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viva [34]
3 years ago
14

Wine goes bad soon after opening because the ethanol CH3CH2OH dissolved in it reacts with oxygen O2 gas to form water and aqueou

s acetic acid CH3COOH, the main ingredient in vinegar. Calculate the moles of ethanol needed to produce 0.095mol of water. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.
Chemistry
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer:

We need 0.095 moles of ethanol

Explanation:

Step 1: Data given

Number of moles water = 0.095 moles

Step 2: The balanced equation

CH3CH2OH + O2 ⇒ H2O + CH3COOH

Step 3: Calculate moles of ethanol

For 1 mol ethanol we need 1 mol oxygen to produce 1 mol water and 1 mol acetic acid

For 0.095 moles water, we need 0.095 moles ethanol and 0.095 moles  oxygen

We need 0.095 moles of ethanol

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Ionic compound is defined as the compound which is formed by complete transfer of electrons from one atom to another atom.

The atom which looses the electron is known as electropositive atom and the atom which gains the electron is known as electronegative atom. This bond is usually formed between a metal and a non-metal.

Magnesium is the 12th element of the periodic table having electronic configuration of 1s^22s^22p^63s^2

This element will loose 2 electrons to form Mg^{2+} ion

Chlorine is the 17th element of the periodic table having electronic configuration of 1s^22s^22p^63s^23p^5

This element will gain 1 electron to form Cl^{-} ion

So, for every 1 atom of magnesium, 2 atoms of chlorine are required. Thus, the chemical formula becomes MgCl_2

Hence, the ionic compound formed is magnesium chloride having formula MgCl_2

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How many hydrogen atoms are in 1 molecule of C3H7O
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Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.
Katyanochek1 [597]

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.

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