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Scrat [10]
4 years ago
7

the red line observed in the line spectrum for hydrogen has 3.03x10^-19J, what is the wavelength, in nm, of this?

Chemistry
1 answer:
djverab [1.8K]4 years ago
6 0

Answer:

λ = 6.5604 x 1016 nm

Explanation:

Given Data:

The energy of the red line in Hydrogen Spectra = 3.03 x 10-19

Formula to calculate Wave length

E= hv

Where E is Energy

h is Planks Constant = 6.626 x 10–34 J s

v is frequency

In turn

v= c/ λ

where c is speed of light = 3.00 x 108 m s–1

λ is wavelength = to find

Solution:

Formula to be Used:

E= hv………………………… (1)

Putting the value v in equation 1

               E= h c/ λ…………………… (2)

Put the value in equation 2

3.03 x 10-19 J  = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) / λ ……………………….(3)

By rearranging equation 3

λ = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) /3.03 x 10-19  J

λ = 6.5604 x 107 m

The answer is in “m”

So we have to convert it into nm

So for this to convert “m” to “nm” multiply the answer with 109

λ = 6.5604 x 107 x 109

λ = 6.5604 x 1016 nm

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Answer:

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1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

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The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

Explanation:

A molecular equation is a balanced chemical equation which shows the reacting species as molecules rather than as componenet ions in their compounds with subscripts written beside the molecules to indicate the state in which they occur in the chemical reaction.

An ionic equation expresses the reacting species as components ions in a chemical reation. All the ions and molecules reacting are shown.

In a net ionic equation, the ions which remain in the ionic state also known as spectator ions are not written as part of the equation.

From the given attachment;

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

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