17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.
First, we will convert 17.8 g of NaClO₄ to moles using its molar mass (122.44 g/mol).

Next, we will convert 0.145 moles to molecules of NaClO₄ using Avogadro's number; there are 6.02 × 10²³ molecules in 1 mole of molecules.

NaClO₄ is a strong electrolyte that dissociates according to the following equation.
NaClO₄ ⇒ Na⁺ + ClO₄⁻
The molar ratio of NaClO₄ to Na⁺ is 1:1. The number of Na⁺ in 8.73 × 10²² molecules of NaClO₄ is:

The molar ratio of NaClO₄ to ClO₄⁻ is 1:1. The number of ClO₄⁻ in 8.73 × 10²² molecules of NaClO₄ is:

The molar ratio of ClO₄⁻ to Cl is 1:1. The number of Cl in 8.73 × 10²² ions of ClO₄⁻ is:

The molar ratio of ClO₄⁻ to O is 1:1. The number of O in 8.73 × 10²² ions of ClO₄⁻ is:

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.
You can learn more Avogadro's number here: brainly.com/question/13302703
Answer:
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Answer: this is all of it
Explanation: i know this because its all of it
Answer:
The correct answer is option C. The unknown solution definitely has Hg22+ present.
Explanation:
In the analysis of group 1 metal cation, the unknown solution is treated with sufficient quantity of 6 M HCl solution and if group 1 metal cations are present then white precipitate of Agcl, PbCl2 or Hg2Cl2 is formed. The precipitate of PbCl2 is soluble in hot water but the other two remains insoluble after treating with hot water. Precipitate of AgCl disappears upon treatment of NH3 solution but Hg2Cl2 becomes black in the reaction with NH3. The black Colour appears due to the formation of metallic Hg.
Balanced chemical equation of the reation is -
Hg2Cl2 + 2NH3 ---------> HgNH2Cl (white ppt.) + Hg (black ppt.) + NH4Cl
Therefore, from the given information the conclusion which can be drawn is that the unknown solution definitely has Hg22+ present.