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4 years ago

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Reduction is a reaction which results in a 1 in electrons and a 2____ in positive charge of the atom or ion

1 answer:

mestny [16]4 years ago

7 0

Hello! Your answer would be a gain of electrons and a decrease in positive charge.

Contrary to how it sounds, reduction is actually a gain of electrons. It is part of a set of reactions known as a redox reaction, reduction being a gain of electrons and oxidation being a loss.

With the gain of electrons, the element would become more negative as electrons bring with them a negative charge. Therefore, this would decrease the positive charge.

Hope this helped!

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A 0.327-g sample of azulene (C10H8) is burned in a bomb calorimeter and the temperature increases from 25.20 °C to 27.60 °C. The

liq [111]

Explanation:

The given data is as follows.

Molecular weight of azulene = 128 g/mol

Hence, calculate the number of moles as follows.

No. of moles = $\frac{mass}{\text{molecular weight}}$

= $\frac{0.392 g}{128 g/mol}$

= 0.0030625 mol of azulene

Also, $-Q_{rxn} = Q_{solution} + Q_{cal}$

$Q_{rxn} = n \times dE$

$Q_{solution} = m \times C \times (T_{f} - T_{i})$

$Q_{cal} = C_{cal} \times (T_{f} - T_{i})$

Now, putting the given values as follows.

$Q_{solution} = 1.17 \times 10^{3} g \times 10^{3} \times 4.184 J/g^{o}C \times (27.60 - 25.20)^{o}C$

= 11748.67 J

So, $Q_{cal} = 786 J/^{o}C \times (27.60 - 25.20)^{o}C$

= 1886.4 J

Therefore, heat of reaction will be calculated as follows.

$-Q_{rxn}$ = (11748.67 + 1886.4) J

= 13635.07 J

As, $Q_{rxn} = n \times dE$

13635.07 J = $-n \times dE$

dE = $\frac{13635.07 J}{0.0030625 mol}$

= 4452267.75 J/mol

or, = 4452.26 kJ/mol (as 1 kJ = 1000 J)

Thus, we can conclude that $\Delta E$ for the given combustion reaction per mole of azulene burned is 4452.26 kJ/mol.

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