Firstly, you'll need to convert the grams into number of mols.
Since no. of mol = mass/molecular weight
3.55/39.1 = 0.09079 mol
because 1 mol of potassium consists of 6.02x10^23 atoms;
0.09079 mol of potassium will consist of
0.09079x(6.02x10^23)=5.466x10^22 atoms of potassium.
This number is approx. 18.10e+23 atoms.
The number of atoms is 18,066422571.10e23.
Since molecules of potassium contain only single potassium atoms, molecules of iodine contain two atoms, and moles of potassium iodide contain one atom of each element, 2.5 moles of iodine are needed to react completely with 5 moles of potassium.
Two times Avogadro's number (6.022×10 to the 23)
15 moles3 moles potassium3 moles nitrogen9 moles oxygen15 X 6.022 X 10 to the 23rd power
A mole of potassium consists of Avogadro's number of atoms. Therefore 6.85 X 1025 atoms comprises (6.85 X 1025)/(6.022 X 1023) or 114 moles, to the justified number of significant digits.
0.1 moles
The answer is 12, 044 280.1023 atoms.
The number of atoms is 101,341.10e23.
6.81 X 1024 atoms potassium (1 mole K atoms/6.022 X 1023) = 11.3 moles of potassium atoms ========================
Since molecules of potassium contain only single potassium atoms, molecules of iodine contain two atoms, and moles of potassium iodide contain one atom of each element, 2.5 moles of iodine are needed to react completely with 5 moles of potassium.
Two times Avogadro's number (6.022×10 to the 23)
3KNO3, so 9 oxygen atoms.
15 moles3 moles potassium3 moles nitrogen9 moles oxygen15 X 6.022 X 10 to the 23rd power
10.0 moles K2SO4 (6.022 X 1023/1 mole K2SO4) = 6.02 X 1024 atoms of potassium sulfate ==========================
0.3 moles K (6.022 X 10^23/1mol K) = 1.8 X 10^23 atoms of K
2 moles of ANY element contains 2x6.02x10^23 atoms of that element = 1.2x10^24 atoms.
0.0384 moles K x 6.02x10^23 atoms/mole = 2.31x10^22 atoms
A mole of potassium consists of Avogadro's number of atoms. Therefore 6.85 X 1025 atoms comprises (6.85 X 1025)/(6.022 X 1023) or 114 moles, to the justified number of significant digits.