What would happen if the Earth was tilted at 45⁰ rather than 23.5⁰?

Chemistry

1 answer:

denis23 [38]3 years ago

4 0

Answer:

it would cause global warming

Explanation: hope this helps

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An atom of which element has the greatest attraction for the electrons in a bond with a hydrogen atom?

Nadusha1986 [10]

The question can be changed into a new form:
Which element has the most negative electron affinity, or attraction for electrons? halogens have the highest electron affinities, and thus are more attracted to the electrons in the Hydrogen atom than any element in their respective periods.
In this case all the following choices are in the same period, thus Cl or Chlorine is the answer as it is a halogen.

7 0

3 years ago

A 2.5 mg sample of magnesium powder is ignited with 2 mg oxygen in a sealed container. All of the magnesium is consumed and 4.15

lys-0071 [83]

Answer:

Total mass of the reactant = 2+2.5 =4.5 mg

Total mass of product = 4.15 mg

therefore, mass of unreacted oxygen = 4.50-4.15 = 0.35 g

7 0

3 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.

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