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finlep [7]
3 years ago
8

Methyl orange can change color by transitioning from one chromophore to another. When added to a clear solution and the solution

turns red, it is determined to be a(n) __________ in its __________ stable form.
Chemistry
1 answer:
Anvisha [2.4K]3 years ago
4 0

Answer:

acidic titration in its stable form

Explanation:

Methyl orange can change its color in titration solution. The yellow color is towards alkaline solution and red color is towards acidic solution. The Ph value of solution will change during this chemical process.

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If ice is warmed and becomes a liquid, which type of process is it?
ser-zykov [4K]
If ice is warmed and becomes a liquid, the process is endothermic.

The process requires heat in order to proceed. If ice stays in a very cold place, it will not melt unless it's heated. If ice is placed outside where it melts on its own, it gets the heat from the surroundings.
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As a chemical reaction occurs, the thermometer in the container records a drop in temperature. What is true of the reaction?
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<span>When a chemical reaction occurs and the thermometer in the container records a drop in temperature t</span>he reaction is exothermic because heat was released by the reaction.

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4 0
3 years ago
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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
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Some of the reactants or the products are in the gaseous phase.
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Referring to the experiment in which the scientists studies how long it
pshichka [43]

Answer:

different sizes of the parachute

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this is what is being changed throughout the experiment

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