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Whitepunk [10]
3 years ago
9

For a reversible reaction, what would a large equilibrium constant indicate? Question 9 options: A) At equilibrium, the concentr

ation of the reactants will be much higher than the concentration of the products. B) At equilibrium, the concentration of the products will be about the same as the concentration of the reactants. C) At equilibrium, the concentration of the products will be much higher than the concentration of the reactants. D) At equilibrium, there will be no reactants left because they will all have been turned into products.
Chemistry
1 answer:
jek_recluse [69]3 years ago
3 0
Hey what do you want me to pick out today or tomorrow if not I’ll just get my car I need you a
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It’s neutral with a pH of 7. It’s probably water.
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The<br> is the continuous movement of water between the<br> atmosphere, the land and the oceans.?
vekshin1

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hydrologic cycle

Explanation:

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Nitrous oxide (N2O) is used as an anesthetic (laughing gas) and in aerosol cans to produce whipped cream. It is also a potent gr
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Answer:

five half lives

Explanation:

Half-life is the time required for a quantity to reduce to half of its initial value.

How many half lives it would take to reach 3.13% form 100% of it's initial concentration:

100% - 50% : First Half life

50% - 25%: Second Half life

25% - 12.5%: Third Half life

12.5% - 6.25%: Fourth Half life

6.25% - 3.125%: Fifth Half life

This means it would take five half lives to get to 3.125% (≈ 3.13%) of it's original concentration.

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3 years ago
Density is an example of a​
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A rock is obviously more dense than a crumpled piece of paper of the same size. A styrofoam cup is less dense than a ceramic cup.


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7 0
3 years ago
The photodissociation of ozone by ultraviolet light in the upper atmosphere is a first-order reaction with a rate constant of 1.
atroni [7]

Answer:

[O₃]= 8.84x10⁻⁷M  

Explanation:

<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)

<em>where [O₃](t): is the ozone concentration in the elapsed time and [O₃]₀: is the initial ozone concentration</em>

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
4 years ago
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