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rusak2 [61]
3 years ago
11

Which electrons are the valence electrons in a neutral atom of fluorine (F)? Write the orbital configuration of just the valence

electrons.
Chemistry
1 answer:
ICE Princess25 [194]3 years ago
4 0

Answer:

2s² 2p⁵

There are seven valance electrons in fluorine.

Explanation:

The elements of group 17 are called halogens. These are six elements Fluorine, Chlorine, Bromine, Iodine, Astatine. Halogens are very reactive these elements can not be found free in nature. Their chemical properties are resemble greatly with each other.

Properties of fluorine:

1. it is yellow in color.

2. it is flammable gas.

3. it is highly corrosive.

4. fluorine has pungent smell.

5. its reactions with all other elements are very vigorous except neon, oxygen, krypton and helium.  

Electronic configuration of fluorine:

F₉ = 1s² 2s² 2p⁵

Valance electrons in fluorine are 2s² 2p⁵.

Valance Orbital configuration:

2s² 2p⁵

There are seven valance electrons in fluorine.

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3. CuO(s)+2HCl(aq)\rightarrow CuCl_2(aq)+H_2O(l) : Double displacement

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Cu(s)+4HNO_3(aq)rightarrow Cu(NO_3)_2(aq)+2NO_2(g)+2H_2O(l)

Double displacement reaction is defined as the reaction where exchange of ions takes place. Double displacement reaction in which one of the product remain in solid form are represented by (s) after their chemical formulas. Such double displacement reaction are called as precipitation reaction.

Cu(NO_3)_2(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+2NaNO_3(aq)

Double displacement reaction is defined as the reaction where exchange of ions takes place.

CuO(s)+2HCl(aq)\rightarrow CuCl_2(aq)+H_2O(l)

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5. Durante un estudio de la velocidad de la reacción A2(g) + 3B2(g)  2 AB3(g), se observa que en un recipiente cerrado que cont
weqwewe [10]

Answer:

a) Speed of the reaction = 0.002083 mol/L.s

b) The rate of disappearance of A₂ during this period of time = 0.002083 mol/L.s

c) The rate of appearance of AB₃ = 0.004167 mol/L.s

Explanation:

English Translation

During a study of the reaction rate

A₂ (g) + 3B₂ (g) → 2 AB₃ (g),

it is observed that in a closed container containing a certain amount of A₂ and 0.75 mol / L of B₂, the concentration B₂ decreases to 0.5 mol / L in 40 seconds.

a) What is the speed of the reaction?

b) What is the rate of disappearance of A₂ during this period of time?

c) What is the rate of appearance of AB₃?

Solution

The rate of a chemical reaction is defined as the time rate at which a reactant is used up or the rate at which a product is formed.

It is the rate of change of the concentration of a reactant (rate of decrease of the concentration of the reactant) or a product (rate of increase in the concentration of the product) with time.

Mathematically, for a balanced reaction

aA → bB

Rate = -(1/a)(ΔA/Δt) = (1/b)(ΔB/Δt)

The minus sign attached to the change of the reactant's concentration indicates that the reactant's concentration decreases.

And the coefficients of each reactant and product in the balanced reaction normalize the rate of reaction for each of them

So, for our given reaction,

A₂ (g) + 3B₂ (g) → 2 AB₃ (g)

Rate = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt) = (1/2)(ΔAB₃/Δt)

a) Speed of the reaction = Rate of the reaction

But we are given information on the change of concentration of B₂

Change in concentration of B₂ = ΔB₂ = 0.50 - 0.75 = -0.25 mol/L

Change in time = Δt = 40 - 0 = 40 s

(ΔB₂/Δt) = (-0.25/40) = -0.00625 mol/L.s

Rate of the reaction = -(1/3)(ΔB₂/Δt) = (-1/3) × (-0.00625) = 0.002083 mol/L.s

b) The rate of disappearance of A₂ during this period of time

Recall

Rate = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt)

-(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt)

Rate of disappearance of A₂ = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt) = (-1/3) × (-0.00625) = 0.002083 mol/L.s

c) The rate of appearance of AB₃

Recall

Rate = -(1/3)(ΔB₂/Δt) = (1/2)(ΔAB₃/Δt)

(1/2)(ΔAB₃/Δt) = -(1/3)(ΔB₂/Δt)

(ΔAB₃/Δt) = -(2/3)(ΔB₂/Δt)

rate of appearance of AB₃ = (ΔAB₃/Δt) = -(2/3)(ΔB₂/Δt) = (-2/3) × (-0.00625) = 0.004167 mol/L.s

Hope this Helps!!!

3 0
3 years ago
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