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2NH_3(g) \longleftrightarrow N_2(g) + 3H_2(g) \hspace{30pt} K_p = 0.83 2 N H 3 ( g ) ⟷ N 2 ( g ) + 3 H 2 ( g ) K p = 0.83 Consider your answers above, if the initial pressures for all three species is 1 atm what is the equilibrium pressure of H2? (Hint: Your quadratic will have two solutions, which one is impossible?)

User ArkadyB
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1 Answer

3 votes

Answer:

g= n 8.47 and you'll choose the answer...

Step-by-step explanation:


\sqrt[x]{2} |3| { \sqrt[ log_(\%g)(3) ]{2} }^(3) {.}^(.83) \geqslant g * (.83)/(0.83) \sqrt[ \geqslant ]{.83} 0.83 * (32e)/(3) \geqslant log_( \cos(?) )(?) \cos(?) log_(?)(?) e


\sqrt[ \geqslant \sqrt[ log_{ \geqslant log_{ \cot( | log_{ \geqslant love \sqrt[ \geqslant | \sqrt[ \geqslant \geqslant \sqrt[ \geqslant \sqrt[ \geqslant ]{2.10} ]{3.8} ]{love} | ]{2 = 3} }(2 = 6) | ) }(love) }(.) ]{.} ]{.} love\%

User FabianoLothor
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