The reaction of K₃[Fe(CN)₆] with freshly prepared FeSO₄ solution produces a dark blue precipitate called Turnbull's blue. Reaction of K₄[Fe(CN)₆] with the FeSO₄ solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the FeSO₄ solution with NaNO₃, followed by a slow addition of concentrated H₂SO₄ through the side of the test tube produces a brown ring.

Precipitate X is

The reaction of K₃[Fe(CN)₆] with freshly prepared FeSO₄ solution produces a dark blue precipitate called Turnbull's blue. Reaction of K₄[Fe(CN)₆] with the FeSO₄ solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the FeSO₄ solution with NaNO₃, followed by a slow addition of concentrated H₂SO₄ through the side of the test tube produces a brown ring.

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

One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed by II, as shown below. If the work done by the gas in the two processes are same, the value of $$\ln {{{V_3}} \over {{V_2}}}$$ is _________.


(U : internal energy, S : entropy, p : pressure, V : volume, R : gas constant)

(Given : molar heat capacity at constant volume, C_V,m of the gas is $${5 \over 2}$$R)

Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the velocity (in cm s^$$-$$1) of He atom after the photon absorption is __________.

(Assume : Momentum is conserved when photon is absorbed.

Use : Planck constant = 6.6 $$\times$$ 10^$$-$$34 J s, Avogadro number = 6 $$\times$$ 10²³ mol^$$-$$1, Molar mass of He = 4 g mol^$$-$$1)