Calculate the number of ions present in 5gof copper 2nitrate?

To calculate the number of ions present in 5g of copper (II) nitrate, we first need to determine the molar mass of copper (II) nitrate.

The molar mass of copper (II) nitrate (Cu(NO3)2) can be calculated as follows:

Copper (Cu): 1 atom x 63.55 g/mol = 63.55 g/mol
Nitrogen (N): 2 atoms x 14.01 g/mol = 28.02 g/mol
Oxygen (O): 6 atoms x 16.00 g/mol = 96.00 g/mol

Total molar mass = 63.55 + 28.02 + 96.00 = 187.57 g/mol

Next, we need to calculate the number of moles of copper (II) nitrate in 5g:

Number of moles = mass / molar mass
Number of moles = 5g / 187.57 g/mol ≈ 0.0266 mol

Since copper (II) nitrate dissociates into 1 copper ion (Cu2+) and 2 nitrate ions (NO3-) in solution, the total number of ions present in 5g of copper (II) nitrate can be calculated as follows:

Number of ions = (1 + 2) x number of moles
Number of ions = 3 x 0.0266 mol ≈ 0.08 mol

Therefore, there are approximately 0.08 moles of ions present in 5g of copper (II) nitrate.