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Dahasolnce [82]
3 years ago
5

Could a mixture be made up of only elements and no compounds? explain.

Chemistry
1 answer:
katrin [286]3 years ago
3 0
Yes, you could have a mixture noble gases. They are inert so they will remain in their elemental form and still form a mixture without reacting together.
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What is the molarity of a solution made by diluting 0.025L of 6.0 M HCl until
Nadusha1986 [10]

Answer:

Explanation:

Use the dilution equation: M1V1 = M2V2

M1 = 6.0 M

V1 = 0.025 L

M2 = ?

V2 = 1.75 L

(6.0 M)(0.025 L) = M2(1.75 L)

Solve for M2 --> M2 = 0.086 M HCl

3 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
3 years ago
17) Chlorine and Fluorine react to form gaseous chlorine trifluoride. You start with 1.75 mole of chlorine and 3.68 moles of flu
Mama L [17]

Answer:

a) Cl_2 + 3F_2 \rightarrow 2ClF_3

b) F_2 is the limiting reagent

c) Moles of ClF_3 produced = 2.45 mol

b) Moles of Cl_2 left = 1.75 -1.22 = 0.53

Explanation:

a) Balanced reaction:

Cl_2 + 3F_2 \rightarrow 2ClF_3

b)

No. of mole of Cl_2 = 1.75\ mol

No. of mole of F_2 = 3.68\ mol

Cl_2 + 3F_2 \rightarrow 2ClF_3

As, it is clear from the reaction that,

1 mol of Cl_2 requires 3 moles of F_2

1.75 mol of Cl_2 require = 1.75 × 3 = 5.25 mol of F_2

As, only 3.68 mol of F_2 is present, so F_2 is the limiting reagent.

c)

3 moles of F_2 form 2 moles of ClF_3

3.68 moles of F_2 will form = \frac{2}{3}  \times 3.68 = 2.45\ mol\ of\ ClF_3

d)

Cl_2 is present in excess.

3 moles of F_2 requires 1 mol of Cl_2

3.68 moles of F_2 will require = \frac{1}{3} \times 3.68 = 1.22\ mol\ of\ Cl_2

Cl_2 left = 1.75 -1.22 = 0.53 mol

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