>Cell having 3rd and 4th half reactions:</u>

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,

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

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> 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
What I know about the Alveoli and Gas exchange
Reptile [31]

Explanation:

What happens during gas exchange in the alveoli?

These are called alveoli. They inflate when a person inhales and deflate when a person exhales. During gas exchange oxygen moves from the lungs to the bloodstream. At the same time carbon dioxide passes from the blood to the lungs.

What is the role of alveoli in gas exchange?

The alveoli are where the lungs and the blood exchange oxygen and carbon dioxide during the process of breathing in and breathing out. Oxygen breathed in from the air passes through the alveoli and into the blood and travels to the tissues throughout the body.

7 0
2 years ago
Help me pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeee this 10 points answer the ones i didnt answer
xxMikexx [17]
42cubic centimeter are in the block
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4 0
3 years ago
The chemical formula for beryllium oxide is BeO.A chemist determined by measurements that 0.045 moles of beryllium participated
blondinia [14]

Answer: 0.405g

Explanation:

Molar Mass of Be = 9g/mol

Number of mole of Be = 0.045mol

Mass conc. Of Be = 0.045 x 9 = 0.405g

8 0
3 years ago
Sample A: 300 mL of 1M sodium chloride
yarga [219]

Answer:

sample B contains the larger density

Explanation:

Given;

volume of sample A, V = 300 mL = 0.3 L

Molarity of sample A, C = 1 M

volume of sample B, V = 145 mL = 0.145 L

Molarity of sample B, C = 1.5 M

molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol

Molarity is given as;

C = \frac{moles \ of \ solute, \ mol}{liters \ of \ solvent} \\\\Moles \ of \ solute \ for \ sample \ A = 1 \times 0.3 = 0.3 \ mol\\\\Moles \ of \ solute \ for \ sample \ B = 1.5 \times 0.145 = 0.2175 \ mol

The reacting mass for sample A = 0.3mol x  58.5 g/mol = 17.55 g

The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g

The density of sample A  = \frac{mass}{volume} = \frac{17.55}{0.3} = 58.5 \ g/L

The density of sample B = \frac{mass}{volume} = \frac{12.72}{0.145} = 87.72 \ g/L

Therefore, sample B contains the larger density

5 0
2 years ago
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