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3 years ago

Explain the process of how James Chadwick found the neutron.

Chemistry

1 answer:

Andrews [41]3 years ago

3 0

Answer:

Discovery of neutron

Explanation:

Until 1932, the atom was believed to be composed of a positively charged nucleus surrounded by negatively charged electrons, after successful discovery of electron by J.J. Thompson and proton by Ernest Rutherford.

In 1932, James Chadwick bombarded beryllium atoms with alpha particles.

As Chadwick had previously seen in experiment conducted by Frédéric and Irène Joliot-Curie, they had bombarded a sample of beryllium with alpha radiation.

Chadwick repeated the experiment and observed an unknown radiation, this showed no deflection what-so-ever.

He repeated this experiment by bombarding other materials, like helium, nitrogen and lithium,

Finally, Chadwick found that this radiation was composed of particles with a neutral electrical charge and had approximate mass of a proton.

This particle was then known as the neutron.

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Answer:

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Explanation:

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2 years ago

A student with an unknown code of Q3A conducts a qualitative analysis of her unknown. Her results indicated the formation of a y

insens350 [35]

Answer:

Grey precipitate implies the presence of silver ions

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Explanation:

Qualitative analysis provides us a quick method of identifying ions present in a sample by chemical reactions involving simple reagents. Precipitates having a unique colour is formed. The identity of ions in the sample is deduced from the colour of precipitate obtained when particular reagents are added.

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3 years ago

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curium

−

243

252

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2 years ago

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2 years ago

n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph

olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = 0.111M

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = 0.2501 g of acetic acid

Now, assuming density of solution as 1.00g/mL, 37.54mL weights 37.54g.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)

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2 years ago