I cannot see your question to help you... sorry
I Would Think that the answer is 55.845 u ± 0.002 u
Or you could just do<u> 55.85 Grams</u>
Answer:
The answer is N203 Dinitrogen trioxide.
To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:
N₂ + O₂ → 2 NO
Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.
1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present
Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>
