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arlik [135]
3 years ago
5

An experiment using alpha particles to bombard a thin sheet of gold foil indicated that most of the volume of the atoms in the f

oil is taken up by ...
Chemistry
2 answers:
mars1129 [50]3 years ago
6 0
Empty space. We now know that this is the gap between the electrons and nucleus.
antiseptic1488 [7]3 years ago
3 0

The experiment that used alpha particles to bombard a thin sheet of gold foil indicated that most of the volume of the atoms in the foil is taken up by \boxed{{\text{empty space}}}

Further Explanation:

Rutherford’s experiment:

Rutherford's model of atom is the classic model of the atom instead of having many limitations. He designed an experiment that used alpha particles emitted by radioactive elements as objects that can demonstrate the structure of the atom. Rutherford showed his own physical model for the subatomic structure as a result of the experimental observations.

The postulates of Rutherford’s model are as follows:

1. An atom consists of a positive charge in a very small volume. Most of its mass is concentrated in a very small region of the atom and this region was termed as the nucleus of the atom.

2. The nucleus of the atom is surrounded by the negatively charged particles which were called electron. These electrons were supposed to revolve around the atomic nucleus in a circular path at a very high speed. This circular path is called orbit.

3. A very strong electrostatic force of attraction holds together the negatively charged electrons revolving around the nucleus positively charged concentrated in the nucleus.

Limitations of Rutherford’s model:

1. It did not tell anything about the distribution of electrons in various orbits.

2. This model failed to explain Maxwell’s theory of electromagnetic radiation.

3. It was unable to explain the stability of an atom.

Therefore most of the volume of an atom is taken up by empty space in the atom.

Learn more:

1. The major contribution of Antoine Lavoisier to chemistry: brainly.com/question/2500879

2. Example of physical change: brainly.com/question/1119909

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Atomic Structure

Keywords: Rutherford, atom, volume, nucleus, orbit, postulates, limitations, Maxwell, electromagnetic radiations, distribution of electrons, stability of atom.

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Question 6
hodyreva [135]

Answer:

0.00688 moles

Explanation:

The molarity ratio looks like this:

Molarity = moles / volume (L)

After you convert mL to L, you can plug the values into the equation and simplify to find moles.

27.5 mL / 1,000 = 0.0275 L

Molarity = moles / volume                         <----- Molarity ratio

0.250 M = moles / 0.0275 L                      <----- Insert values

0.00688 = moles                                       <----- Multiply both sides by 0.0275

4 0
2 years ago
How many grams of hydrogen are in 46 g of CH40?<br>​
Romashka [77]

Answer:

5.8 grams of hydrogen

Explanation:

Your welcome

6 0
2 years ago
Read 2 more answers
What is the orbital hybridization of a central atom that has one lone pair and bonds to:
svlad2 [7]

sp^3d is the type of orbital hybridization of a central atom that has one lone pair and bonds to four other atoms.

<h3>What is orbital hybridization?</h3>

In the context of valence bond theory, orbital hybridization (or hybridisation) refers to the idea of combining atomic orbitals to create new hybrid orbitals (with energies, forms, etc., distinct from the component atomic orbitals) suited for the pairing of electrons to form chemical bonds.

For instance, the valence-shell s orbital joins with three valence-shell p orbitals to generate four equivalent sp3 mixes that are arranged in a tetrahedral configuration around the carbon atom to connect to four distinct atoms.

Hybrid orbitals are symmetrically arranged in space and are helpful in the explanation of molecular geometry and atomic bonding characteristics. Usually, atomic orbitals with similar energies are combined to form hybrid orbitals.

Learn more about hybridization

brainly.com/question/22765530

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5 0
1 year ago
A compound contains only change and n combustion of 35.0mg of the compound produces 33.5mg co2 and 41.1mg h2o. What is the empir
Viefleur [7K]

Answer:

The empirical formula is CH6N2

Explanation:

A compound containing only C, H, and N yields the following data. Complete combustion of 35.0 mg of the compound produced 33.5 mg of CO2 and 41.1 mg of H2O. What is the empirical formula of the compound

Step 1: Data given

Mass of the compound = 35.0 mg = 0.035 grams

Mass of CO2 = 33.5 mg = 0.0335 grams

Mass of H2O = 41.1 mg = 0.0411 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 0.0335 grams / 44.01 g/mol

Moles CO2 = 7.61 *10^-4 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 7.61 *10^-4 moles CO2 we have 7.61 *10^-4 moles C

Step 4: Calculate mass C

Mass C = 7.61 *10^-4 moles * 12.01 g/mol

Mass C = 0.00914 grams = 9.14 mg

Step 5: Calculate moles H2O

Moles H2O = 0.0411 grams / 18.02 g/mol

Moles H2O = 0.00228 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.00228 moles H2O we have 2* 0.00228 = 0.00456 moles H

Step 7: Calculate mass H

Mass H = 0.00456 moles * 1.01 g/mol

Mass H = 0.00461 grams = 4.61 mg

Step 8: Calculate mass N

Mass N = 35.0 mg - 9.14 - 4.61 = 21.25 mg = 0.02125 grams

Step 9: Calculate moles N

Moles N = 0.02125 grams / 14.0 g/mol

Moles N = 0.00152 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moes

C: 0.000761 moles / 0.000761 moles= 1

H:  0.00456 moles / 0.000761 moles = 6

N: 0.00152 moles  / 0.000761 moles = 2

For every C atom we have 6 H atoms and 2 N atoms

The empirical formula is CH6N2

5 0
2 years ago
What is the concentration of bromide, in ppm, if 12.41 g MgBr2 is dissolved in 2.55 L water.
pav-90 [236]

Answer:

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

Explanation:

ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.

In our case we have:

mass of MgBr₂ = 12.41 g

volume of water (which is equal to the final solution volume) = 2.55 L

Now we devise the following reasoning:

if         12.41 g of MgBr₂ are dissolved in 2.55 L of water

then         X g of MgBr₂ are dissolved in 1 L of water

X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂

if in         184 g (1 mole) of MgBr₂ we have 160 g of Br⁻

then in   4.867 g of MgBr₂ we have Y g of Br⁻

Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)

4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

7 0
3 years ago
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