<span>A. Exact ecological footprints are often difficult to calculate, but estimates can be useful in comparing populations.
</span>Which of the following could be said about ecological footprints? <u /> <u>Exact ecological footprints are often difficult to calculate, but estimates can be useful in comparing populations.</u><u />
NOT:
b. Ecological footprints can't be used to determine carrying capacity.
C. Ecological footprints don't take into account resources needed to absorb and manage wastes.
<span>D. The average ecological footprints for various countries are nearly identical.</span>
The number of formula units in 2.50 mol of the compound is 15.1 * 10^23.
The question is unclear whether NaNO2 or NaNO3 is implied. However,in either case, the solution applies equally.
6.02 * 10^23 formula units of the compound are contained in 1 mole
x formula units are contained in 2.5 moles of the compound
x = 6.02 * 10^23 formula units * 2.5 moles/ 1 mole
x = 15.1 * 10^23 formula units of the compound.
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Answer:
Explanation:Although the term is quite new, our connection to nature is not. We depend on nature for our survival - without healthy ecosystems, our drinking water isn’t clean nor is the air we breathe. We also enjoy nature... studies show that people who spend time in nature tend to be happier than those that don’t. It can even act as a natural anti-depressant. With industry and urban sprawl expanding at unprecedented rates, Ecosystem Services attempt to translate the benefits we receive from nature into economic terms so we can better understand the trade-offs we are making between nature and industrial development.
Answer:
for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: 
![\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where 
represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
![\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20192.45%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20191.61%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20130.684%5Cfrac%7BJ%7D%7Bmol.K%7D%5D%3D-99.4J%2FK)
So, for the given reaction is -99.4 J/K
The formula for force is F=ma. Because weight is a measure of force, then we can substitute the weight of the meteor, 3204 N, for F in the the equation for force. We also know that the acceleration of gravity on Earth is 9.8 m/s^2. To find the mass, simply divide both sides of the equation by the value of acceleration to get
Therefore, the value of the mass of the meteor is 326.9 kg.
