im not sure which one to answer, and i can hardly see the text.
hey there!:
2HgO (s) => 2Hg (l) + O2 (g)
2 moles of HgO decompose to form 2 moles of Hg and 1 mole of O2 according to the reaction mentioned in the question.
So 4.00 moles of HgO must give 4 moles of Hg and 2 moles of O2 theoretically.
603 g of Hg = 603 / 200.6 = 3 moles
Percent yield = ( actual yield / theoretical yield) * 100
= ( 3/4) * 100
= 75 %
Hope this helps!
Answer:
Amy's question is not testable and too broad.
