What is an example of an abiotic factor in an environment?

1 answer:

3 0

In the Terrestrial ecosystem, examples might include the following such as tempature, light and water. In the Ocean Ecosystem, abiotic factors would include salinity and ocean currents

You might be interested in

Can someone help I’m stuck on this question

RideAnS [48]

Answer:

It is  propyl amine.

as amine =H-N-H

and it contain 3C =prop 

8 0

3 years ago

Photochemical smong result from the interaction of pollution in the presence of

Goshia [24]

Photochemical smong result from the interaction of pollution in the presence of sunlight. The photochemical smog is result of the chemical reaction in the air between the sunlight, nitrogen oxides and volatile organic compounds.
It is noticed as a brown haze above cities. The negative effects are: production of chlorofluorocarbons (CFCs) increased UV radiation, irritation to the eyes, nose, and throat.

6 0

4 years ago

Climate change is a threat to biodiversity on Earth. True False

barxatty [35]

True is the correct answer

7 0

3 years ago

Read 2 more answers

4. A 0.100 M solution of NaOH is used to titrate an HCl solution of unknown concentration. To neutralize the solution, an averag

PtichkaEL [24]

Answer:

C. 0.191 M

Explanation:

Our goal for this question, is to calculate the concentration of the HCl solution. For this, in the experiment, a solution of NaOH was used to find the moles of HCl. Therefore, our first step is to know the reaction between HCl and NaOH:

$HCl~+~NaOH~->~NaCl~+~H_2O$

The "titrant" in this case is the NaOH solution. If we know the concentration of NaOH (0.100M) and the volume of NaOH (38.2 mL=0.0382 L), we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

$0.100~M=\frac{mol}{0.0382~L}$

$mol=0.100~M*0.0382~L=0.0382~mol~of~NaOH$

Now, in the reaction, we have a 1:1 molar ratio between HCl and NaOH (1 mol of HCl is consumed for each mole of NaOH added). Therefore we will have the same amount of moles of HCl in the solution:

$0.0382~mol~of~NaOH\frac{1~mol~HCl}{1~mol~NaOH}=0.0382~mol~HCl$