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marta [7]
3 years ago
14

What is the ΔG given the following? ΔH = 20 kJ/mol T = 15°C ΔS = .101 kJ/molK

Chemistry
1 answer:
DedPeter [7]3 years ago
6 0
∆G = ∆H-T∆S
=20×10^3-(15+273)(.101×10^3)
=20000-(288)(101)
=20000-29088
=-9088 joule = -9.088 kj
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To neutralized 1.65g LiOH, how much .150M HCl would be needed?
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4 years ago
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atroni [7]

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<u>The photodissociation of ozone by UV light is given by:</u>

O₃ + hν → O₂ + O (1)

<u>The first-order reaction of the equation (1) is:</u>

rate = k [O_{3}] = - k \frac{\Delta [O_{3}]}{\Delta t} (2)

<em>where k: is the rate constant and Δ[O₃]/Δt: is the variation in the ozone concentration with time, and the negative sign is by the decrease in the reactant concentration </em>    

<u>We can get the following expression of the </u><u>first-order integrated law</u><u> of the reaction (1), by resolving the equation (2):</u>

[O_{3}]_{t} = [O_{3}]_{0} \cdot e^{-kt} (3)

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We can calculate the initial ozone concentration using equation (3):  

[O_{3}]_{t} = 5.0 \cdot 10^{-3}M \cdot e^{-(1.0\cdot 10^{-5}s^{-1}) (\frac{10d \cdot 24h \cdot 3600 s}{1d \cdot 1h})} = 8.84 \cdot 10^{-7}M

So, the ozone concentration after 10 days is 8.84x10⁻⁷M.

I hope it helps you!                    

3 0
3 years ago
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