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kifflom [539]
3 years ago
5

How did Ernest Rutherford change the atomic model?

Chemistry
1 answer:
nadya68 [22]3 years ago
6 0

Answer:

D. He showed that most of an atom's mass was located in the atom's

nucleus.

Explanation:

Ernest Rutherford changed the atomic model because of his experiment which was the gold foil experiment. A beam of alpha particles was aimed at a piece of gold foil, most particles passed through but some were scattered backward which showed that the middle of an atom (nucleus) is the where most of the mass is located.

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A 0.500 L bottle of water contains 8.82 × 10–7 mL of benzene. What is the concentration of benzene in ppmv?
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0.00176 ppmv

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Exactly 10.0 L of air -25°C is heated to 100.0°C. What is the new volume if the pressure is kept constant?
zepelin [54]

Answer: V2= 15.0403226 Liters

Explanation:

Use V1/T1=V2/T2

Make sure you change the degrees Celsius to Kelvin. (Kelvin = degrees Celsius +273)

10.0L / 248 K = V2/ 373 K

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Which are antiaromatic? check all that apply. the cycloheptatrienyl cation the cycloheptatrienyl anion the cyclopropenyl cation
MrMuchimi

The cyclo-heptatrienyl anion will be considered as antiaromatic compound.

A cyclic compound with a greater energy electron system due to the existence of a 4n delocalized electron is what makes up an anti-aromatic chemical.

Due to its eight pi electrons, the cycloheptatrienyl anion (also known as the tropylium anion) should be antiaromatic; yet, the extra lone pair on one carbon could enable that carbon to become sp^{3} hybridized, placing the extra electrons in an sp^{3} orbital. It would become non-planar as well as non-aromatic as a result.

Therefore, the cyclo-heptatrienyl anion will be considered as antiaromatic compound.

To know more about antiaromatic compound.

brainly.com/question/12790305

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2 years ago
Brainliest will be given to correct answer :)
Lady_Fox [76]

Answer: \Delta G for the reaction   is -90kJ

Explanation:

The balanced chemical reaction is,

2H_2S(g)+SO_2(g)\rightarrow 3S_{rhombic}(s)+2H_2O(g)

The expression for Gibbs free energy change is,

\Delta G=[n\times G_{products}]-[n\times G_{reactants}]

Putting the values we get :

\Delta G=[3\times G_f{S,rhombic}+2\times G_f{H_2O}]-[2\times G_f{H_2S}+1\times G_f{SO_2}]

\Delta G=[(3\times 0kJ/mol)+(2\times -229kJ/mol)]-[(2\times -34kJ/mol) +(1\times -300kJ/mol)]

\Delta G=-90kJ

Thus \Delta G for the reaction2H_2S(g)+SO_2(g)\rightarrow 3S_{rhombic}(s)+2H_2O(g)   is -90kJ

5 0
3 years ago
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