If the reaction is run at STP and the yield is 54%, what is the maximum volume o

Question

If the reaction is run at STP and the yield is 54%, what is the maximum volume of N_2 formed (in liters)? 36.5 liters 19.7 liters 21.5 liters 39.8 liters The density of gold is 19.3 grams/cm^3. Which equation below represents the conversion of this density to units of nanograms/m^3? Density (ng/m^3) = 19.3 Times 10^-9 Times 10^6 Density (ng/m^3) = 19.3 Times 10^-9 Times 10^2 Density (ng/m^3) = 19.3 Times 10^9 Times 10^2 Density (ng/m^3) = 19.3 Times 10^9 Times 10^6 Consider the molecular orbital diagram of O_2 with a bond order of 2. How many electric must be added to the O_2 molecule to reduce the bond order to zero? 4 electrons 3 electrons 2 electrons 1 electron

Explanation / Answer

1) At STP, 1 mole of volume of N2 =22.4 Liters, but yield is =54%, so volume of N2= 0.54*22.4=12.1 L ( but if is 1 mole of N2 only). No correct answer is given.

2. ng= 109gm and cc =10-6 m3

19.3 g/cc= 19.3*109 ng/ 10-6 m3 =19.3*109*106 ng/m3 ( D is correct)

3. When 4 eelctrons are added , the bond order gets reduced to zero. while generating two oxide anions

O2+ 4 e-2 O2- ( a is correct)

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