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madreJ [45]
3 years ago
12

How did the development of the earliest idea about atoms differ from the later work of scientists?

Chemistry
2 answers:
IgorLugansk [536]3 years ago
8 0
The earliest idea of the atom is that it is made up of tiny, invisible particles, called the atoms, is believed to have originated from the Greek philosopher Leucippus of Miletus and his student, Democritus of Abdera, around 5th century B.C. However, it wasn't until centuries later that the atomic theory was taken seriously by the society, thanks to the scientist John Dalton by 1808. Hence, the development of the Atomic Theory began until the modern age. Modern scientist has more technological advancements than the early age. Therefore, the modern scientist can view the atom more detailed than before.

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Igoryamba3 years ago
7 0

Answer:

The smallest unit of matter, which possesses the characteristics of an element is known as an atom. It comprises a dense core known as the nucleus and alignment of outer shells inhabited by orbiting electrons.  

The early concepts of the atom postulate that the atom is a solid or a hollow body comprising nothing within. However, the discoveries and work done by scientists afterward have proved that the atoms comprise the subatomic constituents, that is, the protons, electrons, and neutrons.  

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I NEED HELP PLEASE, THANKS! :)
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Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

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3 years ago
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