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Virty [35]
3 years ago
10

Ion Concentration 1. Which solution has the greatest [SO42-]: a) 0.075 M H2SO4 b) 0.15 M Na2SO4 c) 0.080 M Al2(SO4)3?

Chemistry
1 answer:
soldi70 [24.7K]3 years ago
8 0

Answer:

c) 0.080 M Al₂(SO₄)₃

Explanation:

Ion [SO₄²⁻] concentration of each solution is:

a) 0.075 M H₂SO₄: <em>[SO₄²⁻] = 0.075M</em>. Because 1 mole of H₂SO₄ contains 1 mole of SO₄²⁻

b) 0.15 M Na₂SO₄: <em>[SO₄²⁻] = 0.15M</em>. Also, 1 mole of Na₂SO₄ contains 1 mole of SO₄²⁻

c) 0.080 M Al₂(SO₄)₃ [SO₄²⁻] = 0.080Mₓ3 =<em> 0.240M</em>. Because 1 mole of Al₂(SO₄)₃ contains 3 moles of SO₄²⁻.

<h3>Thus, the soluion that has the greatest [SO₄²⁻] is 0.080 M Al₂(SO₄)₃</h3>
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Water molecules are polar because the?
disa [49]

Answer:

Water (H2O) is polar due to the  bent shape of the molecule. The shape means most of the negative charge from the oxygen is one one side of the molecule and the positive charge of the hydrogen atoms is on the other side of the molecule. This is an example of polar covalent chemical bonding.

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3 years ago
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A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc
Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

<em>2.5x10⁻⁵ = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺]</em>

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

<h3>pH = 2.83</h3>

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3 years ago
Use this oxidation-reduction reaction to answer questions about half-reactions:
Advocard [28]

Answer:

N, O, R

Explanation:

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2 years ago
For each of the following sublevels, give the n and l values and the number of orbitals: (a) 5s; (b) 3p; (c) 4f
OverLord2011 [107]

Answer:

(a) 5s. n = 5. Sublevel s, l = 0. Number of orbitals = 1

(b) 3p. n = 3. Sublevel p, l = 1. Number of orbitals = 3

(c) 4f. n =4. Sublevel f, l = 3. Number of orbitals = 7

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Sublevel number, 0 ≤ l ≤ n − 1

So,

(a) 5s. n = 5, shell number 5. Sublevel s, l = 0. Number of orbitals = 2l +1 = 1

(b) 3p. n = 3, shell number 3. Sublevel p, l = 1. Number of orbitals = 2l +1 = 3

(c) 4f. n =4, shell number 4. Sublevel f, l = 3. Number of orbitals =  2l +1 = 7

6 0
3 years ago
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