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Verdich [7]
3 years ago
10

describe how the discoveries of scientist over time have contributed to our current understanding of atom

Chemistry
1 answer:
Olenka [21]3 years ago
5 0

Answer:

In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the most basic form of matter. He demonstrated that cathode rays could be deflected, or bent, by magnetic or electric fields, which indicated that cathode rays consist of charged particles (Figure  2.2.2 ). More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; these particles are now called electrons. Most relevant to the field of chemistry, Thomson found that the mass-to-charge ratio of cathode rays is independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.

Explanation:

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Describe what happened when sodium chloride (salt) was placed in water in the video What happens when stuff dissolves?"
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Answer:

-

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6 0
2 years ago
Convert the following Grams: 0.200 moles of H2S
ANEK [815]

Answer:

6.82 g H₂S

General Formulas and Concepts:

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

0.200 mol H₂S

<u>Step 2: Identify Conversions</u>

Molar Mass of H - 1.01 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

<u>Step 3: Convert</u>

  1. Set up:                    \displaystyle 0.200 \ mol \ H_2S(\frac{34.09 \ g \ H_2S}{1 \ mol \ H_2S})
  2. Multiply:                  \displaystyle 6.818 \ g \ H_2S

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

6.818 g H₂S ≈ 6.82 g H₂S

7 0
3 years ago
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