Dinitrogen tetraoxide and nitrogen dioxide are two gases that exist in equilibrium at a range of temperatures. no2 is a reddish brown gas while n2o4 is at high temperature the red color is strong. at low temperature the gas has less color. we represent the equilibrium as: 2 no2(g) doublearrow. gifn2o4(g) we can conclude that: 1. this reaction is: a. exothermicb. endothermicc. neutrald. more information is needed to answer this question.2. when the temperature is decreased the equilibrium constant, k: a. increasesb. decreasesc. remains the samed. more information is needed to answer this question.3. when the temperature is decreased the equilibrium concentration of no2: a. increasesb. decreasesc. remains the samed. more information is needed to answer this question.

1-C

2-A

3-B

Explanation:

K=[N204]/[NO2]

In the first question the reaction is at equilibrium. When the temperature is decreased, more N2O4 if formed and K increases. When the temperature is increased, more NO2 is formed and K decreases.

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