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Yakvenalex [24]
3 years ago
6

Solution of the Schrödinger wave equation for the hydrogen atom results in a set of functions (orbitals) that describe the behav

ior of the electron. Each function is characterized by 3 quantum numbers: n, l, and ml ... n is known as the quantum number. ... l is known as the quantum number. ... ml is known as the quantum number. ... n specifies l specifies ml specifies ... A.The orbital orientation. B.The energy and average distance from the nucleus. C.The subshell - orbital shape.
Chemistry
1 answer:
kvv77 [185]3 years ago
5 0

Answer :

'n' specifies  → (B) The energy and average distance from the nucleus.

'l' specifies   → (C) The subshell orbital shape.

'ml' specifies → (A) The orbital orientation.

Explanation :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. 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...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as m_l. The value of this quantum number ranges from (-l\text{ to }+l). When l = 2, the value of

Spin Quantum number : It describes the direction of electron spin. This is represented as m_s. The value of this is +\frac{1}{2} for upward spin and -\frac{1}{2} for downward spin.

As per question we conclude that,

'n' specifies  → The energy and average distance from the nucleus.

'l' specifies   → The subshell orbital shape.

'ml' specifies → The orbital orientation.

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A 6.000L tank at 19.2°C is filled with 18.0g of carbon monoxide gas and 10.6g of chlorine pentafluoride gas. You can assume both
Jobisdone [24]

Answer:

Total pressure: 2.89 atm

Mole fraction CO: 0.88

Partial pressure CO: 2.56 atm

Mole fraction ClF₅: 0.12

Partial pressure ClF₅: 0.33 atm

Explanation:

We should apply the Ideal Gases Law to solve this:

P . V = n . R . T

We need n, which is the total moles for the mixture

Total moles = Moles of CO + Moles of ClF₅

Moles of CO = mass of CO / molar mass CO → 18 g/28 g/mol = 0.643 mol

Moles of ClF₅ = mass of ClF₅ / molar mass ClF₅ → 10.6g/ 130.45 g/m = 0.0812 mol

0.643 mol + 0.0812 mol → 0.724 moles in the mixture

So we have the total moles so with the formula we would know the total pressure.

P . 6L = 0.724 mol . 0.082L.atm/mol.K . 292.2K

P = ( 0.724 mol . 0.082L.atm/mol.K . 292.2K) / 6L

P = 2.89 atm

Mole fraction is defined as the quotient between the moles of gas over total moles, and it is equal to partial pressure of that gas over total pressure

Moles of gas X /Total moles = Partial pressure of gas X/Total pressure

(Moles of gas X / Total moles) . Total pressure = Partial pressure of gas X

Mole fraction CO = 0.643 / 0.724 = 0.88

Partial pressure CO = 0.88 . 2.89 atm → 2.56 atm

Mole fraction ClF₅ = 0.0812 / 0.724 = 0.12

Partial pressure ClF₅ = 0.12 . 2.89 atm → 0.33 atm

5 0
3 years ago
How many moles of gas occupy 98L at a pressure of 2.8atm and a temperature of 292k? and what law would you use?
algol13
Let's assume that the gas has ideal gas behavior.

Then we can use ideal gas equation,
PV = nRT

Where, P is Pressure of the gas (Pa), V is volume of the gas (m³), n is the number of moles of gas (mol), R is the Universal gas constant (8.314 J mol⁻¹ K⁻¹) and T is the temperature in Kelvin (K)

The given data for the gas is,
P = 2.8 atm = 283710 Pa
V = 98 L = 98 x 10⁻³ m³
T = 292 K
R = 8.314 J mol⁻¹ K⁻¹
n = ?

By applying the formula,
283710 Pa x 98 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 292 K
                                       n = 11.45 mol

Hence,moles of gas is 11.45 mol.
8 0
3 years ago
I need this answer quick please show work
Ainat [17]

Answer:

The answer to your question is 25.2 g of acetic acid.

Explanation:

Data

[Acetic acid] = 0.839 M

Volume = 0.5 L

Molecular weight = 60.05 g/mol

Process

1.- Calculate the number of moles of acetic acid

    Molarity = moles / volume

-Solve for moles

    moles = Molarity x volume

-Substitution

    moles = (0.839)(0.5)

-Result

    moles = 0.4195

2.- Calculate the mass of acetic acid using proportions and cross multiplications

                   60.05 g ----------------------- 1 mol

                        x        ----------------------- 0.4195 moles

                        x = (0.4195 x 60.05) / 1

                        x = 25.19 g

3.- Conclusion

25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M

               

4 0
3 years ago
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