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andrew11 [14]
1 year ago
13

What was John Dalton's contribution to the development of the atomic theory?

Chemistry
1 answer:
Lorico [155]1 year ago
6 0

John Dalton made some hypothesis about the structure of atom. He proposed that matter is composed of great number of indivisible particles called atoms they can neither be destroyed nor be created.

<h3>What is atomic theory?</h3>

There are different theories regarding the structure and electronic properties of an atom. Many scientists contributed to the modern theory of atomic structure in which John Dalton was first to mention the word atom.

According to Dalton' theory, matter is composed of indivisible particles called atoms. Atoms can neither be created nor be destroyed. All the atoms of the same element are identical in all aspects.

Atoms of different elements are different and the compounds are formed by the combination of atoms.  Dalton's theory provided a sound basis for the laws of chemical combination and also several properties of gases and liquids known at that time.

However, he could not explain the reason for chemical combination of atoms and did not give any idea about the existence of isotopes and isobars.

Hence, the main aspects of Dalton's theory was the indivisibility of atoms and the chances of chemical combination.

To learn more about Dalton's theory, find the link below:

brainly.com/question/11855975

#SPJ1

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3.0 M HI Unknown NaOH initial burette reading 0.2 ml 0.7 ml final burette reading 47.6 ml 37.5 ml What is the concentration of t
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This is a neutralization, which means we are mixing a base with an acid until the mixture becomes neutral. We add more HI to the NaOH until the number of acid equivalents is equal to the number of base equivalents. We can calculate the acid equivalents using normality and volume, and the same with base equivalents.

A = acid equivalents = normality*volume (in the acid solution)

B = base equivalents = normality*volumen (in the base solution)

A = B

NA*VA = NB*VB

HI is an acid which releases only 1 acid equivalent per molecule, so its molarity is equal to its normality.

NaOH is a base which releases only 1 base equivalent per formula unit, so its molarity is equal to its normality.

MA*VA = MB*VB

We’re trying to find out NaOH molarity, which is equal to the NaOH normality.

MB = MA*VA/VB

DATA:

MA = 3.0 M

<span>The volumen of HI used can be calculated by subtracting the final volume of the burette to its initial volume (the final volume is smaller, as we have taken some volume away)</span>

VA = V0-Vf = 0.7 ml - 0.2 ml = 0.5 ml

VB = V0-Vf = 47.6 ml - 37.5 ml = 10.1 ml

MB = 3.0M*0.5M/10.1M = 0.149 M

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