Answer:
The purpose of this experiment was to verify Hess’s law associated with three reactions. Our experiment confirmed it by combining the enthalpy changes of two reactions to equal of that another single reaction (Reaction 2 in our case). Styrofoam cup was used as a calorimeter to measure the change in temperature, which is a necessary variable for further calculations. We discovered that sum of enthalpy changes for reactions one and three equaled the enthalpy change in reaction two. This experiment was possible because if one of the reactions is the same as the combination of other two, the heat of the reaction of the single reaction should be equal to the heat of other two reaction combined.
Explanation: when you see the reactions that the Hess's law to find the enthalpy, which we did from this lab.
Answer:
Assuming Volume of dolution = 1 liter . The Normality is:

Explanation:
<u>Normality : </u>It is the gram equivalent of solute per liter of solution. It is represented by N.

Number of equivalents weight =

n= acidity of base or basicity of acid
for salts , n = charge present on cation
The relation between Normality and Molarity is :

n = charge present on cation (<u>not moles)</u>
= ferric sulphate
Here <u><em>Fe is cation and its oxidation state is = + 3 so n= 3</em></u>
1. First , calculate the molarity(M)
Molar Mass of ferric sulfate = Fe2(SO4)3
2(mass of Fe)+3 (mass of S) + 12(mass of O)
= 2(56)+3(32)+12(16)
= 400 grams


Molar mass = 400 gram
mass = 50 mg = 0.05 grams



let volume = 1 liter


<u><em>2. Multiply the molarity with n </em></u>


n = 3


Answer: 208g/mol
Explanation:
First of all we have to write the balance equation for the reaction.
BaCl2 + Fe2(SO4)3____> BaSO4 + FeCl3
After balancing we have.
3BaCl2 +Fe2(SO4)3_____> 3BaSO4 +2FeCl3
Looking at the equation, we find out that 3 moles of barium chloride reacts with 1 mole of iron iii sulfate
Therefore we have
3moles of BaCl2 _____> 400g/mole of iron iii sulfate
Xmole of BaCl2 _____> 200g/mole of iron iii sulfate
X = 2 * 200g/mol divide by 400g/mol
X = 1mole
1 mole of BaCl2 will be need to react with 200g/mol of iron iii sulfate.
This 1 mole of BaCl2 is equivalent to 208g/mol of BaCl2.
Therefore the gram of barium chloride that must be present is = 208g/mol//
It is C that is the most testable hypotheses