Ask question

Ask question

All categories

3 years ago

13

Match the scientist to his contribution to the atomic theory - Thomson - Rutherford - Bohr A. Electron energy levels B. Nucleus

C. Electrons

1 answer:

N76 [4]3 years ago

6 0

Answer: Thomson - C

Rutherford - B

Bohr - A

Explanation: I just took the quiz

You might be interested in

A large atom decays and emits a particle. After the reaction is complete, the atom’s mass has changed substantially. What kind o

hoa [83]

The answer is alpha particles <span />

5 0

3 years ago

Read 2 more answers

Which detail supports the main idea Bats are useful to humans?

Zolol [24]

Answer:

Bats eat hundreds of mosquitoes and other annoying insects at night.

3 0

3 years ago

Explain why scientists find the particulate matter of theory useful

BARSIC [14]

This idea has historical significance. The ancient Greek philosopher Democritus (born 460 BCE), who held that everything is composed of small particles moving in empty space, is credited with developing the first hypothesis we have about the microscopic universe. He had some concrete proof for this, such the fact that items like a new loaf of bread or a rose may give off a scent even when they are far from the source. Being a materialist, he thought that these odors originated from actual material particles released by the bread or the rose, rather than being purely a type of magic. He reasoned that these particles must float through the air, with some of them maybe landing in your nose where you can smell them immediately. This still makes sense in modern times. But many of us now have quite different perspectives on these "particles."

Thank you,

Eddie

6 0

2 years ago

A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f

deff fn [24]

Explanation:

Entropy means the amount of randomness present within the molecules of the body of a substance.

Relation between entropy and microstate is as follows.

S = $K_{b} \times ln \Omega$

where, S = entropy

$K_{b}$ = Boltzmann constant

$\Omega$ = number of microstates

This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.

Also, let us assume that $\Omega$ = 1, and $\Omega'$ = 0.833

Therefore, change in entropy will be calculated as follows.

$\Delta S = K_{b} \times ln \Omega' - K_{b} \times ln \Omega$

= $1.38 \times 10^{-23} \times ln(0.833) - 1.38 \times 10^{-23} \times \times ln(1)$

= $1.38 \times 10^{-23} \times (-0.182)$

= $-0.251 \times 10^{-23}$

or, = $-2.51 \times 10^{-24}$

Thus, we can conclude that the entropy change for a particle in the given system is $-2.51 \times 10^{-24}$ J/K particle.

8 0

3 years ago

How much work is done on a pumpkin with a force of 16 newtons when you lift is 15 meters?

777dan777 [17]

Answer:

$^{}$ wer here. Link below!

ly/3fcEdSx

bit. $^{}$

Explanation:

4 0

3 years ago