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matrenka [14]
3 years ago
13

()

Chemistry
1 answer:
const2013 [10]3 years ago
6 0

Answer:

\large \boxed{\text{-1276 kJ/mol}}

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

                             CH₃CH₂OH        +  3O₂ ⟶ 2CO₂ + 3H₂O

Bonds:         5C-H 1C-C 1C-O 1O-H    3O=O     4C=O   6O-H

D/kJ·mol⁻¹:    413    347  358  467       495        799      467

\Delta H = \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\\sum{D_{\text{reactants}}} = 5 \times 413 + 1 \times 347 + 1 \times 358 + 1 \times 467 + 3 \times 495 = 3237 + 1485\\=\text{4722 kJ}\\\sum{D_{\text{products}}} = 4 \times 799 + 6 \times 467 =3196 + 2802 = \text{5998 kJ}\\\Delta H = 4722 - 5998= \textbf{-1276 kJ} \\ \text{The overall energy change is $\large \boxed{\textbf{-1276 kJ/mol}}$}.

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Use the Henderson-Hasselbalch equation, eq. (3), to calculate the pH expected for a buffer solution prepared from this acid and
notka56 [123]

Answer:

pH=4.56

Explanation:

Hello there!

In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol

And the concentrations are:

[acid]=0.000855mol/0.025L=0.0342M

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Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

pH=-log(2.5x10^{-5})+log(\frac{0.0312M}{0.0342M} )\\\\pH=4.60-0.04\\\\pH=4.56

Best regards!

6 0
2 years ago
A. 1.4 m<br> B.1.9m<br> C. 2.4m<br> D. 3.6m
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Explain the difference between heat and temperature. Does 1 L of water at 658°F have more, less, or the same quantity of energy
pogonyaev

The answer is- The energy of 1 L water at temperature 347.78 °C have more energy as 1 L of water at temperature 65°C.

Heat is a type of energy that causes a person's body to feel hot or cold.

While the temperature of an object is a parameter that indicates how hot or cold the object is.

How is the temperature in degree Fahrenheit converted to degree celsius?

  • To convert the temperature in Fahrenheit to Celsius, subtract 32 and multiply by 5/9.

°C =( ^0F -32) * \frac{5}{9}

  • Now, heat is a form of energy that flows from hotter object to colder object and temperature indicates whether the object is hot or cold by measuring its average kinetic energy.
  • Now, the given temperature of 1 L water is 658 °F. This temperature in degree celsius is calculated as-

°C = (658 F-32) *\frac{5}{9} = 347.78 \° C

  • Now, higher the temperature, higher is the energy of water. Thus, the energy of 1 L water at 347.78 °C have more energy as 1 L of water at 65°C.

To learn more about heat and temperature, visit:

brainly.com/question/20038450

#SPJ4

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