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3 years ago

Question 22 Write the empirical formula for at least four ionic compounds that could be formed from the following ions:

Chemistry

1 answer:

Ivenika [448]3 years ago

7 0

Answer:

We can use the criss-cross method to work out the formulas.

Explanation:

The steps are

Write the symbols of the anion and cation.
Criss-cross the numbers of the charges to become the subscripts of the other ion.
Write the formula with the new subscripts.
Divide the subscripts by their highest common factor.
Omit all subscripts that are 1.

When we use this method, the formula for the compound formed from NH₄⁺ and PO₄³⁻ becomes (NH₄)₃PO₄.

The table below lists the formulas for the possible combinations.

$\begin{array}{ccc}& \textbf{PO}_{4}^{3-} & \textbf{ClO}_{3}^{-}\\\textbf{NH}_{4}^{+} & \text{(NH$_{4}$)$_{3}$PO$_{4}$} & \text{NH$_{4}$ClO}_{3}\\\textbf{Fe}^{3+} & \text{FePO}_{4} & \text{Fe(ClO$_{3}$)}_{3}\\\end{array}$

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SF₆ in the air at a concentration of 1.0 ppb, exerts a partial pressure of 1.0 × 10⁻⁹ atm. At this concentration, 2.3 × 10¹⁰ molecules of SF₆ are contained in 1.0 cm³ of air at 46 °C.

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where,

pSF₆: partial pressure of SF₆

P: total pressure of air (we will assume it is 1 atm)

ppb: concentration of SF₆ in parts per billion

$pSF_6 = P \times \frac{ppb}{10^{9} } = 1 atm \times \frac{1.0 ppb}{10^{9} } = 1.0 \times 10^{-9} atm$

Then, we will convert 1.0 cm³ to L using the following conversion factors:

1 cm³ = 1 mL
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$1.0 cm^{3} \times \frac{1mL}{1cm^{3}} \times \frac{1L}{1000 mL} = 1.0 \times 10^{-3} L$

Next, we will convert 46 °C to Kelvin using the following expression.

$K = \° C + 273.15 = 46 + 273.15 = 319 K$

Afterward, we calculate the moles (n) of sulfur hexafluoride using the ideal gas equation.

$P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{1.0 \times 10^{-9} atm \times 1.0 \times 10^{-3} L}{(0.082 atm.L/mol.K) \times 319 K} = 3.8 \times 10^{-14} mol$

Finally, we will convert 3.8 × 10⁻¹⁴ mol to molecules using Avogadro's number.

$3.8 \times 10^{-14} mol \times \frac{6.02 \times 10^{23} molecules }{mol} = 2.3 \times 10^{10} molecules$

You can learn more about partial pressure here: brainly.com/question/13199169

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