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ella [17]
3 years ago
14

Describe what Rutherford would have seen in his gold foil experiment if J.J Thompson's Plum Pudding model was completly accurate

Chemistry
1 answer:
nata0808 [166]3 years ago
6 0
Answer is: Rutherford demonstrate that <span>J.J Thompson's Plum Pudding model was not accurate.

</span><span>Rutherford theorized that atoms have their charge concentrated in a very small nucleus.
</span>This was famous Rutherford's Gold Foil Experiment: he bombarded thin foil of gold with positive alpha particles (helium atom particles, consist of two protons and two neutrons).
Rutherford observed the deflection of alpha particles on the photographic film and notice that most of alpha particles passed straight through foil.
That is different from Plum Pudding model<span>, because it shows that most of the atom is empty space.</span><span>

</span>
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Alkaline earth metals electronegativity
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The radius of a circle is 7.31. How should the circle area be reported?
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I'm not sure but it should be 7.31 times 2 times 3.14 or pi

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If you need to round to the nearest tenth it will be 45.9

If you need to round to the nearest hundredth it will be 45.91

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A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica
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Answer: The empirical formula for the given compound is CH_5N

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.

For Carbon = \frac{3.23}{3.23}=1

For Hydrogen  = \frac{16.2}{3.23}=5.01\approx 5

For Oxygen  = \frac{3.24}{3.23}=1.00\approx 1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is C_1H_5N_1=CH_5N

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