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faust18 [17]
3 years ago
7

Which is the correct interpretation of the formula mgf2? the lattice contains 1 magnesium atom for every 2 fluorine atoms. the l

attice contains 1 magnesium ion for every 2 fluoride ions. the lattice contains 2 magnesium atoms for every fluorine atom. the lattice contains 2 magnesium ions for every fluoride ion.
Physics
1 answer:
mixer [17]3 years ago
7 0

Answer: the lattice contains 1 magnesium ion for every 2 fluoride ions.

MgF_2  is an ionic compound. It contains positive ions of Magnesium and  and negative ions of Fluorine. To balance the +2 positive charge of one magnesium ion, two (-1) negatively charged Fluorine ions are present. Hence, The correct interpretation of MgF_2  is that the lattice contains 1 Magnesium ion for every 2 Fluoride ions.

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Что сделать чтоб предмет облодал магнитными свойствами?​
Brut [27]
What is the question?
4 0
3 years ago
For a given Prandtl-Meyer expansion, the upstream Mach number is 3 and the pressure ratio across the wave is P2/P1 = 0.4. Calcul
loris [4]

Answer:

The angle for the forward Mach line is 19.47°

The angle for the rearward Mach line is 5.21°

Explanation:

From table A-1 (Modern Compressible Flow: with historical perspective):

\frac{P_{o} }{P_{1} } =36.73 (M₁ = 3)

If Po₁ = Po₂

\frac{P_{o2} }{P_{2} } =\frac{P_{o1} }{P_{1} } *\frac{P_{1} }{P_{2} } =36.73*\frac{1}{4} =91.825

Table A-1:

\frac{P_{o2} }{P_{2} } =91.825,M_{2} =3.63

Table A-5:

v₁ = 49.76°

μ₁ = 19.47°

v₂ = 60.55°

μ₂ = 16°

θ = 60.55 - 49.76 = 10.79°

The angle for the forward Mach line is:

μ₁ = 19.47°

The angle for the rearward Mach line is:

θr = μ₂ - θ = 16 - 10.79 = 5.21°

3 0
3 years ago
The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2
Vlada [557]

Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x10^{-11} e^{-255/T}  and 2x10^{-12} e^{-940/T}  

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]  

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]

Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the  reaction with OH

Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x10^{-11} e^{-255/T}  and 2x10^{-12} e^{-940/T}  

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]  

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]

Than, considering the concentration of Cl increases by a factor of 100 to about 4 × 10^{5} molecules cm^{-3}, the result will be that the reaction with OH will be 535 + (100 to about 4 × 10^{5} molecules cm^{-3}) times faster than the  reaction with Cl

Explanation:

4 0
3 years ago
Two particles are located on the x axis. particle 1 has a mass m and is at the origin. particle 2 has a mass 2m and is at x = +l
wlad13 [49]

The solution would be like this for this specific problem:

<span>
The force on m is:</span>

<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 1

The force on 2m is:</span>

<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 2

From (1), you’ll get M = 2mx^2 / L^2 and from (2) you get M = m(L - x)^2 / L^2 

Since the Ms are the same, then 

2mx^2 / L^2 = m(L - x)^2 / L^2 

2x^2 = (L - x)^2 

xsqrt2 = L - x 

x(1 + sqrt2) = L 

x = L / (sqrt2 + 1) From here, we rationalize. 

x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1) 

x = L(sqrt2 - 1) / (2 - 1) 


x = L(sqrt2 - 1) </span>

 

= 0.414L

 

<span>Therefore, the third particle should be located the 0.414L x axis so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles.</span>

8 0
3 years ago
Propose an experiment that could be used to demonstrate that friction is a contact force?
vesna_86 [32]
Maybe push or pull an object with a large amount of mass? you are force a (pushing through object) aka making contact. i hope i helped not good with physics :)
3 0
3 years ago
Read 2 more answers
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