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scoray [572]
3 years ago
8

Describe thomson and millikan contributions to atomic theory

Chemistry
2 answers:
AVprozaik [17]3 years ago
6 0

Thomson's and Millikan's contributions to atomic theory:

Thomson is the first person, who suggest that the theory of the atom contains positive and negative particles and demonstrated the latter which are called "electrons".

Millikan: Millikan discovered that the weight of an electron is 1840 times smaller than a hydrogen atom that is having atomic mass 1.

padilas [110]3 years ago
6 0

Answer:

Thompson discovered the electron and measuring of its charge to mass ratio. Millikan contributed the measurement of electrons charge.

Explanation:

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7 0
2 years ago
How many grams in 6.20 x 10^25 atoms of bromine (Br) ? image attached , will give brainliest
Deffense [45]

Answer:

8239.2g

Explanation:

Given parameters:

Number of atoms in Br  = 6.2 x 10²⁵atoms

Unknown:

Mass of Br = ?

Solution:

From mole concepts, we know that:

       1 mole of a substance contains 6.02 x 10²³ atoms/mol

 Molar mass of Br  = 80g/mol

6.2 x 10²⁵atoms  x \frac{1}{6.02 x 10^{23} } \frac{mol}{atoms} x  80 x \frac{g}{moles}  

          = 8239.2g

8 0
3 years ago
Which structural formula represents an unsaturated hydrocarbon?
bija089 [108]

Answer:

4

Explanation:

your chosen answer is right as unsaturated mean contains double bond and 1st and 3rd is Wrong as Carbon only can have 4 bonds and 2nd is Wrong as it is saturated.

hope this helps :)

5 0
2 years ago
How many moles of Na are needed to produce 4 moles of NaCl in the reaction below?
IgorC [24]

Answer:

4 moles of Na are needed.

8 0
2 years ago
The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio
iren2701 [21]

Answer : The correct option is, (B) CO_2

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

R_2 = rate of effusion of oxygen gas = 14.0\text{ mL }min^{-1}

M_1 = molar mass of unknown gas  = ?

M_2 = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

M_1=44.2g/mole

The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide (CO_2)

5 0
3 years ago
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