Question 1Ernest Rutherford completed his famous gold foil experiment in 1911. In this experiment, alpha particles were fired at a thin sheet of gold foil. He observed that most of the alpha particles passed straight through the gold foil unimpeded, but a small number of alpha particles were deflected. Which of the following conclusions about atomic structure were made from Rutherford’s gold foil experiment?

Answers

Answer 1
Answer:

The conclusions drawn by Ruthford after the experiment are that most of the atom is empty and that the nucleus of the atom is positively charged.

What is the atom?´

  • It is the particular mentor of the matter.
  • It is the element that makes up matter.

After the discovery of the atom, many scientists sought to understand this element more specifically, especially in relation to its composition. These discoveries were strengthened over time, and Rutherford was the one who established how the atom really is, with a positive charge in the nucleus and an electrosphere around it.

Complete question:

Ernest Rutherford completed his famous gold foil experiment in 1911. In this experiment, alpha particles were fired at a thin sheet of gold foil. He observed that most of the alpha particles passed straight through the gold foil unimpeded, but a small number of alpha particles were deflected. Which of the following conclusions about atomic structure were made from Rutherford’s gold foil experiment?

Most of the atom is empty.

The nucleus is positively charged.

The atom is a massive sphere.

The atom is indivisible.

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Answer 2
Answer:

Answer: gold foil. It's in the experiment's name


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The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K. A sample of C8H18 is placed in a closed, evacuated 537 mL container at a temperature of 339 K. It is found that all of the C8H18 is in the vapor phase and that the pressure is 68.0 mm Hg. If the volume of the container is reduced to 338 mL at constant temperature, which of the following statements are correct?a. No condensation will occur.b. Some of the vapor initially present will condense.c. The pressure in the container will be 100. mm Hg.d. Only octane vapor will be present.e. Liquid octane will be present.
Calculate the mass percent of oxygen in KMnO4.
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By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask and diluted to the calibration mark. Determine the molarity of the resulting solution. A stock solution of potassium permanganate (KMnO4) was prepared by dissolving 13.0g KMnO4 with DI H2O in a 100.00-mL volumetric flask and diluting to the calibration mark. Determine the molarity of the solution Molarity= O.822 M

Combustion of Solid Fuel. A fuel analyzes 74.0 wt % C and 12.0% ash (inert). Air is added to burn the fuel, producing a flue gas of 12.4% CO2, 1.2% CO, 5.7% O2, and 80.7% N2. Calculate the kg of fuel used for 100 kg mol of outlet flue gas and the kg mol of air used. (Hint: First calculate the mol O2 added in the air, using the fact that the N2 in the flue gas equals the N2 added in the air. Then make a carbon balance to obtain the total moles of C added.)

Answers

Answer:

Kg of fuel used = 220.54Kg

Explanation:

The concept of material balance is applied here by taking into consideration the percentage composition of each flue gas in the atmosphere. This is used with each of the percentage composition of the flue gas that burn in the fuel and the detailed steps is as shown in the attachment.

What do lactate dehydrogenase, aspartate aminotransfcrase, and creatine kinase all have in common? a. they all are allosteric enzymes b. they are all zymogens c, they are all used to diagnose medical conditions d. they all function at abeornally high temperatures

Answers

Answer:

c. they are all used to diagnose medical conditions

Explanation:

Lactate dehydrogenase, aspartate aminotransfcrase, and creatine kinase all are used to diagnose medical conditions.

A gummie bear was tested through a flame-calorimeter test. the bear had a mass of 1.850 grams and the temperature of 100.0 milliliters of water increased by 15.0 degrees celsius. how many calories were in the gummie bear? show all of your calculations.

Answers

Answer:- 1500 calories

Solution:- mass of bear = 1.850 g

volume of water = 100.0 mL

Density of water is 1.00 g/moL. So, mass of water would be 100.0 g.

delta T for water = 15.0 degree C

specific heat capacity for water is 1 cal/(g* degree C)

q = m x c x delta T

where, q is the heat energy, m is mass, c is specific heat capacity and delta T is change in temperature.

for water, q = 100.0 x 1 x 15.0

q = 1500 calorie

heat gained by water = heat lost by bear

So, the 1.850 g bear has 1500 cal or 1.50 Cal.

(Where, 1 Cal = 1000 cal)

Which of the following statements is true about exothermic reactions?

Answers

Answer:

An Exothermic Reaction , gives off more heat, and a little energy to its surroundings.

this can helps us figure out that the answer is , C, More heat is given off into its products.

Explanation:

Joan has four containers. The chart below shows the mass and volume of each of the containers. Two of the containers are filled with solids, one is filled with a liquid, and one is filled with a gas.

Answers

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, it should be noted that if the containers are compared with an equal average volume, the containers having solids will have larger masses than that containing liquid which will also have a larger mass than that containing gas. This is because solids have there molecules touching each other in compact manner which makes the molecule exert a certain combined force/mass. The molecules of liquid are also close to one another but are not compact like the solids and are hence exerting a lesser force/mass than solids. Gases have free molecules that are far apart and thus are usually the lightest when they occupy the same volume as liquids and solids.

how does the presence of a strong electrolyte in solution affect the colligative properties of a solution when compared to the same number of moles of a nonelectrolyte solute?

Answers

Answer:

  • lowered vapor pressure
  • higher boiling point

Explanation:

The colligative properties of a given solution can be defined as the properties of that solution that are dependent on the concentration of the molecules or ions of the solute in the solution, and not on the type or identity of that solute. Examples include:

1. vapor pressure lowering

2. boiling point elevation

3. freezing point depression

4. Osmotic pressure

In this case, vapor pressure would be lowered because with an electrolyte introduced into a solution, the number of solute particles would be larger because the solute particles dissociate into ions, thereby competing with the solvent molecules at the surface of the solution, which in turn reduces the rate at which the solvent evaporates and condenses. Vapor pressure is lower compared to a solution with the same number of moles of  nonelectrolyte solute.

The higher the number of ions in the solution, the greater the colligative properties of the solution will be impacted.

Final answer:

The presence of a strong electrolyte in solution affects the colligative properties differently than a non-electrolyte solute. Strong electrolytes dissociate into ions, increasing the number of particles in solution. This affects colligative properties such as vapor pressure, boiling point elevation, freezing point depression, and osmotic pressure.

Explanation:

The presence of a strong electrolyte in solution affects the colligative properties differently than the same number of moles of a non-electrolyte solute. This is because strong electrolytes dissociate into ions when dissolved in solution, while non-electrolytes do not. The dissociation of strong electrolytes increases the total number of particles in solution, which affects colligative properties such as vapor pressure, boiling point elevation, freezing point depression, and osmotic pressure.

For example, let's compare a solution of 1 mole of sodium chloride (NaCl) to a solution of 1 mole of sucrose (C12H22O11). The sodium chloride will dissociate into Na+ and Cl- ions, which means there are now 2 particles in solution (1 Na+ and 1 Cl-) instead of just 1 molecule of sucrose. This higher particle concentration will result in a greater depression of the freezing point and elevation of the boiling point compared to the sucrose solution.

In summary, the presence of a strong electrolyte increases the number of particles in solution, leading to greater deviations in colligative properties compared to the same number of moles of a non-electrolyte solute.

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