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2 years ago

Write a full set of quantum numbers for the following:

Chemistry

1 answer:

bogdanovich [222]2 years ago

8 0

It is known that the maximum value of ml is equal to the value of l. But the minimum value of n is as follows.

n = l + 1

where n = principal quantum number

l = azimuthal quantum number

Values of n can be 1, 2, 3, 4, and so on. Whereas the values of l are 0 for s, 1 for p, 2 for d, 3 for f, and so on.

Also, "m" is known as a magnetic quantum number whose values can be equal to -l and +l.

The electronic configuration of Li is. So here, n = 2, l = 0, m = 0 and s = ±.

Electronic configuration of is. So here, n = 4, l = 1, m = -1, 0, +1, and s = ± .

Electronic configuration of is. So here, n = 5, l = 1, m = -1, 0, +1, and s = ± The electronic configuration of B is. So here, n = 2, l = 1, m = -1, 0, +1, and s = ±.

<h3>What is quantum number?</h3>

Quantum numbers in quantum physics and chemistry explain the values of conserved quantities in a quantum system's dynamics.
Quantum numbers are quantities that can be precisely known at the same time as the system's energy and are related to the eigenvalues of operators that commute with the Hamiltonian and their corresponding eigenspaces.
A base state of a quantum system is fully described by the specification of all of its quantum numbers, which can theoretically be measured collectively.

<h3>What purpose do quantum numbers serve?</h3>

Because they can be used to determine an atom's electron configuration and the likely placement of its electrons, quantum numbers are significant.
The atomic radius and other properties of atoms, such as ionization energy, are also understood using quantum numbers.

Learn more about quantum numbers here:

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820 g Li2SO4 is dissolved into 2500 mL of solution. What is the molar concentration?

Andreas93 [3]

Molarity is defined as the moles of solute per liter of solution. $M=\frac{n}{V}$ . Where M is molarity, n is the number of moles and V is the volume. First we must find the molar mass of $Li_2SO_4$ which is 109.98 g/mol

$Li_2SO_4= 2 \times Li_{Ar} + S_{Ar} + 4 \times O_{Ar}\\= (2 \times 6.941 + 32.1 + 4 \times 16.0) = 109.98 g/mol$

$820\ g\ Li_2SO_4 \times \frac{mol}{109.98\ g}= 7.46\ mol\ Li_2SO4$

Then we find the molarity using above equation

$M= \frac{n}{V} = \frac{7.46 mol}{2500ml} \times\frac{1000ml}{L} = 2.98\ M$

6 0

4 years ago

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×

Anuta_ua [19.1K]

Answer:

$Kc=~1.49x10^3^4}$

Explanation:

We have the reactions:

A: $N_2_(_g_) + O_2_(_g_) 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9$

Our <u>target reaction</u> is:

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)$

We have $NO_(_g_)$ as a reactive in the target reaction and $NO_(_g_)$ is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

Then if we add reactions A and B we can obtain the target reaction, so:

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9$

For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}$

$Kc=~1.49x10^+^3^4}$

3 0

3 years ago

Consider the following reaction:C2H4(g) + F2(g) -----------> C2H4F2(g) Delta H = -549 kJEstimate the carbon-fluorine bond ene

frutty [35]

Answer:

Bond energy of carbon-fluorine bond is 485 kJ/mol

Explanation:

Enthalpy change for a reaction, is given as:

$\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]$

Where $(E_{bond})_{i}$ and $(E_{bond})_{j}$ represents average bond energy in breaking "i" th bond and forming "j" th bond respectively. $n_{i}$ and $n_{j}$ are number of moles of bond break and form respectively.

In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed

So, $-549kJ=(1mol\times 614kJ/mo)+(4mol\times E_{C-H})+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(4mol\times E_{C-H})-(2mol\times E_{C-F})$

or, $-549kJ=(1mol\times 614kJ/mo)+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(2mol\times E_{C-F})$

or, $E_{C-F}=485kJ/mol$

So bond energy of carbon-fluorine bond is 485 kJ/mol

8 0

3 years ago

What were the four elements that ancient Greeks believed to exist

ruslelena [56]

Answer:

The ancient Greeks believed in the four elements of air, earth, fire, and water. The Greek philosopher Empedocles is known for originating the theory of these classic four elements being the basis of all things.

4 0

3 years ago

A chemist titrates 150.0 mL of a 0.2653 M carbonic acid (H2CO3) solution with 0.2196 M NaOH solution at 25 °C. Calculate the pH

xxTIMURxx [149]

Answer:

9.3

Explanation:

This is long and complicated so get ready

We are going to use the conjugate base of carbonic acid with water to make carbonic acid and OH- (Na is simply a spectator ion and is irrelavent here)

Let the conjugate base be A- and Carbonic acid be HA

A- + H20 ⇄ HA + OH-

To find the concentration of A- we must find the concentration of the reactants given. We know this will be equal because it is a strong base and all of it disassociates.

to get moles of acid we take the concentration and multiply by liters to cancel

.2653 x .150 = .039795 mol HA

Because it is at equivalence point we know the moles will be equal. We are given the concentration so we only have to solve for liters

We plug it into the equation and found: .181 L

Now use moles and combined volums to fins concentrarion which is .120 M

Now plug that use the Ka converted to Kb to find the cincentrations of HA and OH-

Ka is (10^-3.60) = 2.4E-4

Kb x Ka is 10^-14

Kb = 3.98E-11

Now we know Kb = [HA] [OH] / [A-]

Solve for this through algebra by using x for the values you dont know

youll find x^2 = 3.3E-10

X = 1.8 E -5

this is the OH- concentration

-log [oh] = pOH

pOH = 4.73

We know 14-pOH = ph so pH= 9.3

6 0

4 years ago