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3 years ago

Consider hydrogen atom in an excited state of 3s1. What is the energy of its electron

Chemistry

1 answer:

SVEN [57.7K]3 years ago

8 0

Given:

Excited state electron configuration of H atom = 3s¹

To determine:

Energy of the electron in this state

Explanation:

The energy 'E' of an electron in a hydrogen atom occupying a level, 'n' is given by the formula:

En = -E₀/n²

where, E₀ is the ground state energy = 13.6 eV

n = principle quantum number i.e. n = 1,2,3...

Therefore,

En = -13.6/n² eV

In this case for a 3s¹ configuration, n = 3

E = -13.6/(3)² = -1.51 eV

Ans: Energy of the electron in a 3s1 state is -1.51 eV

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I PLEASE HELP! How many liters of a 2.25 molar hydrobromic acid (HBr) solution would be needed to react completely with 100.0 gr

PtichkaEL [24]

Moles of calcium metal used = 100/40.1=2.5
Moles of HBr need to react = 5 moles
As the molar ratio is 1 is to 2 among them
so
Moles=molarity x volume
5=2.25 x volume
volume=2.22 litres of HBr required for this reaction
ANSWER IS 2.22 LITRES

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3 years ago

H2(g) + Br2(l) ⇄ 2HBr(g) Kc = 4.8 × 108

elena-14-01-66 [18.8K]

Answer:

1.5 × 10⁻⁹M

Explanation:

1. Equilibrium equation

H₂(g) + Br₂(l) ⇄ 2HBr(g)

2. Equilibrium constant

The liquid substances do not appear in the expression of the equilibrium constant.

$k_c=\dfrac{[HBr(g)]^2}{[H_2]}=4.8\times 10^8M$

3. ICE table.

Write the initial, change, equilibrium table:

Molar concentrations:

H₂(g) + Br₂(l) ⇄ 2HBr(g)

I 0.400 0

C - x +2x

E 0.400 - x 2x

4. Substitute into the expression of the equilibrium constant

$4.8\times 10^8=\dfrac{(2x)^2}{0.400-x}$

5. Solve the quadratic equation

192,000,000 - 480,000,000x = 4x²

x² + 120,000,000x - 48,000,000 = 0

Use the quadratic formula:

$x=\dfrac{-120,000,00\pm\sqrt{(120,000,000)^2-4(1)(-48,000,000}}{2(1)}$

The only valid solution is x = 0.39999999851M

Thus, the final concentration of H₂(g) is 0.400 - 0.39999999851 ≈ 0.00000000149 ≈ 1.5 × 10⁻⁹M

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3 years ago

Choose all the answers that apply. Which of the following has kinetic energy? rolling ball, moving car, book on a table, spinnin

Virty [35]

Spinning top, moving car, and rolling ball have kinetic energy I believe

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3 years ago

In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691

andre [41]

Explanation:

Let us assume that the value of $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s$

Also at 1500 K, $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s$

$K_{r} = 400.613 m^{6}/mol^{2}s$

Relation between $K_{p}$ and $K_{c}$ is as follows.

$K_{p} = K_{c}RT$

Putting the given values into the above formula as follows.

$K_{p} = K_{c}RT$

$0.003691 = K_{c} \times 8.314 \times 1500$

$K_{c} = 2.9 \times 10^{-7}$

Also, $K_{c} = \frac{K_{f}}{K_{r}}$

or, $K_{f} = K_{c} \times K_{r}$

= $2.9 \times 10^{-7} \times 400.613$

= $1.1617 \times 10^{-4} m^{6}/mol^{2}s$

Thus, we can conclude that the value of $K_{f}$ is $1.1617 \times 10^{-4} m^{6}/mol^{2}s$ .

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3 years ago

4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60

Bingel [31]

464 g radioisotope was present when the sample was put in storage

<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

t = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

Half-life of Co-60 = 5.3 years

Input the value :

$\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}$

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3 years ago