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3 years ago

Determine the number of unpaired electrons in the octahedral coordination complex [fex6]3–, where x = any halide.

1 answer:

3 0

There will be four unpaired electrons
The metal complex is [FeX₆]³⁻
X being the halogen ligand
X = F, CL, Br, and I
The oxidation of metal state is +3
The ground state configuration is
₂₆Fe =Is² 2s²2p⁶ 3s² 3p⁶ 3d⁶ 4s²
Metal, Fe(III) ion electron configures
₂₆Fe³⁺ = Is2 2s² 2p⁶ 3s² 3p⁶ 3d⁵

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A 69.5-kg person throws a 0.0475-kg snowball forward with a ground speed of 31.5 m/s. A second person, with a mass of 57.5 kg, c

Leno4ka [110]

Answer:

- After throwing the snow, velocity of the thrower is 2.33 m/s

- the velocity of the receiver is 0.026 m/s

Explanation:

Given the data in the question;

Using conservation of momentum,

Initial thrower has a momentum of mv; $m_{total$ v

(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s

Now, When he throws it at 31.5 m/s, these constitutes a momentum of;

(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s

hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s

To get his velocity, we say;

161.94035 = mv

{ he lost weight of the snow ball so, m = 69.5 kg )

161.94035 = 69.5 × v

v = 161.94035 / 69.5

v = 2.33 m/s

Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s

Next is the Receiver;

the receiver will gain momentum of 1.49625 kg.m/s

he has no momentum initially and after he catches the snow ball;

1.49625 kg.m/s = mv

1.49625 kg.m/s = ( 57.5 kg + 0.0475 kg ) × v

1.49625 kg.m/s = 57.5475 kg × v

v = ( 1.49625 kg.m/s ) / 57.5475 kg

v = 0.026 m/s

Therefore, the velocity of the receiver is 0.026 m/s

3 0

3 years ago

Assume that the average mass of each of the approximately 1 billion people in China is 55 kg.Assume that they all gather in one

Harman [31]

Answer: $5.9(10)^{-8} m$

Explanation:

The equation to calculate the center of mass $C_{M}$ of a particle system is:

$C_{M}=\frac{m_{1}r_{1}+m_{1}r_{1}+...+m_{n}r_{n}}{m_{1}+m_{2}+...+m_{n}}$

In this case we can arrange for one dimension, assuming the geometric center of the Earth and the ladder are on a line, and assuming original center of mass located at the Earth's geometric center:

$C_{M}=\frac{m_{E}(0 m) + m_{p} r_{E-p}}{m_{E}+m_{p}}$

Where:

$m_{E}=5.9(10)^{24} kg$ is the mass of the Earth

$m_{p}=55(10)^{9} kg$ is the mass of 1 billion people

$r_{E}=6371000 m$ is the radius of the Earth

$r_{E-p}=6371000 m- 2m=6370998 m$ is the distance between the center of the Earth and the position of the people (2 m above the Earth's surface)

$C_{M}=\frac{m_{p}55(10)^{9} kg (6370998 m)}{5.9(10)^{24} kg+55(10)^{9} kg}$

$C_{M}=5.9(10)^{-8} m$ This is the displacement of Earth's center of mass from the original center.

8 0

3 years ago

The period of rotation of Mars is 1 day and 37 minutes. Determine its frequency of rotation in Hertz.

Sholpan [36]

The frequency of rotation of Mars is 0.0000113 Hertz.

Given the following data:

Period = 1 day and 37 minutes.

To find the frequency of rotation in Hertz:

First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.

Conversion:

1 day = 24 hours

24 hours to minutes = $60$ × $24$ = $1440$ minutes