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olya-2409 [2.1K]
2 years ago
15

In Rutherford experiment some alpha particles fired at a gold foil bounced backward as a result of ... A.reflection from the sur

faces of gold atoms B.electrostatic repulsion by gold nuclei C.electrostatic repulsion by electronsD.all of the aboveE.none of the above
Chemistry
2 answers:
IRINA_888 [86]2 years ago
6 0

Answer: The correct option is B (electrostatic repulsion by gold nuclei).

Explanation:

In the Rutherford's experiment, he used positively charged particles called alpha particles to bombard an atom in order to find out what is inside the atom. Together with two other scientists, Geiger and Marsden, they used a narrow beam of alpha particles emitted from a radioactive source to bombard a thin gold foil. The scattering of the particles from the gold foil was detected by a movable zinc sulphide screen which could be rotated to various positions around the foil.

Each time an alpha particle hit the screen, a visible flash of light or scintillation was produced. This was observed by a microscope attached to the screen. It was then observed that some of the particles followed a straight path through the gold foil while a few where scattered in a backward direction. This was as a result of electrostatic repulsion by gold nuclei which occurs due to the greater part of the mass of the atom was concentrated in a minute nucleus with positive charge.

Amanda [17]2 years ago
3 0

Answer:

B.electrostatic repulsion by gold nuclei

Explanation:

According to Rutherford's experiment, a thin gold foil was bombarded with alpha particles. Some of the particles passed through the foil undeviated, some were scattered through large angles while some bounced backwards.

It follows that the particles that bounced backwards must have encountered a massive particle of like charge.

The atom is composed of a nucleus which contains positively charged particles. Some of the alpha particles which are positively charged particles bounced back when they encountered the positively charged particles in the nucleus.

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3 years ago
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Which species is classified as a Arrenhius base? 1. CH3OH 2. LiOH 3. PO43- 4. CO32-
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Answer:

2. LiOH

Explanation:

An Arrhenius base is a substance or chemical compound which increases  the number of OH- ions when added in water.

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3 years ago
An air tight freezer measures 4 mx 5 m x 2.5 m high. With the door open, it fills with 22 °C air at 1 atm pressure.
____ [38]

Answer:

(a) Density of the air = 1.204 kg/m3

(b) Pressure = 93772 Pa or 0.703 mmHg

(c) Force needed to open the door =  15106 N

Explanation:

(a) The density of the air at 22 deg C and 1 atm can be calculated using the Ideal Gas Law:

\rho_{air}=\frac{P}{R*T}

First, we change the units of P to Pa and T to deg K:

P=1 atm * \frac{101,325Pa}{1atm}=101,325 Pa\\\\ T=22+273.15=293.15^{\circ}K

Then we have

\rho_{air}=\frac{P}{R*T}=\frac{101325Pa}{287.05 J/(kg*K)*293.15K} =1.204 \frac{kg}{m3}

(b) To calculate the change in pressure, we use again the Ideal Gas law, expressed in another way:

PV=nRT\\P/T=nR/V=constant\Rightarrow P_{1}/T_{1}=P_{2}/T_{2}\\\\P_{2}=P_{1}*\frac{T_{2}}{T_{1}}=101325Pa*\frac{7+273.15}{22+273.15}=101,325Pa*0.9254=93,772Pa\\\\P2=93,772 Pa*\frac{1mmHg}{133,322Pa}= 0.703 mmHg

(c) To calculate the force needed to open we have to multiply the difference of pressure between the inside of the freezer and the outside and the surface of the door. We also take into account that Pa = N/m2.

F=S_{door}*\Delta P=2m^{2} *(101325Pa - 93772Pa)=2m^{2} *7553N/m2=15106N

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