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3 years ago

Find out how many atoms will fit in one apple . Adius of an apple is about 3.5 cm and the radius of an atom is 1.5 angstrom

Chemistry

1 answer:

e-lub [12.9K]3 years ago

6 0

Answer:

1.3 × 10²⁵ atoms

Explanation:

Step 1: Calculate the volume of an apple

The radius of an apple is 3.5 cm (3.5 × 10⁻² m). If we assume it is a sphere, we can calculate its volume using the following expression.

V = 4/3 × π × r³

V(apple) = 4/3 × π × (3.5 × 10⁻² m)³ = 1.8 × 10⁻⁴ m³

Step 2: Calculate the volume of 1 atom

The radius of 1 atom is 1.5 A (1.5 × 10⁻¹⁰ m). If we assume it is a sphere, we can calculate its volume using the following expression.

V = 4/3 × π × r³

V(atom) = 4/3 × π × (1.5 × 10⁻¹⁰ m)³ = 1.4 × 10⁻²⁹ m³

Step 3: Calculate how many atoms are in 1 apple.

V(apple)/V(atom) = 1.8 × 10⁻⁴ m³/1.4 × 10⁻²⁹ m³ = 1.3 × 10²⁵

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Mashcka [7]

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$ .

(a) Relation between $K_{a}$ and $pK_{a}$ is as follows.

$pK_{a} = -log (K_{a})$

Putting the values into the above formula as follows.

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

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Also, relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log\frac{[conjugate base]}{[acid]}$

= $3.347+ log \frac{0.15}{0.12}$

= 3.44

Therefore, pH of the buffer is 3.44.

(b) No. of moles of HCl added = $Molarity \times volume$

= $11.6 M \times 0.001 L$

= 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form $HNO_{2}$

Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

= 0.15 mol

And, no. of moles of $HNO_{2}$ = $0.12 M \times 1.0 L$

= 0.12 mol

On the other hand, after the reaction :

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

= (0.15 - 0.0116) mol

= 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

= (0.12 + 0.0116) mol

= 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = $pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

= 3.369

= 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

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Answer:

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