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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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15
Vinil7 [7]

Answer:

barium and silicon has same valence electrons

Explanation:

barium-2,8,18,18,8,2

neon-2,8

silicon-2,8,2,2

carbon-2,4

5 0
2 years ago
A sample of a compound contains 3.21 g of sulfur and 11.4 g of fluorine. Which of the following represents the empirical formula
Sergeeva-Olga [200]

Answer:

The empirical formula is SF6 (option E)

Explanation:

Step 1: Data given

Mass of sulfur = 3.21 grams

Mass of fluorine = 11.4 grams

Molar mass sulfur = 32.065 g/mol

Molar mass fluorine = 19.00 g/mol

Step 2: Calculate moles

Moles = mass /molar mass

Moles sulfur = 3.21 grams / 32.065 g/mol

Moles sulfur = 0.100 moles

Moles fluorine = 11.4 grams / 19.00 g/mol

Moles fluorine = 0.600 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.100 / 0.100 = 1

F : 0.600 / 0.100 = 6

The empirical formula is SF6 (option E)

7 0
3 years ago
El ácido sulfúrico H2SO4 es uno de los compuestos que se utiliza para la producción de fertilizantes como el nitrosulfato amónic
Serjik [45]

Explanation:

12 hours ago

El ácido sulfúrico H2SO4 es uno de los compuestos que se utiliza para la producción de fertilizantes como el nitrosulfato amónico. Si disponemos de 8 mL de H2SO4 al 37 %P/P (d=1,26 g /mL), los cuales se disolvieron hasta alcanzar un volumen de solución de 400 mL, con una densidad de 1,08 g/mL. (La densidad del soluto es corresponde a 1,83 g/cm³)

5 0
2 years ago
Which planet has a distinctive red spot? A. Mercury B. Venus C. Jupiter D. Mars
Firdavs [7]
Jupiter------------------------
3 0
3 years ago
Read 2 more answers
A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch
ivanzaharov [21]

Answer:

The reactions free energy \Delta G = -49.36 kJ

Explanation:

From the question we are told that

      The pressure of (NO) is P_{NO} = 9.20 \ atm

      The  pressure of  (Cl) gas is  P_{Cl} = 9.15 \ atm

       The  pressure of nitrosly chloride (NOCl) is P_{(NOCl)} = 7.70 \ atm

The reaction is

              2NO_{(g)} + Cl_2 (g)    ⇆   2 NOCl_{(g)}

 From the reaction we can  mathematically evaluate the \Delta G^o (Standard state  free energy ) as

                    \Delta G^o = 2 \Delta G^o _{NOCl} -   \Delta G^o _{Cl_2}  - 2 \Delta G^o _{NO}

The Standard state  free energy for NO is  constant with a value  

                 \Delta G^o _{NO} = 86.55 kJ/mol

 The Standard state  free energy for Cl_2 is  constant with a value                  

             \Delta G^o _{Cl_2} = 0kJ/mol

 The Standard state  free energy for NOCl is  constant with a value

         \Delta G^o _{NOCl} =66.1kJ/mol

Now substituting this into the equation

        \Delta G^o = 2 * 66.1 - 0 - 2 * 87.6

                = -43 kJ/mol

The pressure constant is evaluated as

         Q =  \frac{Pressure \ of  \ product }{ Pressure  \ of \ reactant }

Substituting  values  

        Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}

           = 0.0765

The free energy for this reaction is evaluated as

           \Delta  G  =  \Delta  G^o  + RT ln Q

Where R is gas constant with a value  of  R = 8.314 J / K \cdot mol

          T is temperature in K  with a given value of  T = 25+273 = 298 K

   Substituting value

                \Delta  G  = -43 *10^{3} + 8.314 *298 * ln [0.0765]

                       = -43-6.36

                      \Delta G = -49.36 kJ

4 0
3 years ago
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