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N^(3-),O^(2-),F^-,Na^+, Mg^(2+),Al^(3+)Arrange them in the order of increasing ionic radii.

N_(2(s))+3H_(2(g)) hArr 2NH_3state the condition for industrial production of ammonia.

N_(2)O_(5) is more acidic than N_(2)O_(3).

N_(2)O_(4)(g)hArr2NO_(2)(g),K_(C)=4. This reversible reaction is studied graphically as shown in figure. Select the correct statements.

N_2O_(4(g)) hArr 2NO_(2(g)) of K_p = 0.15 atm (298 K) , so calculate K in torr and mol/L. (1 atm = 760 torr, R= 0.0821 L atm mol^(-1) K^(-1))

N_(2)O_(4)(g) hArr 2NO_(2)(g) The equilibrium reaction shown is endothermic as written. Which change will increase the amount of NO_(2) at equilibrium?

N_(2)O_(4) is dissociated to 33% and 50% at total pressure P_(1) and P_(2)atm respectively. The ratio of (P_(1))/(P_(2)) is :

N_(2)O_(4) harr 2NO_(2), N_(2)O_(4(g)) dissociates until the partial pressures of N_(2)O_(4) and NO_(2) become equal, initial pressure of N_(2)O_(4) is 9 atmosphere. What is K_(P) ?

N_(2)O_(4) hArr 2NO_(2).At 300K N_(2)O_(4) at 6 atmosphere is heated to 600K where equilibrium is established witha total pressure of 16 atmospheres. What isK_(p) at 600K ?

N_2O_4 at an initial pressure of 2 atm and 300K dissociates to an extent of 20% at the same temperature by the time equilibrium is established. K_p for the reaction 2NO_2 harr N_2O_4 is

N_(2) O_(2) hArr 2NO, K_(1) , (1)/(2) N_(2) + (1)/(2) O_(2) hArr NO. K_(2) , 2NO hArr N_(2) + O_(2). K_(3) , NO hArr (1)/(2) N_(2) + (1)/(2) O_(2). K_(4)

N_(2)O is polar even though it is linear-why?

N_(2)H_(4) loses 10 mol e^(-) and form new compound Y number of N does not change so what is the oxidation number of N in Y atom ?

N_(2(g)) + O_(2(g)) hArr 2NO_((g)) , Delta_rH^(ө)=180 kJ .To increase the temperature of this reaction, what is the effect on products and value of K?

N_(2(g)) + 3H_(2(g)) underset"Fe"overset"200 Atm"hArr 2NH_(3(g)) , At which temperature this reaction will be possible ?

N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)), The K_p of this reaction is 35 at 500 K temp. Calculate K_p of following reaction at this temp. (i)4NH_(3(g)) hArr 2N_(2(g)) + 6H_(2(g)) (ii)1/2N_(2(g)) +3/2H_(2(g)) hArr NH_(3(g))

N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)). If some HCl gas is passed into the reaction mixture at the equilibrium of this R reaction,

N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)) Ifargon introduce in this reaction than what happen ?

N_(2(g)) + 3H_(2(g)) harr 2NH_(3(g)) is a gaseous phase equilibrium reaction taking place at 400K in a 5 L flask. For this

N_(2(g))+3H_(2(g)) hArr 2NH_(3(g)) , For this equilibrium reaction at given temperature find relation between K_p and K_c

N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)), DeltaH=-92.38 "kJ mol"^(-1) the reaction is exothermic. So at lowertemperature more product is obtained still why the reaction is carried out at high temperature ?

N_(2)3H_(2)hArr2NH_(3) K=4xx10^(6) "at" 298 K=41 "at" 400 K Which statements is correct?

N_(2)has higher order than NO. Explain .

N_2 gas is present in one litre flask at a pressure of 7.6 xx 10^(-10)mm of Hg. The number of N_2 gas molecules in the flask at 0^@C are

N_(2)gas is present in one litre flask at a pressure of 7.6 xx 10^(-10)mm of Hg. The number of N_(2)gas molecules in the flask at 0°C are

N_(2) gas id added to the reaction equilibrium PCl_(5) (g) hArr PCl_(3) (g) + Cl_(2) (g) at constant temperature . If pressure is kept constant, equilibrium constant will …… and equilibrium will shift in the ………. direction.

N_(2(g)) + 2O_(2(g)) rarr 2NO_(2) + xkJ, 2NO_((g)) + O_(2(g)) rarr 2NO_(2(g)) + ykJ. The enthalpy of formation of NO is

N_2and O_2 are converted into monoanions N_(2)^(-)and O_(2)^(-) respectively, which of the following statements is wrong?

N_2+ 3H_2 hArr 2NH_3+ Energy, in this equilibrium reaction what is the following when increase the total pressure ?

N_(0)//2 atoms of X_((g)) are converted into X_((g))^(o+) by energy E_(1), N_(0)//2 atoms of X_((g)) are converted inot X_((g))^(ɵ) by energy E_(2). Hence ionisation potential and electron affinity of X_((g)) per atom are

In the formation of a chemical bond between Na^(+) and Cl^(-),they attain the stable configuration of ……………….

n-Propylmagnesium bromide on hydrolysis gives propane. Is there another Grignard reagent which also gives propane ? If so, give its name , structure and equation for the reaction.

n-propylamine containing no secondary and tertiary amines as impurities is prepared by

n - Propyl chloride and isopropyl chloride mixture is subjected to the Wurtz reaction, which one of the following compounds is not formed

n-Propyl alcohol on treatment with excess hot conc. H_(2)SO_(4) gives compounds X. X on treatment with HBr in the presence of benzoyl peroxide give Y. Y when dissolved in ether and treated with sodium gives

n-Propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent

n-Octane when heated to 773 K under a pressure of 10-20 atm and in presence of a mixture of Cr_2O_3, V_2O_5 and Mo_2O_3 supported over Al_2O_3 as catalyst, gives

n' number of alkenes yields 2,2,3,4,4- pentamethyl-pentane on catalytic hydrogenation and 'm' number report your answer as (n+m)

n-Octane in presence of Cr_2O_3 // Al_2O_3 ultimately changes into

n mole each of H_(2)O(g),H_(2)(g) "and" O_(2)(g) are mixed at a suitable high temperature to attain the equilibrium 2H_(2)O(g)hArr2H_(2)(g)+O_(2)(g). If y and mole of H_(2)O(g) are the dissociated and the total pressure maintained is P, calculate the K_(P).

'n' moles of an ideal gas at temperature ‘T’ occupy ‘V’ litres of volume, exerting a pressure of ‘p’ atmospheres. What is its concentration in mole lit^(-1) (R = gas constant)

n, l and m values of the 2p_(z) orbital are

n hexane passsed over chromic oxide supported on alumina at 873 k will give ……………..

n-Hexane on heating to 773 K at 10-20 atmospheric pressure in the presence of oxides of vanadium supported over alumina, yields

n-hexadecane has cetane number

N has fractional oxidation number in

n g of substance X reacts with mg of substance Y to form p g of substance R and q g of substance S . This reaction can be represented as , X + Y = R + S . The relation which can be establish in the amounts of the reactants and the products will be

n-factors for Cu_(2)S and Cus when they react with KMnO_4 in acidic medium are (neglecting the further oxidation of released SO_2

N-factor for the following reaction is FeS_(2)toFe_(2)O_(3)+SO_(2)

n-factors for Cu_(s)S and CuS when they react with KMnO_(4) in acidic medium are