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const2013 [10]
3 years ago
7

In Rutherford's Gold foil experiment, were the alpha particles directed to different areas of the gold foil or only the same spo

t? If the same spot, how did the particles get deflected in different directions sometimes and go through most of the time
Chemistry
1 answer:
MrRissso [65]3 years ago
8 0

Answer:

See explanation

Explanation:

In the Rutherford experiment, alpha particles were directed at the same spot on a thin gold foil.

As the alpha particles hit the foil, most of the alpha particles went through the foil. In Rutherford's interpretation, most of the particles went through because the atom consisted largely of empty space.

However, some of the alpha particles were deflected through large angles, in Rutherford's interpretation, the deflected alpha particles had hit the dense positive core of the atom which he called the nucleus.

This accounted for their scattering through large angles throughout the foil in all directions.

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Give the mathematical relationship for an unsaturated solution in comparing Q with K sp:__________.
e-lub [12.9K]

Answer:

Q < Ksp

Explanation:

The general equilibrium of a constant product solubility, ksp, is:

AB ⇄ A⁺ + B⁻

<em>Where Ksp is defined as:</em>

Ksp = [A⁺] [B⁻]

When [A⁺] [B⁻] = Ksp, the solution is saturated or oversaturated because there are the maximum amount of ions that solution can dissolve.

When the solution is oversaturated, AB is produced.

Now, in a unsaturated solution, the [A⁺] [B⁻] is less than the maximum amount that can be dissolved. That means:

[A⁺] [B⁻] = Q < Ksp

Q is defined in the same way than Ksp, just in Q the system is not in equilibrium.

Right answer is:

<h3>Q < Ksp</h3>
3 0
3 years ago
The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is add
JulijaS [17]

Answer:

a)The Ksp was found to be equal to 13.69

Explanation:

Terminology

Qsp of a dissolving ionic solid — is the solubility product of the concentration of ions in solution.

Ksp however, is the solubility product of the concentration of ions in solution at EQUILIBRIUM with the dissolving ionic solid.

Note that if Qsp > Ksp , the solid at a certain temperature, will precipitate and form solid. That means the equilibrium will shift to the left in order to attain or reach equilibrium (Ksp).

Step-by-step solution:

To solve this: 

#./ Substitute the molar solubility of KCl as given into the ion-product equation to find the Ksp of KCl.

#./ Find the total concentration of ionic chloride in each beaker after the addition of HCl. We pay attention to the amount moles present at the beginning and the moles added.

#./ Find the Qsp value to to know if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.

a) The equation of solubility equilibrium for KCL is thus;

KCL_(s) ---> K+(aq) + Cl- (aq)

The solubility of KCl given is 3.7 M.

Ksp= [K+][Cl-] = (3.7)(3.7) =13.69

The Ksp was found to be equal to 14.

In pure water KCl

Ksp =13.69 KCl =[K+][Cl-]

Let x= molar solubility [K+],/[Cl-] :. × , x

Ksp =13.69 = [K+][Cl-] = (x)(x) = x²

x= √ 13.69 = 3.7 M moles of KCl requires to make 100mL saturated solutio

37M moles/L

The Ksp was found to be equal to 14.

4.0 M HCl = KCl =[K+][Cl-]

Let y= molar solubility :. y, y+4

Ksp =13.69= [K+][Cl-] = (y)(y*+4)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)

13.69=4y:. y= 3.42 moles/100mL

y= 34.2moles/L

8 M HCl = KCl =[K+][Cl-]

Let b= molar solubility :. B, b+8

Ksp =13.69= [K+][Cl-] = (b)(b*+8)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)

13.69=8b:. b= 1.71 moles/100mL

17.1 moles/L

Therefore in a solution with a common ion, the solubility of the compound reduces dramatically.

8 0
3 years ago
Read 2 more answers
Calculate the pressure of 2.50 Liters of a gas at 25.0oC if it has a volume of 4.50 Liters at
MariettaO [177]

Answer:

P_2 =0.51  atm

Explanation:

Given that:

Volume (V1) = 2.50 L

Temperature (T1) = 298 K

Volume (V2) = 4.50 L

at standard temperature and pressure;

Pressure (P1) = 1 atm

Temperature (T2) = 273 K

Pressure P2 = ??

Using combined gas law:

\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\ \\ \dfrac{1 *2.5}{298} = \dfrac{P_2*4.5}{273}

0.008389261745 \times 273 = 4.5P_2

P_2 =\dfrac{0.008389261745 \times 273 }{4.5}

P_2 =0.51 \ atm

4 0
3 years ago
Identify the name of the compound that goes with this chemical compound
postnew [5]
What is the chemical compound?
8 0
3 years ago
199.5 grams unrefined dark crystalline sugar to cups
Dmitry [639]

Answer:

0.9975 cup  

Step-by-step explanation:

"Unrefined dark crystalline sugar" is what non-chemists call "brown sugar."

200.0 g brown sugar = 1 cup

 199.5 g brown sugar = 199.5× 1/200 .0

 199.5 g brown sugar = 0.9975 cup

A standard measuring cup is not capable of this precision and, furthermore, the mass of brown sugar you can get into a cup depends on how tightly you pack it.

Your Mole Day cake will be fine if you use 1 cup of brown sugar as usual.

4 0
3 years ago
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