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dimulka [17.4K]
1 year ago
13

Is each of the following very soluble in water? Explain. (b) Sodium hydroxide

Chemistry
1 answer:
vova2212 [387]1 year ago
7 0

The answer is highly soluble

<h3>How does sodium hydroxide dissolve in water?</h3>

When hydroxide (NaOH) dissolved in water, it divides into positively and negatively charged sodium ions (cations) and hydroxide ions (anions). These ions move freely and independently in water, while cations prefer to be surrounded more closely by anions and vice versa.

Although sodium hydroxide is more soluble in hot than in cold water, the process of dissolving sodium hydroxide in water is exothermic.

The unknown compound's insolubility in water might be attributed to its big hydrocarbon group. Because it is dissolved in 5% Naoh, its substituent may be acidic because the interaction with NaOH produces an anion that is soluble in aqueous solution.

learn more about sodium hydroxide refer

brainly.com/question/13041783

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What is Y? 222/86 Rn A/z + 4/2He
kondaur [170]

^{222}_{\phantom{2}86}\text{Rn} \to ^{218}_{\phantom{2}84}\text{Po}+ ^{4}_{2}\text{He}

Y is Po-218.

  • A = 218
  • Z = 84.
<h3>Explanation </h3>

^{222}_{\phantom{2}86}\text{Rn} \to ^{A}_{Z}\text{Y}+ ^{4}_{2}\text{He}

Here's the symbol of a particle in a nuclear reaction. ^{A}_{Z}\text{Y}.

A stands for mass number. Z stands for atomic number. Both numbers shall conserve in a nuclear reaction.

  • The mass number on the left hand side is 222.
  • The two mass numbers on the right hand side add up to A + 4.
  • 222 = A + 4.
  • A = 218

So is the case for the atomic number. Try figure out the atomic number of Y using the same approach.

  • The atomic number on the left hand side is 86.
  • The two atomic number on the right hand side add up to __ + __.
  • 86 = __ + __.
  • Z = 84.

What element is Y? The atomic number of Y is 84. Refer to a periodic table. Element 84 corresponds to Po (polonium). Y is Polonium-218. The symbol of Y should be written as ^{218}_{\phantom{2}84}\text{Po}. Hence the equation:

^{222}_{\phantom{2}86}\text{Rn} \to ^{218}_{\phantom{2}84}\text{Po}+ ^{4}_{2}\text{He}.

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