ration 33 A1273 K and I am, I L of N04) decomposes to NO E) given N0. NO, librium, original volume is 25% less than the existing volume. Percentage decomposition of NO. (b) 50% (e) 33.33% (d) 56.16 Solution

At 273 K and I atm, I0 L of N_(2)O_(4(g)) decomposes to NO_(2(g)) as given, N_(2)O_(4(g)) 2NO_(2(g)) , At equilibrium . original volume is 25% lessthan the exisiting volume percentage decomposition of N_(2)O_(4(g)) is thus,

A1 273 K and I atm, I Lof N_(2)O_(4(g)) decomposes to NO_(2(g)) as given, N_(2)O_(4(g)) 2NO_(2(g)) , At equilibrium . original volume is 25% lessthan the exisiting volume percentage decomposition of N_(2)O_(4(g)) is thus,

At 273 K one atm, 'a' litre of N_(2)O_(4) decomposes to NO_(2) as : N_(2)O_(4)(g) hArr 2NO_(2)(g) . To what extent has the decomposition proceeded when he original volume is 25% less then that of existing volume ? [Report your answer up to decimal places.]

If x is 25% less than y, then what percent is y more than x?16(1)/(2)% b.29% c.33(1)/(3)% d. none of these

If the radius of a sphere is increased by 10%, then the volume will be increased by (a) 33.1% (b) 30% (c) 50% (d) 10%

At 273K and 1 atm a L of N_(2)O_(4) decomposes to NO_(2) according to equation N_(2)O_(4)(g) hArr 2NO_(2)(g) . To what extent ha the decomposition proceeded when the original volume is 25% less than that of exisiting volume?

At 273 K and 1atm , 10 litre of N_(2)O_(4) decompose to NO_(4) decompoes to NO_(2) according to equation N_(2)O_(4)(g)hArr2NO_(@)(G) What is degree of dissociation (alpha) when the original volume is 25% less then that os existing volume?

At 273 K and 1atm , 10 litre of N_(2)O_(4) decompose to NO_(4) decompoes to NO_(2) according to equation N_(2)O_(4)(g)hArr2NO_(@)(G) What is degree of dissociation (alpha) when the original volume is 25% less then that os existing volume?

A is 20% more than B, B is 25% more than C, C is 50% less than D and D is 10% more than E. Which of the following is true?