What would be the change in pressure in a sealed 10.0 l vessel due to the formation of n2 gas when the ammonium nitrite in 1.20 l of 0.500 m nh4no2 decomposes at 25.0°c?

First, let's find the number of moles of ammonium nitrite.

0.5 mol/L * 1.20 L = 0.6 mol

The decomposition reaction is

NH₄NO₂ → N₂ + 2 H₂O

Let's calculate the amount of N₂ gas produced.

0.6 mol * 1 mol N₂/1 mol NH₄NO₂ = 0.6 mol N₂

Then, let's compute P through the ideal gas equation:
P = nRT/V
P = (0.6)(0.0821 L-atm/mol-K)(25+273 K)/10 L
P = 1.47 atm

Answer) by studying proton collisions hope this : )