Answer:
The yield of the product in gram is 
Explanation:
Given that:
the molecular mass weight of the product = 96.2 g/mol
the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g
given that the millimoles of the reagent = 2,7 millimoles = 
We know that:
Number of moles = mass/molar mass
Then:


mass = 0.697
Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100
i.e
Theoretical yield = 
where;
and 
Theoretical yield = 
Given that the theoretical yield = 100%
Then:



where,
= derived weight of the product
the molecular mass of the derived product
the molecular mass of the reagent
= weight in a gram of the reagent



Mass of CaCl₂ = 0.732 g
<h3>Further explanation</h3>
The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)
pH = 2.88 ==>
![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
=

= 0.001

The change in Concentration Δ
![[CH_{3}COOH]](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D)
= 0.001

CH3COOH H+ CH3COOH
Initial

0 0
Change

-0.001 +0.001 +0.001
Equilibrium

- 0.001 0.001 0.001
Since the

value is so small, the assumption
![[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D_%7Binitial%7D%20%3D%20%5BCH_%7B3%7DCOOH%5D_%7Bequilibrium%7D)
can be made.
![k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}](https://tex.z-dn.net/?f=%20k_%7Ba%7D%20%3D%20%5Btex%5D%3D%201.8%2A10%5E%7B-5%7D%20%20%3D%20%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%20%3D%20%20%5Cfrac%7B0.001%5E%7B2%7D%7D%7Bx%7D%20)
Solve for x to get the required concentration.
note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.
2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind
Hope this helps!
The molarity of the diluted solution is 0.33 M
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 0. 5 M
Volume of stock solution (V₁) = 100 mL
Volume of diluted solution (V₂) = 100 + 50 = 150 mL
<h3>Molarity of diluted solution (M₂) =? </h3>
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
<h3>M₁V₁ = M₂V₂</h3>
0.5 × 100 = M₂ × 150
50 = M₂ × 150
Divide both side by 150
M₂ = 50 / 150
<h3>M₂ = 0.33 M</h3>
Therefore, the molarity of the diluted solution is 0.33 M
Learn more: brainly.com/question/24625656