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Anna71 [15]
4 years ago
6

How did rutherford's experimental evidence lead to the development of a new atomic number

Chemistry
1 answer:
borishaifa [10]4 years ago
8 0
Rutherford's experiment was the gold foil experiment.

The gold foil experiment was him shooting alpha particles (you could think of this as a Helium atom without its electrons) into a gold foil. The whole experiment was surrounded with something called Zinc Sulfide which sparked when the alpha particles hit it.

Most of the alpha particles went through, approximately 1 in 8000 alpha particles deflected at a large angle (almost right back to where it was shot).

This constant ratio caused him to conclude that:-the atom was mostly empty space (since most alpha particles went through)-there was something very positive in the atom (the proton)-the proton was very dense (since it made something going light speed deflect back at a large angle)-The proton was also very small (since only 1 in 8000 hit it)

Prior to the discovery of the proton, John Dalton's periodic table was used. Having "elements" such as soda and potash. Now that we have discovered the proton and found out that each atom's number of protons is unique, we used that to classify each element's identity.
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castortr0y [4]
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3 years ago
The half-life of radioactive element krypton-91 is 10 seconds. if 16 grams of krypton-91 are initially present, how many grams a
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6 0
4 years ago
Drag each label to the correct location on the equation. Each label may be used more than once.
elena-14-01-66 [18.8K]
HCl is an acid, C5H5N is a base, Cl is an acid, and HC5H5N is an acid.

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8 0
2 years ago
What is the oh- in a solution with a poh of 5.71
Rudik [331]

Answer:- The hydroxide ion concentration of the solution is 1.95*10^-^6 .

Solution:- The formula used to calculate pOH from hydroxide ion is:

pOH=-log[OH^-]

When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:

[OH^-]=10^-^p^O^H

pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:

[OH^-]=10^-^5^.^7^1

[OH^-] = 0.00000195 or 1.95*10^-^6

So, the hydroxide ion concentration of the solution is 1.95*10^-^6 .



3 0
3 years ago
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