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elena-s [515]
2 years ago
12

In a Lewis structure, why must pairs of unshared of lone pairs if dots be placed around the outside of some of the atoms

Chemistry
1 answer:
Maksim231197 [3]2 years ago
6 0

Answer:

In a Lewis symbol, the symbol for the element is used to represent the atom and its core electrons. Dots placed around the atom are used to indicate the valence electrons. ... In 1916, Gilbert Newton Lewis, an American chemist, suggested that molecules were formed when atoms shared pairs of outer electrons.

Explanation:

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You have a different rock with a volume of 30cm and a mass of 60g, What is its density?
Galina-37 [17]
Density = mass/volume

Therefore,
Density = 60g/30cm
8 0
3 years ago
Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.
Firlakuza [10]

Answer:

  • [Ag⁺] = 1.3 × 10⁻⁵M
  • s = 3.3 × 10⁻¹⁰ M
  • Because the common ion effect.

Explanation:

<u></u>

<u>1. Value of [Ag⁺]  in a saturated solution of AgCl in distilled water.</u>

The value of [Ag⁺]  in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                    0

C          -s                      +s                  +s

E         X - s                   s                     s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × s
  • Ksp = s²

By substitution:

  • 1.8 × 10⁻¹⁰ = s²
  • s = 1.34 × 10⁻⁵M

Rounding to two significant figures:

  • [Ag⁺] = 1.3 × 10⁻⁵M ← answer

<u></u>

<u>2. Molar solubility of AgCl(s) in seawater</u>

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                  0.54

C          -s                      +s                  +s

E         X - s                   s                     s + 0.54

The new equation for the Ksp equation will be:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × ( s + 0.54)
  • Ksp = s² + 0.54s

By substitution:

  • 1.8 × 10⁻¹⁰ = s² + 0.54s
  • s² + 0.54s - 1.8 × 10⁻¹⁰ = 0

Now you must solve a quadratic equation.

Use the quadratic formula:

     

     s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

<u>3. Why is AgCl(s) less soluble in seawater than in distilled water.</u>

AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.

This is known as the common ion effect.

By LeChatelier's principle, you know that an increase in the concentrations of one of the substances that participate in the equilibrium displaces the reaction to the direction that minimizes this efect.

In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.

3 0
2 years ago
Yusef adds all of the values in his data set and then divides by the number of values in the set. What is Yusef most likely find
Zanzabum

He is most likely finding the mean of the data.

7 0
2 years ago
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1. The definition of the relative atomic mass for an element, X, is:
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In a combustion experiment, it was found that 12.096 g of hydrogen molecules combined with 96.000 g of oxygen molecules to form
Mkey [24]
The mass change, or the mass defect, can be calculated by the formula that is very known to be associated with Albert Einstein. 

E = Δmc²
where
E is the energy gained or released during the reaction
c is the speed of light equal to 3×10⁸ m/s
Δm is the mass change

(1.715×10³ kJ)(1,000 J/1 kJ) = Δm(3×10⁸ m/s)²
Δm = 1.91×10⁻¹¹ kg
5 0
3 years ago
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