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Anna71 [15]
3 years ago
6

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

Chemistry
1 answer:
borishaifa [10]3 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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I need to make sure of my answer
choli [55]

Answer:

A) W and X

Explanation:

They are both non metals

3 0
2 years ago
A compound containing sodium, chlorine, and oxygen is 25.42% sodium by mass. A 3.25 g sample gives 4.33×1022 atoms of oxygen. Wh
Tcecarenko [31]
step  one 
calculate  the  %  of  oxygen
from  avogadro  constant
1moles =  6.02  x  10  ^23  atoms
what  about    4.33  x10^22  atoms
= ( 4.33  x  10^ 22 x 1 mole )  /  6.02  10^23=   0.0719 moles
mass=  0.0719  x16=  1.1504   g
% composition   is therefore= ( 1.1504/3.25)  x100 = 35.40%
 step  two
calculate the  %  composition  of  chrorine
100-  (25.42  +  35.40)=39.18%

step  3
calculate the  moles   of  each  element
that   is  
Na  =  25.42  /23=1.1052  moles
Cl=  39.18  /35.5=1.1037moles
O=  35.40/16=  2.2125   moles
step  4
find  the  mole  ratio  by  dividing  each  mole  by  1.1037  moles
that  is
Na  =  1.1052/1.1037=1.001
Cl= 1.1037/1.1037=  1
0=2.2125 = 2
therefore  the  empirical  formula= NaClO2
8 0
3 years ago
Which of the following is an oxidation-reduction reaction? Fe2O3 3CO Right arrow. 2Fe 3CO2 CuSO4 2NaOH Right arrow. Cu(OH)2 Na2S
Damm [24]

The reaction, Fe2O3 + 3CO------> 2Fe + 3CO2 is an oxidation-reduction reaction.

An oxidation-reduction reaction is a reaction in which there is a change in oxidation number from left to right in the reaction. This is because, a specie is oxidized and another specie is reduced.

In the reaction;  Fe2O3 + 3CO------> 2Fe + 3CO2, we can see that the oxidation number of iron decreased from +3 on the left hand side to zero on the right hand side. The oxidation number of carbon was increased from + 2 to +4.

Learn more: brainly.com/question/10079361

6 0
2 years ago
What is the volume of a 5.30g piece of aluminum? The density for aluminum is 2.70mL
Makovka662 [10]

Answer:

1.96mL

Explanation:

Density = mass/volume, and rearranged to solve for volume, volume = mass/density.

So:

volume = 5.30g/2.70g/mL = 1.96mL (assuming your unit was g/mL for density)

7 0
2 years ago
The pressure of a sample of argon gas was increased from 3.14 atm to 7.98 at a constant temperature. If the final volume of argo
noname [10]

Answer:

<h2>36.09 L</h2>

Explanation:

The initial volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume.

Since we're finding the initial volume

V_1 =  \frac{P_2V_2}{P_1}  \\

We have

V_1 =  \frac{7.98 \times 14.2}{3.14} =   \frac{113.316}{3.14}  \\  = 36.0878...

We have the final answer as

<h3>36.09 L</h3>

Hope this helps you

8 0
2 years ago
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