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Mila [183]
1 year ago
15

calculate the quantity of 0.001 m aq naoh needed to neutralize the hcl produced by complete solvolysis

Chemistry
1 answer:
abruzzese [7]1 year ago
5 0

The quantity of 0.001 m aq naoh needed to neutralize the hcl produced by complete solvolysis is 200 ml.

Solvolysis is a type of nucleophilic substitution or elimination wherein the nucleophile is a solvent molecule. function of SN1 reactions, solvolysis of a chiral reactant provides the racemate.

Calculation :-

using the titration equation,

M₁V₁ = M₂V₂

substituting values

V₁ = M₂ V₂/M₁

  = 0.2 × 10 ml /0.01

= 2/0.01 ml

= 200 ml.

A reaction in which the solvent is a reactant, and turns into part of the response product. Hydrolysis of tert-butyl chloride; solvent = water. Fischer esterification reaction; solvent = methanol. related phrases: Alcoholysis, aminolysis.

Learn more about solvolysis here:-brainly.com/question/29555642

#SPJ4

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<u>Disclaimer:- your question is incomplete, please see below for the complete question.</u>

Calculate the quantity of 0.01M aq NaOH needed to neutralize the HCl produced by complete solvolysis of the t-BuCl in 10ml of 0.2M t-BuCl in acetone.

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7 0
3 years ago
A compound has a molar mass of 100 g/mol and the percent composition (by mass) of 65.45% C, 5.45% H, and 29.09% 0. Determine the
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Answer:

B) C3H3O and C6H6O2

Explanation:

Given data:

Molar mass of compound = 100 g/mol

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Percentage of carbon = 65.45%

Percentage of oxygen = 29.09%

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of H = 5.45 / 1.01 = 5.4

Number of gram atoms of O = 29.09/ 16 = 1.8

Number of gram atoms of C = 65.45 / 12 = 5.5

Atomic ratio:

            C                      :      H            :         O

           5.5/1.8              :     5.4/1.8     :        1.8/1.8

            3                      :        3          :        1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = 12×3 + 1.01 ×3 + 16×1 = 55.03  

n = 100 / 5503

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 (C₃H₃O)

Molecular formula = C₆H₆O₂

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