Answer:
The correct statement is:
E - The entropy of the products is greater than the entropy of the reactants.
Explanation:
C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O
As glucose is a large molecule and then it is transformed into many molecules of water and carbon dioxide, the entropy of the system increases. If the number of molecules increases, the disorder increases.
Initial state: 7 molecules (1 glucose + 6 oxygen)
Final state: 12 molecules (6 carbon dioxide + 6 water)
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.
H2SO4(aq)+NaOH(s)----->NA2SO4 + H20
10 cubic inches
We will use Boyle's law that states that for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional.
P1 V1 = P2 V2
Where
P1 is initial pressure = 5 psi
V1 is initial volume = 20 cubic inch
P2 is final pressure = 10 psi
V2 is final volume = unknown
V2 = P1,V1 / P2
V2 = 20 × 5 / 10
V2 = 100/10
V2 = 10 cubic inches
No beginning or end I believe
