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Zarrin [17]
3 years ago
10

Which of these locations is NOT a main-oil producing region for the United States?

Chemistry
2 answers:
Makovka662 [10]3 years ago
5 0

I believe the answer is D

trapecia [35]3 years ago
3 0

Answer:

D

Explanation:

i looked at the other answer

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Some one please help me name the following compound CsCrO4​
lozanna [386]

Answer:

Cs is cesium and CrO₄ is chromate so CsCrO₄ is cesium chromate.

8 0
3 years ago
Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid (H2CO3), causing the pH of clean, unpolluted ra
anastassius [24]

Answer:

The range of [H⁺] is from 2.51 x 10⁻⁶ M to 6.31 x 10⁻⁶ M,

Explanation:

To answer this problem we need to keep in mind the <u>definition of pH</u>:

  • pH = -log [H⁺]

So now we <u>calculate [H⁺] using a pH value of 5.2 and of 5.6</u>:

  • 5.2 = -log [H⁺]

-5.2 = log [H⁺]

10^{-5.2} = [H⁺]

6.31 x 10⁻⁶ M = [H⁺]

  • 5.6 = -log [H⁺]

-5.6 = log [H⁺]

10^{-5.6} = [H⁺]

2.51 x 10⁻⁶ M = [H⁺]

6 0
3 years ago
The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process
denis23 [38]

Answer:

0.1066 hours

Explanation:

A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.

t1/2 = ln2/k

t1/2 = ln2/6.5 h⁻¹

t1/2 = 0.1066 h

The half-life of the pesticide is 0.1066 hours.

7 0
3 years ago
What causes the Moon to revolve around Earth?
Mama L [17]

Answer:

A, Earth Gravity

Explanation:

7 0
2 years ago
Read 2 more answers
The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0
3 years ago
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