When the football is on its way up, Kinetic energy is being transformed into Gravitational Potential energy.
Molar mass NaHCO₃ = 23 + 1 + 12 + 16 x 3 = 84 g/mol
1 mole ---------- 84 g
? mole ---------- 110 g
moles NaHCO₃ = 110 . 1 / 84
moles NaHCO₃ = 110 / 84
= 1.309 moles
hope this helps!
1) d
2) b because the independent variable is the thing you change/control in an experiment
3) c because the dependent variable is the thing being measured in an experiment
4)hmm it might be d, as c and a are both correct as different sized feeders would make it an unfair test and different types of food would as well
5) c
6) a
7) b obviously because if he activated them at different times then the ones activated last would have an advantage
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.
