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Anika [276]
3 years ago
10

Who was the first person to discover the existence of electrons​

Chemistry
2 answers:
nignag [31]3 years ago
6 0

Answer:

JJ Thomson was the first person to discover the existence of electrons. Although Julius plucker in 1859 observed electrons he couldn't specify his observations instead Thomson was the one who further studied plucker observations and specified it as electrons..

k0ka [10]3 years ago
4 0
Electron was discovered by J. J. Thomson in 1897 when he was studying the properties of cathode ray.
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Energy released from cell respiration is know as ATP
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3 years ago
If one mole of a substance has a mass of 56.0 g, what is the mass of 11 nanomoles of the substance? Express your answer in nanog
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Answer:

616,0 ng is the right answer.

Explanation:

You should know that 1 mole = 1 .10^9 nanomoles

Get the rule of three.

1 .10^9 nanomoles ...................... 56.0 gr

11 nanomoles .....................

(11 x 56) / 1 .10^9 nanomoles = 6.16 x 10^-7 gr

Let's convert

6.16 x 10^-7 gr x 1 .10^9 = 616 ngr

8 0
3 years ago
Which of the following mixture types is characterized by the settling of particles?
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The answer is B. Suspension. Suspension mixtures are composed of two or more materials mixed together wherein the solute particles are usually larger than those found in a solution or colloid. In cases of solid-fluid suspension mixtures, the solid solute particles tend to settle at the bottom of the mixture after some time.
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3 years ago
When Mendeleev started his work on organizing the elements, about how many had
Talja [164]

Answer:

About 60.

Explanation:

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8 0
3 years ago
What is the maximum mass of P2I4 that can be prepared from 8.80g of P4O6 and 12.37g of iodine according to the reaction:
Dvinal [7]
The balanced chemical reaction would be as follows:

<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10

We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:

8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2

Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:

0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
8 0
3 years ago
Read 2 more answers
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