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Lera25 [3.4K]
2 years ago
12

What is the frequency of electromagnetic radiation needed to excite an electron in a hydrogen atom from the n = 3 orbit to the n

= 6 orbit? How much energy does 800. millimoles of these photons have?
Chemistry
1 answer:
Andreas93 [3]2 years ago
5 0

The energy of 800 milimoles of photons is 1.38 * 10^-19 J.

<h3>What is excitation of electrons?</h3>

We know that an electron can move from a lower to higher energy level according to the formula;

1/λ = RH(1/ni^2 - 1/nf^2)

RH = 1.097 * 10^7m-1

ni = 3

nf = 6

Thus;

1/λ = 1.097 * 10^7(1/3^2 - 1/6^2)

1/λ = 1.097 * 10^7(1/9 - 1/36)

1/λ = 1.097 * 10^7(0.11 - 0.03)

λ =1.14 * 10^-6 m

v = λf

f = v/λ

f = 3 * 10^8 m/s/1.14 * 10^-6 m

f = 2.6 * 10^14 Hz

Now

E = hf

E= 6.63 * 10^-34 * 2.6 * 10^14 Hz

E = 1.72 * 10^-19 J

Now;

1 mole of photons has 1.72 * 10^-19 J of energy

800 * 10^-3 moles of photons has 800 * 10^-3 moles * 1.72 * 10^-19 J/ 1 mole

= 1.38 * 10^-19 J

Learn more about energy of photons:brainly.com/question/2393994

#SPJ1

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Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress
ElenaW [278]

Answer:

(a) W_{isoentropic}=8.125\frac{kJ}{mol}

(b) W_{polytropic}=7.579\frac{kJ}{mol}

(c) W_{isothermal}=5.743\frac{kJ}{mol}

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant k should be computed for air as an ideal gas by:

\frac{R}{Cp_{air}}=1-\frac{1}{k}  \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\

0.2856=1-\frac{1}{k}\\\\k=1.4

Next, we compute the final temperature:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K

Thus, the work is computed by:

W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}

(b) In this case, since n is given, we compute the final temperature as well:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K

And the isentropic work:

W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}

Regards.

8 0
3 years ago
There are two containers: one container with a 0.15 molar solution of compound A and another with 0.20 molar solution of compoun
Margaret [11]

Answer:

The container that has 0.20 moles of the compound has more moles compared to the container that has 0.15 moles of compound A

Explanation:

The molarity is defined as the concentration of a substance in solution, expressed as the number of moles of a solute per liter of solution.

From the given information.

In compound A

Molarity of the solution = 0.15 molar = 0.15 m/L

It implies that, In compound A, there are 0.15 moles present in 1 L of the solution.

Similarly;

In compound B

Molarity of the solution = 0.20 molar = 0.20 m/L

It implies that, In compound B, there are 0.120 moles present in 1 L of the solution.

Thus, we can conclude that:

The container that has 0.20 moles of the compound has more moles compared to the container that has 0.15 moles of compound A

6 0
3 years ago
Which of the following solutions has the greater buffer capacity: (a) 100 mL of 0.30M HNO2 - 0.30 M NaNO2 or (b) 100 mL of 0.10
never [62]

Answer:

100mL of 0.10M HNO2 and 0.10M NaNO2

Explanation:

because solution has the greatest buffering capacity when the concentration of the weak acid is = at the concentration of its conjugate base.

4 0
4 years ago
Help please answer show proof
slavikrds [6]

Answer:

d or c

Explanation:

5 0
3 years ago
What is the enthalpy for reaction 1 reversed? reaction 1 reversed: 2co2 + 3h2o→c2h5oh + 3o2 express your answer numerically in k
Rama09 [41]

Answer: The enthalpy of the given reaction is 1234.8kJ/mol.

Explanation: Enthalpy change of the reaction is the amount of heat released or absorbed in a given chemical reaction.

Mathematically,

\Delta H_{rxn}=\Delta H_f_{(products)}-\Delta H_f_{(reactants)}

For the given reaction:

2CO_2(g)+3H_2O(g)\rightarrow C_2H_5OH(l)+3O_2(g)

H_f_{(CO_2)}=-393.5kJ/mol

H_f_{(H_2O)}=-241.8kJ/mol

H_f_{(C_2H_5OH)}=-277.6kJ/mol

H_f_{(O_2)}=0kJ/mol

\Delta H_{rxn}=\Delta H_f_{(C_2H_5OH)}+3\Delta H_f_{(O_2)}-[2\Delta H_f_{(CO_2)}+3\Delta H_f_{(H_2O)}]

Putting values in above equation, we get:

\Delta H_{rxn}=[-277.6+3(0)]-[2(-393.5)+3(-241.8)]kJ/mol

\Delta H_{rxn}=1234.8kJ/mol

7 0
3 years ago
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