Specific heat is the quantity of heat required to change the temperature of 1 gram of a substance by 1 degree Celsius. It is the amount per unit mass that is required to raise the temperature by one degree Celsius. Every substance has its own specific heat and each has its own distinct value. The units of specific heat are joules per gram-degree Celsius (J/f C) and sometimes J/Kg K may also be used.
Answer:
V₂ =279.9 cm³
Explanation:
Given data:
Initial volume = 360 cm³
Initial temperature = 50°C
Initial pressure = 700 mmHg
Final volume = ?
Final temperature = 273 k
Final pressure = 1 atm
Solution:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Solution:
<em>We will convert the mmHg to atm.</em>
700/760 = 0.92 atm
<em>and °C to kelvin.</em>
50+273 = 323 K
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 0.92 atm × 360 cm³ × 273 K / 323 K ×1 atm
V₂ = 290417.6 atm .cm³. K / 323 k. atm
V₂ =279.9 cm³
Answer:
it cannot !!
Explanation:
a chemical reaction MUST occur to separate a compound! :)
<h3>
Answer:</h3>
P₂ = 0.67 atm
<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>Gas Laws</u>
Boyle's Law: P₁V₁ = P₂V₂
- P₁ is pressure 1
- V₁ is volume 1
- P₂ is pressure 2
- V₂ is volume 2
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] P₁ = 2.02 atm
[Given] V₁ = 4.0 L
[Given] V₂ = 12.0 L
[Solve] P₂
<u>Step 2: Solve</u>
- Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
- [Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
- [Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
- [Pressure] Rewrite: P₂ = 0.673333 atm
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>
0.673333 atm ≈ 0.67 atm
