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3 years ago

The elements fluorine and chlorine can form minerals in the _____ group

Physics

2 answers:

navik [9.2K]3 years ago

5 0

The elements fluorine and chlorine can form minerals in the "halide"group .

Halide refers to the group containing halogens. The halogens turn out to be less responsive as you go down the group. Fluorine, at the highest point of the group, is the most responsive or reactive halogen. It is to a great degree risky, causing serious synthetic consumes on contact with skin.

Halogens also act as bleaching agents.They will evacuate the shade of colors. Chlorine is utilized to bleach wood mash to influence white paper.

Dovator [93]3 years ago

4 0

I would say the Halide group

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Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

SSSSS [86.1K]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

$\rho(r) = ar^2 - br^3$

r is the radius of the spherical shell

dr is the thickness

volume of shell

$dV = 4 \pi r^2 dr$

mass of shell

$dM = \rho(r)dV$

$\rho = \rho_0 - br$

now,

$dM = (\rho_0 - br)(4 \pi r^2)dr$

integrating both side

$M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr$

$M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})$

$M = \pi R^3(\dfrac{\rho_0}{3}+\rho)$

we know,

$a = \dfrac{GM}{R^2}$

$a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}$

$a =\pi RG(\dfrac{\rho_0}{3}+\rho)$

$a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)$

a = 9.94 m/s²

7 0

2 years ago

sound of thunder is heard 5s after the flash of lighting is seen. calculate the distance of nearest point of lighting if the spe

Karolina [17]

Answer:

formula

speed = distance / time

distance = speed x time

distance = 346 x 5

distance = 1730 m

The distance of the lightening strike would be 1730 m or 1.7 km

Explanation:

6 0

3 years ago

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What is the cause of the beats

mestny [16]

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2 years ago

You use an electron microscope in which the matter wave associated with the electron beam has a wavelength of 0.0173 nm. What is

Serggg [28]

Answer:

The kinetic energy of an electron in the beam is 5.04 keV.

Explanation:

We need to find the velocity of the electron by using the De Broglie wavelength:

$\lambda = \frac{h}{mv}$

Where:

λ: is the wavelength = 0.0173 nm

v: is the velocity

m: is the electron's mass = 9.1x10⁻³¹ kg

h: is the Planck constant = 6.62x10⁻³⁴ J.s

$v = \frac{h}{m\lambda} = \frac{6.62 \cdot 10^{-34} J.s}{9.1 \cdot 10^{-31} kg*0.0173 \cdot 10^{-9} m} = 4.21 \cdot 10^{7} m/s$

Now, we can find the kinetic energy:

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}9.1 \cdot 10^{-31} kg*(4.21 \cdot 10^{7} m/s)^{2} = 8.06 \cdot 10^{-16} J*\frac{1 eV}{1.6 \cdot 10^{-19} J} = 5038 eV = 5.04 keV$

Therefore, the kinetic energy of an electron in the beam is 5.04 keV.

I hope it helps you!

3 0

2 years ago

The value of gravitational acceleration of a body on Earth is 9.8 meters/second2. The gravitational potential energy for a 1.00

solong [7]

PE = mgh₁ - mgh₂= 12.5J

h₁ = ?
h₂ = 0

h₁ = PE / mg = 12.5 / 9.8

6 0

2 years ago

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