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pishuonlain [190]
3 years ago
9

How did Rutherford’s atomic model fix the shortcomings of Thomson’s atomic model?

Chemistry
1 answer:
dangina [55]3 years ago
7 0

Answer: The correct options are b and e.

Explanation:

Thomson's model was generally known as plum pudding model which says that the electrons are distributed in an atom which is similar to the plum present in pudding.

Rutherford's model was known as the gold-foil experiment. In this experiment, he bombarded alpha particles to an atom. He thought that the particles will directly pass through the atom because it has a lot of empty space in it. But to his surprise, he found that some of the particles reflected their path and some directly reflected their way. As the alpha particles carry +2 charge and positive charges repel each other. Hence, he concluded that the atom has a positive charge which is concentrated in a small area.

Hence, the correct statements are b and e.

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Help pls :) I am stuck on this chemistry question about percentage yields!
Charra [1.4K]
You just switch them around 
4 0
3 years ago
This is Science, if you can answer this, much appreciated :) Both questions a and b, thank you!! I need this asap :)
Andrej [43]
Asap means as soom as possible i blieb
7 0
4 years ago
The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How
BARSIC [14]
<h3>Answer:</h3>

              6.21 × 10²² Carbon Atoms

<h3>Solution:</h3>

Data Given:

                 Mass of Butane (C₄H₁₀)  =  1.50 g

                 M.Mass of Butane  =  58.1 g.mol⁻¹

Step 1: Calculate Moles of Butane as,

                 Moles  =  Mass ÷ M.Mass

Putting values,

                 Moles  =  1.50 g ÷ 58.1 g.mol⁻¹

                 Moles  =  0.0258 mol

Step 2: Calculate number of Butane Molecules;

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Butane Molecules can be written as,

            Moles  =  Number of C₄H₁₀ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Butane molecules,

             Number of C₄H₁₀ Molecules  =  Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

     Number of C₄H₁₀ Molecules  =  0.0258 mol × 6.022 × 10²³ Molecules.mol⁻¹

                 Number of C₄H₁₀ Molecules  =  1.55 × 10²² C₄H₁₀ Molecules

Step 3: Calculate Number of Carbon Atoms:

As,

                            1 Molecule of C₄H₁₀ contains  =  4 Atoms of Carbon

So,

          1.55 × 10²² C₄H₁₀ Molecules will contain  =  X Atoms of Carbon

Solving for X,

 X =  (1.55 × 10²² C₄H₁₀ Molecules × 4 Atoms of Carbon) ÷ 1 Molecule of C₄H₁₀

X  =  6.21 × 10²² Atoms of Carbon

5 0
4 years ago
Read 2 more answers
What does the triple point on a phase diagram describe
cluponka [151]
It describes the point at which the element is a solid liquid and has at a certain temperature and pressure
7 0
3 years ago
What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su
anastassius [24]

Answer:

V=27992L=28.00m^3

Explanation:

Hello,

In this case, the combustion of methane is shown below:

CH_4+2O_2\rightarrow CO_2+2H_2O

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3

Best regards.

8 0
4 years ago
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