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3 years ago

Which of the following is NOT true regarding Rutherford's Gold Foil experiment?

Chemistry

1 answer:

erastova [34]3 years ago

8 0

Answer:

The area around the nucleus must be of low mass.

Explanation:

Rutherford`s experiment showed that there are some positive charges in the center of the atoms, and because they are all together, they will give a great mass to the atom.

It was quite different from Thomson`s experiment, in which it was thought that the negative charges were mixed with the positive charges, around the atom (like a Pudding Model). In Rutherford`s experiment, because the direction of beta particles, it was the prediction of the positive nucleus.

Hope this info is useful.

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AysviL [449]

Answer:

Yes

Explanation:

Isotopes are atoms of the same element that have different numbers of neutrons in their nuclei. Everything else about them is the same.(If you want more explanation tell me).

5 0

1 year ago

Read 2 more answers

What involves 2 non metals ionic or covalent bonds

Juli2301 [7.4K]

Answer:

A bond forms between two non metals is covalent

Explanation:

For example Chlorine Cl is a non metal bond in Cl - Cl is covalent

8 0

2 years ago

Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.

Elenna [48]

The given concentration of boric acid = 0.0500 M

Required volume of the solution = 2 L

Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.

Calculating the moles of 0.0500 M boric acid present in 2 L solution:

$2 L * \frac{0.0500 mol B(OH)_{3} }{1 L} = 0.100 mol B(OH)_{3}$

Converting moles of boric acid to mass:

$0.100 mol B(OH)_{3} * \frac{61.83 g}{mol B(OH)_{3}} = 6.183 g$

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.

5 0

3 years ago

A 100.0 mL sample of 0.200 M HCl is mixed with a 100.0 mL sample of 0.205 M NaOH in a coffee cup calorimeter. If both solutions

Bas_tet [7]

I just need more points that's why I'm doing this

8 0

3 years ago

In the laboratory you dissolve 19.1 g of ammonium fluoride in a volumetric flask and add water to a total volume of 375 . mL. Wh

11111nata11111 [884]

Answer:

1.376 M

Explanation:

The following data were obtained from the question:

Mass of ammonium fluoride (NH₄F) = 19.1 g

Volume of solution = 375 mL

Molarity of ammonium fluoride (NH₄F) =?

Next, we shall convert 375 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

375 mL = 375 mL × 1 L / 1000 mL

375 mL = 0.375 L

Next, we shall determine the number of mole in 19.1 g of ammonium fluoride (NH₄F). This can be obtained as follow:

Mass of NH₄F = 19.1 g

Molar mass of NH₄F = 14 + (4×1) + 19

= 14 + 4 + 19

Molar mass of NH₄F = 37 g/mol

Mole of NH₄F =?

Mole = mass /Molar mass

Mole of NH₄F = 19.1 / 37

Mole of NH₄F = 0.516 mole

Finally, we shall determine the molarity of the solution. This can be obtained as follow:

Volume of solution = 0.375 L

Mole of NH₄F = 0.516 mole

Molarity of NH₄F =?

Molarity = mole /Volume

Molarity of NH₄F = 0.516 / 0.375

Molarity of NH₄F = 1.376 M

Therefore, the molarity of the ammonium fluoride (NH₄F) is 1.376 M

3 0

2 years ago