The ecosystem services provided by the each of the given examples are as follows:
- A cornfield in Kansas provides provisioning services.
- Bacteria that decompose waste along the Gulf Coast provides regulating services.
- Ocean currents that keep Pacific Northwest air cool and moist provides regulating services.
- Flower garden at a national landmark provides cultural services.
- Lumber from an oak tree provides provisioning services.
- Animals that eat seeds and then spread the seeds through their waste regulating services.
<h3>Ecosystem services</h3>
Ecosystem services are defined as the benefits derived by man from the surroundings ecosystems.
<h3>Categories of ecosystem services</h3>
The four categories of ecosystem services are:
- regulating services,
- provisioning services,
- cultural services, and
- supporting services
A cornfield in Kansas provides provisioning services.
Bacteria that decompose waste along the Gulf Coast provides regulating services.
Ocean currents that keep Pacific Northwest air cool and moist provides regulating services.
Flower garden at a national landmark provides cultural services.
Lumber from an oak tree provides provisioning services.
Animals that eat seeds and then spread the seeds through their waste regulating services.
Learn more about ecosystem services at: brainly.com/question/2191258
Answer:
2.895*10^24
Explanation:
mass of Oxygen give = 153.9g
molar mass of O2 molecule = 16*2=32g/mol
n= mole
To find the mole
n= mass/ molar mass
n= 153.9/32
n=4.81mol.
To find the number of molecules of o
Nm= number of molecule
Nn = Number of mole
NA = number of Avogados
Nm= Nn * NA
Nm= 4.81 *6.02*10^23
Nm= 2.895*10^24
Answer:
- The limiting reactant is lead(II) nitrate.
- 7.20 g of precipitate are formed.
- 1.9 g of the excess reactant remain.
Explanation:
The reaction that takes place is:
- Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)
With a percent yield of 87.5%.
To determine the limiting reactant, first we <u>convert the masses of each reactant to moles</u>, using their molar mass:
- 9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂
- 5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl
Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.
To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:
- 0.0296 mol Pb(NO₃)₂
*
* 87.5/100 = 7.20 g PbCl₂
- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:
- 0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂
Now we <u>convert that amount to moles of KCl and finally into grams of KCl</u>:
- 0.0259 mol Pb(NO₃)₂
*
= 3.86 g KCl
3.86 g of KCl would react, so the amount remaining would be:
false
Explanation:
..................................................................