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torisob [31]
2 years ago
6

Función que describe el movimiento armónico simple

Chemistry
1 answer:
n200080 [17]2 years ago
3 0
Sorry I don’t understand
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What do we call compounds that will not dissolve in water
denis-greek [22]

Answer:

covalent compounds

Explanation:

8 0
3 years ago
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Which planets' orbits is the belt located between the main asteroid belt?
hichkok12 [17]
<span>circumstellar is the answer</span>
3 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
A drinking water plant adds 500 grams of fluoride to a water tank containing 500,000 liters of drinking water. what is the conce
nlexa [21]
Answer is: <span>concentration of fluoride in the water in parts-per-million is 1 ppm.
</span>Parts-per-million (10⁻⁶) is<span> present at one-millionth of a </span>gram per gram of sample solution, f<span>or example mg/kg.
</span>m(fluoride) = 500 g · 1000 mg/g = 500000 mg.
m(water) = d(water) · V(water).
m(water) = 1 kg/L · 500000 L.
m(water) = 500000 kg.
arts-per-million = 500000 mg ÷ 500000 kg = 1 mg/kg = 1 ppm.
5 0
2 years ago
The liquid dispensed from a burette is called ___________.
11Alexandr11 [23.1K]

The liquid that is been dispensed during titration as regards this question is Titrant.

  • Titration can be regarded as  common laboratory method that is been carried out during quantitative chemical analysis.
  • This analysis helps to know the concentration of an identified analyte.
  • Burette can be regarded as laboratory apparatus.

It is used in the in measurements of variable amounts of liquid ,this apparatus helps in dispensation   of liquid, especially when performing titration.

  • The specifications is been done base on their volume, or resolution.
  • The liquid that comes out of this apparatus is regarded as Titrant, and this is gotten during titration process, which is usually carried out during volumetric analysis.

Therefore, burrete is used in volumetric analysis.

Learn more at:

brainly.com/question/2728613?referrer=searchResults

8 0
3 years ago
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