Answer:
Ir(NO2)3
Explanation:
The molar mass is 330.2335, in case that's also required.
The percentage of calcium carbonate that reacted is 2.5%. The reaction in question allows us to determine the equilibrium Kp: Kp = the partial pressure of carbon dioxide, since the other components are solids. We'll apply the ICE table to the provided equilibrium. At the start, we have 0.2 for calcium carbonate with no initial moles of other substances. As the reaction progresses, we set the changes to be -x for calcium carbonate, +x for carbon dioxide, and +x for the other product, leading us to an equilibrium of 0.2-x for calcium carbonate while both other products are at x. Using Kp = Kc(RT)ⁿ, where n represents the mole difference of gaseous products and reactants, we find n to equal 1 for this reaction. With R as the gas constant (8.314 J/mol K) and the temperature at 800 °C (1073 K), we substitute the values accordingly. Upon calculation, we find x = 0.005, which indicates the amount of calcium carbonate that dissociated or reacted, leading us to the reacted percentage.
A total of 1.505×10^23 lead atoms
In the lungs, the volume of lead equals the total lung volume, which is 5.60L
1 mole corresponds to 22.4L
Thus, 5.6L of lead converts to 5.6/22.4 = 0.25 moles
According to Avogadro's law
1 mole of lead contains 6.02×10^23 lead atoms
Thus, 0.25 moles of lead equates to 0.25×6.02×10^23 = 1.505×10^23 lead atoms
