Answer:
The correct answer is " As you increase altitude, Earth's gravity is not acting as greatly as when closer to Earth's surface so the weight of air molecules decrease, which also in turn decreases air pressure."
Explanation:
As altitude increases air becomes rarefied. At higher altitudes there is less Earth's gravity acting on the air molecules than at areas close the earth surface. Air pressure is the weight of air acting per unit surface area. At higher altitudes as the weight of air decreases due to reduced gravitational pull the air pressure also decreases.
According the VSEPR theory the molecular geometry for CH3+ is triagonal planar
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)
Answer:
The mass stays the same only volume changes, the volume decreases
Explanation:
The ice shrinks (decreases volume) and becomes more dense. The weight will not (and cannot) change.
