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3 years ago

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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Read 2 more answers

A 13 g sample of P4010 contains how many

Alex777 [14]

Answer:

$\large \boxed{5.5 \times 10^{22}\text{ molecules of P$_{2}$O}_{5}}$

Explanation:

You must calculate the moles of P₄O₁₀, convert to moles of P₂O₅, then convert to molecules of P₂O₅.

1. Moles of P₄O₁₀

$\text{Moles of P$_{4}$O}_{10} = \text{13 g P$_{4}$O}_{10} \times \dfrac{\text{1 mol P$_{4}$O}_{10}}{\text{283.89 g P$_{4}$O}_{10}} = \text{0.0458 mol P$_{4}$O}_{10}$

2. Moles of P₂O₅

P₄O₁₀ ⟶ 2P₂O₅

The molar ratio is 2 mol P₂O₅:1 mol P₄O₁₀

$\text{Moles of P$_{2}$O}_{5} = \text{0.0458 mol P$_{4}$O}_{10} \times \dfrac{\text{2 mol P$_{2}$O}_{5}}{\text{1 mol P$_{4}$O}_{10}} = \text{0.0916 mol P$_{2}$O}_{5}$

3. Molecules of P₂O₅

$\text{No. of molecules} = \text{0.0916 mol P$_{2}$O}_{5} \times \dfrac{6.022 \times 10^{23}\text{ molecules P$_{2}$O}_{5}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{5.5 \times 10^{22}}\textbf{ molecules P$_{2}$O}_{5}\\\text{There are $\large \boxed{\mathbf{5.5 \times 10^{22}}\textbf{ molecules of P$_{2}$O}_{5}}$}$

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