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2 years ago

When a boron atom loses its valence electrons, what is the formula for the resulting ion?

Chemistry

2 answers:

kirill [66]2 years ago

8 0

Answer:

option b. B3+

Explanation:

Boron takes the 5th position on the periodic table, therefore it has 5 electrons....2 on the inside and 3 on the outside. when it lost it 3 external electrons, it become positively charged with the amount of electron it loses.

ololo11 [35]2 years ago

4 0

b. B 3+

..................

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Describe the kinetic energy/ movement of the particles in plasma.

exis [7]

kinetic energy is a moving object potential energy is stored energy

8 0

3 years ago

Read 2 more answers

A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a

Jet001 [13]

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

BaC₂O₄ ⇌ Ba²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 5.0 × 10⁻⁵ c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

ZnC₂O₄ ⇌ Zn²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 2.0 × 10⁻⁷ c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

Ag₂C₂O₄ ⇌ 2Ag⁺ + C₂O₄²⁻

E/mol·L⁻¹: 3.0 × 10⁻⁵ c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028 mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

4 0

3 years ago

What is the net ionic equation for 2Sb(OH)3 (s) + 3Na2S (aq) = Sb2S3 + 6NaOH

tatuchka [14]

Answer:

2Sb^(+3) (aq) + 3S^(-2) (aq) = Sb_2•S_3

Explanation:

First of all, let us balance the equation to give;

2Sb(OH)3 (s) + 3Na2S (aq) = Sb2S3 + 3NaOH

Now, we can observe the presence of positive Sodium ions (Na+) and negative hydroxyl ions (OH-) on both left and right sides of the equation.

Now, the two ions will cancel out. These ions are not really involved in the overall reaction and thus do not require being written in the overall equation. Hence, the overall net ionic reaction can now be written as:

2Sb^(+3) (aq) + 3S^(-2) (aq) = Sb_2•S_3

6 0

3 years ago

Select correct examples of linear molecules for five electron groups. Select correct examples of linear molecules for five elect

Shalnov [3]

Explanation:

Formula to calculate hybridization is as follows.

Hybridization = $\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

So, hybridization of $BeCl_{2}$ is as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[2 + 2]$

= 2

Hybridization of $BeCl_{2}$ is sp. Therefore, $BeCl_{2}$ is a linear molecule. There will be only two electron groups through which Be is attached.

Similarly, hybridization of $XeF_{2}$ is calculated as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[8 + 2]$

= 5

Therefore, hybridization of $XeF_{2}$ is $sp^{3}d. Therefore, [tex]XeF_{2}$ is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.

Thus, we can conclude that out of the given options $XeF_{2}$ is the correct examples of linear molecules for five electron groups.

7 0

2 years ago

Write a balanced equation for rusting. (Assume rust becomes Fe+2)

Stells [14]

Answer:

4Fe + 3O2 + 6H2O → 4Fe(OH)3

Explanation:

The chemical formula for rust is Fe2O3 and is commonly known as ferric oxide or iron oxide. The final product is a series of chemical reactions simplified below as- The rusting of the iron formula is simply 4Fe + 3O2 + 6H2O → 4Fe(OH)3. The rusting process requires both the elements of oxygen and water.

8 0

2 years ago