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Sedaia [141]
2 years ago
12

What did Josef Loschmidt and Amedeo Avogadro Contribute to our understanding of basic molecular numbers, sizes, and reaction rat

ios.
Chemistry
1 answer:
mario62 [17]2 years ago
4 0

From Avogadro we obtained a physical constant of matter which is Avogadro's number, and from both scientists we understand that elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.

<h3>What is Avogadro's number?</h3>

Avogadro's number, or Avogadro's constant, is the number of particles found in one mole of a substance.

The Avogadro's number is given as 6.02 x 10²³.

Summary of Josef Loschmidt and Amedeo Avogadro Contribution to chemistry.

  • Equal volumes of gas contain equal numbers of molecules,
  • Elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.

Thus, from Avogadro we obtained a physical constant of matter which is Avogadro's number, and from both scientists we understand that elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.

Learn more about Avogadro's here: brainly.com/question/1581342

#SPJ1

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What are the 5 principles of kinetic molecular theory?
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8 0
3 years ago
How do you solve this ??
kotegsom [21]
Answer : Option A) 2.00 eV

Explanation : The conversion of J to eV is done with the following formula;

E_{eV} = E_{J} X (6.241 X 10^{18})

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So, on substituting we get,
E_{eV} = 3.2 X 10^{-19}  X  (6.241 X 10^{18} )


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3 0
3 years ago
5 Questions to answer please
ryzh [129]
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4. Chemical reactions follows the law of conservation of mass. Therefore mass of reactant = mass of product = 4 grams.
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5 0
2 years ago
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Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

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