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4 years ago

Which combinations of n and l represent real orbitals, and which do not exist

Chemistry

2 answers:

Korolek [52]4 years ago

6 0

Pretty sure 1d is not a real thing

Crank4 years ago

3 0

Answer : The combinations of n and l don not exist is, (d) 1d

Explanation :

Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

(a) For 2s :

n=2 and l can have value 0 to (n-1) i.e. 0 and 1 only. For s, l=0 which is possible.

(b) For 3p :

n=3 and l can have value 0 to (n-1) i.e. 0, 1 and 2 only. For p, l=1 which is possible.

(c) For 5s :

n=1 and l can have value 0 to (n-1) i.e. 0, 1, 2, 3 and 4 only. For s, l=0 which is possible.

(d) For 1d :

n=1 and l can have value 0 to (n-1) i.e. 0 only. For d, l=2 which is not possible.

Hence, the combinations of n and l don not exist is, (d) 1d

You might be interested in

What are products of the light reactions of photosynthesis that are utilized in the calvin cycle?

Bumek [7]

Answer:

ATP and NADPH

Explanation:

The photosynthesis involves two stages: the light-dependent reactions and the Calvin cycle.

The first stage is called the light-dependent reactions, in which the chlorophylls convert the light energy from the sun into two storage molecule energy, namely ATP (Adenosine triphosphate) and NADPH (energy-carrying molecule). In this stage, the water is converted to oxygen gas.

The second stage is known as the Calvin cycle, in which the transformation of the carbon from the carbon dioxide into glucose requires the use of ATP and NADPH from the light-dependent reactions.

Therefore, the products of the light reactions of photosynthesis that are utilized in the Calvin cycle are ATP and NADPH.

I hope it helps you!

8 0

3 years ago

*explain your answer for brainliest

Over [174]

Answer:

3 mol Cl₂/2 mol AlCl₃ (three over two)

Step-by-step explanation:

Start with the balanced equation"

2Al + 3Cl₂⟶ 2AlCl₃

The steps in the calculation are

mass of AlCl₃ ⟶ moles of AlCl₃⟶ moles of Cl₂ ⟶ mass of Cl₂

The critical step is the conversion of moles.

You multiply the moles of AlCl₃ by a conversion factor to get moles of Al:

Moles of AlCl₃ × conversion factor = moles of Al.

The conversion factor is the molar ratio, and it uses the coefficients of the formulas in the balanced equation.

It is either (2 mol AlCl₃/3 mol Cl₂) or (3 mol Cl₂/2 mol AlCl₃).

You choose the one that has the desired units of the answer in the numerator.

We choose the second option, because it has the correct units.

For example,

$\text{0.2 mol AlCl}_{3} \times \frac{3\textbf{ mol Cl}_{2} }{ 2\text{ mol AlCl}_{3} } = 0.3\textbf{ mol Cl}_{2})$

Notice how the units "mol AlCl₃" cancel and the correct units appear in the answer.

If we had used the other conversion factor, we would have gotten the wrong units.

3 0

3 years ago

What is an important step to determine the number of grams of SO3 that can be produced by reacting 6 moles of SO2 with oxygen in

vagabundo [1.1K]

Answer:

480.40 g.

Explanation:

According to the balanced equation:

2 SO₂(g) + O₂(g) → 2 SO₃(g),

it is clear that 2.0 moles of SO₂ react with 1.0 mole of oxygen to produce 2.0 moles of SO₃.

We can get the no. of moles of SO₃ produced:

∵ 2.0 moles of SO₂ produce → 2.0 moles of SO₃, from the stichiometry.

∴ 6.0 moles of SO₂ produce → 6.0 moles of SO₃.

Then, we can get the mass of the produced 6.0 moles of SO₃ using the relation:

mass = no. of moles x molar mass of SO₃ = (6.0 moles)(80.066 g/mol) = 480.396 g ≅ 480.40 g.

3 0

3 years ago

What is the correct name of CUCIO2 Chemistry

Deffense [45]

Answer:

Copper(II) Chlorite = Cu(ClO2)

Explanation:

how you write the compound is very important as well

5 0

3 years ago

A 581.2 mL sample of carbon dioxide was heated to 347 K.

svetlana [45]

Answer:

The temperature at 581.2 mL volume was 242.8 K.

Explanation:

Using Charle's law

$\frac {V_1}{T_1}=\frac {V_2}{T_2}$

Given ,

V₁ = 581.2 mL

V₂ = 830.5 mL

T₁ = ?

T₂ = 347 K

Using above equation as:

$\frac {581.2\ mL}{T_1}=\frac {830.5\ mL}{347\ K}$

$T_1=\frac{581.2\times 347}{830.5}\ K=242.8\ K$

The temperature at 581.2 mL volume was 242.8 K.

3 0

4 years ago