Answer:
The rule is especially applicable to carbon, nitrogen, oxygen, and the halogens, but also to metals such as sodium or magnesium. ... All four of these electrons are counted in both the carbon octet and the oxygen octet, so that both atoms are considered to obey the octet rule.
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
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Answer:
34g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
H2S + 2AgNO3 —> 2HNO3 + Ag2S
Next, we shall determine the number of mole of H2S required to react with 2 moles of AgNO3.
This is illustrated below:
From the balanced equation above,
We can see that 1 mole of H2S is required to react completely with 2 moles of AgNO3.
Finally, we shall convert 1 mole of H2S to grams. This is shown below:
Number of mole H2S = 1 mole
Molar mass of H2S = (2x1) + 32 = 34g/mol
Mass = number of mole x molar Mass
Mass of H2S = 1 x 34
Mass of H2S = 34g
Therefore, 34g of H2S is needed to react with 2 moles of AgNO3.
