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3 years ago

The electron cloud model describes which of the following?

Physics

2 answers:

son4ous [18]3 years ago

7 0

Answer:

1.B.The trials left by an electron as it moves around the nucleus .

2.8

3.Its mass is lowered, but it is still the same element.

Explanation:

1.Electron cloud model: It says that we cannot know exactly where an electron is at any given time, but the electrons are more likely to be in specific areas.The electron model says that electron have no fixed position so, it traces their path.

Therefore, the trials left by an electron as it moves around the nucleus .

Option B is true.

2.We are given that

Number of protons=8

Number of neutrons=10

We have to find the atomic number

We know that atomic number=Number of protons=8

Hence, atomic number of oxygen=8

3.When an atom loses a neutron then mass decreases but the element does not change.Because when mass number different and atomic number are same then the element are called isotopes of an element.In isotopes , neutron numbers are different.

If an atom loses an neutron then atomic number remain same and element remain same.

Answer:Its mass is lowered, but it is still the same element.

GalinKa [24]3 years ago

3 0

1. I think it's the trails left by an electron as it moves around the nucleus.

2. The atomic number is the number of protons so it is 8.

3. It's mass is lowered but it is still the same element.

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Answer:

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Explanation:

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2 years ago

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3 years ago

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

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3 years ago

A 60kg student traveling in a 1000kg car with a constant velocity has a kinetic energy of 1.2 x 10^4 J. What is the speedometer

777dan777 [17]

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

$K = \frac{1}{2}mv^{2}$

By substituting the values

$1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}$

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, $v = \left (\frac{4.9}{1000} \right )\times \left ( \frac{3600}{1} \right )$

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

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3 years ago

The measurement 0.025563 g should be reported as?

Tomtit [17]

The given mass is 0.025563 g.

Examine the given choices.
a. 0.026 g
This uses 2 significant digits, with rounding to the 3rd decimal place.

b. 2.5 x 10² g = 250 g.
It is incorrect.

c. 0.025 g.
This uses 2 significant digits. It is inaccurate because it does not use rounding to the 3rd decimal place.

d. 0.02 g
This uses one significant digit. It is incorrect for representing the given data.

Answer: a. 0.026 g

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3 years ago