Answer:
The correct answer is B. exogenous
Explanation:
Let us try to describe exogenous and endogenous variables an exogenous variable value is influenced only by factors outside a model or system and is forced onto the model, while a change in an exogenous variable is known as an exogenous change. Also an endogenous variable is one whose value is influenced only by the system or model under study.
Cold blooded animal? it’s a bit vague sorry
Answer:
kp= 3.1 x 10^(-2)
Explanation:
To solve this problem we have to write down the reaction and use the ICE table for pressures:
2SO2 + O2 ⇄ 2SO3
Initial 3.4 atm 1.3 atm 0 atm
Change -2x - x + 2x
Equilibrium 3.4 - 2x 1.3 -x 0.52 atm
In order to know the x value:
2x = 0.52
x=(0.52)/2= 0.26
2SO2 + O2 ⇄ 2SO3
Equilibrium 3.4 - 0.52 1.3 - 0.26 0.52 atm
Equilibrium 2.88 atm 1.04 atm 0.52 atm
with the partial pressure in the equilibrium, we can obtain Kp.
Answer:
16.5 dm³
Explanation:
Data Given:
no. moles of O₂ = 0.735 moles
volume of O₂ = ?
Solution:
Now
we have to find volume of O₂ gas
Formula used for this purpose
No. of moles = Volume / molar volume
where
molar volume at STP for Oxygen (O₂) = 22.4 dm³/mol
No. of moles O₂ = Volume of O₂ / 22.4 dm³/mol . . . . . .(1)
Put values in equation 1
0.735 = Volume of O₂ / 22.4 dm³/ mol
rearrange above equation
Volume of O₂ = 0.735 x 22.4 dm³/ mol
Volume of O₂ = 16.5 dm³
So,
the volume of O₂ at STP is 16.5 dm³
