The littoral zone of a lake is the area closest to the shore. It has very little biological activity but includes a lot of oxygen. The water in the lake's littoral zone is freshwater, free of living organisms such as plants and fish.
Answer: yes they do.
Explanation:
I'm tryna get more points because my other account got blocked for the next 48 hours.
(Don't be mad at me please)
Answer:
a) ΔHvap=35.3395 kJ/mol
b) Tb=98.62 °C
Explanation:
Given the reaction:
C₇H₁₆ (l) ⇔ C₇H₁₆ (g)
Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.
When
T₁=50°C=323.15K ⇒P₁=0.179
T₂=86°C=359.15K ⇒P₂=0.669
The Clasius-Clapeyron equation is:
ΔHvap=35339.5 J/mol=35.3395 KJ/mol
Normal boiling point ⇒ P=1 atm
Hence, we find the normal boiling point where:
T₁=323.15K
P₁=0.179 atm
P₂=1 atm
T₂=371.77 K= 98.62 °C
The answer is 4.45 × 10²⁴ units.
To calculate this, we will use Avogadro's number which is the number of units (atoms, molecules) in 1 mole of substance:
6.02 × 10²³ units per 1 mole
So, we need a proportion:
If 6.02 × 10²³ units are in 1 mole, how many units will be in 7.40 moles:
6.02 × 10²³ units : 1 mole = x : 7.40 moles
After crossing the products:
1 mole * x = 7.40 moles * 6.02 × 10²³ units
x = 7.40 * 6.02 × 10²³ units
x = 44.5 × 10²³ units = 4.45× 10²⁴ unit
