Cambiar dos meses a segundos

The first ionization energy of na(g) is 0.50 mj/mole. can a photon with a wavelength of 500 nm ionize a sodium atom? explain.

Sonbull [250]

<h3>Answer:</h3>

No, a photon with a wavelength of 500 nm can't ionize a sodium atom.

<h3>Explanation:</h3>

Step 1: Find out Energy of given wavelength:

The relation between wavelength and frequency is as follow,

υ = c / λ

where

υ = frequency = ?

c = velocity of light = 3.0 × 10⁸ ms⁻¹

λ = wavenumber = 500 nm = 500 × 10⁻⁹ m

Putting the given values,

υ = 3.0 × 10⁸ ms⁻¹ / 500 × 10⁻⁹ m

υ = 6.0 × 10¹⁴ s⁻¹

Also we know that,

E = h υ

Where;

E = Energy

h = Plank's Constant = 6.6262 × 10⁻³⁴ Js

Putting value of frequency,

E = 6.6262 × 10⁻³⁴ Js × 6.0 × 10¹⁴ s⁻¹

E = 3.97 × 10⁻¹⁹ J

Step 2: Compare given energy and calculated energy:

As we know,

0.5 mJ = 500000 J

Also,

500000 J > 3.97 × 10⁻¹⁹ J

<h3>Conclusion:</h3>

The energy of calculated wavelength of photon is less than 0.5 mJ hence, it is unable to ionize sodium atom.

4 0

3 years ago

Read 2 more answers

Suppose that coal of density 1.5 g/cm^3 is pure carbon. (It is, in fact, much more complicated, but this is a reasonable first a

NISA [10]

Answer:

q = -6464.9 kJ

Explanation:

We are given that the heat of combustion is ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.

vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³

m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g

mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol

q = −394 kJ /mol C x 16.41 mol C = -6464.9 kJ

7 0

3 years ago

Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and

rusak2 [61]

Answer:

$4Zn_(_s_)~+~7OH^-~_(_a_q_)~+~NO_3^-_(_a_q_)~+~6H_2O_(_l_)~-->4Zn(OH)_4^-^2_(_a_q_)~+~NH_3_(_g_)$

-) Oxidizing agent: $NO_3^-_(_a_q_)$

-) Reducing agent: $Zn_(_s_)$

Explanation:

The first step is separate the reaction into the <u>semireactions</u>:

A. $Zn~->Zn(OH)_4^-^2$

B. $NO_3^-~->~NH_3$

If we want to balance in <u>basic medium </u>we have to follow the rules:

1. We adjust the oxygen with $OH^-$

2. We adjust the H with $H_2O$

3. We adjust the charge with $e^-$

Lets balance the first semireaction A. :

$Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-$

Now, lets balance semireaction B:

$NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-$

Finally, we have to add the two semireactions:

_________________________________________

$8~(Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-)$

$2~(NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-)$

_________________________________________

$(8Zn~+~32OH^-~->8Zn(OH)_4^-^2~+~16e^-)$

$(2NO_3^-~+~16e^-~+~12H_2O~->~2NH_3~+~18OH^-)$

Cancel out the species on both sides:

$8Zn~+~14OH^-~+~2NO_3^-~+~12H_2O~-->8Zn(OH)_4^-^2~+~2NH_3$

Simplifying the equation :

$4Zn~+~7OH^-~+~NO_3^-~+~6H_2O~-->4Zn(OH)_4^-^2~+~NH_3$

The $Zn_(_s_)$ is <u>oxidized</u> therefefore is the <u>reducing agent</u>. The $NO_3^-_(_a_q_)$ is<u> reduced</u> therefore is the <u>oxidizing agent</u>.

4 0

4 years ago

PLEASE HELP!!!!!!!!!!

Olegator [25]

I believe the answer is 3

8 0

3 years ago

Write a balanced equation for the oxidation-reduction reaction that occurs when hydrogen peroxide reacts with ferrous ion

Alex17521 [72]

H₂O₂ + 2FeSO₄ + H₂SO₄ → Fe₂(SO₄)₃ + 2H₂O

H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O k=1
Fe²⁺ → Fe³⁺ + e⁻ k=2

H₂O₂ + 2H⁺ + 2Fe²⁺ → 2H₂O + 2Fe³⁺

7 0

3 years ago