The configuration is 1s2 . Hope I answered and helped out! :)
Answer:
there are 4 hydrogen so
A.the mass of Hydrogen in the reactant side of the equation above is 1×4=4 amu.
B.the mass of Hydrogen on the product side of the equation above =1×4=4 amu.
<u>Note</u><u>:</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>reactant</u><u> </u><u>=</u><u>mass</u><u> </u><u>of</u><u> </u><u>product</u><u>.</u>
1.1 moles of C10H8 is the limiting reagent in the reaction between reaction C10H8 and O2
.
C10H8 + 12O2 ----> 10CO2 + 4H2O
C10H8 is the limiting reagent since 1.1 moles of C10H8 is totally consumed during the reaction
I think the answer is c but I’m not for sure