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ankoles [38]
3 years ago
10

Consider this question: What is the molarity of HCL if 35.23 mL of a solution of HCL contains 0.3366 g of HCL?

Chemistry
1 answer:
Crazy boy [7]3 years ago
6 0

<u>Answer:</u> The molarity of HCl solution is 0.262 M

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Given mass of HCl = 0.3366 g

Molar mass of HCl = 36.5 g/mol

Volume of the solution = 35.23 mL

Putting values in above equation, we get:

\text{Molarity of HCl}=\frac{0.3366g\times 1000}{36.5g/mol\times 35.23mL}\\\\\text{Molarity of HCl}=0.262M

Hence, the molarity of HCl solution is 0.262 M.

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For the following reaction, identify whether the compound in bold is behaving as an acid or a base.
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Explanation: it dissolves in water to produce hydronium ion as the only positive ion

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Which type of substance is an electron-pair acceptor?
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Answer:

Explanation:

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27. Sample of neon gas has a pressure of 4.1 atm and argon gas with a pressure of 3.2 atm are added to a
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Answer:

Option C. 13.5 atm

Explanation:

From the question given above, the following data were obtained:

Pressure of Neon (Pₙₑ) = 4.1 atm

Pressure of Argon (Pₐᵣ) = 3.2 atm

Pressure of nitrogen (Pₙ₂) = 6.2 atm

Total pressure (Pₜ) =?

The total pressure in the container can be obtained by adding the pressure of the individual gases. This is illustrated below:

Pₜ = Pₙₑ + Pₐᵣ + Pₙ₂

Pₜ = 4.1 + 3.2 + 6.2

Pₜ = 13.5 atm

Therefore, the total pressure in the container is 13.5 atm

8 0
3 years ago
The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori
Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0
3 years ago
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