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oksano4ka [1.4K]
3 years ago
8

14) The central iodine atom in the ICl4- ion has __________ nonbonded electron pairs and __________ bonded electron pairs in its

valence shell.
Chemistry
1 answer:
masha68 [24]3 years ago
7 0

Answer:

Two non bonded electron pairs and four bonded electron pairs

Explanation:

An image of the compound as obtained from chemlibretext is attached to this answer.

The ion ICl4- ion, is an AX4E2 ion. This implies that there are four bond pairs and two lone pairs of electrons. As expected, the shape of the ion is square planar since the lone pairs are found above and below the plane of the square. This is clear from the image attached.

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Digiron [165]
<h3>1</h3>

Species shown in bold are precipitates.

  • Ca(NO₃)₂ + 2 KOH → Ca(OH)₂ + 2 KNO₃
  • Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2 NaNO₃
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  • Zn(NO₃)₂ + Na₂C₂O₄ → ZnC₂O₄ + 2 NaNO₃

<h3>2</h3>

A double replacement reaction takes place only if it reduces in the concentration of ions in the solution. For example, the reaction between Ca(NO₃)₂ and KOH produces Ca(OH)₂. Ca(OH)₂ barely dissolves. The reaction has removed Ca²⁺ and OH⁻ ions from the solution.

Some of the reactions lead to neither precipitates nor gases. They will not take place since they are not energetically favored.


<h3>3</h3>

Compare the first and last row:

Both Ca(NO₃)₂ and Zn(NO₃)₂ react with KOH. However, between the two precipitates formed, Ca(OH)₂ is more soluble than Zn(OH)₂.

As a result, add the same amount of KOH to two Ca(NO₃)₂ and Zn(NO₃)₂ of equal concentration. The solution that end up with more precipitate shall belong to Zn(NO₃)₂.


<h3>4</h3>

Compare the second and third row:

Cu(NO₃)₂ reacts with KI, but Ni(NO₃)₂ does not. Thus, add equal amount of KI to the two unknowns. The solution that forms precipitate shall belong to Cu(NO₃)₂.

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3 years ago
You need to make 10.0 L of 1.2 M KNO3. What molar ( concentration) would the potassium nitrate solution need to be if you were t
solniwko [45]

Answer:

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=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution

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