Biotic and abiotic factors are the environmental conditions that the organisms have to face to live in a specified environment.
-Abiotic factors-
Abiotic factors are the physical and chemical conditions of an environment. For example : heat, salinity, pressure, light, wind, pH ...
-Biotic factors-
Biotic factors are all the biological conditions of an environment for a specie/taxa. It can include prey and predator abundance, available food amount, available space, intra and interspecific competition...
The development of organims is under the control of abiotic factors. Some are adapted to heat, cold etc ... The abiotic factors will define which organisms are able or not to live in a specified place.
The living organisms will constitute the biotic factors, which define if and how can an organism live in a specified environment.
So, the abiotic factors are controling the biotic factors of an environment.
Hope it helps you !
Answer:
B) As you move across the row, the number of electrons increases and reactivity also increases.
Explanation:
The periodic table is arranged in a way that if you go across a period, the number of protons, neutrons, and electrons in an element increases. In terms of reactivity, the most reactive elements are the ones which have a high electronegativity. The electronegativity of the elements increases as you travel to the right and upwards on the periodic table.
Answer:
Transition elements are elements which have partially filled d-orbitals and form at least one or more stable ions.
Answer:
the conversion factor is f= 6 mol of glucose/ mol of CO2
Explanation:
First we need to balance the equation:
C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)
the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:
f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction
f = 6 moles of CO2 / 1 mol of glucose = 6 mol of glucose/ mol of CO2
f = 6 mol of CO2/ mol of glucose
for example, for 2 moles of glucose the number of moles of CO2 produced are
n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2
Answer:
a)23.2 L
b)68.3kPa
c)7.5 atm
d)60.5L
e)1.67 atm
Explanation:
From Boyle's law:
P1V1=P2V2
P1= 748mmHg
P2=725mmHg
V1= 22.5L
V2??
V2= P1V1/P2= 748×22.5/725= 23.2 L
b)
V1=4.0L
P1= 205×10^3Pa
V2= 12.0L
P2=???
P2= P1V1/V2= 205×10^3×4/12
P2= 68.3×10^3 Pa or 68.3kPa
c)
P1= 1 atm
V1= 196.0L
P2= ??
V2= 26.0L
P2= P1V1/V2=1×196.0/26.0
P2= 7.5 atm
d)
V1= 40.0L
P1= 12.7×10^3Pa
V2=???
P2= 8.4×103Pa
V2= P1V1/P2= 12.7×10^3×40.0/8.4×103
V2=60.5L
e)
V1= 100mL
P1= 1atm
V2= 60mL
P2=???
P2= P1V1/V2= 1×100/60
P2= 1.67 atm