Chemistry, 18.11.2019 20:31 sierravick123owr441

Nitric oxide is formed in automobile exhaust when nitrogen and oxygen in air react at high temperatures. n2(g) + o2(g) 2no(g)the equilibrium constant kp for the reaction is 0.0025 at 2127�c. if a container is charged with 8.00 atm of nitrogen and 5.00 atm of oxygen and the mixture is allowed to reach equilibrium, what will be the equilibrium partial pressure of nitrogen? a) 0.16 atm b) 0.31 atm c) 3.1 atm d) 7.7 atm e) 7.8 atm

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Answer from: kezionhoward13

Answer : The correct option is, (E) 7.8 atm

Explanation :

The partial pressure of N_2 = 8.00 atm

The partial pressure of O_2 = 5.00 atm

K_p = 0.0025

The balanced equilibrium reaction is,

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

Initial pressure 8.00 5.00 0

At eqm. (8.00-x) (5.00-x) 2x

The expression of equilibrium constant K_p for the reaction will be:

$K_p=\frac{(p_{NO})^2}{(p_{N_2})(p_{O_2})}$

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

$0.0025=\frac{(2x)^2}{(8.00-x)\times (5.00-x)}$

By solving the terms, we get:

x=0.15atm

The equilibrium partial pressure of N_2 = (8.00 - x) = (8.00 - 0.15) = 7.8 atm

Therefore, the equilibrium partial pressure of N_2 is 7.8 atm.

Answer from: Quest

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. $hbr$ completely dissociates to $h^+$ and $br-$ ions. a is incorrect because $h_2s$ does not fully ionize to $h^+$ and $hs^-$ ions and some remain as the compound $h_2s$ . c and d are incorrect because, $li_2o$ and $libr$ do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise, if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both $li_2o$ and $libr$ have a complete octet so can not accept any electrons which disqualifies them as acids.

Answer from: Quest

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

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