Chemistry, 19.05.2020 23:48, ashlee7877
Una de las maneras de eliminar el NO en las emisiones de humo es hacerle reaccionar con amoniaco
4NH3(g) +6NO(g) = SN2+6H2O
Suponiendo que el rendimiento de la reacción sea del 100%:
A) ¿Cuántos litros de N2 medidos en condiciones normales sr obtendran apartir de 17g de NH3?
B) ¿Cuántos gramos de nitrógeno y de agua se obtendrán a partir de 180g de NO y 180g de NH3?
Answers: 3
Chemistry, 22.06.2019 04:00, mgnbrnne
Two nitro no2 groups are chemically bonded to a patch of surface. they can't move to another location on the surface, but they can rotate (see sketch at right). it turns out that the amount of rotational kinetic energy each no2 group can have is required to be a multiple of ε, where =ε×1.010−24 j. in other words, each no2 group could have ε of rotational kinetic energy, or 2ε, or 3ε, and so forth — but it cannot have just any old amount of rotational kinetic energy. suppose the total rotational kinetic energy in this system is initially known to be 32ε. then, some heat is removed from the system, and the total rotational kinetic energy falls to 18ε. calculate the change in entropy. round your answer to 3 significant digits, and be sure it has the correct unit symbol.
Answers: 2
Una de las maneras de eliminar el NO en las emisiones de humo es hacerle reaccionar con amoniaco
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