The path of energy flow from the sun to the humpback whale is as follows:
- Sun---> Plankton ---> Small fishes ---> Humpback whale.
<h3>What is energy?</h3>
Energy is the ability to do work.
The primary source of energy on the earth is the sun.
The energy from the sun is used by producers to produce food on which other organisms depend on.
The energy from the sun gets to the humpback whale through producers such as plankton.
The path of energy flow from the sun to the humpback whale is as follows:
- Sun---> Plankton ---> Small fishes ---> Humpback whale.
Learn more about energy flow at: brainly.com/question/21786633
Answer:
The answer is
<h2>20 %</h2>
Explanation:
The percentage error of a certain measurement can be found by using the formula

From the question
actual volume = 35 mg
error = 35 - 28 = 7
The percentage error is

We have the final answer as
<h3>20 %</h3>
Hope this helps you
Answer ; The question is missing in some details, but here are he details ;
The two naturally occurring isotopes of bromine are
81Br (80.916 amu, 49.31%) and
79Br (78.918 amu, 50.69%).
The two naturally occurring isotopes of chlorine are
37Cl (36.966 amu, 24.23%) and
35Cl (34.969 amu, 75.77%).
Bromine and chlorine combine to form bromine monochloride, BrCl.
Explanation:
The detaile calculation is as shown in the attachment.
Answer:
The light emitted by a light bulb is a form of radiation that occurs when the filament heats up and its thermal emission gains enough energy to move into the visible spectrum.
Explanation:
Light bulbs contain a filament which is heated up electrically. When this filament is heated up,energy in the form of heat is imparted to the electrons in the filament.
This thermal excitation of electrons ultimately leads to emission of light in the viable spectrum. This light is now radiated through a light bulb.
Answer:
H+(aq) + OH-(aq) → H2O(l)
Explanation:
Step 1: Data given
nitrious acid = HNO3
sodium hydroxide = NaOH
Step 2: The unbalance equation
HNO3(aq) + NaOH(aq) →NaNO3(aq) + H2O(l)
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side (Ba^2+ and Br-), look like this:
H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq) →Na+(aq) +NO3(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)