Explanation:
The given data is as follows.
= 98.70 kPa = 98700 Pa,
T = 
= (30 + 273) K = 303 K
height (h) = 30 mm = 0.03 m (as 1 m = 100 mm)
Density = 13.534 g/mL = 
= 13534 
The relation between pressure and atmospheric pressure is as follows.
P = 
Putting the given values into the above formula as follows.
P =
= 
= 102683.05 Pa
= 102.68 kPa
thus, we can conclude that the pressure of the given methane gas is 102.68 kPa.
B. slows down is your answer, obviously as it approaches carrying capacity, there would be less available space to find in the place of inhabitance, so less and less population units would be able to find the place of inhabitance suitable for living, or can't find enough space to live in.
Answer:
C6H12O6 —> 2C2H5OH + 2CO2
Explanation:
The equation for the reaction is given below:
C6H12O6 —> C2H5OH + CO2
We can balance the equation above as follow:
There are 12 atoms of H on the left side and 6 atoms of the right side. It can be balance by putting 2 in front of C2H5OH as shown below:
C6H12O6 —> 2C2H5OH + CO2
There are 6 atoms of C on the left side and 5 atoms on the right side. It can be balance by putting 2 in front of CO2 as shown below:
C6H12O6 —> 2C2H5OH + 2CO2
Now the equation is balanced.
Unlike solid matter, where particles are tightly packed and slightly vibrating, or gas, where particles go around everywhere and are extremely loose, a liquid has particles that are loosely packed but are still in slight contact with each other. Hope that's good enough
Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
