Answer:
I think its a.The hockey puck loses heat molecule's.
The pressure exerted by 0.400 moles of carbon dioxide in a 5.00 Liter container at 25 °C would be 1.9563 atm or 1486.788 mm Hg.
<h3>The ideal gas law</h3>
According to the ideal gas law, the product of the pressure and volume of a gas is a constant.
This can be mathematically expressed as:
pv = nRT
Where:
p = pressure of the gas
v = volume
n = number of moles
R = Rydberg constant (0.08206 L•atm•mol-1K)
T = temperature.
In this case:
p is what we are looking for.
v = 5.00 L
n = 0.400 moles
T = 25 + 273
= 298 K
Now, let's make p the subject of the formula of the equation.
p = nRT/v
= 0.400 x 0.08206 x 298/5
= 1.9563 atm
Recall that: 1 atm = 760 mm Hg
Thus:
1.9563 atm = 1.9563 x 760 mm Hg
= 1486.788 mm Hg
In other words, the pressure exerted by the gas in atm is 1.9563 atm and in mm HG is 1486.788 mm Hg.
More on the ideal gas law can be found here: brainly.com/question/28257995
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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)
