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
Answer: 0.0220275 M
Explanation:
So, we are given the following data or parameters which are going to help in solving this particular Question/problem.
=> Averagely, we have the volume = 5.0 L of blood in human body .
=> Mass of sugar eaten = 37.7 g of sugar (sucrose, 342.30 g/mol).
Therefore, the molarity of the blood sugar change can be calculated as below:
The molarity of the blood sugar change = (1/ volume) × mass/molar mass.
Thus, the molarity of the blood sugar change = (1/5) × 37.7/342.30 = 0.0220275 M.
Answer: 4.1 g of barium precipitated.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
To calculate the moles, we use the equation:
Given : moles of barium = 0.030
Molar mass of barium = 137 g/mol
x= 4.1 g
Thus there are 4.1 g of barium that precipitated.
First one is False. The second is true.
