I put down all the info that I found but I cant figure out the questions 1 & 2 at the bottom!!

Empty Calorimeter- 11.6 g
Room temp - 21 degrees celcius
Three different experiments of 50mL 1(M) HCl reacts with 0.15 g Mg, 0.25g Mg, and 0.35g Mg respectively.
The maximum temperatures after reaction were, 29.27 degrees Celcius, 34.78 degrees Celcius and 40.29 degrees celcius. Which makes deltaT 8.27, 13.78, and 19.29.
The weight after reaction minus the empty calorimeter is: 49.957g, 50.049g, and 50.141g.

1.Calculate the number of moles reacted in the three experiments (MW= 24.305 g/mol)
2.Calculate the heat released into the solution for the 3 reactions, according to:
3. Find the molar heat of reaction for each experiment in units of Joules / (mole of Mg) by dividing the heat of reaction by the moles of Mg used.

4. Calculate and record the average molar heat of reaction from the three results.

q(reaction) = Ccal * Delta T + mass(contents) * Cp (contents) * deltaT
assume that calorimeter constant is 0

Note the correct spelling of celsius.

1. moles Mg = grams Mg/molar mass Mg.
2. q = heat released = mass water x specific heat water x delta T. Note that the answer to 2 is note at bottom (after question 4) "q(reaction) = ...._"

To answer questions 1 and 2, we need to use the given information and apply the relevant formulas. Let's break down each question step by step:

1. To calculate the number of moles reacted in the three experiments, we can use the formula:

moles = mass / molar mass

For the first experiment:
moles = 0.15 g / 24.305 g/mol
moles = 0.00617 mol (rounded to five decimal places)

For the second experiment:
moles = 0.25 g / 24.305 g/mol
moles = 0.01028 mol (rounded to five decimal places)

For the third experiment:
moles = 0.35 g / 24.305 g/mol
moles = 0.01440 mol (rounded to five decimal places)

Therefore, the number of moles reacted in the three experiments are approximately 0.00617 mol, 0.01028 mol, and 0.01440 mol.

2. To calculate the heat released into the solution for the three reactions, we can use the formula:

q(reaction) = Ccal * Delta T + mass(contents) * Cp (contents) * deltaT

Since the calorimeter constant is assumed to be 0, the formula simplifies to:

q(reaction) = mass(contents) * Cp (contents) * deltaT

For the first experiment:
q(reaction) = 49.957 g * Cp (contents) * 8.27 degrees Celsius

For the second experiment:
q(reaction) = 50.049 g * Cp (contents) * 13.78 degrees Celsius

For the third experiment:
q(reaction) = 50.141 g * Cp (contents) * 19.29 degrees Celsius

Please note that we are missing the specific heat capacity (Cp) value, which should be provided in the question. The specific heat capacity depends on the substance involved in the reaction.

To find the specific heat capacity value, you may need to refer to a reference table or look up the value online. Once you have the specific heat capacity, you can substitute it into the above calculations to find the heat released into the solution for each reaction.

I hope this helps you understand how to approach questions 1 and 2. If you have any further questions, feel free to ask!