1) Compund Ir (x) O(y)
2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g
3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g
4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g
5) Convert grams to moles
moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles
moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131
6) Find the proportion of moles
Divide by the least of the number of moles, i.e. 0.00656
Ir: 0.00656 / 0.00656 = 1
O: 0.0131 / 0.00656 = 2
=> Empirical formula = Ir O2 (where 2 is the superscript for O)
Answer: Ir O2
Molarity is expressed as:
Molarity = moles / liter
Given that the cell is rod-shaped, its volume is calculated using the formula for a cylinder's volume:
V = πr²L
V = π * (0.6)² * 4.9
V = 5.54 μm³
1 Liter = 10³ mm³
1 mm = 10³ μm
1 mm³ = 10⁹ μm³
1 liter = 10¹² μm³
So the volume in liters is:
5.54 x 10⁻¹² L
Moles = molarity * liters
Moles = 0.0029 * 5.54 x 10⁻¹²
Moles = 1.61 x 10⁻¹⁴
To get the number of molecules, we multiply the moles by Avagadro's number
Number of molecules = 1.61 x 10⁻¹⁴ * 6.02 x 10²³
There are 9.69 x 10⁹ molecules in the cell
I’m not to sure but let me figure it out hold up
