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Rom4ik [11]
3 years ago
10

The hybrid orbitals used for bonding by the sulfur atom in the sf4 molecule are ________ orbitals.

Chemistry
1 answer:
mote1985 [20]3 years ago
3 0
The hybrid orbitals used for bonding by the sulfur atom in the sf4 molecule are sp^3d orbitals. Hybrid orbitals are the reaction of a model which incorporate atomic orbitals on a single atom in ways that lead to a new set of orbitals that have geometries appropriate.
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Compare the de Broglie wavelength of an alpha particle moving at 3.40×107 miles per hour (1.52×107 m/s) to that of a baseball mo
denis-greek [22]

Answer:

1.74 x 10⁻²⁰(baseball) : 1.90 x 10⁴ (electron) : 1 (alpha particle)

baseball not detected

electron x ray

alpha particle beyond gama rays

Explanation:

The strategy for us here is to utilize deBroglie relation of wavelength and momentum:

λ = h/ mv

where  λ is wavelength,  h isPlanck´s constant, m is the mass of the particle, and v its velocity.

For determining the regions of spectrum the wavelengths fall into, we need to consult charts that describe the wavelengths vs region.

So  using the above  equation we will compute the wavelengths for the alpha particle, baseball,  and an electron.

Notice we are missing the masses of the alpha particle, and that of the electron, and baseball

The mass of the alpha particle is 4 times  amu  (4 x 1.66 x 10⁻²⁷ Kg ) since the alpha particle is esentially a helium nucleus which has atomic number 4.

The mass of the electron is 9.11 x 10⁻³¹ kg.

Mass of baseball per rules around 0.145 kg.

Notice we are working in the metric system, so use the velocities in m/s.

Now that we have all the data required, lets proceed to calculate the respective wavelengths.

λ ( alpha particle ) = 6.626 x 10⁻³⁴ J·s / ( 4 x 1.66 x 10⁻²⁷ kg x 1.52 x 10⁷ m/s )

= 6.57 x 10⁻¹⁵ m

λ ( baseball ) = 6.626 x 10⁻³⁴ J·s / ( 0.145 kg  x 40.2 m/s ) = 1.14 x 10⁻³⁴ m

λ ( electron ) = 6.626 x 10⁻³⁴ J·s / ( 9.11 x 10⁻³¹ kg x 5.81 x 10⁶ m/s )

= 1.25 x 10⁻¹⁰ m

Comparing the wavelengths from largest to smallest we have:

λ ( baseball) : λ ( electron )   : λ ( alpha particle )

1.14 x 10⁻³⁴  : 1.25 x 10⁻¹⁰ :  6.57 x 10⁻¹⁵

1.74 x 10⁻²⁰(baseball) : 1.90 x 10⁴ (electron) : 1 (alpha particle)

We can see the wavelength  of the baseball is very, very small compared to that of an electron and an alpha particle. For this small wavelength  we are not going to see effects such as diffraction or interference.This is the reason that for everyday macroscopic objects we do not talk about its associated wavelength. The wavelength can only be appreciated in  very small microscopic particles as exemplified for the cases of the electron and alpha particle in this question.

The wavelength of the baseball cannot be detected so a placement in the electromagnetic spectrum is undefined.

The wavelength of the electron in this question will fall into the x ray region of the spectrum (   region  around 10⁻⁹ to 10⁻¹² ) and the alpha particle will fall beyond the gamma rays ( that is wavelength shorter than 10⁻¹² m ).

3 0
3 years ago
during a rising tide, ocean waves often become larger. if the amplitude of a wave increases by a factor of 1.1, by how much does
Sergio039 [100]

Answer:

1.21 times

Explanation:

The energy of a wave is proportional to the square of the amplitude of the wave.

Mathematically:

E\propto A^2

where

E is the energy of the wave

A is its amplitude

In this problem, the amplitude of the wave increases by a factor of 1.1; it means that the new amplitude can be written as

A'=1.1 A

Therefore, this means that the energy of the wave increases by a factor:

E'\propto A'^2=(1.1 A)^2 = 1.1^2 A^2 =1.21 A^2 = 1.21 E

Therefore, the energy of the wave increases by a factor 1.21.

8 0
3 years ago
What is the meaning of plate tectonics
hodyreva [135]

Answer:

A tectonic plate (also called lithospheric plate) is a massive, irregularly shaped slab of solid rock, generally composed of both continental and oceanic lithosphere. Plate size can vary greatly, from a few hundred to thousands of kilometers across; the Pacific and Antarctic Plates are among the largest.

7 0
3 years ago
Read 2 more answers
How many significant figures are in $10,000,210<br> 7<br> 3
nevsk [136]

Answer:

7

Explanation:

7 0
3 years ago
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Ammonia (NH3) clouds are present around some planets. Calculate the number of grams of ammonia produced by the reaction of 5.4 g
Komok [63]

Answer:

30.4 g. NH3

Explanation:

This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.

But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:

5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)

Now, we can use the relationship between H2 and NH3.

2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3

Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.

1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)

5 0
2 years ago
Read 2 more answers
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