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Ymorist [56]
3 years ago
5

Os etapas que devem sem compridas para obter separadamente os componentes (a as substâncias ) de uma mistura de água ,sal e arei

a ,são respectivamente : A) filtração e sifonação B) decantação e sifonação C) filtração e destilação simples D) catação e decantação
Chemistry
1 answer:
SashulF [63]3 years ago
4 0

Answer:

The steps that must be long to obtain separately the components (to the substances) of a mixture of water, salt and sand, are respectively: A) filtration and siphoning B) decanting and siphoning C) simple filtration and distillation D) collection and decantation.

Simple filtration and Siphoning is the method. When an insoluble solid is present in a liquid. Filtration is the ideal method of separation,The sand is filtered by the filtered paper as water is allowed to pass through the filter paper into a beaker where the funnel which hold the filter paper emptied into.

Water molecules is Siphoned, pull by force of gravity through the sand, past  the filter paper to emerge as clean in the beaker below.

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Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4. Calculate the pH of a solution containing a caffeine concentration of 41
RSB [31]

Answer:

pH → 7.47

Explanation:

Caffeine is a sort of amine, which is a weak base. Then, this pH should be higher than 7.

Caffeine + H₂O  ⇄  Caffeine⁺  +  OH⁻      Kb

1 mol of caffeine in water can give hydroxides and protonated caffeine.

We convert the concentration from mg/L to M

415 mg = 0.415 g

0.415 g / 194.19 g/mol = 2.14×10⁻³ mol

[Caffeine] = 2.14×10⁻³  M

Let's calculate pH. As we don't have Kb, we can obtain it from pKb.

- log Kb = pKb → 10^-pKb = Kb

10⁻¹⁰'⁴ = 3.98×10⁻¹¹

We go to equilibrium:

Caffeine + H₂O  ⇄  Caffeine⁺  +  OH⁻      Kb

Initially we have 2.14×10⁻³ moles of caffeine, so, after the equilibrium we may have (2.14×10⁻³ - x)

X will be the amount of protonated caffeine and OH⁻

     Caffeine     +    H₂O  ⇄  Caffeine⁺  +  OH⁻      Kb

   (2.14×10⁻³ - x)                         x                x

We make the expression for Kb:

3.98×10⁻¹¹ = x² / (2.14×10⁻³ - x)

We can missed the -x in denominator, because Kb it's a very small value.

So: 3.98×10⁻¹¹ = x² / 2.14×10⁻³

√(3.98×10⁻¹¹ . 2.14×10⁻³) = x → 2.92×10⁻⁷

That's the [OH⁻].  - log [OH⁻] = pOH

- log 2.92×10⁻⁷ = 6.53 → pOH

14 - pOH = pH → 14 - 6.53 = 7.47

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