MCQ
On heating $NaNO_3$ gives
  • $O_2$
  • B
    $NO_2$
  • C
    $O_2 + NO_2$
  • D
    $Na_2O$

Answer

Correct option: A.
$O_2$
a
$NaN{O_3}\xrightarrow{\Delta }NaN{O_2} + \frac{1}{2}{O_2}$

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Similar questions

The reaction of $K _3\left[ Fe ( CN )_6\right]$ with freshly prepared $FeSO _4$ solution produces a dark blue precipitate called Turnbull's blue. Reaction of $K _4\left[ Fe ( CN )_6\right]$ with the $FeSO _4$ solution in complete absence of air produces a white precipitate $X$, which turns blue in air. Mixing the $FeSO _4$ solution with $NaNO _3$, followed by a slow addition of concentrated $H _2 SO _4$ through the side of the test tube produces a brown ring.

Precipitate $X$ is

$(A)$ $Fe _4\left[ Fe ( CN )_6\right]_3$

$(B)$ $Fe \left[ Fe ( CN )_6\right]$

$(C)$ $K _2 Fe \left[ Fe ( CN )_6\right]$

$(D)$ $KFe \left[ Fe ( CN )_6\right]$

Among the following, the brown ring is due to the formation of

$(A)$ $\left[ Fe ( NO )_2\left( SO _4\right)_2\right]^{2-}$

$(B)$ $\left[ Fe ( NO )_2\left( H _2 O \right)_4\right]^{3+}$

$(C)$ $\left[ Fe ( NO )_4\left( SO _4\right)_2\right]$

$(D)$ $\left[ Fe ( NO )\left( H _2 O \right)_5\right]^{2+}$

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