Answer:
The value of the heat capacity of the Calorimeter
= 54.4 
Explanation:
Given data
Heat added Q = 4.168 KJ = 4168 J
Mass of water
= 75.40 gm
Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c
From the given condition
Q =
ΔT +
ΔT
Put all the values in above equation we get
4168 = 75.70 × 4.18 × 11.24 +
× 11.24
611.37 =
× 11.24
= 54.4 
This is the value of the heat capacity of the Calorimeter.
Answer:
2.335 J/g*degrees celcius
Explanation:
<h3>
Answer:</h3>

<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2C + O₂ → 2CO₂
[Given] 0.25 moles O₂
[Solve] moles CO₂
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol O₂ → 2 mol CO₂
<u>Step 3: Stoichiometry</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:
