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lesya [120]
3 years ago
12

Suppose you were preparing 1.0 L of a bleaching solution in a volumetric flask, and it calls for 0.21 mol of NaOCl. If all you h

ad available was a jug of bleach that contained 0.82 M NaOCl, what volume of bleach would you need to add to the volumetric flask before you added enough water to reach the 1.0 L line?
Chemistry
2 answers:
Ilya [14]3 years ago
7 0

Answer:

V₁ = 0.26 L

Explanation:

The final solution requires 0.21 moles in  V₂ = 1.0 L, which means a concentration of C₂ = 0.21 mol / 1.0 L = 0.21 M. The concentration of the initial solution is C₁ = 0.82 M. We can find the volume V₁ required using the dilution rule.

C₁ × V₁ = C₂ × V₂

0.82 M × V₁ = 0.21 M × 1.0 L

V₁ = 0.26 L

yulyashka [42]3 years ago
3 0

Answer:

0.256 L  

Explanation:

We should use the following formula:

concentration (1) × volume (1) =  concentration (2) × volume (2)

concentration (1) = 0.82 M NaOCl

volume (1) = ?

concentration (2) = 0.21 M NaOCl

volume (2) = 1 L

volume (1) = [concentration (2) × volume (2)] / concentration (1)

volume (1) = [0.21 / 1] / 0.82 = 0.256 L

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The question is in the picture below
Rus_ich [418]

Answer:

\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

\bf{\text{A} \rightarrow 2\text{B}}                  (Δ\text{H}_1)  -----(1)

Since we have 2B, multiply the whole of II. by 2:

\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}       (2Δ\text{H}_2) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is \text{A} \rightarrow 2\text{C} +2\text{D}.

Reversing III. gives us a negative enthalpy change as such:

\bf{2\text{D} \rightarrow \text{E}}                  (-Δ\text{H}_3) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of \text{A} \rightarrow 2\text{C} +\text{E}, which is also the equation of interest.

Adding all three together:

\text{A} \rightarrow 2\text{C}+\text{E}            (\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 })

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

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