Calculate the number of potassium ions present in 7.45g of potassium oxide (atomic mass of K=39u)?
K2O potassium oxide
moles K2O=mass/formula mass= 7.45/(2*39+16)=.079moles
number K ions= 2*.079*avagradros number
To calculate the number of potassium ions present in 7.45 grams of potassium oxide (K2O), you'll need to follow these steps:
Step 1: Determine the molar mass of potassium oxide (K2O).
The molar mass is the sum of the atomic masses of all the atoms in a molecule. In this case, 1 molecule of K2O contains 2 potassium (K) atoms and 1 oxygen (O) atom.
Molar mass of K2O = (2 x atomic mass of K) + (1 x atomic mass of O)
= (2 x 39u) + (1 x 16u)
= 78u + 16u
= 94u
Step 2: Calculate the number of moles of potassium oxide (K2O).
To calculate the number of moles, use the formula:
moles = mass / molar mass
moles of K2O = mass of K2O / molar mass of K2O
= 7.45g / 94u
≈ 0.079 moles
Step 3: Determine the number of potassium ions.
Since 1 molecule of K2O contains 2 potassium (K) ions, and 1 mole is equal to Avogadro's number (6.022 x 10^23) of particles, we can use the following conversion:
number of potassium ions = moles of K2O x 2 x Avogadro's number
number of potassium ions = 0.079 moles x 2 x 6.022 x 10^23
= 9.53 x 10^22 potassium ions
Therefore, there are approximately 9.53 x 10^22 potassium ions present in 7.45 grams of potassium oxide.