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3 years ago

A chemical bond formed when atoms share three pairs of electrons is __________.

Physics

1 answer:

nikdorinn [45]3 years ago

5 0

Coordinator bond ! the three bond pair makes triple bond !

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Why does a comet's tail point away from the Sun? 1Choice 1 The solar wind blows the tail away from the Sun. 2 It is being pulled

AnnyKZ [126]

Choice 1

The Sun's radiation and solar wind cause the dust and gas around the comet (coma) to stretch the coma. The solar wind electromagnetically blows the ions in the coma away.

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3 years ago

What happens when you dip open bottle in bucket full of water?why? chapter=pressure

Degger [83]

Answer: It goes inside filling most of it .

Explanation: This happens because the bottle is empty and the bucket is full of water.

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3 years ago

C. If you knew the pressure, volume, and temperature of the air in the

densk [106]

Answer:

Explanation:

First, let's review the ideal gas law, PV = nRT. In this equation, 'P' is the pressure in atmospheres, 'V' is the volume in liters, 'n' is the number of particles in moles, 'T' is the temperature in Kelvin and 'R' is the ideal gas constant (0.0821 liter atmospheres per moles Kelvin).

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2 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

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3 years ago

An electric grinder uses a grinding wheel

luda_lava [24]

(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s²

(B) 
(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev 

(C) 
(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s

3 0

2 years ago

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