Rutherford’s gold foil experiment gave evidence that an atom is mostly empty space. true false

Chemistry

1 answer:

Sholpan [36]2 years ago

4 0

The given statement is true .

<h3>What is Rutherford’s gold foil experiment?</h3>

A piece of gold foil was hit with alpha particles, which have a favorable charge. Most alpha particles went right around. This showed that the gold particles were mostly space.

The Rutherford gold leaf investigation supposed that most (99%) of all the mass of an atom is in the middle of the atom, that the nucleus is very small (105 times small than the length of the atom) and that is positively captured.

For the distribution experiment, Rutherford enjoyed a metal sheet that could be as thin as practicable. Gold is the most malleable of all known metals. It can easily be converted into very thin sheets. Hence, Rutherford established a gold foil for his alpha-ray scattering experimentation.

To learn more about Rutherford’s gold foil experiment, refer to:

brainly.com/question/4113533

#SPJ4

You might be interested in

Identify each number that is given is scientific notation

Neko [114]

Answer:

Need more information.

Explanation:

I believe you either left out info or explained the question confusingly.

5 0

3 years ago

How many total moles of ions are released when the following sample dissolves completely in water? Enter your answer in scientif

Katena32 [7]

<u>Answer:</u> The number of moles of strontium bicarbonate is $7.5\times 10^{-9}mol$

<u>Explanation:</u>

Formula units are defined as lowest whole number ratio of ions in an ionic compound. It is calculate by multiplying the number of moles by Avogadro's number which is $6.022\times 10^{23}$

We are given:

Number of formula units of $Sr(HCO_3)_2=4.55\times 10^{15}$

As, $6.022\times 10^{23}$ number of formula units are contained in 1 mole of a substance.

So, $4.55\times 10^{15}$ number of formula units will be contained in = $\frac{1}{6.022\times 10^{23}}\times 4.55\times 10^{15}=7.5\times 10^{-9}mol$ of strontium bicarbonate.