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4 years ago

What did Ernest Rutherford s gold foil experiment demonstrate about atoms?

1 answer:

mote1985 [20]4 years ago

6 0

Rutherford performed gold foil experiment to understand that how negative and positive particles could Co exist in an atom. He bombarded alpha particles on a 0.00004 cm thick gold foil.

He proposed a planetary model of the atom and concluded following results and demonstrated that,
1. An atom produces a line spectrum.
2. An Electron revolves around the nucleus without any orbits.
3. Since most of the particles passed through the foil undeflected it means that most of the volume occupied by an atom is empty.
4. An Atom as a whole is neutral.
5. The deflection of few particles on the foil suggested that there is center of positive particles in an atom called the nucleus of the atom.
6. The complete rebounce of few particles on the gold foil suggested that the nucleus is very dense and hard.

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True

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3 years ago

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The Haber process can be used to produce ammonia (NH3) from hydrogen gas (H2) and nitrogen gas (N2). The balanced equation for t

kotykmax [81]

Answer:

Explanation:

We are given that ammonia can be produced from hydrogen gas and nitrogen gas according to the equation:

$\displaystyle 3\text{H$_2$} + \text{N$_2$} \longrightarrow 2\text{NH$_3$}$

We want to determine the mass of hydrogen gas that must have reacted if 0.575 g of NH₃ was produced.

To do so, we can convert from grams of NH₃ to moles of NH₃, moles of NH₃ to moles of H₂, and moles of H₂ to grams of H₂.

We are given that the molar masses of NH₃ and H₂ are 17.03 g/mol and 2.0158 g/mol, respectively.

From the equation, we can see that two moles of NH₃ is produced from every three moles of H₂.

With the initial value, perform dimensional analysis:

$\displaystyle \begin{aligned} 0.575\text{ g NH$_3$}& \cdot \frac{1\text{ mol NH$_3$}}{17.03\text{ g NH$_3$}} \cdot\frac{3\text{ mol H$_2$}}{2\text{ mol NH$_3$}} \cdot \frac{2.0158\text{ g H$_2$}}{1\text{ mol H$_2$}} \\ \\ & = 0.102\text{ g H$_2$}\end{aligned}$

*Assuming 100% efficiency.

Our final answer should have three significant figures. (The first term has three, the second term has four (the one is exact), the third term is exact, and the fourth term has five. Hence, the product should have only three.)

In conclusion, our answer is B.

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3 years ago

When NEM is added to a purified solution of creatine kinase, Cys 278 is alkylated, but no other Cys residues in the protein are

USPshnik [31]

Answer:

The answer is given below

Explanation:

Cys 278 residue is the only available cysteine which is alkylated by the addition of N-Ethylmaleimide or NEM (alkylating agent). It works by only alkylating the sulfhydryls. In this case, Cys 278 residue is the only one which has exposed cysteine residue.

While the other residues have their sulfhydryls group either involved in the synthesis of disulfide bonds of proteins or their Cys residues are intrinsically placed in the proteins and cannot be alkylated with NEM.

NEM cannot alkylate if its protein is not available in the free form or it is in bounded form. For NEM to alkylate Cys 278, it should be free and should have sulfhydryls available for alkylation.

Alkylation: it is the transfer of alkyl groups. Alkyl groups contain Hydrogen and Carbon in their structure.

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4 years ago

How is nitrogen converted in the nitrogen cycle

castortr0y [4]

Answer:

nitrogen fixing bacteria

Explanation:

These bacteria made the nitrogen suitable for plant use.

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3 years ago