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klasskru [66]
4 years ago
7

If the plum pudding model of the atom was correct, what should the results of Rutherford’s experiment be?

Chemistry
1 answer:
shusha [124]4 years ago
6 0

Answer:most of the positively charge particles should be bounce back at a range of angles as they collide with the atoms in the foil; only a few should pass straight through the foil

Explanation:

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Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
3 years ago
The freezing point of a nonelectrolyte solution containing 30.0 g of a solute dissolved in 250.0 g of water is observed to be -2
scZoUnD [109]

Answer:

Molar mass of solute is 89.28 g/m

Explanation:

Colligative property of freezing point depression to solve this:

ΔT = Kf . m . i

i = number of particles, dissolved in solution. In this case, it is a nonelectrolyte, so i = 1.

m = molalilty (mol of solute/1kg of solvent

ΔT = T° freeze pure solvent - T° freeze solution

0°C - (-2.50°C) = 1.86 °C/m . m

2.50°C / 1.86 m/°C = m

1.34 mol solute/kg solvent = m

This means, that in 1000 g of solvent, we have 1.34 moles but we have 250 g of solvent, so let's make a rule of three.

1000 g ____ 1.34 moles

250 g _____(2.50 . 1.34) / 1000 = 0.336 moles

To find the molar mass, we divide mass / moles

30 g/ 0.336 moles = 89.28 g/m

7 0
3 years ago
3. A metal tank contains three gases: oxygen, helium, and nitrogen. If the partial pressures of the three
VikaD [51]

Answer:

35.0 atm + 5.0 atm + 25.0 atm = 65 atm

Explanation:

Partial pressure + Partial pressure + Partial pressure = Total pressure

5 0
3 years ago
The atomic mass of magnesium is the weighted average of the atomic masses of
mojhsa [17]
Of... its isotopes multiplied by each of their respective abundances
6 0
4 years ago
What do calories have to do with combustion
kumpel [21]

Answer:

A calorie is the amount of energy needed to raise the temperature of 1 gram of water 1 degree Celsius. The complete combustion of a large kitchen match, for example, gives you about one kilocalorie of heat

7 0
3 years ago
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