Explanation
Though the individual speeds are changing, the distribution of speeds remains constant at a particular temperature.
$$K.E_{ O_{ 2 } }=\dfrac { \dfrac { 3 }{ 2 } \times \dfrac { N }{ 32 } \times R\times 150 }{ \dfrac { 3 }{ 2 } \times \dfrac { { N }' }{ 32 } \times R\times 300 } $$
$$=\dfrac { x }{ 2x } $$
Given
Colorless gas - $$X$$
Brown gas - $$Y$$
The bromine does not change its color at a given temperature, but nitrogen dioxide and dinitrogen tetraoxide change their color at a given temperature and give the same color as given in the question.
Heat dinitrogen tetroxide
$${N_2}{O_4}$$ $$\xrightarrow{{{{160}^{\rm O}}C}}$$ $$2N{O_2}$$ $$\xrightarrow{{{{600}^{\rm O}}C}}$$ $$2NO$$ $$ + $$ $${O_2}$$
In this reaction,
$${N_2}{O_4}$$ - dinitrogen tetroxide (colorless)
$$2N{O_2}$$ - nitrogen dioxide (brown)
$$2NO$$ - Nitric oxide (colorless)
Therefore, the name of $$Y$$ gas is dinitrogen tetroxide that is brown.
Hence, option (C) is the correct answer.
An LPG cylinder can withstand pressure difference of 15 atm across its boundaries. If at room temp $$({ 27 }^{ o })$$ it is filled with 2 atm pressure. Determine the temperature at which it will explode?
$$18.0\ g$$ of glucose$$\left( {{C_6}{H_{12}}{O_6}} \right)$$ is added to $$178.2\ g$$ of water. The vapour pressure of water for this aqueous solution at $${100^o}\ C$$ is :
Given,
heat dinitrogen tetraoxide
in this reaction,
$${N_2}{O_4}$$ - dinitrogen tetraoxide (colorless)
Therefore, the name of $$X$$gas is nitric oxide that is colorless.
Hence, option(d) is the correct answer.
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