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3 years ago

Draw the model of an atom with an atomic number of 7 and a mass number of 15. How many valence electrons are in this atom.

Physics

1 answer:

o-na [289]3 years ago

7 0

Answer:

There are 5 valence electrons

Explanation:

Valence electrons are electrons found in the outermost shell of an atom.

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A snowball starting at rest rolls down a hill and reaches 5 m/s. If the hill is

Lostsunrise [7]

Answer:

The acceleration of the snowball is 0.3125

Explanation:

The initial speed of the snowball up the hill, u = 0

The speed the snowball reaches, v = 5 m/s

The length of the hill, s = 40 m

The equation of motion of the snowball given the above parameters is therefore;

v² = u² + 2·a·s

Where;

a = The acceleration of the snowball

Plugging in the values, we have;

5² = 0² + 2 × a × 40

∴ 2 × 40 × a = 5² = 25

80 × a = 25

a = 25/80 = 5/16

a = The acceleration of the snowball = 5/16 m/s².

The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .

4 0

3 years ago

B. How could you determine the instantaneous speed during the trip? Be specific.

alex41 [277]

Answer:

Speedometer.

Explanation:

If you look at the speedometer in a car, you will see your instantaneous speed. Speedometers tell you your speed of that moment and thats what instantaneous speed.

6 0

3 years ago

Which type of energy warms the eartha surface

cestrela7 [59]

B. Thermal energy is the energy that warms the earths surface.

5 0

3 years ago

Read 2 more answers

In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum

gulaghasi [49]

Answer:

Ratio of length will be $\frac{L_2}{L_1}=1.108$

Explanation:

We have given time period of the pendulum when length is $L_1$ is $T_1=0.95sec$

And when length is $L_2$ time period $T_2=1sec$

We know that time period is given by

$T=2\pi \sqrt{\frac{L}{g}}$

So $0.95=2\pi \sqrt{\frac{L_1}{g}}$ ----eqn 1

And $1=2\pi \sqrt{\frac{L_2}{g}}$ -------eqn 2

Dividing eqn 2 by eqn 1

$\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}$

Squaring both side

$\frac{L_2}{L_1}=1.108$

8 0

4 years ago

A series of lines involving a common level in the spectrum of atomic hydrogen lies at 656.46 nm, 486.27 nm, 434.17 nm, and 410.2

julsineya [31]

Answer:

The wavelength of next line in the series will be 397.05 nm

Explanation:

From Rydberg equation;

$\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})$

where;

λ is the wavelength

n lines in the series

RH is Rydberg constant = 1.097 x 10⁷ m⁻¹

Also at a given maximum wavelength, we can determine the first line n₁ in the series

$\frac{1}{\lambda_{max}R_H} = \frac{1}{n_1^2} -\frac{1}{(n_1 +1)^2} \\\\$

$\frac{1}{\lambda_{max}R_H} = \frac{2n_1+1}{n_1^2(n_1 +1)^2} \\\\\lambda_{max}R_H} = \frac{n_1^2(n_1 +1)^2}{2n_1+1}$

Given;

maximum wavelength = 656.46 nm

$\lambda_{max}R_H} = \frac{n_1^2(n_1 +1)^2}{2n_1+1}\\\\656.46 *10^{-9}*1.097*10^7 = \frac{n_1^2(n_1 +1)^2}{2n_1+1}\\\\7.2 = \frac{n_1^2(n_1 +1)^2}{2n_1+1}$

Now, test for different values of n that will be equal to 7.2

let n₁ = 1

$\frac{n_1^2(n_1 +1)^2}{2n_1+1} = \frac{(1)^2(1 +1)^2}{2(1)+1} = 1.3\\$

n₁(1) ≠ 7.2

Again, let n₁ = 2

$\frac{n_1^2(n_1 +1)^2}{2n_1+1} = \frac{(2)^2(2 +1)^2}{2(2)+1} = 7.2\\$

∴ n₁(2) = 7.2

For the least wavelength given as 410.29 nm, n = ?

$\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})\\\\\frac{1}{410.29*10^{-9}} = 1.097*10^7(\frac{1}{2^2} -\frac{1}{n_2^2})\\\\\frac{1}{4.5} =\frac{1}{4} -\frac{1}{n_2^2}\\\\\frac{1}{n_2^2} =\frac{1}{4} -\frac{1}{4.5} \\\\\frac{1}{n_2^2} = 0.0277778\\\\n_2^2 = \frac{1}{0.0277778} \\\\n_2^2 = 36\\\\n_2= \sqrt{36}\ = 6$

next line in the series will be 7

The wavelength of next line in the series will be;

$\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})\\\\\frac{1}{\lambda} = 1.097*10^7(\frac{1}{2^2} -\frac{1}{7^2})\\\\\frac{1}{\lambda} = 1.097*10^7(0.22959)\\\\\frac{1}{\lambda} = 2518602.3\\\\\lambda = 397.05 \ nm$

Therefore, the wavelength of next line in the series will be 397.05 nm

5 0

3 years ago