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2 years ago

Which of the following locations would experience the highest rate of sublimation?

Chemistry

2 answers:

g100num [7]2 years ago

6 0

I think its the arctic ice caps

Evgen [1.6K]2 years ago

6 0

I Think it is Sahara Desert

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Which of these is a benefit of nuclear fusion?

dybincka [34]

Answer:

B. There is a lower risk of runaway chain reactions.

Explanation:

Ap Ex Confirmed

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1 year ago

How many molecules are in 20.2 moles of CO₂?

mrs_skeptik [129]

Answer:

1.22 x 10²⁵ molecules CO₂

Explanation:

To find the amount of molecules, you need to multiply the number of moles by Avogadro's Number. Avogadro's Number is a ratio which represents the amount of molecules per every 1 mole. It is important to arrange this ratio in a way that allows for the cancellation of units (since you are going from moles to molecules, moles should be in the denominator). The final answer should have 3 sig figs like the given value.

Avogadro's Number:

6.022 x 10²³ molecules = 1 mole

20.2 moles CO₂ 6.022 x 10²³ molecules
--------------------------- x -------------------------------------- = 1.22 x 10²⁵ molecules
1 mole

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3 months ago

A gaseous mixture of o2 and n2 contains 35.8 % nitrogen by mass. what is the partial pressure of oxygen in the mixture if the to

lubasha [3.4K]

The partial pressure is the amount of linguistic compound there is which makes the lagitude of the element 64.663

5 0

2 years ago

Please someone answer and help :)

algol [13]

13. D
14. A
15. C
16. B
17. F
18. E

7 0

2 years 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

6 0

1 year ago