Which of the following choices is a sign that a glacier was present at one point in history?

A thin metallic spherical shell of radius 41.6 cm has a total charge of 8.55 μC uniformly distributed on it. At the center of th

Paul [167]

Answer:

E = 1.2443*10⁶ N/C

Explanation:

R = 41.6 cm = 0.416 m

Q₁ = 8.55 μC = 8.55*10⁻⁶C

Q₀ = 4.43 μC = 4.43*10⁻⁶C

r = 17.9 cm = 0.179 m

K = 9*10⁹ N*m²/C²

Since r < R we can apply Gauss's Law as follows

E = K*Q₀ / r²

⇒ E = (9*10⁹ N*m²/C²)*(4.43*10⁻⁶C) / (0.179 m)²

⇒ E = 1.2443*10⁶ N/C

5 0

3 years ago

Two moles of gas A spontaneously convert to 3 moles of gas b in a container where the temperature and pressure are held constant

erica [24]

At constant temperature and pressure, If the amount of gas increases to the given value, its volume also increases to 20.85L.

<h3>What is Avogadro's law?</h3>

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."

It is expressed as;

V₁/n₁ = V₂/n₂

Given the data in the question;

Initial amount of gas n₁ = 2moles

Initial volume v₁ = 13.9L

Final amount of gas n₁ = 3moles

Final volume v₂ = ?

V₁/n₁ = V₂/n₂

V₁n₂ = V₂n₁

V₂ = V₁n₂ / n₁

V₂ = (13.9L × 3moles) / 2moles

V₂ = 41.7molL / 2mol

V₂ = 20.85L

At constant temperature and pressure, If the amount of gas increases to the given value, its volume also increases to 20.85L.

Learn more about Avogadro's law here: brainly.com/question/15613065

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4 0

1 year ago

Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th

lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

"Your question is not complete, it seems to be missing the diagram of the emission spectrum"

the diagram of the emission spectrum has been added.

From the given chart;

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

c is the speed of light = 3 x 10⁸ m/s
f is the frequency of the wave
λ is the wavelength

$f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz$

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ Js

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928