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puteri [66]
3 years ago
6

Recall all the models you described in task 1. Think about the results each model would predict for Thomson’s experiment. Which

atomic models does Thomson’s experimental evidence support? Explain why these models are compatible with the experimental results.
Models in task one were: Dalton's atomic model, Thomson's atomic model, Rutherford's atomic model, and Bohr's atomic model.
Chemistry
2 answers:
svetlana [45]3 years ago
7 0

<u>Answer:</u>

The very early discovery of particles that were subatomic was proposed by Dalton, which was then negated by the postulates suggested by Thomson that proved Dalton's theory.

<u>Explanation:</u>

The structural arrangement of atoms were approved by Thomson and Rutherford using the technique of air condition in x-ray showing the arrangement of protons and neutrons within Centre.

A nucleus and the negative particles that we called electrons moving around which were also shown to be much lighter than the particles present in the centre.

padilas [110]3 years ago
4 0

Answer:

Thomson’s cathode ray tube experiment gives results that support the atomic models proposed by Thomson, Rutherford, and Bohr. In each of these models, the atom contains negatively charged electrons. These electrons can be released from the atom if there is a sufficiently strong electric field

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Answer:

0.1066 hours

Explanation:

A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.

t1/2 = ln2/k

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t1/2 = 0.1066 h

The half-life of the pesticide is 0.1066 hours.

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1. Find the molar mass of the compounds<br>a. K2Cro4​
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194.1903

Explanation:

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How many km are in 5.6mm? 5.6x103 5.6x10-6 5.6x10-3 5.6x106​
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Hey there :)

<em>Q</em><em>u</em><em>e</em><em>s</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em><em> </em><em>How many km are in 5.6mm? </em>

<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>3 </em>

<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>-6 </em>

<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>-3 </em>

<em>=</em><em>></em><em> </em><em>5.6x10</em><em>^</em><em>6</em>

<em>A</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em>-</em>

5.6 \times 10^{ - 6}

<em>E</em><em>x</em><em>p</em><em>l</em><em>a</em><em>n</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em><em>-</em>

By using the formula-

1millimeter =  \frac{1}{1000000}

As 1 with 6 zeros, we convert it into exponential form.

=  >  \frac{1}{10^{6} }

As this above value is fraction type, we can do the reciprocal, thus, the exponent gets a negative value.

=  > 10^{ - 6}

Now combine with given question.

=  > 5.6 \times 10^{ - 6}

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