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:
hope it helps.
<h3>stay safe healthy and happy.<u>.</u><u>.</u></h3>
For the answer to the questions above,
a) Ag2CO3(s) => Ag2O(s)+CO2(g)
<span>b) Cl2(g)+2(KI)(aq) => I2(s)+2(KCl)(aq) (coefficients are for balanced equation) </span>
<span>net ionic is Cl2(g)+2I- => I2(s)+2Cl-(aq) </span>
<span>c) I2(s)+3(Cl2)(g)=>2(ICl3)
</span>I hope I helped you with your problem
As the question tells you, you need to use the formula
% mass= mass of solute/ mass of solution x 100
mass solute= 30.0 g
mass of solution= 30.0 + 270.0= 300.0 g
% mass= 30.0/ 300.0 x 100= 10%
answer is B
