Answer:
the number of possible configurations of the locations and energies of the atoms or molecules that comprise a system
Explanation:
Ludwig Boltzmann was the first to suggest that the concept of entropy could be calculated by examining the positions and energies of molecules. This was developed into an equation, known as the Boltzmann equation, which relates entropy to the number of microstates (W):
S = k ln W
where k is the Boltzmann constant (1.38 x 10-23 J/K), and W is the number of microstates.
Microstates was used to imply the number of different possible arrangements of molecular position and kinetic energy at a particular thermodynamic state. Therefore any process that gives an increase in the number of microstates therefore increases the entropy. Hence the answer.
Answer:
Phase C - Liquid State
Phase E - Gaseous State
Explanation:
Usually, in phases of water, we have the following;
When temperature is less than zero, it is said to be in its solid phase as ice.
When temperature is between 0 to 100, we can say it is in the liquid phase as water.
When temperature is above 100°C, It is said to be in the gaseous phase as vapour.
From the diagram;
Phase C is the only liquid state because it falls between temperature of 0°C and 100°
Also, only phase E is in the gaseous phase because the temperature is above 100°C.
All of them are right except the neutron one. The neutrons is actually 176 because you subtract the number of protons from the atomic mass to find this.
Answer:
200 g C₆H₁₂O₆
General Formulas and Concepts:
<u>Chemistry - Solutions</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Molarity = moles of solute / liters of solution
Explanation:
<u>Step 1: Define</u>
1 M C₆H₁₂O₆
1 L of solution
<u>Step 2: Identify Conversions</u>
Molar Mass of C - 12.01 g/mol
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
<u>Step 3: Find moles of solute</u>
1 M C₆H₁₂O₆ = x mol C₆H₁₂O₆ / 1 L
x = 1 mol C₆H₁₂O₆
<u>Step 4: Convert</u>
<u />
= 180.18 g C₆H₁₂O₆
<u>Step 5: Check</u>
<em>We are given 1 sig figs. Follow sig fig rules and round.</em>
180.18 g C₆H₁₂O₆ ≈ 200 g C₆H₁₂O₆
Whichever one has a negative enthalpy value.
