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WINSTONCH [101]
2 years ago
10

PLEASE HELP!!!! Draw a representation of John Dalton’s atomic model.

Chemistry
1 answer:
larisa [96]2 years ago
7 0

Answer:

John Dalton's Atomic Model Below ⬇

Explanation:

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PLEASE HELP MY TEACHER IS GOING TO SEE ME AFTER CLASS
Crazy boy [7]

Answer:

The Nucelous helps prevent bad cells in

Explanation:

6 0
2 years ago
32. List three examples of substances. Explain why each<br> is a substance. (3.1)
Alex

Answer: Chemical X H3 and f1

Explanation:

8 0
2 years ago
What is the wavenumber of the radiation emitted when a hydrogen
LUCKY_DIMON [66]

Answer: Wavenumber of the radiation emitted  is 0.08\times 10^{8}m^{-1}

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

E=\frac{hc}{\lambda}

where,

E = energy of the radiation = 1.634\times 10^{-18}J

h = Planck's constant  = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength of radiation = ?

Putting values in above equation, we get:

1.634\times 10^{-18}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}\\\\\lambda=12.16\times 10^{-8}m

\bar {\nu}=\frac{1}{\lambda}=\frac{1}{12.16\times 10^{-8}}=0.08\times 10^{8}m^{-1}

Thus wavenumber of the radiation emitted  is 0.08\times 10^{8}m^{-1}

8 0
3 years ago
Why do we use a double arrow in the dissociation equation for a weak acid answer?
Zepler [3.9K]
Answer is: because weak acids do not dissociate completely.

The strength of an Arrhenius acid determines percentage of ionization of acid and the number of H⁺ ions formed. <span>
Strong acids completely ionize in water and give large amount ofhydrogen ions (H</span>⁺), so we use only one arrow, because reaction goes in one direction and there no molecules of acid in solution.

For example hydrochloric acid: HCl(aq) → H⁺(aq) + Cl⁻(aq).

<span> Weak acid partially ionize in water and give only a few hydrogen ions (H</span>⁺), in the solution there molecules of acid and ions.

For example cyanide acid: HCN(aq) ⇄ H⁺(aq) + CN⁻(aq).


4 0
3 years ago
A certain first order reaction has a half-life of 54. 3 s. How long will it take (in s) for the reactant concentration to decrea
Maksim231197 [3]

Answer:

82.4 s

Explanation:

Find the NUMBEr of half lives...then multiply by 54.3

2.27 = 6.5 (1/2)^n

log (2.27/6.5) / log (1/2) = n = 1.52 half lives

1.52 * 54.3 = 82.4 s

8 0
1 year ago
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