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gizmo_the_mogwai [7]
3 years ago
6

Choose two of the following scientists: Anton Lavoisier, John Dalton, JJ Thomson, Robert Millikan, Ernest Rutherford, James Chad

wick, & Niels Bohr. Describe their experiments and discoveries. Tell how that changed the atomic theory.
Chemistry
1 answer:
elixir [45]3 years ago
6 0

Answer:

John Dalton:

John Dalton was the scientist who introduced atomic theory in the field of chemistry. Dalton worked on different gases and formulated this theory. The main points of Dalton's theory are:

  • Every element present is made up of atoms.
  • Atoms of an elements are have the same same properties whereas these properties are different for each element.
  • According to his theory, an atom could not be broken down.
  • Different atoms combine or get separated from each other during a chemical reaction.

Ernest Rutherford:

Ernest Rutherford is known as the father of nuclear physics due to his impressing research work on radioactivity of atoms. Rutherford was the first scientist to discover the nucleus of an atom and prove that the nucleus was charged. He also described that the electrons circle around the nucleus of an atom.

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What are the different between alpha , beta , gamma rays ​
vovikov84 [41]

Answer:

Alpha is when you are the leader of the pack. Beta is the weaker but can become alpha by killing th ealpha.

Explanation:

4 0
3 years ago
Rank these acids according to their expected pKa values.
givi [52]

Answer:

According to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH  is the weakest acid, so the highest pKa value.

Explanation:

The pKa values are the negative logarithm of dissociation constant. It represents the relative strengths of the acids. Stronger acids show smaller pKa values and weak acids present larger pKa value. The stronger the acid, the weaker it's the conjugate base. The larger the pKa of the conjugate base, the stronger the acid. The strength of an acid is inversely related to the strength of its conjugate.

Conjugate bases are the substance that has one less proton than the parent acid. The conjugate base of the acid presented in the problem are:

ClCH2COOH -> ClCH2COO-  + H+

ClCH2CH2COOH -> ClCH2CH2COO- + H+

CH3CH2COOH -> CH3CH2COO- + H+

Cl2CHCOOH -> Cl2CHCOO - + H+

Cl2CHCOOH. The negative charge presented on its conjugate base is by resonance and inductive effect. This is the strongest acid.

ClCH2COOH. A negative charge is stabilized by resonance and electron-withdrawing but only one atom is present. So this acid is less strong than the first one.

ClCH2CH2COOH. The negative charge is stabilized by resonance and electron-withdrawing atom but the effect is less compared to the two acids showed previously.

CH3CH2COOH. The negative charge is stabilized by resonance and destabilized due to CH3 group. This is the weakest acid among the problem.

Stronger acids have smaller pKa values and weak acids have larger pKa values. Due to the information present in this problem, Cl2CHCOOH is the strongest acid and the lowest pKa. CH3CH2COOH is the weakest acid, so the highest pKa value.

Finally, we can conclude that according to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH  is the weakest acid, so the highest pKa value.

3 0
3 years ago
In the molecules below, areas that have a partial negative charge are pink and areas that have a partial positive charge are blu
Kaylis [27]

Answer:

Dipole-dipole interactions

Step-by-step explanation:

Each molecule consists of <em>two different elements</em>.

Thus, each molecule has permanent <em>bond dipoles</em>.

The dipoles do not cancel, so the attractive forces are dipole-dipole attractions.

"Covalent bonds" is <em>wrong,</em> because there are no bonds between the two molecules.

There are dipole-induced dipole and London dispersion forces, but they are much weaker than the dipole-dipole attractions.

4 0
3 years ago
HELP MEEE PLEASEEEEEEEEEEEEEE
Nesterboy [21]

I think its the sea cucumber and sea urchin

but it could be the star too

5 0
3 years ago
Sodium hydride racts with excess water to produce
romanna [79]

Answer:

0.96g of sodium hydride

Explanation:

Equation of reaction:

NaH + H20 = NaOH + H2

Mass of hydrogen gas produced (m) = PVM/RT

P = 765torr - 28torr = 737torr = 737/760 = 0.97atm, V = 982mL = 982cm^3, M = 2g/mol, R = 82.057cm^3.atm/gmol.K, T = 28°C = 28 + 273K = 301K

m = (0.97×982×2)/(82.057×301) = 0.08g of hydrogen gas

From the equation of reaction

1 mole (24g) of sodium hydride produced 1 mole (2g) of hydrogen gas

0.08g of hydrogen gas would be produced by (24×0.08)/2 = 0.96g of sodium hydride

8 0
3 years ago
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