The answers that fit the blanks are SMALL and LITTLE, respectively. The particles or molecules or fas are small which makes it loose and easily moves around, and these only exert little attraction for other gas particles. The answer for this would be option D.
<em>Five household items that </em><em>contain</em><em> the same chemicals as cigarettes :</em><em>-</em>
- <em>Perfume</em>
- <em>Household</em><em> </em><em>cleaning</em><em> </em><em>products</em>
- <em>Beauty</em><em> </em><em>Products</em>
- <em>Sunscreen</em>
- <em>Water</em><em> </em><em>bottles</em>
<em>They</em><em> </em><em>contain</em><em> </em><em>toxic</em><em> </em><em>substances</em><em>!</em>
<em>hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>!</em>
Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
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2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
