Gaseous indium dihydride is formed from the elements at elevated temperature:

ln(g)+H2(g)?lnH2(g)Kp = 1.48 at 973 K

Partial pressures measured in a reaction vessel are:
PIn = 0.0650atm , PH2 = 0.0330atm , PInH2 = 0.0790atm

1. Calculate Qp
2. Determine the direction of reaction to attain equilibrium
3. Determine the equilibrium partial pressure of In
4. Determine the equilibrium partial pressure of H2
5. Determine the equilibrium partial pressure of InH2

For 1: The value of for above reaction is 36.83

For 2: The value of for above reaction is 36.83

For 3: The equilibrium partial pressure of Indium is 0.126 atm

For 4: The equilibrium partial pressure of hydrogen gas is 0.094 atm

For 5: The equilibrium partial pressure of Indium dihydrogen is 0.018 atm

Explanation:

We are given:

Partial pressure of Indium gas = 0.0650 atm

Partial pressure of hydrogen gas = 0.0330 atm

Partial pressure of Indium dihydride = 0.0790 atm

The given chemical equation follows:

                     

Initial:               0.065     0.033           0.079

At eqllm:       0.065-x   0.033-x       0.079+x

For 1:

The expression of for above reaction follows:

Putting values in above equation, we get:

Hence, the value of for above reaction is 36.83

For 2:

We are given:

of the reaction = 1.48

There are 3 conditions:

When ; the reaction is product favored.When ; the reaction is reactant favored.When ; the reaction is in equilibrium.

As, for the given reaction, the reaction is reactant favored.

Hence, the reaction proceed in the backward direction to attain equilibrium

For 3:

The expression of for above reaction follows:

Putting values in above equation, we get:

Neglecting the value of x = 0.835 because the reaction is going backwards. So, by taking this value, the pressure of the reactants will decrease

So, equilibrium partial pressure of Indium = (0.065 - x) = [0.065 - (-0.061)] = 0.126 atm

For 4:

The equilibrium partial pressure of hydrogen gas = (0.033 - x) = [0.033 - (-0.061)] = 0.094 atm

For 5:

The equilibrium partial pressure of Indium dihydrogen = (0.079 + x) = [0.079 + (-0.061)] = 0.018 atm

Answer: abeexplanation:

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

D. most models represent a real world object in a simpler way