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bekas [8.4K]
3 years ago
8

What are bonds in metals like? Please Answer - In full sentences (3)

Chemistry
2 answers:
mylen [45]3 years ago
8 0

Metallic bonding involves the sharing of valence electrons among a number of metal ions. The valence electrons stay delocalized and are not held tightly by any one metal cation. The nature of the metallic bonding is such that it attributes all metallic properties such as ductility, malleability and more. In metallic bonding there is electrostatic attraction between the delocalized electrons and the metallic cation.

NARA [144]3 years ago
7 0
Metallic bonds are bonds between 2 metals. Their electrons flow around freely unlike covalent or ionic bonds.
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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
In sience is copper a element or not a element
sukhopar [10]

Answer:

a element

Explanation:

In science copper is a element. It is one of the transition metals found on the periodic table of elements.

Elements are distinct substances that cannot be split-up into simpler substances. Such substances consist of one kind of atom.

There are over a hundred elements that are known till date.

Each elements combines to form a compound

7 0
2 years ago
A chemical reaction occurs between Calcium, Oxygen, and Chlorine. The reaction is demonstrated:
ad-work [718]

Answer:

D.

Double replacement, CaO + Cl2O

Explanation:

8 0
3 years ago
Read 2 more answers
Gold forms face-centered cubic crystals. The atomic radius of a gold atom is 144 pm. Consider the face of a unit cell with the n
kipiarov [429]

Answer:

The length of an edge of this unit cell is 407.294 pm

Explanation:

Face centered cubic structure contains 4 atoms in each unit cell and 12 coordination number, occupying about 74% volume of the total cell. Face centered cubic structure is known for efficient use of space for atom packing.

To determine the edge length, a relationship between the radius of the atom and edge length is used.

X = R√8

Where;

X is the length of an edge of this unit cell

R is the radius of the gold atom = 144 pm = 144 X 10⁻¹² m

X = 144 X 10⁻¹²√8

X = 407.294 X 10⁻¹² m

X = 407.294 pm

Therefore, the length of an edge of this unit cell is 407.294 pm

8 0
2 years ago
Sulfur burns in the air to form 33.6L of sulfur dioxide. (S = 32, O =16, 64g of sulfur dioxide has 22.4L, 32g of oxygen has 22.4
ivolga24 [154]

The mass of oxygen reacted/required in this reaction is obtained as 48g.

<h3>What is stoichiometry?</h3>

The term stoichiometry has to do with mass- volume or mass - mole relationship which ultimately depends on the balanced reaction equation.

Now, we have the reaction; S + O2 ------>SO2

If 1 mole of sulfur dioxide contains 22.4 L

x moles of sulfur dioxide contains 33.6L

x = 1.5 moles of sulfur dioxide.

Since the reaction is 1:1, the number if moles of oxygen required/reacted is 1.5 moles.

Mass of oxygen required/reacted = 1.5 moles * 32 g/mol = 48g

Learn more anout stoichiometry: brainly.com/question/9743981

8 0
2 years ago
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