Set up a hess’s law cycle, and use the following information to calculate δh∘f for aqueous nitric acid, hno3(aq). you will need to use fractional coefficients for some equations. 3no2(g)+h2o(l)→2hno3(aq)+no(g) δh∘ = -137.3 kj 2no(g)+o2(g)→2no2(g) δh∘ = -116.2 kj 4nh3(g)+5o2(g)→4no(g)+6h2o(l) δh∘ = -1165.2 kj nh3(g) δh∘f = -46.1 kj h2o(l) δh∘f = -285.8 kj

-207,4 kJ = ΔH∘fHNO₃(aq)

Explanation:

Ussing Hess's law it is possible to obtain the enthalpy of a reaction or a product by the sum of different half-reactions.

(1) 3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g) ΔH° = -137.3 kJ

(2) 2NO(g) + O₂(g) → 2NO₂(g) ΔH° = -116.2 kJ

(3) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l) ΔH° = -1165.2 kJ

The sum of ²/₃ (1) + (2) gives:

⁴/₃NO(g) + ²/₃H₂O(l) + O₂(g) → ⁴/₃HNO₃(aq) ΔH° = ²/₃ (1) + (2) = -207,7 kJ

This reaction + ¹/₃ (3) produce:

⁴/₃NH₃(g) + ⁸/₃O₂(g) → + ⁴/₃H₂O(l) + ⁴/₃HNO₃(aq) ΔH° = -207,7 kJ +  ¹/₃ (3) = -596,1 kJ

This ΔH° is =  ⁴/₃ΔH∘fH₂O(l) + ⁴/₃ΔH∘fHNO₃(aq) - (⁴/₃ΔH∘fNH₃(g) + ⁸/₃ΔH∘fO₂(g))

As: NH₃(g) ΔH∘f = -46.1 kJ; H₂O(l) ΔH∘f = -285.8 kJ; O₂(g) ΔH∘f =0kJ:

-596,1 kJ = -⁴/₃285,8kJ +⁴/₃ΔH∘fHNO₃(aq)  - (-⁴/₃46,1kJ + 0 kJ)

-276,5 kJ = ⁴/₃ΔH∘fHNO₃(aq)

-207,4 kJ = ΔH∘fHNO₃(aq)

I hope it helps!

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

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