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mario62 [17]
2 years ago
15

2. In 1804, almost a century before the nucleus was discovered, the English scientist John Dalton provided evidence for the exis

tence of the atom. Dalton thought that atoms were the smallest particles of matter, which couldn't be divided into smaller particles. He modeled atoms with solid wooden balls. In 1897, another English scientist, named J. J. Thomson, discovered the electron. It was first subatomic particle to be identified. Because atoms are neutral in electric charge, Thomson assumed that atoms must also contain areas of positive charge to cancel out the negatively charged electrons. He thought that an atom was like a plum pudding, consisting mostly of positively charged matter with negative electrons scattered through it.
The nucleus of the atom was discovered next. It was discovered in 1911 by a scientist from New Zealand named Ernest Rutherford, who is pictured in Figure below. Through his clever research, Rutherford showed that the positive charge of an atom is confined to a tiny massive region at the center of the atom, rather than being spread evenly throughout the “pudding” of the atom as Thomson had suggested.

Question: Why did Thomson think that the atom also had to have a positive charge?

Question: What two particles are found in the nucleus?

3.Question: What particle did Rutherford discover in the nucleus?
4.Question: What did Rutherford predict was also in the nucleus?
5.Question: How did Rutherford describe the size of the nucleus?
6.Question: According to Rutherford, how was Thomson's model of the atom incorrect?
7.Question: How are all atoms of an element identical?
Chemistry
1 answer:
belka [17]2 years ago
3 0

Answer:

The Electron found by J.J Thompson

Explanation:

Hope this helps:)

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Number of moles of NaOH = V(NaOH) * M(NaOH)= 0.150 L * 0.1 moles/L = 0.015 moles

Number of moles of formic acid, HCOOH = V(HCOOH) * M(HCOOH) = 0.200 L * 0.1 moles/L = 0.020 moles

Here, the limiting reagent is NaOH

The reaction is represented as:

HCOOH + NaOH ↔HCOONa + H2O

Moles of HCOONa formed = Moles of the limiting reagent, NaOH = 0.015 moles

Moles of HCOOH remaining = 0.020-0.015 = 0.005 moles

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Therefore: [HCOOH] = 0.005 moles/1 L = 0.005 M

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As per Henderson-Hasselbalch equation

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Structure of the following three isomeric esters with chemical formula C₇H₁₂O₂

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