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3 years ago

All of the orbitals in a given subshell have the same value of the __________ quantum number.

1 answer:

3 0

All of the orbitals in a given subshell have the same value of the "<span>magnetic and principal" quantum number

Hope this helps!</span>

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Reading Check: (Describe) how pressure changes as the velocity of a fluid<br> increases.

juin [17]

Answer:

Bernoulli's equation states mathematically that if a fluid is flowing through a tube and the tube diameter decreases, then the velocity of the fluid increases, the pressure decreases, and the mass flow (and therefore volumetric flow) remains constant so long as the air density is constan

Explanation:

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2 years ago

What molecular geometry would be expected for f2 and hf?

Kaylis [27]

The molecular geometry of both F2 and HF is linear.

There are only two atoms which are covalently bonded and thus, the bonding scheme with the atoms looks like this;

F --- F

H---F

So, both are linear.

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2 years ago

suppose a vacuum cleaner use 120 j of electrical energy. If 45 j are used to pull air into the vacuum cleaner, how efficient is

Mice21 [21]

It depends on how much time 45 j will go, so if you told me that it went through 45 j per minute it will go through for 2 minutes so not efficient unless it went through 45 j per hour then it is efficient.

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3 years ago

How can Magnets cause objects to have kinetic energy?

lianna [129]

Answer:

If there is a system of magnets being held in place, there is potential energy. When you let go, the potential energy converts to kinetic energy and the magnets move. Putting the system of magnets close together which creates an opposing force.

Explanation:

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2 years ago

Read 2 more answers

A wooden object (conically shaped) has a diameter of 8cm and height of 14cm. It floats in oil with 6cm of its height above oil l

baherus [9]

Answer:

(a) The density of the object is 316/343 × the density of the oil

(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume

Explanation:

(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;

$V = \dfrac{\pi r^{2} h}{3}$

The volume of the object is therefore;

$\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3$

Where 6 cm is above the oil level we have;

$\dfrac{\pi \times \left (6 \times \dfrac{4}{14} \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3$ above the oil level

Therefore, volume of the oil displaced = $68\tfrac{116}{147}\pi$ cm³ = 216.11 cm³

The density of the object is thus;

$\dfrac{68\tfrac{116}{147}\pi}{ 74\tfrac{2}{3}\pi} \times Density \ of \ the \ oil = \dfrac{316}{343} \right ) \times Density \ of \ the \ oil$

The density of the object = 316/343 × the density of the oil.

(b) The volume of the oil = 2 × Volume of the object = $2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3$

The fraction of the volume displaced, x, after immersing the object is given as follows;

$x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461$

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil

8 0

3 years ago