Compare Ions: 1 Mol Nickel (II) vs 1 Mol Copper (I)

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SUMMARY

1 mol of Nickel (II) ions results in approximately 1.03 x 1022 ions, while 1 mol of Copper (I) ions yields about 9.48 x 1021 ions. Therefore, Nickel (II) has a greater number of ions compared to Copper (I). The calculations utilize the molar masses of Nickel (58.69 g/mol) and Copper (63.55 g/mol) along with Avogadro's number (6.022 x 1023 ions/mol). Understanding the concept of moles as a counting unit is crucial for these comparisons.

PREREQUISITES
  • Understanding of molar mass (Nickel: 58.69 g/mol, Copper: 63.55 g/mol)
  • Familiarity with Avogadro's number (6.022 x 1023 ions/mol)
  • Basic knowledge of ionic compounds and their properties
  • Concept of moles as a counting unit in chemistry
NEXT STEPS
  • Study the concept of moles in greater detail, focusing on its applications in chemical calculations
  • Learn about the properties and uses of Nickel (II) and Copper (I) ions in various chemical reactions
  • Explore the significance of molar mass in stoichiometry and its impact on chemical equations
  • Investigate the differences between ionic and covalent compounds and their respective behaviors in solutions
USEFUL FOR

Chemistry students, educators, and professionals interested in understanding ionic compounds and their quantitative comparisons.

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Which will have the greater number of ions, 1 mol of nickel (II) or 1 mol of copper (I)?



2. No equations



3. Nickel: 1 mol/58.69 g of Ni * 6.022*10^23 ions/ 1 mol = 1.03 * 10^22 ions.

Copper: 1 mol/ 63.55 g of Cu * 6.022 * 10^23 ions/ 1 mol = 9.48 * 10 ^ 21 ions.

Nickel will have the greater number of ions. Is this correct?
 
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This is like asking "Which bag will have more items, one with a dozen tennis balls or one with a dozen ping pong balls?"

Remember that "moles" is really a number (like dozen is a number) referring to how many of a certain thing.
 
Think your two calculations should have the units of "ions/gm", not just "ions", after you cancel both "mole" units.

This and chemisttree's info should lead you to the correct answer.

Good luck.

Steve
 

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