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REY [17]
3 years ago
5

A chemist has synthesized two new dyes based on the molecular structure of plant-based dyes. The lowest energy absorption line f

or the first dye is light in the visible region at 530 nm. The lowest energy absorption line for the second dye is light in the visible region at 645 nm. Based on this evidence, which molecule has the larger HOMO-LUMO gap?
1. The dye that absorbs at 530 nm
2. There is not enough information given to predict
3. The gap is the same, because they both absorb in visible region
4. The dye that absorbs at 645 nm
Chemistry
1 answer:
baherus [9]3 years ago
7 0

Answer:

1. The dye that absorbs at 530 nm

Explanation:

With a larger HOMO-LUMO gap, there's also a higher absorption energy, so this means that the dye with the higher absortion energy has the larger HOMO-LUMO gap.

The relationship between energy and wavelenght can be expressed by the formula E = hc/λ, this means that the <em>lower</em> the wavelenght, the <em>higher</em> the energy is. So the dye that absorbs at a lower wavelenght has a larger HOMO-LUMO gap.

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For the 52nd electron must pair with one of the<br> electrons in the 5p orbital.
Umnica [9.8K]

Answer:

This question is incomplete

Explanation:

This question is incomplete, however, the element that has 52 electrons only is Tellurium (Te) and when the electronic configuration of elements with more than 52 electrons are written, the 52nd electron is indicated/paired the same way the 52nd electron of Te is indicated/paired. Hence, while writing the electronic configuration of Te, it is written as

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5 0
3 years ago
If 5 grams of hydrogen reacted with 40 games of oxygen to form water, how much of water would be formed?
GREYUIT [131]
2H₂ + O₂ = 2H₂O

n(H₂)=m(H₂)/M(H₂)
n(H₂)=5g/2.0g/mol=2.5 mol

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n(O₂)=40g/32.0g/mol=1.25 mol

H₂ : O₂ = 2 : 1

2.5 : 1.25 = 2 : 1

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3 0
3 years ago
25.0 mL of nitrous acid (HNO2) is titrated with a 1.235 M solution of KOH. The equivalence point (stoichiometric point) is obser
tatuchka [14]

Answer:

0.456 M

Explanation:

Step 1: Write the balanced neutralization equation

HNO₂ + KOH ⇒ KNO₂ + H₂O

Step 2: Calculate the reacting moles of KOH

9.26 mL of 1.235 M KOH react.

0.00926 L × 1.235 mol/L = 0.0114 mol

Step 3: Calculate the reacting moles of HNO₂

The molar ratio of HNO₂ to KOH is 1:1. The reacting moles of HNO₂ are 1/1 × 0.0114 mol = 0.0114 mol.

Step 4: Calculate the initial concentration of HNO₂

0.0114 moles of HNO₂ are in 25.0 mL of solution.

[HNO₂] = 0.0114 mol / 0.0250 L = 0.456 M

3 0
3 years ago
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