The central vacuole stores materials, wastes, and helps give the plant structure and support.
Hope this helps!
Jim, because he ran a greater distance in the same time :)
By the way, this is a maths question
Answer:
252.68 K or -20.46 °C
Explanation:
According to Gay-Lussac's Law, "Pressure and Temperature at given volume are directly proportional to each other".
Mathematically,
P₁ / T₁ = P₂ / T₂ ---- (1)
Data Given:
P₁ = 30.7 kPa
T₁ = 0.00 °C = 273.15 K
P₂ = 28.4 kPa
T₂ = <u>???</u>
Solving equation for T₂,
T₂ = P₂ T₁ / P₁
Putting values,
T₂ = 28.4 kPa × 273.15 K / 30.7 kPa
T₂ = 252.68 K or -20.46 °C
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.
