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Alborosie
3 years ago
8

List three of Rutherford’s major achievements that helped earn him the title “Father of Nuclear Physics.”

Chemistry
2 answers:
labwork [276]3 years ago
8 0
Discovered the necleus
proposed the proton existence
discovered fusion reaction from fission
Nikolay [14]3 years ago
6 0

The correct answer is:

discovered the necleus proposed the proton existence discovered fusion reaction from fission

Explanation:

Rutherford (1871–1937) was subject for an extraordinary array of discoveries in the fields of radioactivity and nuclear physics. He identified alpha and beta rays, set forth the rules of radioactive decay, and recognized alpha particles as helium nuclei.In 1911, Rutherford realized that atoms not detached, they are managed by something else, a nucleus. He noticed that the preponderance of an atom's mass is contemplated at a very tiny point in the center where the nucleus was determined. The nucleus is a positive force enclosed by electrons which are negative free flying particles.

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Naturally occurring silicon has an atomic mass of 28.086 and consists of three isotopes. The major isotope is 28Si, natural abun
Elden [556K]

Answer:

29Si has a natural abundance of 4.68%.

30Si has a relative atomic mass of 29.99288 and a natural abundance of 3.09%.

Explanation:

The atomic mass of silicon is given by:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

Where:

Si: atomic mass of silicon (28.086)

Si²⁸: relative atomic mass of 28Si (27.97693)

A₁: natural abundance of 28Si (92.23%)

Si²⁹: relative atomic mass of 29Si (28.97649)

A₂: natural abundance of 29Si

Si³⁰: relative atomic mass of 30Si

A₃: natural abundance of 30Si

We also know that 30Si natural abundance is in the ratio of 0.6592 to that of 29Si.

We have to set up a system of three equations in three unknowns:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

A₃=0.6592×A₂

A₁+A₂+A₃=1

First, we find substitute the value of A₃ in the third equation and solv teh value of A₂:

A₁+A₂+0.6592×A₂=1

A₁+1.6592×A₂=1

1.6592×A₂=1-A₁

A₂=\frac{1-A₁}{1.6592}=\frac{1-0.9223}{1.6592}=0.0468

Then, we find the value of A₃:

A₃=0.6592×A₂

A₃=0.6592×0.0468=0.0309

Finally, we find the value of Si³⁰ in the first equation:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

28.086=27.97693×0.9223+28.97649×0.0468+Si³⁰×0.0309

28.086=27.15922+Si³⁰×0.0309

28.086-27.15922=Si³⁰×0.0309

\frac{0.92678}{0.0309}=Si³⁰

Si³⁰=29.99288

8 0
3 years ago
Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.
Elenna [48]

The given concentration of boric acid = 0.0500 M

Required volume of the solution = 2 L

Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.

Calculating the moles of 0.0500 M boric acid present in 2 L solution:

2 L * \frac{0.0500 mol B(OH)_{3} }{1 L} = 0.100 mol B(OH)_{3}

Converting moles of boric acid to mass:

0.100 mol B(OH)_{3} * \frac{61.83 g}{mol B(OH)_{3}}   = 6.183 g

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.


5 0
3 years ago
Present value lets us compare dollar values from different time periods.
kramer
True? Not sure what the question is
5 0
3 years ago
Of the following: H2(g); He(g); CO2(g); which would behave least like an ideal gas? Why?
Margarita [4]
Answer:

CO2(g)

Because CO2 is the larges molecule with specific geometric, therefore it is not likely to behave as an ideal gas.
4 0
3 years ago
Which of the following are organic compounds?
KIM [24]
I think methane and propane
3 0
3 years ago
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