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julia-pushkina [17]
3 years ago
6

A certain metal fluoride crystallizes in such a way that the fluoride ions occupy simple cubic lattice sites, while the metal io

ns occupy the body centers of half the unit cells. The formula for the metal fluoride is
Chemistry
1 answer:
Bezzdna [24]3 years ago
3 0

Answer:

MF_2

Explanation:

In a simple cubic lattice lattice, the atoms are present at the eight corners of the cibe.

Since it is mentioned that the fluorine is present at the corners and also 1 corners are shared by 8 unit cells. So, share of atom in one unit cell is:- 8\times \frac{1}{8}=1

Also, the metal, M occupy half of the body centre. A cube has only one body centre. So, share of M in each unit cell:- \frac{1}{2}

Thus, the formula is:-

M_{\frac{1}{2}}F Or simplifying MF_2

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8 0
3 years ago
If the frequency of a wave is 68 hertz then the period of the wave is going to be
tigry1 [53]
The answer is 4.41x10^1 m.

Explanation:

You would use this formula to calculate it
λ = C/f

Where,

λ (Lambda) = Wavelength in meters

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So we have the frequency, 68 Hz, and we have the speed of light. Now we put it into the equation and it will look like this:


λ= (299,792,458 m/s) / (68 Hz)

λ= 4.41x10^1


6 0
3 years ago
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calcium reacts with fluorine to produce calcium fluoride. how does oxidation and reduction take place in this reaction?
Assoli18 [71]

The oxidation is occurring on Calcium ions as it release one electron and reduction will be occurring on fluorine ion as it accepts one electron.

<u>Explanation:</u>

An element will undergo oxidation and form a positive ion on releasing one or more electrons from its valence shell. While reduction is occurred in a chemical reaction, then the element will be forming a negative ion with the acceptance of one or more electrons in its valence shell.

So in the given process of calcium fluoride, the one electron from the valence shell of calcium will be released making it as c a^{+} ions and this is termed as oxidation process. This one electron will be getting accepted by the fluorine ion and thus it will convert to F^{-} ions. This process of acceptance of electrons is termed as reduction.

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8 0
3 years ago
1. How would the series of figures change in the presence of a catalyst?
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No, it won't change the amount of reactants nor the products as a catalyst will only provide an alternative path where lower activation energy is needed for the process to take place.

hope this explains it

If it does, please give it a brainliest :)))

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