Choose all the answers that apply.

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:

The photodissociation of ozone by UV light is given by:

O₃ + hν → O₂ + O (1)

The first-order reaction of the equation (1) is:

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

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

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

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

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

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

Please help me (number which answer is which number)

Klio2033 [76]

1) An endothermic reaction is a reaction where heat is "absorbed" and the surrounding area is cooled. An example of an endothermic reaction is the classic baking soda and vinegar reaction. The mixture gets cold.
An exothermic reaction is one which gives off heat, such as burning, which is itself a chemical reaction.

2) (I don't actually know this one but i hope i helped with number one)

7 0

3 years ago

A model used for the yield Y of an agricultural crop as a function of the nitrogen level N in the soil (measured in appropriate

lisov135 [29]

Answer:

It is 3.

Explanation:

The model written correctly is:

$Y=\frac{kN}{9+N^2}$

This is a mathematical question, instead of a chemistry question, and you should use calculus to find the nitrogen level that gives the best yield, since this is an optimization problem.

The best yield is the maximum yield, and the maximum, provided that it exists, is found using the first derivative and making it equal to zero: Y' = 0

To find Y' you must use the quotient rule.

$Y'=\frac{(kN)'(9+N^2)- (kN)(9+N^2)'}{(9+N^2)^2}\\ \\Y'=\frac{k(9+N^2)-kN(2N)}{(9+N^2)^2}\\ \\Y'=\frac{9k + kN^2 - 2kN^2}{(9 + N^2)^2}=\frac{9k-kN^2}{(9+N^2)^2}$

Now make Y' = 0

The denominator is never equal to zero, because it is always positive and greater than 9.

Make the numerator equal to zero:

9k - kN² = 0

Factor: k (9 - N²) = 0

Since k is a positve constant, it is not equal to zero, and the other factor, 9 - N², must be equal to zero:

9 - N² = 0 ⇒ (3 - N) (3 + N) = 0

⇒ 3 - N = 0 or 3 + N = 0 ⇒ N = 3 or N = -3.

Since N is nitrogen level, it cannot be negative and the only valid answer is N = 3.

You can prove that it is a maximum (instead of a minimum) finding the second derivative or testing some points around 3 (e.g. 2.5 and 3.5).

8 0

3 years ago

Sodium reacts with water to produce sodium hydroxide and hydrogen gas:

Kruka [31]

Answer:

3

Explanation:

3 0

3 years ago

N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?

just olya [345]

Answer:

$\large \boxed{\text{197.4 g}}$

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 28.01 17.03

N₂ + 3H₂ ⟶ 2NH₃

m/g: 240.0

(a) Moles of NH₃

$\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}$

(b) Moles of N₂

$\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}$

(c) Mass of N₂

$\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}$

7 0

3 years ago

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