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wolverine [178]
3 years ago
13

Why are there two fluoride ions in magnesium fluoride but only one fluoride ion in lithium fluoride

Physics
2 answers:
Serjik [45]3 years ago
6 0

Answer:

Idk

Explanation:

postnew [5]3 years ago
5 0
This would be more of a chemistry question. Remember magnesium has a charge of 2+, and would need to hand off its two extra electrons. Fluorine can only take one electron at a time, so there needs to be two fluorines to take one magnesium's 2 electrons. 

With lithium, it has a +1 charge, so it has one extra electron, which it can hand off to just 1 fluorine atom. 

Another way of looking at this is: Mg^{2+} + 2F^{-} = MgF2  (the charges must balance out to zero)

Li^{+} + F^{-} = LiF (the charges balance out to zero)
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Why are most substances denser in the solid state than in the liquid state?
Rom4ik [11]

Answer:

For other liquids, solidification when the temperature drops includes the lowering of kinetic energy, which allows molecules to pack more tightly and makes the solid denser than its liquid form. Because ice is less dense than water, it is able to float at the surface of water.

Explanation:

have a great day:)

3 0
2 years ago
An eagle is flying horizontally 16.4 meters above a lake at a speed of 9.3 m/s, carrying a small pumpkin in its talons. The pump
Dima020 [189]

Answer:

The horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m

Explanation:

Given;

height above the ground, h = 16.4 m

speed of the eagle, v = 9.3 m/s

The time it will take the pumpkin to fall at the given height is calculated as;

t = \sqrt{\frac{2h}{g} }\\\\t =  \sqrt{\frac{2*16.4}{9.8} }\\\\t = 1.83 \ s

The horizontal distance traveled at this time is given by;

x = vt

x = (9.3)(1.83)

x = 17.02 m

Therefore, the horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m

4 0
2 years ago
A certain freely falling object, released from rest, requires 1.95 s to travel the last 23.5 m before it hits the ground. (a) Fi
Ratling [72]

Answer:

(a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

Explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

Put  the value into the formula

-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2

u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}

u=-2.496\ m/s

(b). We need to calculate the total distance the object travels during the fall

Using equation of motion

v = u+gt

Put the value in the equation

-2.496=0-9.8\times t

t =\dfrac{2.496}{9.8}

t=0.254\ sec

The total time is

t'=t+1.95

t'=0.254+1.95

t'=2.204\ sec

We need to calculate the distance

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

Put the value into the formula

s=0+\dfrac{1}{2}\times9.8\times(2.204)^2

s=23.80\ m

Hence, (a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

4 0
2 years ago
Which of the following is a characteristic of a white dwarf star?
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