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Scilla [17]
3 years ago
5

Please help me with chemistry. Thanks

Chemistry
2 answers:
il63 [147K]3 years ago
7 0
Constant:
Test tubes

Independent:
<span>volume of gas

Dependent:
</span>
<span>amount of H2O2 </span>
olga2289 [7]3 years ago
6 0
Hello there!

A constant stays the same so is the test tube.
An independent variable is changed to test the effects so is the volume of gas.
A dependent variable is being tested and so is the amount of H2O2.

Hope This Helps You!
Good Luck :)
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What is the H+ concentration for an aqueous solution with pOH = 4.01 at 25 ∘C? Express your answer to two significant figures an
EleoNora [17]

I believe pH = -log[H+]

Also, 14 = pH + pOH

Therefore pH = 14 - pOH

pH = 14 - 4.01

pH = 9.99

9.99 = -log[H+]

Solve for H+

5 0
3 years ago
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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shepuryov [24]

Answer:

............................................

6 0
2 years ago
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Naddik [55]
Answer is: 0,275 moles of oxygen are consumed.
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3 years ago
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Please identify and specify the question please!
8 0
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