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weqwewe [10]
3 years ago
6

Which of the following conclusions can be drawn from JJ Thomson's cathode ray experiments?A.) Atoms contain electronsB.) Practic

ally all the mass of an atom is contained in its nucleusC.) Atoms contain protons, neutrons, and electronsD.) Atoms have positively charged nuclues surrounded by electron cloud
Chemistry
1 answer:
kari74 [83]3 years ago
3 0

Answer:

A.) Atoms contain electrons

Explanation:

J.J. Thomson conducted an experiment in which he took a gas at low pressure in a discharge tube and applied high voltage current. When he did so, he noticed that there are some particles emitting from cathode going towards the anode. <u>Then he concluded that they are negatively charged particles and coined them electrons.</u>

<u>Hence, out of the options:- A.) Atoms contain electrons is correct.</u>

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3 years ago
Which of the following elements have properties similar to those of astatine?
vekshin1
<h3><u>Answer;</u></h3>

F, Br, l

<h3><u>Explanation;</u></h3>
  • Astatine is a member of the halogen family, elements in Group 17 (VIIA) of the periodic table.
  • It is a highly radioactive element and it is the heaviest known halogen. Astatine is similar to the elements above it in Group 17 (VIIA) of the periodic table, especially iodine.
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3 years ago
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If air contains 78% Nitrogen by volume, what is the solubility of nitrogen in water at 25 degree celisus​
Svet_ta [14]

The solubility of nitrogen in water at 25 °C= 4.88 x 10⁻⁴ mol/L

<h3>Further explanation</h3>

Given

78% Nitrogen by volume

Required

The solubility of nitrogen in water

Solution

Henry's Law states that the solubility of a gas is proportional to its partial pressure  

Can be formulated  

S = kH. P.  

S = gas solubility, mol / L  

kH = Henry constant, mol / L.atm  

P = partial gas pressure  

In the standard 25 C state, the air pressure is considered to be 1 atm, so the partial pressure of N₂ -nitrogen becomes:

Vn / Vtot = Pn / Ptot

78/100 = Pn / 1

Pn = 0.78 atm

Henry constant for N₂ at 25 °c = 1600 atm/mol.L=6.25.10⁻⁴ mol/L.atm

The solubility :

\tt S=6.25.10^{-4}\times 0.78\\\\S=4.88\times 10^{-4}~mol/L

7 0
2 years ago
Part A. Two containers, one at 305 K and the other at 295 K, are placed in contact with each other. 1. 1 J of heat flows from th
posledela

Answer:

0.00011 JK.

The process does NOT violate the second law of thermodynamics

Explanation:

The following parameters are given which are going to help in solving for the change in entropy of the system. The term "entropy'' simply means the degree of disorderliness of a system.

=> The temperature of container A = 305 K, the temperature of container B = 295 K and the amount of heat generated when the containers are placed in contact with each other = 1. 1 J.

The change in entropy of the hot container = -(1/305) = - 0.00328 J/K.

The change in entropy of the cold container = 1/295 = 0.00339 J/K.

Therefore, the change in the entropy of the system = - 0.00328 J/K + 0.00339 J/K = 0.00011 JK.

Note that the change in entropy of the system gives a positive value. Hence, this process does not violate the second law of thermodynamics.

The process does NOT violate the second law of thermodynamics.

7 0
3 years ago
When we experience higher than normal low tides and lower than normal high tides we are experiencing?
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Answer: a full or new moon

Explanation:

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