Ask question

Ask question

All categories

3 years ago

5

The following charges on individual oil droplets were obtained during an experiment similar to Millikan's. Determine a charge fo

r the electron On C, coulombs), and explain your answer: -3.204><10 -19 C; -4.806x10-19 C; ~8.0le10-19 C; - 1.442x10-8 C.

1 answer:

RSB [31]3 years ago

4 0

Answer:

- 1.602 x 10⁻¹⁹coulombs

Explanation:

Charge on individual oil droplet would be multiple of charge on one electron . So we will find out the minimum common factor of given individual charges that is the LCM of all the charges given.

LCM of given charges like 3.204 , 4.806 ,8.01 and 14.42 . We have neglected the power of ten( 10⁻¹⁹) because it is already a common factor to all.

The LCM is 1.602 . So charge on electron is 1.602 x 10⁻¹⁹.

You might be interested in

Ms. Clark added a spoonful of baking soda and then poured a little vinegar into the cup. We observed fizzing and bubbling. Based

leva [86]

Answer:

we know that it was the vinegar and baking soda because gas had been created, the gas was held within the bubbles. the gas that was created was called Carbon Dioxide or CO2

7 0

3 years ago

What is the main source of energy in a clear water stream in an open area

solniwko [45]

Answer:

c.sunlight

Explanation:

4 0

3 years ago

Read 2 more answers

(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe

Dmitry [639]

Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration $C_o$ = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

$time (t) = \dfrac{V}{Q}$

$time (t) = \dfrac{200}{0.3}$

$time (t) = 666.66 \ sec$

$time (t) = \dfrac{666.66 }{3600} hrs$

$time (t) = 0.185 hrs$

$\text{Using First Order Reaction:}$

$\dfrac{dc}{dt}=kc$

where;

$t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)$

$0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)$

$0.942 = \Big ( \dfrac{200}{C_e}- 1 \Big)$

$1+ 0.942 = \Big ( \dfrac{200}{C_e} \Big)$

$\dfrac{200}{C_e} = 1.942$

$C_e = \dfrac{200}{1.942}$

$\mathbf{C_e = 102.98 \ mg/l}$

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

$t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)$

$V_{PFR} =0.196(0.663)*0.3*3600$

$V_{PFR} = 140.34 \ m^3$

The volume of the PFR is ≅ 140 m³

3 0

3 years ago

In the following reaction, if you wanted to produce more hydrochloric acid (HCl), what should you do? (2 points)

Paul [167]

Answer:

Add more H2O

Explanation:

Took the test

3 0

3 years ago

How many moles of chlorine are in 100g of chlorine

cricket20 [7]

No. of moles = mass / molar mass
= 100/35.5

5 0

3 years ago