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0.652xx10^(-2)=?...

`0.652xx10^(-2)=?`

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The correct Answer is:
0.00652
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Calculate the equilibrium constant (K) for the formation of NH^ in the following reaction: N_2(g) + 3H_2(g) At equilibrium, the concentration of NH_3 , H_2 and N_2 are 1.2 xx 10^(-2) ,3.0 xx 10^(-2) and 1.5 xx 10^(-2) M respectively.

Following data are obtained for the reaction: N_(2)O_(5) rarr 2NO_(2) + .^(1//2) O_(2) |{:(t//s,0,300,600),([N_(2)O_(5)]//mol L^(-1),1.6 xx 10^(-2),0.8 xx 10^(-2),0.4 xx 10^(-2)):}| (a) Show that it follows first order reaction (b) Calculate the half-life (Given log 2 = 0.3010 log 4 = 0.6021]