Answer:
Q9. The independent variable in this experiment is the fertilizer. It is independent because she manipulating the variable to compare the growth.
Q10. The dependent variable in this experiment is the amount of growth of the corn. It is this because the growth depends on what the scientist did on the corn.
Q11. The variable controlled in this experiment is the amount of sun and water. These two variables never change so this is why it is the control.
Explanation:
Answer:
Cloud condensation nuclei or CCNs are small particles typically 0.2 µm, or 1/100 the size of a cloud droplet on which water vapor condenses. Water requires a non-gaseous surface to make the transition from a vapour to a liquid; this process is called condensation.
Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen 
if it occupies 31.6 mL at 
.
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.
Temperature = 
As molar mass of 
is 32 g/mol. Hence, the number of moles of are calculated as follows.
Using the ideal gas equation calculate the pressure exerted by given gas as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the value into above formula as follows.
Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen if it occupies 31.6 mL at .
Answer:
Explanation:
To calculate their average atomic masses which is otherwise known as the relative atomic mass, we simply multiply the given abundances of the atoms and the given atomic masses.
The abundace is the proportion or percentage or fraction by which each of the isotopes of an element occurs in nature.
This can be expressed below:
RAM = Σmₙαₙ
where mₙ is the mass of isotope n
αₙ is the abundance of isotope n
for this problem:
RAM of Li = m₆α₆ + m₇α₇
m₆ is mass of isotope Li-6
α₆ is the abundance of isotope Li-6
m₇ is mass of isotope Li-7
α₇ is the abundance of isotope Li-7
