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3 years ago

What did Rutherford's gold foil experiment tell about the atom?

1 answer:

3 0

The alpha particles that were fired at the gold foil were positively charged. ... These experiments led Rutherford to describe the atom as containing mostly empty space, with a very small, dense, positively charged nucleus at the center, which contained most of the mass of the atom, with the electrons orbiting the nucleus.

hope this helped

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Draw the orbital notation for the element Boron

Wewaii [24]

Answer:

I can't draw but you could draw 2 electrons in the first orbit and 3 electrons in the second orbit.

Explanation:

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3 years ago

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How is nucleic acids dna and rna are related to the process by which two cells com from on cell

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3 years ago

What is the edge length of a face-centered cubic unit cell that is made of of atoms, each with a radius of 154 pm

gladu [14]

Answer:

The edge length of a face-centered cubic unit cell is 435.6 pm.

Explanation:

In a face-centered cubic unit cell, each of the eight corners is occupied by one atom and each of the six faces is occupied by a single atom.

Hence, the number of atoms in an FCC unit cell is:

$8*\frac{1}{8} + 6*\frac{1}{2} = 4 atoms$

In a face-centered cubic unit cell, to find the edge length we need to use Pythagorean Theorem:

$a^{2} + a^{2} = (4R)^{2}$ (1)

Where:

a: is the edge length

R: is the radius of each atom = 154 pm

By solving equation (1) for "a" we have:

$a = 2R\sqrt{2} = 2*154 pm*\sqrt{2} = 435.6 pm$

Therefore, the edge length of a face-centered cubic unit cell is 435.6 pm.

I hope it helps you!

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4 years ago

Which property describes a mixture? It cannot be separated by physical methods. It has a single chemical composition. It cannot

natima [27]

Answer:

it cannot be separated by physical methods

Explanation:

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3 years ago

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Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.