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kotykmax [81]
2 years ago
8

Which atomic model was proposed as a result of J.J Thomsons work?

Chemistry
1 answer:
Dafna11 [192]2 years ago
6 0
<span>, J.J. Thomson discovered the electron by experimenting with a Crookes, or cathode ray, tube. </span>
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In the activity series, any metal on the list can be oxidized by the ions of elements_______it.
Aleonysh [2.5K]

Answer:

Below

Nitric acid

Lead

Sodium nitrate

Explanation:

The activity series is an arrangement of metals in order of decreasing reactivity. Metals that are higher up in the series displace metals that are lower in the series from dilute solutions. Hence, when the ion of a metal that is lower in the series reacts with a metal that is higher up in the series, the latter is oxidized.

Dilute acids dissolve metals above hydrogen in the activity series such as as zinc and lead. Platinum is much lower than sodium in the activity series hence platinum does not react with sodium nitrate.

3 0
2 years ago
A) State the Law of Conservation of Matter.​
Mariulka [41]

Answer:

matter can either be created nor destroyed but can be transmitted fromone state to another state

4 0
2 years ago
In 1860, Chemists could make which of the following statements about the known chemical elements?
LenKa [72]

Answer:

b. Some had similar properties

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2 years ago
Topics: Fundamentals of electrochemistry; Electrochemical cells
irina1246 [14]

Answer:

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2 years ago
In one sample of a compound of copper and oxygen, 3.12g of the compound contains 2.50g of copper and the remainder is oxygen. Ca
blsea [12.9K]

Answer:

% composition O = 19.9%

% composition Cu = 80.1%

Explanation:

Given data:

Total mass of compound = 3.12 g

Mass of copper = 2.50 g

Mass of oxygen = 3.12 - 2.50 = 0.62 g

% composition = ?

Solution:

Formula:

<em>% composition = ( mass of element/ total mass)×100</em>

% composition Cu = (2.50 g / 3.12 g)×100

% composition Cu = 0.80 ×100

% composition Cu = 80.1%

For oxygen:

<em>% composition = ( mass of element/ total mass)×100</em>

% composition O = (0.62 g / 3.12 g)×100

% composition O = 0.199 ×100

% composition O = 19.9%

5 0
2 years ago
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