The answer can be explained when you burn something cleanly (with a very hot item) or not. With a candle lots of Carbon dioxide is producted but when using a bunson burner hardly any CO2 is produced.
<span>Double displacement reaction is your answer.</span>
Answer:
ΔH°_rxn = -195.9 kJ·mol⁻¹
Explanation:
4NH₃(g) + 3O₂(g) ⟶ 2N₂(g) +6H₂O(g)
ΔH°_f/(kJ·mol⁻¹): -45.9 0 0 -241.8
The formula relating ΔH°_rxn and enthalpies of formation (ΔH°_f) is
ΔH°_rxn = ΣΔH°_f(products) – ΣΔH°_f(reactants)
ΣΔH°_f(products) = -6(241.8) = -1450.8 kJ
ΣΔH°_f(reactants) = -4(45.9) = -183.6 kJ
ΔH°_rxn = (-1450.8 + 183.6) kJ = -1267.2 kJ
Answer:
41.9 g
Explanation:
We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity
m: mass
ΔT: change in temperature
If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.
According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.
Qw + Qs = 0
Qw = -Qs
cw × mw × ΔTw = -cs × ms × ΔTs
(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)
ms = 41.9 g
A Beta particles is emitted when an atom of 85Kr spontaneously decays.
