Which is the first step to take to help reduce pollution

Describe the shapes and relative energies of the SPD & f atomic orbitals

Deffense [45]

Explanation:

The shapes and relative energies of the orbitals s,p,d and f orbitals are given by the principal quantum number and the azimuthal quantum number.

The principal quantum number gives the main energy level and the azimuthal quantum number denotes the shape of the orbitals.

For the principal quantum number, they represent the energy levels in which the orbital is located or the average distance of the orbital from the nucleus. It takes the number n = 1,2,3,4,5,6,7......
The azimuthal quantum number(L) shows the shape of the orbitals in subshells accommodating electrons. The number of possible shapes is limited by the the principal quantum number.

L Name of orbital shape of orbital

0 s spherical

1 p dumb-bell

2 d double dumb-bell

3 f complex

Principal Azimuthal Orbital

Quantum Quantum Designation of

Number (N) Number(l) Sublevel

1 0 1s

2 0 2s

1 2p

3 0 3s

1 3p

2 3d

4 0 4s

1 4p

2 4d

3 4f

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6 0

3 years ago

Can someone help me with this molar mass problem?[It’s the last one]

quester [9]

Answer:

$54.18 \times 10^{23} \ moles$ in 3 mole of $Al_2(SO_4)_3$

Explanation:

It is clear that in the given $1\ mole$ of $Al_2(SO_4)_3$ have $3\ ions$ of $SO_4^2^-$

Therefore 3 moles of $Al_2(SO_4)_3$ will have $3\times3=9 \ ions$ of $SO_4^2^-$

Since 1 ion of anything is equivalent to $6.02\times10^{23} \ moles$

Therefore 3 moles of $Al_2(SO_4)_3$ will have $3\times3=9 \ ions$ of $SO_4^2^-$

Which is equivalent to $9 \times6.02\times10^{23}=54.18\times10^{23} \ moles$

Thus 3 moles of $Al_2(SO_4)_3$ gives $54.18\times10^{23} \ moles$ of $SO_4^2^-$ .

5 0

3 years ago

Reaction of (r)-2-chloro-4-methylhexane with excess nai in acetone gives racemic 2-iodo-4-methylhexane. what is the explanation

saveliy_v [14]

I believe this question has the following five choices to choose from:

>an SN2 reaction has occurred with inversion of configuration

>racemization followed by an S N 2 attack

>an SN1 reaction has taken over resulting in inversion of configuration

>an SN1 reaction has occurred due to carbocation formation

>an SN1 reaction followed by an S N 2 “backside” attack

The correct answer is:

an SN1 reaction has occurred due to carbocation formation

4 0

2 years ago

When sucrose is hydrolyzed, Choose... will be created as the glycosidic linkage is broken. Hydrolyzed sucrose Choose... give a p

Flauer [41]

When sucrose is hydrolyzed, <u>two reducing groups</u> will be created as the glycosidic linkage is broken. Hydrolyzed sucrose <u>will</u> give a positive Benedict's test. Hydrolysis can be achieved by adding <u>strong acids or enzymes.</u>

<h3>How is sucrose hydrolyzed in the body?</h3>

The reaction that breaks bonds and releases energy is called hydrolysis. It is a significant biological process that causes energy to be released from within our bodies. The human small intestine contains the -glucosidase enzyme sucrase, which hydrolyzes sucrose into its component monosaccharides fructose and glucose. In the brush boundary of the upper gastrointestinal system, roughly 10–25% of the fructose is converted to glucose.

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6 0

1 year ago

What is the empirical formula of a vanadium (i) oxide given that 20.38 grams of vanadium combines with oxygen to form 23.58 gram

oksano4ka [1.4K]

V₂O is the empirical formula of a vanadium (i) oxide given that 20.38 grams of vanadium combines with oxygen to form 23.58 grams of the oxide.

The simplest whole number ratio of the atoms of the elements in a specific compound is shown by the empirical formula for that compound. When the mass of each component of a compound or its % composition by mass is known, an empirical formula may typically be computed.

% composition is equal to (mass of an element minus mass of the compound) / 100%.

Given

Vanadium weighs 20.38 g.

The oxide's mass is 23.58 g.

% Vanadium composition: (20.38 g/23.58 g) 100%

= 86.43%

Oxygen mass equals 23.58 g minus 20.38 g

= 3.2 g

100% of oxygen's composition is (3.2/g/23.58 g)

= 13.57%

Atomic mass and number are related by composition percentage.

Vanadium's atomic weight is 50.94 g/mol.

Atomic weight of V = 86.43 / 50.94

= 1.6967

Oxygen has an atomic mass of 16 g/mol.

O atom count is 13.57/16.

= 0.8481

Now on calculation the ratio of Vanadium to the oxygen is

1.6967/0.8481 = 2/1

Ratio of whole numbers is 2:1

Thus V₂O is the empirical formula .

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6 0

4 months ago