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3 years ago

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

1 answer:

3 0

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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A 2.40 kg can of coffee moving at 1.50 m/s in the +x-direction on a kitchen counter collides head-on with a 1.20 kg box of macar

liraira [26]

Answer:

The macaroni travels with a velocity of 1.35 m/s in the +x-direction.

Explanation:

By the conservation of momentum principle and assuming there is no external force (e.g. friction) acting on the system, the total initial momentum is equal to the total final momentum.

We take all velocities in the +x-direction as positive.

Initial momentum of coffee can = (2.40 kg)(1.50 m/s) = 3.60 kg m/s

Initial momentum of macaroni = (1.20 kg)(0.00 m/s) = 0.00 kg m/s

Total initial momentum = (3.60 + 0.00) kg m/s = 3.60 kg m/s

Final momentum of coffee can = (2.40 kg)(0.825 m/s) = 1.98 kg m/s

Final momentum of macaroni = (1.20 kg)(v m/s) = 1.20v kg m/s

Total final momentum = (1.98 + 1.20v) kg m/s

Equating both initial and final total momenta,

3.60 kg m/s = (1.98 + 1.20v) kg m/s

1.20v kg m/s = 1.62 kg m/s

v = (1.62 kg m/s) ÷ (1.20 kg) = 1.35 m/s

Since this is positive, the macaroni travels with a velocity of 1.35 m/s in the +x-direction.

6 0

3 years ago

A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05

Lilit [14]

Answer:

v = 1.08 m/s

Explanation:

What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?

The decrease in PE is

d = 80.0cm * 1 / 1000m = 0.80m

h = 0.80 m /2 = 0.40 m

ΔPE = m*g*h

ΔPE = (0.0500 - 0.0200)kg * 9.8m/s² * 0.400 m

ΔPE = 0.1176 J

The moment of inertia of the assembly is

I = 1/12*m*L² + (m1 + m2)*(L/2)²

I = 1/12*0.390kg*(0.800m)² + 0.0700kg*(0.400m)²

I = 0.032 kg·m²

KE = ½Iω²

0.1176 J = ½ * 0.032kg·m² * ω²

ω = 2.71 rad/s

v = ωr = 2.71 rad/s * 0.400m

The linear velocity

v = 1.08 m/s

3 0

3 years ago

Describe where each subatomic particle is found in an atom.

zysi [14]

The last column in the table lists the location of the three subatomic particles.Protons<span> and </span>neutrons<span> are located in the nucleus, a dense central core in the middle of the atom, while the electrons are located outside the nucleus.</span>

7 0

3 years ago

Read 2 more answers

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago

A force of magnitude 33.73 lb directed toward the right is exerted on an object. What other force must be applied to the object

vova2212 [387]

Answer:

33.73 lb to the left

Explanation:

You need to exert a force with the same magnitude, but opposite direction. You can visualize it in this way: When you push an object, the object will follow your path, but if there is another person opposing the force you are exerting, the object will just not move. If the force that the other person exerts were higher, then the object would move in the opposite direction. So, you need them to have the same magnitude.

7 0

3 years ago