Answer:
487.33 K.
Explanation:
- To calculate the no. of moles of a gas, we can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant.
T is the temperature of the gas in K.
- If n is constant, and have two different values of (P, V and T):
<em>P₁V₁T₂ = P₂V₂T₁</em>
<em></em>
P₁ = 5.4 atm, V₁ = 1.0 L, T₁ = 33°C + 273 = 306 K.
P₂ = 4.3 atm, V₂ = 2.0 L, T₂ =??? K.
<em>∴ T₂ = P₂V₂T₁/P₁V₁</em> = (4.3 atm)(2.0 L)(306 K)/(5.4 atm)(1.0 L) = <em>487.33 K.</em>
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)
Formula: NA2S2O3. Valency: 2
Answer:
The answer is
<h2>250 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
volume of object = 25 mL
Density = 10 g/mL
The mass of the object is
mass = 25 × 10
We have the final answer as
<h3>250 g</h3>
Hope this helps you
