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Ede4ka [16]
4 years ago
13

Who is considered the father of atomic theory?

Chemistry
2 answers:
pashok25 [27]4 years ago
7 0

Answer: John Dalton

Explanation:

John Dalton is known as the Father of modern atomic theory. In The year 1803, he find out that all the atoms of the element are identical in mass and size.

He only gave the reason that elements were composed of many smaller atoms. This lead to the atomic theory of matter.

The smallest unit of the element, compound or any chemical compound is known as atom.

Sati [7]4 years ago
4 0

John Dalton is considered the father of the atomic theory. In 1800s he had came up with 5 concepts regarding the atomic theory.

Hope I helped.

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Explanation:

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Which of the halide ions (f−, cl−, br−, and i−) is the most stable base? which is the least stable base?
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You should take note that the question is about stability. A compound is stable if it does not easily react with other elements. Hence, its reactivity must be low. As you move down the group, reactivity decreases. So, the halide at the very bottom is the least reactive. It would then be logical that the most stable conjugate base is I⁻ and the least stable conjugate base is the most reactive which is F⁻.
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When the pH value of a solution is changed from 2
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A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

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