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:
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Answer:
0.107 mole of SO2.
Explanation:
1 mole of a gas occupy 22.4 L at standard temperature and pressure (STP).
With the above information, we can simply calculate the number of mole of SO2 that will occupy 2.4 L at STP.
This can be obtained as follow:
22.4 L contains 1 mole of SO2.
Therefore, 2.4 L will contain = 2.4/22.4 = 0.107 mole of SO2.
Therefore, 0.107 mole of SO2 is present in 2.4 L at STP.
Answer:
Explanation:
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In this case, the undergoing chemical reaction is:
In such a way, as 7.38 moles of potassium chloride are decomposed, the resulting grams of oxygen are computed considering a 2 to 3 molar relationship in the chemical reaction:
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