SOMEONE PLEASE HELP WITH 7,8&9

Fill the three 3d subshells with the specified number of electrons.

lana66690 [7]

1. 5 electrons.

$3d^{5}$

Therefore, the 3d subshells blanks will be like this:
↑ ↑ ↑ ↑ ↑

2. 6 electrons.

$3d^{6}$

The 3d subshells blanks will be:
↑↓ ↑ ↑ ↑ ↑

3. 7 electrons.

$3d^{7}$

The 3d subshells blanks will be:
↑↓ ↑↓ ↑ ↑ ↑

Hope you could understand.

If you have any query, feel free to ask.

8 0

2 years ago

Arc AB is One-sixth of the circumference of a circle. What is the radian measure of the central angle?

soldi70 [24.7K]

Answer:

?

Explanation:

7 0

2 years ago

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The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘c is 2000 m/s. Note that 1. 0 mol of diatomic hydrogen at

Leno4ka [110]

The rms speed will be 500 m/s

<h3>What is Root mean square speed ?</h3>

Root mean square speed (Vrms) is defined as the square root of the mean of the square of speeds of all molecules.

Root mean square speed (vrms) Root mean square speed (vrms) is defined as the square root of the mean of the square of speeds of all molecules

It is given that

Speed of a diatomic hydrogen molecule,2000 m/s

Mol of diatomic hydrogen,1.0

Temperature,50°C

The rms speed of diatomic molecule will be:

√(3KT)/( m)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

(3/2)KT = (1/2) mv²

v = √(3KT)/m

FOR H₂: v = √(3KT)/m = 2000 m/s

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

Therefore the rms (root-mean-square) speed of a oxygen molecule at 50∘c is 500m/s.

To know more about Root mean square speed

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3 0

2 years ago

A student places a beaker on a hot plate. To the beaker, he adds a clear liquid and a white powder. The mixture instantly begins

Elenna [48]

True, because if it wasn't a chemical reaction it would have proceeded to stay the same. but it begins to bubble.
sorry if this isn't the best answer I'm trying my best.

7 0

3 years ago

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Compound A is an alkene that was treated with ozone to yield only (CH3CH2CH2)2C=O. Draw the major product that is expected when

DochEvi [55]

Answer:

2,2,3,3-tetrapropyloxirane

Explanation:

In this case, we have to know first the alkene that will react with the peroxyacid. So:

What do we know about the unknown alkene?

We know the product of the ozonolysis reaction (see figure 1). This reaction is an oxidative rupture reaction. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If $(CH_3CH_2CH_2)_2C=O$ is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.

What is the product with the peroxyacid?

This compound in the presence of alkenes will produce peroxides. Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product 2,2,3,3-tetrapropyloxirane. (see figure 2)

3 0

3 years ago