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3 years ago

Dinitrogen pentoxide gas is formed by the reaction of nitrogen trioxide gas and nitrogen dioxide gas balanced equation

Chemistry

1 answer:

leonid [27]3 years ago

7 0

Answer:

Balanced chemical equation:

NO₂ + NO₃ → N₂O₅

Explanation:

Chemical equation:

NO₂ + NO₃ → N₂O₅

Balanced chemical equation:

NO₂ + NO₃ → N₂O₅

Nitrogen trioxide gas combine with nitrogen dioxide gas and form nitrogen pentoxide.

This is the simple synthesis reaction in which two substance combine to form a new substance.

Synthesis reaction:

It is the reaction in which two or more simple substance react to give one or more complex product.

General chemical equation:

A + B → AB

A and B are reactants that combine to form AB product.

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The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C

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Explanation:

The given data is as follows.

$P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

$T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

$T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

$P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

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According to Clasius-Clapeyron equation,

$log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

$log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

$log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

$\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

= $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

= 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

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