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2 years ago

Write empirical formula

Chemistry

1 answer:

AURORKA [14]2 years ago

6 0

Answer:

$Fe(MnO_{4})_{2}\\\\ Fe(IO_{3})_{2}\\\\NH_{4}MnO_{4}\\\\NH_{4}IO_{3}$

Explanation:

$Fe^{2+}(MnO_{4}^{-})_{2}--->Fe(MnO_{4})_{2}\\\\Fe^{2+}(IO_{3}^{-})_{2}---> Fe(IO_{3})_{2}\\\\NH_{4}^{+}MnO_{4}^{-}--->NH_{4}MnO_{4}\\\\NH_{4}^{+}IO_{3}^{-}--->NH_{4}IO_{3}$

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Answer:

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Explanation:

Given that:-

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Given the value of pressure = 43,836 Pa

So,

$6.9\times 10^5\ Pa$ = $\frac{6.9\times 10^5}{101325}$ atm

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Volume = $2.3\times 10^{-3}\ m^3$ = 2.3 L ( 1 m³ = 1000 L)

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Using ideal gas equation as:

PV=nRT

where,

P is the pressure

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R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

6.80977 atm × 2.3 L = 2 mol × 0.0821 L.atm/K.mol × T

⇒T = 95.39 K

The expression for the kinetic energy is:-

$K.E.=\frac{3}{2}\times K\times T$

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

T is the temperature

So, $K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 95.39\ J$

$K.E.=1.97\times 10^{-21}\ J$

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