Answer:
Intertidal zone
Neritic zone
Open-ocean zone
Note: the correct questions are found below;
In which zone do you find marshes and mangrove forests?
In which zone are plankton plentiful, providing plenty of food for the fish that live there?
In which zone would you find very little plant or animal life compared to other zones?
Explanation:
The intertidal zone, sometimes called the littoral zone, is the area of the marine shoreline that is exposed to air at low tide, and covered with seawater when the tide is high. Intertidal zonation refers to the tendency of plants and animals to form distinct communities between the high and low tide lines. Some microclimates in the littoral zone are moderated by local features and larger plants such as mangroves.
The neritic zone is the region of shallow water (200 meters depth) above the continental shelf where light penetrates to the sea floor.
Due to the abundant supply of sunlight and nutrients such as plankton in this zone, it is the most productive ocean zone supporting the vast majority of marine life.
The open oceans or pelagic ecosystems are the areas away from the coastal boundaries and above the seabed. It encompasses the entire water column and lies beyond the edge of the continental shelf. It extends from the tropics to the polar regions and from the sea surface to the abyssal depths.
Answer:
T₂ = 150 K
Explanation:
Given data:
Initial volume = 4 L
Initial temperature = 300 K
Final volume = 2 L
Final temperature = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = T₁V₂/V₁
T₂ = 300 K × 2L / 4 L
T₂ = 600 L.K / 4 L
T₂ = 150 K
Answer:
0.256 L
Explanation:
We should use the following formula:
concentration (1) × volume (1) = concentration (2) × volume (2)
concentration (1) = 0.82 M NaOCl
volume (1) = ?
concentration (2) = 0.21 M NaOCl
volume (2) = 1 L
volume (1) = [concentration (2) × volume (2)] / concentration (1)
volume (1) = [0.21 / 1] / 0.82 = 0.256 L
Answer:
(70 x 2 + 16 x 3) x 29/70 x 2 = 38,9428571 g.
Explanation:
