A. Natural Gas is a substance that is a source of hydrocarbons.
Answer:
(2) Adding more O2(g) would shift the equilibrium to the right because a higher concentration of oxygen is offered than its initial position, therefore more products have to be yielded to maintain equilibrium.
Explanation:
The quantity of heat absorbed is 33.4 kJ.
Δ<em>H</em>_fus = 334 J·g⁻¹
<em>q = m</em>Δ<em>H</em>_fus = 100.0 g × 334 J·g⁻¹ = 3.34 × 10⁴ J = 33.4 kJ
Answer:
(a) rate = 4.82 x 10⁻³s⁻¹ [N2O5]
(b) rate = 1.16 x 10⁻⁴ M/s
(c) rate = 2.32 x 10⁻⁴ M/s
(d) rate = 5.80 x 10⁻⁵ M/s
Explanation:
We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration of N₂O₅, so
(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]
(b) rate = 4.82×10⁻³s⁻¹ x 0.0240 M = 1.16 x 10⁻⁴ M/s
(c) Since the reaction is first order if the concentration of N₂O₅ is double the rate will double too: 2 x 1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s
(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to
1.16 x 10⁻⁴ M/s / 2 = 5.80 x 10⁻⁵ M/s
They both turn something on, and the way they are different is the way they turn it off<span>
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