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blondinia [14]
3 years ago
15

Hydrogen atoms in a discharge tube are in their ground state. If a beam of free electrons of energy12.9 eV are fired at the hydr

ogen atoms in the discharge tube, (a) determine the maximum wavelength of the radiation that can be emitted by the hydrogen atoms, AND (b) determine the minimum wavelength of the radiation that can be emitted by the hydrogen atoms.
Chemistry
1 answer:
defon3 years ago
3 0

Answer:

See explanation

Explanation:

-0.7ev = -13.6/n^2

n= 4

4--->1

4--->3, 3---> 2, 2--->1

E4-E1= -13.36(1/4^2 - 1/1^2) = 12.53 eV

E4-E3 = -13.36 (1/4^2 - 1/3^2) = 0.65 ev

E3-E2 = -13.36 (1/3^2 - 1/2^2) = 1.86 eV

E2- E1 = -13.6 ( 1/2^2 - 1/1^2) = 10.02 ev

a)

λ= 6.6 × 10^-34 × 3 × 10^8/12.53 × 1.6 × 10^-19

λ= 19.8 × 10^-26/20.048 ×10^-19

λ= 9.9 × 10^ -8 m

b)

λ= 6.6 × 10^-34 × 3 × 10^8/0.65 × 1.6 × 10^-19

λ= 19.8 × 10^-26/1.04 ×10^-19

λ= 19.0 × 10^ -7 m

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Which elements above will form cations? List them below.
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Many oxidation-reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each
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Answer:

Balanced reaction: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

Reduced: O₂

Oxidized : C₃H₈

Explanation:

For the given reaction:

C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(l)

In the reactants, there are 3 C, and in the products only one, so we multiply CO₂ by 3:

C₃H₈(g) + O₂(g) → 3CO₂(g) + H₂O(l)

In the reactants, there are 8 H, and in the products, there are 2 H, so we multiply H₂O by 4:

C₃H₈(g) + O₂(g) → 3CO₂(g) + 4H₂O(l)

In the reactants are 2 O, and in the products 10 O, so we multiply O₂ by 5:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

The reaction is balanced.

Let's identify the oxidation number (Nox) of the elements in each compound.

C₃H₈:

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x = -8/3

O₂:

Because is a pure compound, the Nox of O is 0.

CO₂:

The Nox of O is fix equal to -2, and the molecule is neutral, so, calling x the Nox of C:

2x -4 = 0

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H₂O:

The Nox of H and O are fixed, respectively, +1 and -2.

So, the carbon in C₃H₈ is oxidized because its Nox is increasing, and oxygen in O₂ is reduced because its Nox is decreasing.

6 0
3 years ago
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Answer:

The concentration of HCl is 0.275 M

Explanation:

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

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Volume of HCl = 35.7 mL

<u>Step 2: </u>The balanced equation:

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<u>Step 3</u>: Calculate moles of Na2CO3

Moles Na2CO3 = mass Na2CO3 / Molar mass Na2CO3

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moles of HCL = 2 * 0.0049 = 0.0098 moles HCl

<u>Step 5:</u> Calculate molarity of HCl

Molarity = number of moles of HCl / Volume (liters)

Molarity = 0.0098 moles / 0.0357 L

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The concentration of HCl is 0.275 M

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