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3 years ago

Which submission doesn’t flow Omaplata Triangle Armbar Leg lock

2 answers:

4 0

((Leg locks)) don't really flow well with the other submissions because leg locks have to do with hyperextending the foot while the other submissions focus on upper body submissions. Omaplata armbar and triangle all transition very well with one another.

Lana71 [14]3 years ago

3 0

Answer: Leg locks

Leg locks don't really flow well with the other submissions because leg locks have to do with hyperextending the foot while the other submissions focus on upper body submissions. Omaplata armbar and triangle all transition very well with one another. It would be difficult to transition to the feet if you are focusing on the upper body especially if you're on the bottom position of full guard.

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For the equilibrium: 2 NO (g) ⇌ N 2(g)+ O 2(g)at 300 K, the equilibrium constant, Kc, is 0.185. If 1.45 moles each of N

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Answer: The concentrations of $N_2,O_2\text{ and }NO$ at equilibrium are 0.112 M, 0.112 M and 0.260 M

Explanation:

Moles of $N_2$ = 1.45 mole

Moles of $O_2$ = 1.45 mole

Volume of solution = 6.00 L

Initial concentration of $N_2$ = $\frac{moles}{Volume}=\frac{1.45mol}{6.00L}=0.242M$

Initial concentration of $O_2$ = $\frac{moles}{Volume}=\frac{1.45mol}{6.00L}=0.242M$

The given balanced equilibrium reaction is,

$2NO(g)\rightleftharpoons N_2(g)+O_2(g)$

Initial conc. 0 M 0.242 M 0.242 M

At eqm. conc. (2x) M (0.242-x) M (0.242-x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[N_2]\times [O_2]}{[NO]^2}$

Now put all the given values in this expression, we get :

$0.185=\frac{(0.242-x)^2}{(2x)^2}$

By solving the term 'x', we get :

x = 0.130

Thus, the concentrations of at equilibrium are :

Concentration of = (0.242-x) M = (0.242-0.130) M = 0.112 M

Concentration of = 2x M = = 0.260 M

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