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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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object carries a charge of -8.1 ÂµC, while another carries a charge of -2.0 ÂµC. How many electrons must be transferred from the

LenKa [72]

Number of electrons transferred: $1.91\cdot 10^{13}$

Explanation:

The charge on the first object is

$Q_1 = -8.1\mu C$

while the charge on the 2nd object is

$Q_2=-2.0 \mu C$

When they are in contact, the final charge on each object will be

$Q=\frac{Q_1+Q_2}{2}=\frac{-8.1+(-2.0)}{2}=-5.05 \mu C$

So, the amount of charge (electrons) transferred from the 1st object to the 2nd object is

$\Delta Q = Q_1 - Q = -8.1 -(5.05)=-3.05 \mu C = -3.05\cdot 10^{-6}C$

The charge of one electron is

$e=-1.6\cdot 10^{-19}C$

Therefore, the number of electrons transferred is

$N=\frac{Q}{e}=\frac{-3.05\cdot 10^{-6}}{-1.6\cdot 10^{-19}}=1.91\cdot 10^{13}$

Learn more about electrons:

brainly.com/question/2757829

#LearnwithBrainly

8 0

3 years ago

What is the difference in the speed of the generator with a small magnet and a generator with a large magnet?

Delvig [45]

With a small magnet with a generator it will be taken up quickly because how small it is while with a big generator it would take more force for it for the generator to attach because the larger the magnet that heavier it will be because it is attached to the North Pole magnet

8 0

3 years ago

Read 2 more answers

All of the noble gases, Group 18, have eight valence electrons in its outer shell (excluding helium which only has two).

mina [271]

C) 0 cause it’s a stable reaction

8 0

4 years ago

Read 2 more answers

Hii:) I need help with qn 1 & please explain if possible too :) , thanks!

andrezito [222]

1. In the first 1.5 seconds, the lift accelerates from 0 to 3m/s. By definition, the acceleration is the ratio between the change in velocity and the time elapsed to change the velocity.

The change in velocity is $\Delta v = v_{\text{final}}-v_{\text{initial}}=3-0=3$ .

The time elapsed is 1.5 seconds, so the acceleration is

$a = \dfrac{3}{1.5}=2$

meters per second squared.

2. We know, from the previous point, that the lift travelled 20m from the first floor. Since it returns to the first floor after the ascent, it must travel again those same 20m, just in reverse (descending instead of ascending). So, the total distance travelled is $20+20=40$ meters.

The displacement, though, is zero, because it measures the distance between the starting and ending point of a certain motion. Since the lift starts and ends its motion at the same place (the first floor), its total displacement is zero.

3 0

3 years ago

A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin

gogolik [260]

Answer:

a) 23 m/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$p_{o} = p_{f} (1)$

The initial momentum p₀, can be written as follows:

$p_{o} = m_{1} * v_{1o} + m_{2}* v_{2o} = 6.0 kg * 5.0 m/s + 2.0 kg * v_{2o} (2)$

The final momentum pf, can be written as follows:

$p_{f} = (m_{1} + m_{2} )* v_{f} = 8.0 kg* (-2.0 m/s) (3)$

Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

$v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg} = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s (4)$

This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.

6 0

4 years ago