All categories

3 years ago

Explain how the gold foil experiment was conducted

1 answer:

4 0

Explanation:

Physicist Ernest Rutherford established the nuclear theory of the atom with his gold-foil experiment. When he shot a beam of alpha particles at a sheet of gold foil, a few of the particles were deflected. He concluded that a tiny, dense nucleus was causing the deflections.

You might be interested in

Why does an atom of 23NA have 23 electrons

Julli [10]

It is actually something standardized more so than theoretical, however in terms of atoms in general the electrons are based on the amount of electronic shells that an atom has and the amount of electrons that atom can accommodate. In the case of sodium, it has three shells with 2 electrons on the first shell which is the maximum, 8 on the second shell which is also the max and 1 on their final shell
So simply put an ATOM of Sodium (Na) has 23 electrons because it has 23 protons which is a fact
<span />

7 0

3 years ago

5. Write word equations for the following reactions

Andrews [41]

Answer:

a) Sulphur + Oxygen → Sulphur dioxide

b) Carbon + Oxygen → Carbon dioxide

c) Sulphur + Iron → Iron sulphide

6 0

3 years ago

Which of the following is not true of alloys?

Scilla [17]

Option d: copper.

Because copper is an element, not a mixture.

7 0

4 years ago

Four grams of magnesium react with an excess of dilute hydrochloric acid. Calculate the volume in liters of hydrogen gas collect

Helga [31]

Sorry I am confuse 3

5 0

3 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is

Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

7 0

4 years ago