Answer:
linear ,trigonal planar,tethradral geometry
Answer:
ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given) (exothermic)
Explanation:
3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g) + 17H₂O(g) + 10CO₂(g)
ΔH⁰(f): 3(-365.6)Kj 1(-301)Kj 14(0)Kj 3(0)Kj 17(-241.8)Kj 10(-393.5)Kj
= -1096.8Kj = -301Kj = 0Kj = 0Kj = -4110.6Kj = -3930.5Kj
ΔHₙ°(rxn) = ∑
(ΔH˚(f)products) - ∑(ΔH˚(f)reactants)
= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]
= [-(8041.1) - (-1397.8)]Kj
= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)
∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)
ΔH°(rxn for 14.11g NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given)
The answer is A I hope it was right
Answer:
39.7 %
Explanation:
magnesium + oxygen ⟶ magnesium oxide
10.57 g 6.96 g 17.53 g
According to the <em>Law of Conservation of Mass</em>, the mass of the product must equal the total mass of the reactants.
Mass of MgO = 10.57 + 6.96
Mass of MgO = 17.53 g
The formula for mass percent is
% by mass = Mass of component/Total mass × 100 %
In this case,
% O = mass of O/mass of MgO × 100 %
Mass of O = 6.96 g
Mass of MgO = 17.53 g
% O = 6.96/17.53 × 100
% O = 0.3970 × 100
% O = 39.7 %
