STP | এ N_(2) এর 1.4 গ্রাম দ্বারা দখলকৃত আয়তন গণনা করুন 10 | মোল কনসেপ্ট এবং স্টোচিওমেট্রি...

The function u_(n) takes on the following values : u_(1) = ( 1)/( 4) , u_(2) = ( 1)/( 4) + ( 1)/( 10)"…....." u_(n) =(1)/( 3+1) + ( 1)/( 3^(2) + 1)+"…........." (1)/( 3^(n) + 1)"…........." Prove that underset( n rarr oo) ( "lim") u_(a) lt (1)/(2)

Henry 's law constant for the solubility of N_(2) gas in water at 298 K is 1.0 xx 10^(5) atm . The mole fraction of N_(2) in air is 0.6 . The number of moles of N_(2) from air dissolved in 10 moles of water at 298 K and 5 atm pressure is a. 3.0 xx 10^(-4) b. 4.0 xx 10^(-5) c. 5.0 xx 10^(-4) d. 6.0 xx 10^(-6)

N_(2)O_(4) is 60% dissociated into NO_(2) at 340K and 1 atm pressure. Find the volume of 10 g N_(2)O_(4) occupy under these condition.

The rate constant of the reaction, 2N_(2)O_(5) to 4NO_(2) + O_(2) at 300 K is 3 xx 10^(-5)s^(-1) . If the rate of the reaction at the same temperaturre is 2.4 xx 10^(-5) mol dm^(-3)s^(-1) , then the molar concentration of N_(2)O_(5) is

The rate constant for the reaction, 2N_(2)O_(5) rarr 4NO_(2) + O_(2) is 3.0 xx 10^(-5) s^(-1) . If the rate is 2.40 xx 10^(-5) mol L^(-1) s^(-1) , then the initial concentration of N_(2)O_(5) (in mol L^(-1) ) is

The rate constant for the reaction 2N_(2)O_(5) rarr 4NO_(2)+O_(2) is 3.0 xx 10^(-5) s^(-1) . If the rate is 2.40 xx 10^(-5) mol L^(-1) s^(-1) , then the concentration of N_(2)O_(5) (in mol L^(-1) ) is