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IceJOKER [234]
3 years ago
9

Classify each of these soluble solutes as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. Solutes Formula Nitric

acid HNO3 Calcium hydroxide Ca(OH)2 Acetic acid H3CCOOH Methyl amine CH3NH2 Potassium chloride KCl Ethanol C2H5OH Glucose C6H12O6
Chemistry
1 answer:
Art [367]3 years ago
7 0

Answer: Strong electrolytes:(HNO_{3}) Nitric acid, (Ca(OH)_{2}))Calcium Hydroxide and (KCl) potassium chloride

Weak electrolytes: (CH_{3}COOH)Acetic acid and (CH_{3}NH_{2}) Methyl amine

Non-electrolytes:(C_{2}H_{5}OH)Ethanol and (C_{6}H_{12}O_{6})Glucose

Explanation: Electrolytes are those compounds which can conduct electricity when dissolved in any polar solvent.

Strong electrolytes are those compounds which completely ionise when dissolved in polar solvent and hence produce ions in solution . So greater the capacity of an compound to ionize itself greater number of ions would be present in solution and hence greater will be the capacity of the solution to conduct electricity.

Ionic compounds like (HNO_{3}) Nitric acid ,(KCl) Potassium chloride and (Ca(OH)_{2}))Calcium hydroxide are completely ionized  when dissolved in polar solvent so these compounds are strong electrolytes.

Weak electrolytes are those  compounds which undergo partial ionization when dissolve in polar solvents . So they are not able to produce more ions in the solution and hence the conductivity of a solution containing weak electrolytes is low.

(CH_{3}COOH)Acetic acid and CH_{3}NH_{2}Methyl amine are partially ionized when dissolved in polar solvent so these electrolytes are weak electrolytes.

Non-electrolytes are those compounds which are not at all ionized in the polar solvent and they remain as molecules itself even if they are dissolved.

(C_{2}H_{5}OH)Ethanol and  (C_{6}H_{12}O_{6})Glucose do not ionize when dissolved in polar solvent and remain as molecules itself so the solutions of these compounds will not have ions and hence they would be unable to conduct electricity.

so

 

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Consider a solution that contains both C6H5NH2 and C6H5NH3+. Calculate the ratio [C6H5NH2]/[C6H5NH3+] if the solution has the fo
Tcecarenko [31]

The question has incomplete information. The pH values are not discriminated. Let's do, as an example, to pH = 4.69.

Answer:

For pH = 4.69, [C₆H₅NH₂]/[C₆H₅NH₃⁺] = 0.708

Explanation:

C₆H₅NH₂ is a base, and, when in aqueous solution, forms a conjugate acid, which is C₆H₅NH₃⁺. The equilibrium between these two species makes a buffer, a solution that prevents the change in pH by the addition of base or acid.

The pH of a buffer can be calculated by the equation of Handerson-Halsebach:

pH = pKa + log[conjugate base]/[acid]

Where pKa = -logKa, and Ka is the equilibrium constant of the acid. For the base, the equilibrium constant, Kb, is equal to 7x10⁻¹⁰, and

Ka*Kb = Kw

Where Kw is the equilibrium constant for water 1.00x10⁻¹⁴, so Ka for the conjugate acid is:

Ka = 1.00x10⁻¹⁴/7.00x10⁻¹⁰

Ka = 1.43x10⁻⁵

pKa = 4.84

pH = 4.84 + log [C₆H₅NH₂]/[C₆H₅NH₃⁺] (By this step, the pH values given by the question must be substituted)

4.69 = 4.84 + log[C₆H₅NH₂]/[C₆H₅NH₃⁺]

log[C₆H₅NH₂]/[C₆H₅NH₃⁺] = - 0.15

[C₆H₅NH₂]/[C₆H₅NH₃⁺] = 10^{-0.15}

[C₆H₅NH₂]/[C₆H₅NH₃⁺] = 0.708

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The chemical reaction in which number of atoms of each element present in the reactant side is equal to the number of atoms of that element in product side, such reactions are said to be a balanced chemical reaction.

The chemical symbol for sodium is Na.

The chemical symbol for fluorine gas is F_2.

The chemical symbol for sodium fluoride is NaF

The sodium fluoride is prepared from the reaction between sodium metal and fluorine gas can be written as:

Na (s)+F_2(g) \rightarrow NaF(s)

The above reaction is not balanced as the number of fluorine atoms are not same on reactant and product side. So, in order to balance the reaction we will multiply Na with 2 on reactant side and NaF with 2 on product side. Thus, the balanced reaction will be:

2Na (s)+F_2 (g)\rightarrow 2NaF(s)

Thus, the balanced chemical equation is 2Na (s)+F_2 (g)\rightarrow 2NaF(s).

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Answer: 0.0624 atm

Explanation:-

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R= Gas constant = 0.0821 atmL/K mol

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