All categories

3 years ago

How did Thomson discover the electron?

1 answer:

4 0

J. J. Thomson discovered electron by performing an experiment using cathode ray tubes. High voltage across is applied across two electrodes at that causes a beam of particles to flow from the the negatively-charged electrode that is cathode to the positively-charged electrode that is anode. Properties of the particles, are tested using two oppositely-charged electric plates around the cathode ray. The cathode ray was deflected away from the negatively-charged electric plate and towards the positively-charged plate. This indicated that the cathode ray was composed of negatively-charged particles. And these negatively charged particles are called electrons.

You might be interested in

A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na

s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f: after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

$Normality=molarity \times n-factor$

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point: $N_1V_1=N_2V_2$

Before equivalence point : $N_1V_1>N_2V_2$

After the equivalence point: $N_1V_1$

$N_1V_1=100\times1=100$

case a: 5.00 mL of 1.00 M NaOH

$N_2V_2=5\times1=5$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

$N_2V_2=100\times1=100$

$N_1V_1=N_2V_2$ hence it is equivalence point

case c: 10.0 mL of 1.00 M NaOH

$N_2V_2=10\times1=10$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

$N_2V_2=150\times1=150$

$N_1V_1$ hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

$N_2V_2=50\times1=50$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

$N_2V_2=200\times1=200$

$N_1V_1$ hence it is after the eqivalence point

7 0

3 years ago

A 225 g sample of an unknown solid is heated 67C and placed into a calorimeter containing 25.6 g of water at 15.6°c. If the fina

Stella [2.4K]

Answer:

= 1.271 J/g°C

Explanation:

Heat released by the metal sample will be equivalent to the heat absorbed by water.

But heat = mass × specific heat capacity × temperature change

Thus;

Heat released by the solid;

= 225 g × c ×(67 -53) , where c is the specific heat capacity of the metal

= 3150 c joules

Heat absorbed by water;

= 25.6 g × 4.18 J/g°C × (53-15.6)

= 4002.0992 joules

Therefore;

3150 c joules = 4002.0992 joules

c =4002.0992/3150

6 0

3 years ago

Which property of water explains why seawater contains over 20 different elements? select one:

Gnesinka [82]

I believe the awnser is a. universal solvent.

8 0

3 years ago

Read 2 more answers

What is the mechanism by which Ozone is formed in the presence of sunshine?

katovenus [111]

Stratospheric ozone is formed naturally through the interaction of UV radiation with molecular oxygen.

not sure if this is what you want but hope it helps!!!

8 0

3 years ago

Which kind of material would best accomplish this task ?

timofeeve [1]

What's the question ?

8 0

3 years ago

Read 2 more answers