Answer:
980C
Explanation:
but it need a scale tell me the scale
Answer:
A
Explanation:
The only factor that would not decrease the carrying capacity of the fur seal's environment is the population of predators.
<em>The carrying capacity is defined as the maximum number of individuals of a species an environment can sustain based on the resources available. Hence, only a fluctuation in the resources available in the environment of the fur seal can affect the carrying capacity. </em>
<u>Pollution </u>and <u>destruction of the habitat of the fur seal</u> would limit the resources available to the animals and as such, limit the carrying capacity. A <u>fluctuation in the population of prey available</u> for the population of fur seals would also affect the carrying capacity.
The only factor that has no bearing on the carrying capacity of the fur seal's environment is the population of predators. Predators have the capacity to reduce the population of fur seals but not affect the carrying capacity for fur seals.
The correct option is, therefore, A.
Answer:
I think the answer is - "is stuck and won't go any further into the wood."
Answer:
1.195 M.
Explanation:
- We can calculate the concentration of the stock solution using the relation:
<em>M = (10Pd)/(molar mass).</em>
Where, M is the molarity of H₂SO₄.
P is the percent of H₂SO₄ (P = 40%).
d is the density of H₂SO₄ (d = 1.17 g/mL).
molar mass of H₂SO₄ = 98 g/mol.
∴ M of stock H₂SO₄ = (10Pd)/(molar mass) = (10)(40%)(1.17 g/mL) / (98 g/mol) = 4.78 M.
- We have the role that the no. of millimoles of a solution before dilution is equal to the no. of millimoles after dilution.
<em>∴ (MV) before dilution = (MV) after dilution</em>
M before dilution = 4.78 M, V before dilution = 250 mL.
M after dilution = ??? M, V after dilution = 1.0 L = 1000 mL.
∴ M after dilution = (MV) before dilution/(V after dilution) = (4.78 M)(250 mL)/(1000 mL) = 1.195 M.
Answer:
3 Hz
The frequency is three waves per second, or 3 Hz.
