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:
from the ideal gas law, P2 /P1 = T2 /T1 for the same volume and quantity of gas. P2 = (758 mmHg) (45 +273 K) / (15 +273K) = 837 mmHg.
Explanation:
Answer: A
Explanation:
heat always passes from a warmer object to a cooler object until all objects are the same temperature. Conduction is how heat travels between objects that are touching. Conduction travels fastest through solids, but liquids and gases can also conduct heat.
<u>Given:</u>
Mass of solvent water = 4.50 kg
Freezing point of the solution = -11 C
Freezing point depression constant = 1.86 C/m
<u>To determine:</u>
Moles of methanol to be added
<u>Explanation:</u>
The freezing point depression ΔTf is related to the molality m through the constant kf, as follows:
ΔTf = kf*m
where ΔTf = Freezing point of pure solvent (water) - Freezing pt of solution
ΔTf = 0 C - (-11.0 C) = 11.0 C
m = molality = moles of methanol/kg of water = moles of methanol/4.50 kg
11.0 = 1.86 * moles of methanol/4.50
moles of methanol = 26.613 moles
Ans: Thus around 26.6 moles of methanol should be added to 4.50 kg of water.
Answer:
183 cg = 0.00183 kg
0.25 kg = 250 g
Explanation:
Use conversion factors. 1kg is equal to 1 x 10^5 cg (100000) and 1 kg is equal to 1 x 10^3 grams (1000 grams).
