Calculate ΔS°rxn (J/k) for 3NO2(g) + H2O(l)LaTeX: \longrightarrow⟶ NO(g) +2HNO3(l) C6H12O6(s) + 6O2(g) LaTeX: \longrightarrow⟶ 6
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Answer:
hydroperoxyethane
Explanation:
tomsFor the Lewis structure we have to remember that all the atoms must have <u>8 electrons</u> (except for hydrogen). In this structure, we have three types of atoms, Carbon, Hydrogen and Oxygen. So, we have to remember the <u>valence electrons</u> for each atom:
-) Carbon : 4 electrons
-) Hydrogen: 1 electron
-) Oxygen: 6 electrons
We can start with the "" part. We can put 3 hydrogen bond arroun the carbon. We can use this same logic with "
". Finally for oxygens, we can put it one bond with
and a bond between oxygens with a final bond with hydrogen to obtain <u>hydroperoxyethane</u>.
See figure 1 for further explanations.
₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb +
Your equation is:
₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + ?
It becomes easier to balance the equation if we replace the "?" with an element symbol _<em>x</em>^<em>y</em>Z.
Then the equation becomes
₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb +_<em>x</em>^<em>y</em>Z
The main point to remember in balancing nuclear equations is that the <em>sums of the superscripts and the subscripts</em> must be the same <em>on each side of the equation</em>.
Sum of superscripts: 210 = 206 + <em>y</em>, so <em>y</em> =4.
Sum of subscripts: 84 = 82 + <em>x</em>, so<em> x</em> = 2.
₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb +
We should recall that is an <em>α particl</em>e.
Thus, the equation represents the α decay of polonium-210 to lead-206.
The nuclear equation is
₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb +