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ludmilkaskok [199]
2 years ago
6

Use the bond energies to answer the question.

Chemistry
1 answer:
Schach [20]2 years ago
4 0

Answer:

D

Explanation:

Hello!

This is a topic I just learned this year so I might not help the best as of explaining. But I will try my best.

To solve this problem, we must first figure the balanced equation.

H₂+O₂=H₂O₂

Now, we have to find the bond energy of the reactants and products. This is how.

Let's find the reactant first.

For the compound H₂, the Lewis dot structure says that it only consists of one singular bond. So the bond energy of H₂ would be 432.

For the compound O₂, the Lewis dot structure that is the best out of the resonance structures form a double bond. Let's use the bond energy 495 for this since there are double bonds and we need the corresponding double bond energy to make the bond energy accurate.

Let's add this amount together. 927

Now, we have to find the bond energy of the products.

H₂O₂ sounds complicated but it's fairly easy. When you write the Lewis dot structure for this specific compound, you will get something along the lines of this:

H-O-O-H (disregarding lone pairs)

In this compound, there are two bonds of H-O and one bond that is a O-O. This might take you a long time to see this, but just think of what atom is touching the atom next door. This might help distinguish the bonds there are between the compound.

Let's take the bond energy of H-O (467) and multiply this by two since we have two. Add that number to the bond energy of O-O (146). We should have the numbers 1080.

Since the problem is asking for the difference in total energy between the reactants and products, let's subtract the total bond energy of the products-reactants.

1080-927=153

Therefore, your final answer of the total bond energy is 153.

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Explanation:

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A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
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Answer:

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Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

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Answer:

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