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vlabodo [156]
3 years ago
8

11.

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

Answer:

(1) -12 Kcal/mol

Explanation:

Our answer options for this question are:

(1) -12 Kcal/mol

(2) -13 Kcal/mol

(3) -15 Kcal/mol

(4) -16 Kcal/mol

With this in mind, we can start with the chemical reaction (Figure 1). In this reaction, <u>two bonds are broken</u>, a C-H and a Br-Br. Additionally, a C-Br and a H-Br are <u>formed</u>.

If we want to calculate the enthalpy value, we can use the equation:

<u>ΔH=ΔHbonds broken-ΔHbonds formed</u>

If we use the energy values reported, its possible to calculate the energy for each set of bonds:

<u>ΔHbonds broken</u>

<u />

C-H = 94.5 Kcal/mol

Br-Br = 51.5 Kcal/mol

Therefore:

105 Kcal/mol + 53.5 Kcal/mol = 146 Kcal/mol

<u>ΔHbonds formed</u>

C-Br = 70.5 Kcal/mol

H-Br = 87.5 Kcal/mol

Therefore:

70.5 Kcal/mol + 87.5 Kcal/mol = 158 Kcal/mol

<u>ΔH of reaction</u>

<u />

ΔH=ΔHbonds broken-ΔHbonds formed=(146-158) Kcal/mol = -12 Kcal/mol

I hope it helps!

<u />

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Explanation:

<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>

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