The decomposition of ammonia is characterized by the following decomposition equation:
2NH₃<span> → N</span>₂ <span> + 3H</span>₂
The mole ratio of N₂ : H₂ is 1 : 3
If the number of moles of N₂ = 0.0351 mol
Then the number of moles of H₂ = 0.0351 mol × 3
= 0.1053 mol
The number of moles of hydrogen gas produced when 0.0351 mol of Nitrogen gas is produced after the decomposition of Ammonia is 0.105 mol (OPTION 3).
Three factors that affect magma viscosity are temperature, composition, and presence of dissolved gases.
I think 25m hopes dis helps
A mole of any gas occupied 22.4 L at STP. So, the number of moles of nitrogen gas at STP in 846 L would be 846/22.4 = 37.8 moles of nitrogen gas.
Alternatively, you can go the long route and use the ideal gas law to solve for the number of moles of nitrogen given STP conditions (273 K and 1.00 atm). From PV = nRT, we can get n = PV/RT. Plugging in our values, and using 0.08206 L•atm/K•mol as our gas constant, R, we get n = (1.00)(846)/(0.08206)(273) = 37.8 moles, which confirms our answer.
