<h3>
Answer:</h3>
0.424 J/g °C
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Chemistry</u>
<u>Thermochemistry</u>
Specific Heat Formula: q = mcΔT
- q is heat (in Joules)
- m is mass (in grams)
- c is specific heat (in J/g °C)
- ΔT is change in temperature
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] m = 38.8 g
[Given] q = 181 J
[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C
[Solve] c
<u>Step 2: Solve for Specific Heat</u>
- Substitute in variables [Specific Heat Formula]: 181 J = (38.8 g)c(11.0 °C)
- Multiply: 181 J = (426.8 g °C)c
- [Division Property of Equality] Isolate <em>c</em>: 0.424086 J/g °C = c
- Rewrite: c = 0.424086 J/g °C
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.424086 J/g °C ≈ 0.424 J/g °C
Answer:
Conclusion
Explanation:
I believe you were asking for the term that best matches with the description given. Typically the conclusion summarizes your experiment in a 1 to 2 paragraph format.
Answer:
C. 1.3 mol
Explanation:
PV = nRT
where P is absolute pressure,
V is volume,
n is number of moles,
R is universal gas constant,
and T is absolute temperature.
Given:
P = 121.59 kPa
V = 31 L
T = 360 K
R = 8.3145 L kPa / mol / K
Find: n
n = PV / (RT)
n = (121.59 kPa × 31 L) / (8.3145 L kPa / mol / K × 360 K)
n = (3769.29 L kPa) / (2993.22 L kPa / mol)
n = 1.26 mol
Round to two significant figures, there are 1.3 moles of gas.
