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andrey2020 [161]
2 years ago
12

The Density of pure carbon in Diamond form is 3.52 g/cm^3. How many cubic inches would 23.7 moles of pure diamond occupy?

Chemistry
1 answer:
Vedmedyk [2.9K]2 years ago
5 0

Answer : The volume of pure diamond is 0.493inch^3

Explanation : Given,

Density of pure carbon in diamond = 3.52g/cm^3

Moles of pure diamond = 23.7 moles

Molar mass of carbon = 12 g/mol

First we have to calculate the mass of carbon or pure diamond.

\text{ Mass of carbon}=\text{ Moles of carbon}\times \text{ Molar mass of carbon}

Molar mass of carbon = 12 g/mol

\text{ Mass of carbon}=(23.7moles)\times (12g/mole)=284.4g

Now we have to calculate the volume of carbon or pure diamond.

Formula used:

Density=\frac{Mass}{Volume}

Now putting all the given values in this formula, we get:

3.52g/cm^3=\frac{284.4g}{Volume}

Volume = 80.8cm^3

As we know that:

1cm^3=0.061inch^3

So,

Volume = 0.061\times 80.8inch^3

Volume = 0.493inch^3

Therefore, the volume of pure diamond is 0.493inch^3

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This problem is asking for the percent by mass of hydrogen in hydrofluoric acid. At the end, the answer turns out to be D. 5%​ as shown below:

<h3>Percent compositions:</h3>

In chemistry, percent compositions are used for us to know the relative amount of a specific element in a compound. In order to do so for hydrogen, we use the following formula, which can also be applied to any other element in a given compound:

\% H=\frac{m_H*1}{M_{HF}}*100\%

Where m_H stands for the atomic mass of hydrogen and M_{HF} for the molar mass of hydrofluoric acid. In such a way, we plug in the atomic masses of hydrogen (1.01 g/mol) and fluorine (19.0 g/mol) to obtain:

\% H=\frac{1.01g/mol*1}{1.01g/mol+19.0g/mol}*100\%\\\\\% H=5\%

Learn more about percent compositions: brainly.com/question/12247957

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Hydrofluoric acid, hf, has a ka of 6.8 × 10−4. what are [h3o+], [f−], and [oh−] in 0.710 m hf?
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[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.

Explanation:

  • For a weak acid like HF, the dissociation of HF will be:

<em>HF + H₂O ⇄ H₃O⁺ + F⁻.</em>

[H₃O⁺] = [F⁻].

<em>∵ [H₃O⁺] = √Ka.C,</em>

Ka = 6.8 x 10⁻⁴, C = 0.710 M.

∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.

<em>∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.</em>

<em></em>

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺]</em> = 10⁻¹⁴/(2.2 x 10⁻²) = <em>4.55 x 10⁻¹³.</em>

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