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horrorfan [7]
3 years ago
10

Write the formula for potassium oxide. why do you not need prefixes in the name ​

Chemistry
1 answer:
Molodets [167]3 years ago
7 0

Potassium oxide: K₂O.

There's no need for prefixes since K₂O is an ionic compound.

<h3>Explanation</h3>

Find the two elements on a periodic table:

  • Potassium- K- on the left end of period four.
  • Oxygen- O- near the right end of periodic two.

Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.

  • Potassium is a metal,
  • Oxygen is a nonmetal.

A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:

  • Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.
  • Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion \text{O}^{2-} when it combines with metals.

The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each \text{O}^{2-} ion is twice that on a \text{K}^{+} ion. Each \text{K}^{+} would pair up with two \text{O}^{2-}. Hence the subscript in the formula: \text{K}_{\bf 2}\text{O}.

There are two classes of compounds:

  • Covalent compounds, which need prefixes, and
  • Ionic compounds, which need no prefix.

Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide \text{CO}_2, the prefix di- indicates that there are two oxygen atoms in the formula \text{CO}_2. However, there's no need for prefix in ionic compounds such as \text{K}_2\text{O}.

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PH of acidic buffer = pKa + log [CH₃COONa - HCl] / [CH₃COOH + HCl]
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the solubility product Ag3PO4 is: Ksp = 2.8 x 10^-18. What is the solubility of Ag3PO4 in water, in moles per liter?
guapka [62]

Answer : The solubility of Ag_3PO_4 in water is, 1.8\times 10^{-5}mol/L

Explanation :

The solubility equilibrium reaction will be:

Ag_3PO_4\rightleftharpoons 3Ag^++PO_4^{3-}

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^{+}]^3[PO_4^{3-}]

K_{sp}=(3s)^3\times (s)

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Given:

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Now put all the given values in the above expression, we get:

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s=1.8\times 10^{-5}M=1.8\times 10^{-5}mol/L

Therefore, the solubility of Ag_3PO_4 in water is, 1.8\times 10^{-5}mol/L

3 0
3 years ago
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