50 points please answer ASAP

What number and type of hybrid orbital(s) form(s) when one p and one s atomic orbital mix?

Harrizon [31]

Two equivalent hybridized orbitals will form from the mixing of one s-orbital and one p-orbital, that is (sp) orbital.

<h3>What are orbitals?</h3>

Orbital is the place around nucleus where mostly the electrons are present. There are four types of orbitals are present, s, p, d, and f.

The orbitals that are formed by the mixing of these orbitals are called hybrid orbitals.

Thus, two equivalent hybridized orbitals will form from the mixing of one s-orbital and one p-orbital, that is (sp) orbital.

Learn more about orbitals

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3 0

2 years ago

Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.

Juliette [100K]

Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = $\frac{mv}{qB}$

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and charge of a proton are:

m = 1.67 * 10^-27 kg

q = 1.6 * 10^-19 C

So, we get that the radius r will be:

r = $\frac{1.67 * 10^-27 kg * 2.2*10^6 m/s}{1.6 * 10^-19 C* 0.7 T}$ = 0.0328 m, or 3.28 cm.

8 0

3 years ago

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (5*10^{-3})(300)^2$

$KE = 225J$

On the other hand the required Energy to heat up t melting point is

$Q_1 = mC_p \Delta T$

$Q_2 = L_f m$

Where,

m = Mass

$C_p =$ Specific Heat

$\Delta T =$ Change at temperature

$L_f =$ Latent heat of fussion

Heat required to heat up to melting point,

$Q = Q_1+Q_2$

$Q = mC_p \Delta T+L_f m$

$Q = 5*0.128*(327-20) + 5*24.7$

$Q = 310J$

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0

3 years ago

Unpolarized light passes through two polarizers whose transmission axes are at an angle # with respect to each other. What shoul

Yuki888 [10]

Answer:

$63.4^{\circ}$

Explanation:

When unpolarized light passes through the first polarizer, the intensity of the light is reduced by a factor 1/2, so

$I_1 = \frac{1}{2}I_0$ (1)

where I_0 is the intensity of the initial unpolarized light, while I_1 is the intensity of the polarized light coming out from the first filter. Light that comes out from the first polarizer is also polarized, in the same direction as the axis of the first polarizer.

When the (now polarized) light hits the second polarizer, whose axis of polarization is rotated by an angle $\theta$ with respect to the first one, the intensity of the light coming out is