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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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Which TWO statements describe the ocean floor giving brainliest pls help explain ur answer
Gelneren [198K]

Answer:

Dude im not 100% sure but I think its b and c im sorry if im wrong its just that im not really sure which ones are.

Explanation:

8 0
3 years ago
Use the ideal gas law to calculate the concentrations of nitrogen and oxygen present in air at a pressure of 1.0 atm and a tempe
notka56 [123]

Answer:

[N₂] = 0.032 M

[O₂] = 0.0086 M

Explanation:

Ideal Gas Law → P . V = n .  R . T

We assume that the mixture of air occupies a volume of 1 L

78% N₂ → Mole fraction of N₂ = 0.78

21% O₂  → Mole fraction of O₂ = 0.21

1% another gases  → Mole fraction of another gases = 0.01

In a mixture, the total pressure of the system refers to total moles of the mixture

1 atm . 1L = n . 0.082L.atm/mol.K . 298K

n = 1 L.atm / 0.082L.atm/mol.K . 298K → 0.0409 moles

We apply the mole fraction to determine the moles

N₂ moles / Total moles = 0.78 → 0.78 . 0.0409 mol = 0.032 moles N₂

O₂ moles / Total moles = 0.21 → 0.21 . 0.0409 mol = 0.0086 moles O₂

4 0
4 years ago
A compound is 80.0% carbon and 20.0% hydrogen by mass. assume you have a 100.-g sample of this compound. the molar mass of the c
ch4aika [34]
Basis of the calculation: 100g
 
For Carbon: 
 Mass of carbon = (100 g)(0.80) = 80 g
  Number of moles of carbon = (80 g)(1 mole / 12g) = 20/3

For Hydrogen:
  Mass of hydrogen = (100 g)(0.20) = 20 g
     Number of moles of hydrogen = (20 g)(1 mole / 1 g) = 20

Translating the answer to the formula of the substance,
     C20/3H20

Dividing the answer,
    CH3

The molar mass of the empirical formula is:
    12 + 3 = 15 g/mol

Since, the molar mass given for the molecular formula is 30.069 g/mol, the molecular equation is,
    C2H6

ANSWER: C2H6

 
4 0
4 years ago
I need help what is s-1/3=7/9???? please help guys and thank you
nalin [4]

Answer:

10/9

Explanation:

First, let's convert 1/3 and 7/9 so that the have the same denominator. To do this let's find the least common multiple of 3 and 9.

List the multiples of 3 and 9:

3: 3, 9

9: 9

They have a least common multiple of 9

We need to convert 1/3 so it has a denominator of 9:

1/3*3/3 (we can multiply it by 3/3 because any number over itself is 1) = 3/9

s-3/9=7/9

Add 3/9 to both sides to isolate s

s=10/9

6 0
3 years ago
Terms that are true of respiration reaction (such C6H12O6 + 6O2 → 6CO2 + 6H2O) include?
scoray [572]
D
Explicating hit brained
7 0
3 years ago
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