Part b - the effect of an electric field (voltage) on a negatively charged oil droplet in the millikan oil droplet experiment, the oil is sprayed from an atomizer into a chamber. the droplets are allowed to pass through the hole into the chamber so that their fall can be observed. the top and bottom of the chamber consist of electrically charged plates. the upper plate is positively charged, and the lower plate is negatively charged. x rays are introduced into the chamber so that when they strike the oil droplets, the droplets will acquire one or more negative charges. the electric field (voltage) is applied to the metal plates. watch the animation and identify the effects of an electric field on the motion of a negatively charged oil droplet. consider the gravitational force as fg and the electric force as fe. all the other forces acting on the oil droplet can be ignored as their effect on the motion of the oil droplet is negligible. check all that apply.

There will be two dominant forces acting on the oil droplets. The first is the weight of the oil droplets, which will cause them to move downwards in the chamber. The other is the force of attraction the droplets will experience due to the positively charged plate and their own negative charge. The magnitude of the resultant force, which will be equivalent to the difference of these forces, will dictate whether the net movement of the droplet is downwards or upwards.

1) The order of the reaction is of FIRST ORDER

2)   Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min                    0         10         20          30             40          ∞

conc B/(mol/L)    0       0.089    0.153     0.200       0.230    0.312

For  a first order reaction:

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

= initial concentration at time t

= final concentration at time t

= concentration at time t

To start with the value of t when t = 10 mins

When t = 20

When t = 30

When t = 40

We can see that at the different time rates, the rate constant of all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴