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andrew11 [14]
11 months ago
13

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

Chemistry
1 answer:
Lorico [155]11 months 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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Convert 732.0 mmHg to atm
enot [183]

Answer:

The answer is

<h2>0.95 atm</h2>

Explanation:

To solve the question we use the following conversion

That's

1 mmHg \cong 0.0013 atm

So we have

If 1 mmHg \cong 0.0013 atm

Then 732 mmHg will be

732 × 0.0013 atm

We have the final answer as

<h3>0.95 atm</h3>

Hope this helps you

5 0
3 years ago
Electrons involved in bonding between atoms are____.
mariarad [96]

Answer:

valence electrons

8 0
3 years ago
What is the “Island of Stability”?
3241004551 [841]

Answer:

The island of stability is a term from nuclear physics that describes the possibility of elements with particularly stable "magic numbers" of protons and neutrons. This would allow certain isotopes of some transuranic elements to be far more stable than others, that is, decay much more slowly.

Explanation:

7 0
2 years ago
If 50.0 grams of HCl were dissolved in enough water to make a solution with a total volume of 5.0 liters, what would be the mola
SVETLANKA909090 [29]

The molarity of a solution is found to be 10 M.

Explanation:

  • Molarity of a solution = <u>moles solute</u>

                                        Litres solution

  • Therefore,M =<u> 50</u>

                        5

  • Molarity of a solution  = 10 M
5 0
2 years ago
When a 12.8 g sample of KCL dissolves in 75.0 g of water in a calorimeter the temp. drops from 31 Celsius to 21.6 Celsius. Calcu
Delicious77 [7]

Answer:

Step 1: Calculate qsur (the surrounding is

usually the water)

qsur = ? J

m = 75.0 g water

c = 4.184 J/g

oC

ΔT = (Tfinal- Tinitial)= (21.6 – 31.0) = -9.4 oC

qsur = m · c · (ΔT)

qsur = (75.0g) (4.184 J/g

oC) (-9.4 oC)

qsur = - 2949.72 J

First, using the information we know that we

must solve for qsur, which is the water. We know

the mass for water, 75.0g, the specific heat of

the water, 4.184 j/g

o

c, and the change in

temperature, 21.6-31.0 = -9.4 oC. Plugging it

into the equation, we solve for qsur.

Step 2: Calculate qsys qsys = - (qsur)

qsys = - (- 2949.72 J)

qsys = + 2949.72

In this case, the qsur is negative, which means

that the water lost energy. Where did it go? It

went to the system. Thus, the energy of the

system is negative, opposite, the energy of the

surrounding.

Step 3: Calculate moles of the substance

that is the system

Given: 12.8 g KCl

Mol system = (g system given)

(molar mass of system)

Mol system = (12.8 g KCl)

(39.10g + 35.45g)

Mol system = 12.8 g KCl

74.55 g

Mol system = 0.172

Here, we solve for the mol in the system by

using the molar mass of the material in the

system.

Step 4: Calculate ΔH ΔH = q sys .

Mol system

ΔH= + 2949.72 J

0.172 mol

ΔH= +17179.81 J/mol or +1.72 x 104

J/mol

i hope this helps

7 0
3 years ago
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