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3 years ago

What did J. J. Thomson observe when he applied electric voltage to a cathode ray tube in his famous experiment?

Chemistry

2 answers:

svet-max [94.6K]3 years ago

8 0

Explanation:

J.J. Thomson conducted an experiment in which he took a gas at low pressure in a discharge tube and applied high voltage current. When he did so, he noticed that there are some particles emitting from cathode going towards the anode. Then he concluded that they are negatively charged particles and coined them electrons.

irinina [24]3 years ago

7 0

QUICK ANSWER

J.J. Thomson's cathode ray experiment was a set of three experiments that assisted in discovering electrons. He did this using a cathode ray tube or CRT. It is a vacuum sealed tube with a cathode and anode on one side.

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Formic acid, which is a component of insect venom, has a ka = 1.8 ´ 10-4. what is the [h3o+] in a solution that is initially 0.1

scoray [572]

Answer is: concentratio of H₃O⁺ ions is 4.2·10⁻³ M.
Chemical reaction: HCOOH(aq) + H₂O(l) ⇄ HCOO⁻(aq) + H₃O⁺(aq).
c(HCOOH) = 0,1 M.
[H₃O⁺] = [HCOO⁻] = x.
[HCOOH] = 0,1 M - x.
Ka = [H₃O⁺] · [HCOO⁻] / [HCOOH].
0,00018 = x² / (0,1 M - x).
Solve quadratic equation: x = [H₃O⁺] = 0,0042 M.

7 0

3 years ago

A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical

Katarina [22]

Answer:

empirical formula: $H_2SO_4$

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02 : 1.02/1.02 : 4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: $H_2SO_4$ .

7 0

3 years ago

Why aren't descriptive investigations repeatable?

Bad White [126]

In descriptive investigations, we still haven't formed any hypothesis yet so we seek information by asking question.

It's not repeatable because repeating the questions over and over again without any clue about what we want to seek is completely waste of time.

Hope this helps xox :)

8 0

3 years ago

Read 2 more answers

Explain the relationship between an individual and a Society

Rufina [12.5K]

Answer:

the relationship between an individual and a society is society doesn't exist independently without an individual .the individual lives and acts within society but society is nothing ,in spite of the combination of individual s for cooperation effort.

6 0

3 years ago

Acid & Base Calculations Calculate the hydronium ion concentration for each. Tell whether it is an acid or a base. 1. pH = 5

Anuta_ua [19.1K]

Explanation:

pH is the negative logarithm of hydronium ion concentration present in a solution.

If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9
If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
The solution having pH equal to 7 is termed as neutral solution.

To calculate the pH of the solution, we use equation:

$pH=-\log[H_3O^+]$ ......(1)