If in acidic medium \(K_{2}Cr_{2}O_{7}\) has relation N=x \(\times\) M then the value of "x" is

5 mol of Fe_(2)(C_(2)O_(4)) is oxidised by x mol of K_(2)Cr_(2)O_(7) in acidic medium, calculate the value of x ?

Acidic K_(2) Cr_(2)O_(7) oxidises KI to

In acidic medium, H_(2)O_(2) changes Cr_(2)O_(7)^(2-) ion to CrO_(5) . The oxidation states of Cr in Cr_(2)O_(7)^(2-) and CrO_(5) are respectively :

The number of moles of \(K_{2}Cr_{2}O_{7}\) required to oxidise one mole of \(Sn^{+2}\) to "1" mole of\(Sn^{+4}\) in acidic medium is "1/x" .The value of x is (A) 1/3 (B) 3 (C) 6 (D) 2

Assertion : In acidic medium, K_2Cr_2O_7 exists as Cr_2O_7^(2-) (orange) while in basic medium it is converted to CrO_4^(2-) (yellow). Reason : K_2Cr_2O_7 is hygroscopic in nature and changes colour on reaction with water.

\(K_{2}Cr_{2}O_{7}\) in alkaline medium with \(30\%\) \(H_{2}O_{2}\) forms red-brown compound (X). Oxidation state of Cr in 'X' is '+a' and number of peroxy oxygens in 'X' is 'b' .Value of (b-a) is