There are 9.88x1023 molecules of O2 available.

Chemistry

1 answer:

blondinia [14]2 years ago

7 0

Answer:

1.64 moles O₂

Explanation:

Part A:

Remember 1 mole of particles = 6.02 x 10²³ particles

So, the question becomes, how many '6.02 x 10²³'s are there in 9.88 x 10²³ molecules of O₂?

This implies a division of given number of particles by 6.02 x 10²³ particles/mole.

∴moles O₂ = 9.88 x 10²³ molecules O₂ / 6.02 x 10²³ molecules O₂ · mole⁻¹ = 1.64 mole O₂

_______________

Part B needs an equation (usually a combustion of a hydrocarbon).

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1. Write the molecular equation, ionic equation, and net ionic equation for the reaction between calcium chloride and ammonium p

aalyn [17]

Answer:

(molecular) 3 CaCl₂(aq) + 2 (NH₄)₃PO₄(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄Cl(aq)

(ionic) 3 Ca²⁺(aq) + 6 Cl⁻(aq) + 6 NH₄⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄⁺(aq) + 6 Cl⁻(aq)

(net ionic) 3 Ca²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s)

Explanation:

The molecular equation includes al the species in the molecular form.

3 CaCl₂(aq) + 2 (NH₄)₃PO₄(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄Cl(aq)

The ionic equation includes all the ions (species that dissociate in water) and the species that do not dissociate in water.

3 Ca²⁺(aq) + 6 Cl⁻(aq) + 6 NH₄⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄⁺(aq) + 6 Cl⁻(aq)

The net ionic equation includes only the ions that participate in the reaction and the species that do not dissociate in water. In does not include <em>spectator ions</em>.

3 Ca²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s)

3 0

3 years ago

What percentage of the mass of hydrofluoric acid (HF) is contributed by hydrogen?

Morgarella [4.7K]

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: