Answer:

Explanation:
Given that:

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

Multiplying (2) with equation (4) ; we have:

From equation (1) ; multiplying (-1) with equation (1); we have:

From equation (2); multiplying (3) with equation (2); we have:

Now; Adding up equation (5), (6) & (7) ; we get:



<u> </u>

<u> </u>
<u />
(According to Hess Law)


Answer:
Field experiments can often have the potential to give scientists opportunities that are not possible in a lab setting.
Explanation:
Having people "act natural" in a lab setting is impossible to truly achieve, as we all know what happens to our mental state and behavior when we are being actively observed. This is also especially helpful when conducting wildlife research, as there are a myriad of different things influencing animal and plant behavior that would be nigh impossible to recreate perfectly in a controlled lab setting.
Field research can have its disadvantages as well, as it is very hard to only have one thing change (make an independent variable) in a wild environment with ever-changing weather, social effects, etc. Also, you, as the researcher, as causing an impact on the very location that you are observing, which can alter your results in unpredictable ways.
The thing to remember is that each type of study has its advantages and disadvantages; if they didn't, then we'd all do the same type of research! You have to weigh your research options and figure out which one works best for your situation :)
Answer:
A.) Air pollutants wash onto land and water bodies from precipitation
C.) Storm water runoff carries pollution
I hope I helped! ^-^
B. The mouse is hungry , because you dont really know that only the mouse does
Answer:
The volume on the tank is 6, 20 L
Explanation:
We use the formula PV=nRT. We convert the units of pressure in kPa into atm and temperature in Celsius into Kelvin:
0°C=273K
101,325kPa---1 atm
275kPa --------x=(275kPax 1 atm)/101,325kPa= 2,71 atm
PV=nRT --> V=nRT/P
V= 0,750 mol x 0,082 l atm /K mol x 273 K/ 2, 71 atm= <em>6, 20 L</em>