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Mandarinka [93]
3 years ago
13

Match the scientist to his contribution to the atomic theory. Match Term Definition J.J. Thomson A) Positively charged nucleus

Chemistry
2 answers:
marissa [1.9K]3 years ago
6 0

Answer is:

J.J. Thomson- B) Plum Pudding Model.

J. J. Thomson discovered the electron in 1897.  

His "plum pudding" model (1904) suggested: the electrons are embedded in the positive charge.  

John Dalton - C) Atoms indivisible.

Dalton stated that atom is matter that can not be divided

Ernest Rutherford - A) Positively charged nucleus.

According to Rutherford model of the atom:

1) Atoms have their charge concentrated in a very small nucleus.

2) Major space in an atom is empty.

3) Atoms nucleus is surrounded by negatively charged particles called electrons.  

4) An atom is electrically neutral.

James Chadwick - D) Discovery of neutrons.

James Chadwick was a British physicist who was awarded the 1935 Nobel Prize in Physics for his discovery of the neutron in 1932.

pantera1 [17]3 years ago
5 0
J.J. Thomson :Plum Pudding Model

James Chadwick: Discovery of neutrons

John Dalton:Atoms indivisible

Ernest Rutherford:Positively charged nucleus


I hope this helps and have a great day!
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The atomic mass of Cu is 63.5. Find its electrochemical equivalent​
FrozenT [24]

Answer:

The electrochemical equivalent of copper, Cu, is 3.29015544 × 10⁻⁷ g/C

Explanation:

The given parameters are;

The element for which the electrochemical equivalent is sought = Copper

The atomic mass of copper = 63.5

The electrochemical equivalent, 'Z', of an element or a substance is the mass, 'm', of the element or substance deposited by one coulomb of electricity, which is equivalent to a 1 ampere current flowing for a period of 1 second

Mathematically, we have;

m = Z·I·t = Z·Q

We have;

Cu²⁺ (aq) + 2·e⁻ → Cu

Therefore, one mole of Cu, is deposited by 2 moles of electrons

The charge carried one mole of electrons = 1 Faraday = 96500 C

∴ The charge carried two moles of electrons, Q = 2 × 96500 C = 193,000 C

Given that the mass of an atom of Cu = 63.5 a.m.u., the mass of one mole of Cu, m = 63.5 g

Z = \dfrac{m}{Q} = \dfrac{63.5 \ g}{193,000 \ C} = 3.29015544 \times 10^{-4} \, g \cdot C^{-1}

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7 0
3 years ago
Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your
k0ka [10]
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.

<span>glucose-1-phosphate⟶glucose-6-phosphate          ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate          ΔG∘=−1.67 kJ/mol
</span>
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glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate 

In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

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Then, the equation to relate ΔG° to the equilibrium constant K is

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lnK = 2.2635
K = e^2.2635
K = 9.62


6 0
3 years ago
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