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Nezavi [6.7K]
3 years ago
9

What is the term for the process that occurs when k+ and br- ions are pulled into solution?

Chemistry
2 answers:
shtirl [24]3 years ago
4 0
KBr i guess, because the + and - charge is 1 so its like that
Katyanochek1 [597]3 years ago
3 0

The Answer is Dissociation, Had to do research for the Gradpoint answer. Next time give options please.

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Explain, in terms of particle behavior why smoke particles cause the detector alarm to sound
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Explain in terms of particle behavior why smoke particles cause the detector alarm to sound


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Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
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Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb)               <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0
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What is the chemical formula for sodium sulfate?
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Hello!

When finding the chemical formula of a compound, we will need to find the charges of each element/bond.

Looking at our period table, sodium has a +1 charge, written as Na 1+, and sulfate has a charge of -2, and it is written as SO4 2-.

Now, we need to make the charges equivalent. To do this, we need to "criss-cross" the charges. This means that sodium will need to additional atoms to make the charges equal, and sulfate will need one.

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Answer:

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