Technically there is only one phase unless you account for a solution where you have a pure liquid with something dissolved in it. Unless you count aqueous as a phase which is just dissolved. Since you are in high school the answer you are looking for is one. <span />
Answer:

Explanation:
Hello!
In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

We plug in the mass of water, temperature change and specific heat to obtain:

Now, this enthalpy of reaction corresponds to the combustion of propyne:

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

Now, we solve for the enthalpy of formation of C3H4 as shown below:

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

Best regards!
Answer:
Technician B.
Explanation:
The claim of technician B that the PCM will add fuel if the IAT indicates that the incoming air temperature is cold is correct.
Answer:
0.123 moles of ammonia, can be produced
Explanation:
First of all, we need to determine the reaction:
Ammonia is produced by the reaction of hydrogen and nitrogen.
3H₂(g) + N₂(g) → 2NH₃(g)
Ratio is 3:2. Let's solve the question with a rule of three:
If 3 moles of hydrogen can produce 2 moles of ammonia
Then, 0.37 moles will produce (0.37 . 2) /3 = 0.123 moles