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ASHA 777 [7]
4 years ago
15

The metallic radius of an aluminum atom is 143 pm. What is the volume of an aluminum atom in cubic meters?

Chemistry
1 answer:
Marina86 [1]4 years ago
8 0

Answer:

6.61 × 10∧-29 m³

Explanation:

Given data:

Atomic radius= 143 pm = 143 × 10∧-12 m

volume = ?

Formula:

r = a/2√2

143 × 10∧-12 m = a/ 2√2

a= 143 × 10∧-12 m × 2√2

a= 404.4 × 10∧-12 m

where a is edge length, so we can calculate the volume by using following formula:

volume= a³

V= (404.4 × 10∧-12 m)³

v= 6.61 × 10∧-29 m³

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The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
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Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

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Q2.  0.254 g of KHP (204 g/mol) titrated against 20.0 mL of unknown NaOH (40.0 g/mol) solution to get the end point of phenolpht
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Answer:

<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>

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1 mole of NaOH - 40g

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⇒0.0498g of NaOH was used during the titration

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<u><em>= 2.49g/L.</em></u>

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