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Bess [88]
3 years ago
12

Which of the following scientists came up with the first widely recognized atomic theory?. . A. John Dalton. . B. Antione Lavois

ier. . C. Robert Millikan. . D. J.J. Thomson.
Chemistry
2 answers:
Yuri [45]3 years ago
8 0
"John Dalton" is the one scientist among the following choices given in the question that <span>came up with the first widely recognized atomic theory. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that has come to your help.</span>
Vedmedyk [2.9K]3 years ago
3 0

Answer:

John Dalton

Explanation:

done the topic

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Meiosis goes through 2 divisions to create four unique haploid cells. true or false
Aleksandr [31]

Answer:

True

Explanation:

True is the ANSWER.

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2 years ago
Suppose you held a lighted match to a solid hunk of wood and another match to a pile of wood shavings. Which form of wood will c
Reptile [31]

Answer:

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Explanation:

0 0
2 years ago
Question 9 (1 point)
Juliette [100K]

Noble gases

Explanation:

     Electronic configuration 1s²  2s²   2p⁶

 The element belongs to the group of the noble gases.

  • The noble gases have complete outer shell configuration of their atoms.
  • we can infer that the configuration above is for an element in the p-block because the last sub-level filled is the p-orbital.
  • The elements therefore belongs to the p-block
  • The block is from group 111A to O
  • Only the halogens and noble gases fits this picture from the option.
  • The outer most p-subshell have three orbitals requiring 6 electrons to fill them up.
  • This makes a complete and stable configuration.
  • The highest energy level of 2 is also made up of 8 electrons, an octet.
  • This is why we can conclude that they are noble gases.

Learn more:

Noble gas brainly.com/question/1781595

#learnwithBrainly

7 0
2 years ago
Determine the distance of closest approach (in fm) before the alpha particle reverses direction. Assume the lead nucleus remains
kiruha [24]

Answer:

Take E(alpha particle energy) = 5.5 MeV (5.5x106x1.6x10-19)

If the charge on the lead nucleus is +82e(atomic number of lead is 82) = +82x1.6x10-19 C and the charge on the alpha particle is +2e = 2x1.6x10-19 C

Using dc = (1/4πεo)qQ/Eα we have

dc = [9x10^9x(2x1.6x10-19x82x1.6x10-19)]/5.5x10-13 = 6.67x10^-13m. = 6.67 x 10^-13 x 10^15 = 6.67 x 10^2fm

Note: 1meter = 10^15fentometer

Explanation:

This is well inside the atom but some eight nuclear diameters from the centre of the lead nucleus.

7 0
3 years ago
What best describes a solution ?
topjm [15]
A solution is the answer to a problem
8 0
3 years ago
Read 2 more answers
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