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2 years ago

How did Ernest Ruth ford's discovery affect the model of the atom?

1 answer:

5 0

<span>By 1911 the components of the atom had been discovered. The atom consisted of subatomic particles called protons and electrons. However, it was not clear how these protons and electrons were arranged within the atom. J.J. Thomson suggested the"plum pudding" model. In this model the electrons and protons are uniformly mixed throughout the atom: Rutherford tested Thomson's hypothesis by devising his "gold foil" experiment. Rutherford reasoned that if Thomson's model was correct then the mass of the atom was spread out throughout the atom. Then, if he shot high velocity alpha particles (helium nuclei) at an atom then there would be very little to deflect the alpha particles. He decided to test this with a thin film of gold atoms. As expected, most alpha particles went right through the gold foil but to his amazement a few alpha particles rebounded almost directly backwards. These deflections were not consistent with Thomson's model. Rutherford was forced to discard the Plum Pudding model and reasoned that the only way the alpha particles could be deflected backwards was if most of the mass in an atom was concentrated in a nucleus. He thus developed the planetary model of the atom which put all the protons in the nucleus and the electrons orbited around the nucleus like planets around the sun. </span>

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When a sound source approaches you, the pitch you hear is

IgorC [24]

Higher <span>C. higher than when the source is stationary</span>

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3 years ago

Calculate the density of air at 100 Deg C and 1 bar abs. Use the Ideal Gas Law for your calculation and give answer in kg/m3. Us

madam [21]

Answer:

$d=0.92\frac{kg}{m^{3}}$

Explanation:

Using the Ideal Gas Law we have $PV=nRT$ and the number of moles n could be expressed as $n=\frac{m}{M}$ , where m is the mass and M is the molar mass.

Now, replacing the number of moles in the equation for the ideal gass law:

$PV=\frac{m}{M}RT$

If we pass the V to divide:

$P=\frac{m}{V}\frac{RT}{M}$

As the density is expressed as $d=\frac{m}{V}$ , we have:

$P=d\frac{RT}{M}$

Solving for the density:

$d=\frac{PM}{RT}$

Then we need to convert the units to the S.I.:

$T=100^{o}C+273.15$

$T=373.15K$

$P=1bar*\frac{0.98atm}{1bar}$

$P=0.98atm$

$M=28.9\frac{kg}{kmol}*\frac{1kmol}{1000mol}$

$M=0.0289\frac{kg}{mol}$

Finally we replace the values:

$d=\frac{(0.98atm)(0.0289\frac{kg}{mol})}{(0.082\frac{atm.L}{mol.K})(373.15K)}$

$d=9.2*10^{-4}\frac{kg}{L}$

$d=9.2*10^{-4}\frac{kg}{L}*\frac{1L}{0.001m^{3}}$

$d=0.92\frac{kg}{m^{3}}$

5 0

3 years ago

Using the Hoffman apparatus for electrolysis, a chemist decomposes 2.3 moles of water into its gaseous elements. How many grams

solong [7]

Answer:

2.318032g

Explanation:

-The electrolysis equation of water is written as below:

$2H_2O_{(l)}->2H_2_{(g)}+O_2_{(g)}$

-The mole ratio of Water to the hydrogen formed is 1:1, therefore 2.3 moles of hydrogen gas is produced.

-Hydrogen's molar mass is 1.00784 grams:

$Mass=moles\times molar \ mass\\\\\=2.3\times 1.00784\\\\=2.318032\ g$

Hence, 2.318032 grams of hydrogen is produced.

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3 years ago

Which list only includes terms that describe water?

Oxana [17]

D! It’s a pure substance and a compound

4 0

2 years ago

You are examining the DNA sequences that code for the enzyme phosphofructokinase in skinks and Komodo dragons. You notice that t

jek_recluse [69]

Answer: D. Mutation in coding sequences are more likely to be deleterious to the organism than mutations in noncoding sequences.

Explanation: It was not likely to be that the coding sequences are replicated more often. The only possible explanation is that the mutations in coding is more likely to be deleterious to the organism than mutations because it is in a non coding sequence.

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2 years ago