Answer:
-24.76 kJ/g; -601.8 kJ/mol
Explanation:
There are two heat flows in this experiment.
Heat from reaction + heat absorbed by calorimeter = 0
q1 + q2 = 0
mΔH + CΔT = 0
Data:
m = 0.1375 g
C = 3024 J/°C
ΔT = 1.126 °C
Calculations:
0.1375ΔH + 3024 × 1.126 = 0
0.1375ΔH + 3405 = 0
0.1375ΔH = -3405
ΔH = -24 760 J/g = -24.76 kJ/g
ΔH = -24.76 kJ/g ×24.30 g/mol = -601.8 kJ/mol
Answer:
Cobalt (III) Iodide
General Formulas and Concepts:
<u>Chemistry - Compounds</u>
- Reading a Periodic Table
- Writing Compounds
- Naming Compounds
Explanation:
<u>Step 1: Define</u>
CoI₃
<u>Step 2: Identify</u>
Co is a Transition Metal, has either a charge of 2+ or 3+
I is a Halogen, has a charge of 1-
<u>Step 3: Write name</u>
We need to balance the charges, and in order to do that, we must have a Co³⁺
Cobalt (III) Iodide
Answer: 0.4578 g
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
where,
n= moles of solute
= volume of solution in ml = 25.0 ml
Thus 0.4578 g of nickel(II) nitrate is to be dissolved in 25.0 mL of 0.100 M 
solution.
