Ernest Rutherford
J. J Thomson
Explanation:
<u>Ernest Rutherford</u>
In 1911, Ernest Rutherford, a New Zealand chemist performed the gold foil experiment where he gave the modelling of the atom a boost.
Experiment
In his experiment, he bombarded a thin gold foil with alpha particles generated from a radioactive source. He found that most of the alpha particles passed through the gold foil while a few of them were deflected back.
Discovery and reflection on the atomic theory
To account for his observation, Rutherford suggested an atomic model in which an atom has small positively charged center where nearly all the mass is concentrated.
<u>J. J Thomson</u>
Experiment
In 1897 J.J Thomson performed experiments using the gas discharge tube that led to the discovery of the electrons. He called them cathode rays because they originate from the cathode and exits at the anode.
Discovery and reflection on the atomic theory
From his experiment on the gas discharge tube, Thomson was able determine the properties of cathode rays some of which are:
- they move in a straight line
- they possess kinetic energy
- they attract positive charges and repels negative charges
Using his observation, he proposed the plum pudding model of the atom where it is made up of entirely electrons.
learn more:
Rutherford brainly.com/question/1859083
#learnwithBrainly
Answer:
Option C= A hydrogen bond formed between a polar side chain and a hydrophobic side chain.
Explanation:
All three given options a, b and d have common mechanism to accommodate the polar amino acid.
A= A hydrogen bond forms between two polar side chains.
B= A hydrogen bond from between a polar side chain and protein back bone.
D = hydrogen bond form between polar side chains and a buried water molecules.
All these are use to accommodate the polar amino acid.
While option C is not used. which is:
A hydrogen bond formed between a polar side chain and a hydrophobic side chain.
Answer:
see explanation below
Explanation:
First to all, this is a redox reaction, and the reaction taking place is the following:
2KMnO4 + 3H2SO4 + 5H2O2 -----> 2MnSO4 + K2SO4 + 8H2O + 5O2
According to this reaction, we can see that the mole ratio between the peroxide and the permangante is 5:2. Therefore, if the titration required 21.3 mL to reach the equivalence point, then, the moles would be:
MhVh = MpVp
h would be the hydrogen peroxide, and p the permanganate.
But like it was stated before, the mole ratio is 5:2 so:
5MhVh = 2MpVp
Replacing moles:
5nh = 2MpVp
Now, we just have to replace the given data:
nh = 2MpVp/5
nh = 2 * 1.68 * 0.0213 / 5
nh = 0.0143 moles
Now to get the mass, we just need the molecular mass of the peroxide:
MM = 2*1 + 2*16 = 34 g/mol
Finally the mass:
m = 0.0143 * 34
m = 0.4862 g
Answer:
There are 1.4754246675000002e+24 atoms of Hydrogen within the measurement of 2.45 moles of hydrogen!
Explanation:
Water would have a much lower boiling point much like its other hydrides and it would loss its ability to dissolve polar substances plus it couldn't form water columns so no more cohesion between water molecules
