Answer:
the tertiary animals the bigger animals that feed on smaller fish and crustaceans. These include predators like sharks, barracuda and tuna snapper. these fish that are commercially fished at unsustainable levels.
Explanation:
Answer:
0.2193 μm
Explanation:
The reaction showing the Photodissociation of ozone (O3) is given below as:
O₃ + hv --------------------------> O₂ + O⁺
H° (142.9) (0) (438kJ/mol).
The first thing to do here is to determine the change in the enthalpy of the total reaction, this can be done by subtracting the change in the enthalpy of the reactant from the change in enthalpy in the product. Hence, we have:
ΔH° = [438 kJ/mol + 247.5 kJ/mol] - (142.9) = 542.6 KJ/mol.
This value, that is 542.6 KJ/mol will then be used in the determination of the value for the maximum wavelength that could cause this photodissociation.
Therefore, the maximum wavelength could cause this photodissociation ≤ h × c/ E = [ 1.199 × 10⁻⁴]/ 542.6 = 2.193 × 10⁻⁷ = 0.2193 μm
<h2>Answer:</h2>
The correct answer is option C which is cell can be a plant cell or an animal cell.
Explanation:
- The statement is is that cell has cell membrane. Cell membrane is the outer covering membrane which is present in both cells. While cell wall is a external hard membrane present only in plant cells.
- In the statement D, the presence in the large molecule usually in the center of cell is the indication of plant cell. It gives the upright direction in plants and it stores large amount of water and food.
- Hence the correct answer is option C.
Answer:
45.4 L
Explanation:
Using Ideal gas equation for same mole of gas as
Given ,
V₁ = 27.9 L
V₂ = ?
P₁ = 732 mmHg
P₂ = 385 mmHg
T₁ = 30.1 ºC
T₂ = -13.6 ºC
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (30.1 + 273.15) K = 303.25 K
T₂ = (-13.6 + 273.15) K = 259.55 K
Using above equation as:
Solving for V₂ , we get:
<u>V₂ = 45.4 L</u>
