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3 years ago

How many d and f orbitals total are present in the n = 4 shell of a hydrogen atom? Enter your answer as a whole number.

Chemistry

1 answer:

sashaice [31]3 years ago

4 0

Answer:

12 orbitals

Explanation:

The suborbital = d orbital.

n = 4

The orbital is 4d.

There are 5 degenerate orbitals of d orbitals. Thus m value, -2, -1, 0, +1, +2 .

Thus, number of orbitals = 5

The suborbital = f orbital.

n = 4

The orbital is 4f.

There are 7 degenerate orbitals of f orbitals. Thus m value, -3, -2, -1, 0, +1, +2, +3

Thus, number of orbitals = 7

Total number of orbitals = 7 + 5 = 12

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Be sure to answer all parts.The equilibrium constant, Kc, for the formation of nitrosyl chloride from nitric oxide and chlorine,

djverab [1.8K]

<u>Answer:</u> The reaction proceeds in the forward direction

<u>Explanation:</u>

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

Relation of $K_p\text{ with }K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta n_g}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $6.5\times 10^4$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $35^oC=[35+273]K=308K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=6.5\times 10^4\times (0.0821\times 500)^{-1}\\\\K_p=1583.43$

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

The expression of $Q_p$ for above equation follows:

$Q_p=\frac{(p_{NOCl})^2}{p_{Cl_2}\times (p_{NO})^2}$

We are given:

$p_{NOCl}=1.76atm$

$p_{NO}=1.01atm$

$p_{Cl_2}=0.42atm$

Putting values in above equation, we get:

$Q_p=\frac{(1.76)^2}{0.42\times (1.01)^2}=7.23$

We are given:

$K_p=1583.43$

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium

As, $K_p>Q_p$ , the reaction will be favoring product side.

Hence, the reaction proceeds in the forward direction

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