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malfutka [58]
3 years ago
11

How much energy is required to move the electron of the hydrogen atom from the 1s to the 2s orbital?

Chemistry
1 answer:
Alja [10]3 years ago
7 0

Answer:

1.63425 × 10^- 18 Joules.

Explanation:

We are able to solve this kind of problem, all thanks to Bohr's Model atom. With the model we can calculate the energy required to move the electron of the hydrogen atom from the 1s to the 2s orbital.

We will be using the formula in the equation (1) below;

Energy, E(n) = - Z^2 × R(H) × [1/n^2]. -------------------------------------------------(1).

Where R(H) is the Rydberg's constant having a value of 2.179 × 10^-18 Joules and Z is the atomic number= 1 for hydrogen.

Since the Electrons moved in the hydrogen atom from the 1s to the 2s orbital,then we have;

∆E= - R(H) × [1/nf^2 - 1/ni^2 ].

Where nf = 2 = final level= higher orbital, ni= initial level= lower orbital.

Therefore, ∆E= - 2.179 × 10^-18 Joules× [ 1/2^2 - 1/1^2].

= -2.179 × 10^-18 Joules × (0.25 - 1).

= - 2.179 × 10^-18 × (- 0.75).

= 1.63425 × 10^- 18 Joules.

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3 years ago
In an aqueous solution of a certain acid the acid is 0.10% dissociated and the pH is 4.06. Calculate the acid dissociation const
Kruka [31]

Answer : The acid dissociation constant Ka of the acid is, 8.7\times 10^{-5}

Explanation :

First we have to calculate the concentration of hydrogen ion.

pH=-\log [H^+]

Given: pH = 4.06

4.06=-\log [H^+]

[H^+]=8.71\times 10^{-5}M

The dissociation of acid reaction is:

                         HA\rightarrow H^++A^-

Initial conc.      c        0         0

At eqm.           c-cα    cα       cα

Given:

Degree of dissociation = α = 0.10 % = 0.001

[H^+]=c\alpha

8.71\times 10^{-5}=c\times 0.001

c=0.0871M

The expression of dissociation constant of acid is:

K_a=\frac{[H^+][A^-]}{[HA]}

K_a=\frac{(c\times \alpha)\times (c\times \alpha)}{(c-c\alpha)}

Now put all the given values in this expression, we get:

K_a=\frac{(0.0871\times 0.001)\times (0.0871\times 0.001)}{(0.0871-0.0871\times 0.001)}

K_a=8.7\times 10^{-5}

Thus, the acid dissociation constant Ka of the acid is, 8.7\times 10^{-5}

7 0
4 years ago
If the pressure of a gas is 2.43 atm what is it’s pressure in kpa
Viktor [21]

The pressure of the gas is 246 kPa.

<em>p</em> = 2.43 atm × (101.325 kPa/1 atm) = 246 kPa  

8 0
3 years ago
In Experiment 9, a 1.05 g sample of a mixture of NaCl (MW 58.45) and NaNO2 (MW 69.01) is reacted with excess sulfamic acid. The
Kaylis [27]

Answer:

There is 76.6 mL of nitrogen collected

Explanation:

<u>Step 1: </u>Data given

Mass of the sample = 1. 05 grams

The sample is 40.00% by mass NaNO2

MW of NaCl = 58.45 g/mol

MW of NaNO2 = 69.01 g/mol

Temperature = 22.0 °C

Pressure = 750.0 mmHg = (750/760) atm

The vapor pressure of water at 22.0°C is 19.8 mm Hg = 0.02605 atm

<u>Step 2:</u> Calculate mass of NaNO2

(40/100)*1.05  = 0.42 grams

<u>Step 3:</u> Calculate moles of NaNO2

Moles NaNO2 = 0.42 grams / 69.01 g/mol

Moles NaNO2 = 0.00608 moles

<u>Step 4:</u> Calculate moles of N2

For 2 moles of NaNO2 we'll get 1 mol of N2

For 0.00608 moles of NaNO2 we'll get 0.00608/2 = 0.00304 moles

<u>Step 5:</u> Calculate pressure of N2

P = 750.0 - 19.8 = 730.2 mmHg = (730.2/760)atm = 0.96079 atm = 97352 Pa

<u>Step 6:</u> Calculate volume of N2

PV = nRT

⇒ P = the pressure of N2 =  0.96079

⇒ V = the volume of N2 = TO BE DETERMINED

⇒ n = moles of N2 = 0.00304 moles

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 22°C = 295 Kelvin

V = (0.00304*0.08206*295)/0.96079

V = 0.0766 L = 76.6 mL

There is 76.6 mL of nitrogen collected

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Answer:

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Explanation:

5 0
3 years ago
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