Answer:
-162,5 kJ/mol
Explanation:
Cl(g) + 2O2(g) --> ClO(g) + O3(g) ΔH = 122.8 kJ/mol (as we used the reaction in the opposite direction, it will turn the enthalpy from exothermic to endothermic)
2O3(g) --> 3O2(g) ΔH = -285.3 kJ/mol
Cl(g) + O2(g) --> ClO(g) + O3(g) ΔH = 122.8 kJ
+ 2O3 (g) --> 3O2(g) ΔH = - 285.3 kJ
O3(g) + Cl(g) --> ClO(g) + 2O2(g) ΔH = 122.8 + (-285.3) = -162,5 kJ
The correct answer is option C, that is, the universal law of gravitation.
According to Newton's law of universal gravitation, every particle fascinates every other particle in the universe with a force that is directly equivalent to the product of their masses and inversely proportional to the square of the distance between their centers. This is a usual physical law obtained from empirical observations by what Isaac Newton knew as inductive reasoning.
Answer:
Occluded Front
Explanation:
"Occluded Front Forms when a warm air mass gets caught between two cold air masses. The warm air mass rises as the cool air masses push and meet in the middle. The temperature drops as the warm air mass is occluded, or “cut off,” from the ground and pushed upward."
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A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O
Let the total mass of compound = 100g
The mass of Aluminum = 22.10 g
Moles of Al = mass / atomic mass = 22.10 /23= 0.96
The mass of P = 25.40 g
Moles of P = 25.40 / 31 = 0.819
The mass of oxygen = 52.50 g
Moles of oxygen = 52.50 / 16= 3.28
The mole ratio of the two elements will be
Al = 0.96/0.819 = 1.17
P = 0.819/0.819 = 1
O = 3.28 / 0.819 = 4
The whole number ratio will be 1 : 1 : 4
So formula will be AlPO4
Answer:
28 grams CaH₂(s) is required for production of 15L H₂(g) at 25°C and 825Torr.
Explanation:
CaH₂(s) + H₂O(l) => Ca(OH)₂(s) + H₂(g)
Using ideal gas law, PV = nRT
=> moles H₂(g) = PV/RT = [(825/760)Atm](15L)/(0.08206L·Atm·mol⁻¹·K⁻¹)(298K) = 0.6659 mol H₂(g)
From stoichiometry of given equation,
=> 0.6659 mol H₂(g) requires 0.6659 mole CaH₂(s)
Converting moles to grams, multiply by formula weight,
=> 0.6659 mole CaH₂(s) = 0.6659 mole CaH₂(s) x 42g/mole = 27.966 grams CaH₂(s) ≅ 28 grams CaH₂(s) (2 sig. figs.)
