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3 years ago

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Match the element with its number of valence electrons. Column A 1. Boron : Boron 2. Carbon : Carbon 3. Fluorine : Fluorine 4. S

odium : Sodium Column B a. 4 b. 1 c. 3 d. 7

1 answer:

Paha777 [63]3 years ago

6 0

Boron 3
Carbon 4
Fluorine 7
Sodium 1

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Write a complete, balanced chemical equation where tin metal reacts with aqueous hydrochloric acid to produce tin(II) chloride a

AleksAgata [21]

Answer:

1. The balanced equation is given below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

2a. H is oxidized.

2b. Sn is reduced.

Explanation:

1. Balanced equation for the reaction between tin (Sn) metal and aqueous hydrochloric acid (HCl) to produce tin(II) chloride (SnCl₂) and hydrogen gas (H₂).

This is illustrated below:

Sn (s) + HCl (aq) –> SnCl₂ (aq) + H₂ (g)

There are 2 atoms of Cl on the right side and 1 atom on the left side. It can be balance by putting 2 in front of HCl as shown below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

Now, the equation is balanced

2. Determination of the element that is oxidize and reduced.

This can be obtained as follow:

We shall determine the change in oxidation number of each element.

NOTE:

a. The oxidation number of H is always +1 except in hydrides where it is –1.

b. The oxidation state of Cl is always –1.

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

For Tin (Sn):

Sn = 0

SnCl₂ = 0

Sn + 2Cl = 0

Cl = – 1

Sn + 2(–1) = 0

Sn – 2 = 0

Collect like terms

Sn = 0 + 2

Sn = +2

Therefore, the oxidation number of Tin (Sn) changes from 0 to +2

For H:

H = +1

H₂ = 0

The oxidation number of H changes from +1 to 0

For Cl:

Cl is always –1. Therefore no change.

Summary:

Element >>Change in oxidation number

Sn >>>>>>>From 0 to +2

H >>>>>>>>From +1 to 0

Cl >>>>>>>No change

Therefore,

Sn is reduced since its oxidation number increased from 0 to +2.

H is oxidized since it oxidation number reduced from +1 to 0

4 0

2 years ago

A certain half-reaction has a standard reduction potential E0red = +0.13V . An engineer proposes using this half-reaction at the

Ivan

Answer:

a. 1.23 V

b. No maximum

Explanation:

Required:

a. Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?

b. Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have?

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

If E°cell must be at least 1.10 V (E°cell > 1.10 V),

E°red, cat - E°red, an > 1.10 V

E°red, cat - 0.13V > 1.10 V

E°red, cat > 1.23 V

The minimum standard reduction potential is 1.23 V while there is no maximum standard reduction potential.

4 0

3 years ago

The reaction between carbon tetrachloride, CCl4, and water, H2O, to form carbon dioxide, CO2, and hydrogen chloride, HCl, has a

xxTIMURxx [149]

Answer:

High activation energy is the reason behind unsuccessful reaction.

Explanation:

There are two types of reaction: (1) thermodynamically controlled reaction and (2) kinetically controlled reaction.

Thermodynamically controlled reaction are associated with change in enthalpy during reaction. More negative the enthalpy change, more favored will be the reaction.

Kinetically controlled reaction are associated with activation energy of a reaction. The lower the activation energy value, the more rapid will be the reaction.

Here, reaction between $CCl_{4}$ and $H_{2}O$ is thermodynamically favored due to negative enthalpy change but the high activation energy does not allow the reaction to take place by simple mixing.

6 0

3 years ago

Which statement best describes a physical property?

ArbitrLikvidat [17]

The first I believe

3 0

3 years ago

Given the chemical reaction: Hg2+(aq) + Cu(s) + Hg(s) + Cu2+(aq)

Pavel [41]

Answer: When the reaction reaches equilibrium, the cell potential will be 0.00 V

Explanation:

Equilibrium state is the state when reactants and products are present but the concentrations does not change with time.

The equilibrium is dynamic in nature and the reactions are continuous in nature. Rate of forward reaction is equal to the rate of backward reaction.

The standard emf of a cell is related to Gibbs free energy by following relation:

$\Delta G=-nFE^0$

The Gibbs free energy is related to equilibrium constant by following relation:

$\Delta G=-2.303RTlog K$

For equilibrium $\Delta G=0$

Thus $0=-nFE^0$

$E^0=0$

Thus When the reaction reaches equilibrium, the cell potential will be 0.00 V

5 0

3 years ago