Using the equations 2 Fe (s) + 3 Cl₂ (g) → 2 FeCl₃ (s) ∆H° = -800.0 kJ/mol
Si(s) + 2 Cl₂ (g) → SiCl₄ (s) ∆H° = -640.1 kJ/mol

Determine the enthalpy (in kJ/mol) for the reaction
3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s)

The enthalpy for the formation of 4 moles of has been 320.3 kJ/mol.

The enthalpy of formation of 1 mole at standard temperature and pressure has been termed the standard enthalpy of formation.

The reaction enthalpy to be calculated has reactants of 3 moles of , and 4 moles of Fe.

The enthalpy for the formation of has been -640.1 kJ/mol.

The enthalpy for the dissociation of has been the reverse of the formation reaction = +640.1 kJ/mol.

The enthalpy for the dissociation of 3 moles of has been = 3 640.1 kJ/mol.

The enthalpy for the dissociation of 3 moles of has been = 1920.3 kJ/mol.

The enthalpy for the formation of 2 moles of = -800 kJ/mol

The enthalpy for the formation of 4 moles of = -1600 kJ/mol.

The final reaction has 3 moles of  , and 4 moles of .

Thus, the final enthalpy of reaction will be: 1920 kJ/mol + (-1600 kJ/mol)

The final enthalpy of reaction will be = 320.3 kJ/mol.

For more information about the enthalpy of reaction, refer to the link:

link

The enthalpy : 320.3 kJ/mol

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

Based on the principle of Hess's Law,

the change in enthalpy of a reaction will be the same even though it is through several stages or ways  

Reaction

2Fe (s) + 3Cl₂ (g) → 2FeCl₃ (s) ∆H° = -800.0 kJ/mol  x 2

4Fe (s) + 6Cl₂ (g) → 4FeCl₃ (s) ∆H° = -1600.0 kJ/mol

Si(s) + 2 Cl₂ (g) → SiCl₄ (s) ∆H° = -640.1 kJ/mol

Reverse

SiCl₄ (s) → Si(s) + 2 Cl₂ (g) ∆H° = 640.1 kJ/mol x 3

3SiCl₄ (s) → 3Si(s) + 6Cl₂ (g) ∆H° = 1920.3 kJ/mol

+

3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s) ∆H° = 320.3 kJ/mol

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.