Complete question:
Consider the hypothetical reaction 4A + 2B → C + 3D
Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?
Answer:
the final concentration of A is 0.992 M.
Explanation:
Given;
time of reaction, t = 4.0 s
rate of change of the concentration of B = -0.0760 M/s
initial concentration of A = 1.600 M
⇒Determine the rate of change of the concentration of A.
From the given reaction: 4A + 2B → C + 3D
2 moles of B ---------------> 4 moles of A
-0.0760 M/s of B -----------> x
⇒Determine the change in concentration of A after 4s;
ΔA = -0.152 M/s x 4s
ΔA = -0.608 M
⇒ Determine the final concentration of A after 4s
A = A₀ + ΔA
A = 1.6 M + (-0.608 M)
A = 1.6 M - 0.608 M
A = 0.992 M
Therefore, the final concentration of A is 0.992 M.
The positively charged atmosphere attracts negatively charged spider silk, might electrostatic force play in spider dispersal, according to a recent study.
Answer: Option C
<u>Explanation:</u>
The positive charge present in upper of the atmosphere and the negative charge on planet’s surface. During cloudless skies days, the air possesses a voltage of nearly around 100 volts for each and every meter from above the ground.
Ballooning spiders process within this planetary electric field. When their silk relieve their bodies then it picks up a negative charge. This oppose the similar negative charges on the surfaces on which the spiders settles and create sufficient force to lift them into the air. And spiders can hike those forces by climbing onto blades of grass,twigs, or leaves.
The transformation of energy that takes place when a slingshot launches a stone is elastic potential energy to kinetic energy.
Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5
Explanation: In order to solve this problem we have to use the gaussian law in the mentioned regions.
Region 1; 0<r<2
∫E.ds=Qinside the gaussian surface/ε0
inside of the solid conducting sphere the elevctric field is zero because the charge is located at the surface on this sphere.
Region 2; 2<r<4;
E.4*π*r^2=8,84/ε0
E=8,84/(4*π*ε0*r^2)
Region 3; 4<r<5
E=0 because is inside the conductor.
Finally
Region 4; r>5
E.4*π*r^2=(8,84-2.02)/ε0