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andreyandreev [35.5K]
2 years ago
12

Lines in one spectral series can overlap lines in another.

Chemistry
1 answer:
Kipish [7]2 years ago
4 0

Lines in one spectral series can overlap lines in another is rydberg

In several chemical elements, the Rydberg formula is used to compute the wavelengths of spectral lines.

<h3>What causes lines in a line spectrum?</h3>

A spectral line is similar to a fingerprint in that it can be used to identify the atoms, elements, or molecules in a star, galaxy, or cloud of interstellar gas. When we use a prism to split incoming light from a heavenly source, we often observe a rainbow of colours crossed by definite lines. It's worth noting that spectral lines can appear in other parts of the electromagnetic spectrum, however we can't use a prism to assist us identify them.

Emission lines are distinct coloured lines that appear on a black backdrop and correspond to certain wavelengths of light emitted by an item.

Absorption lines are visible as black bars.

learn more about line spectrum refer:

brainly.com/question/16934091

#SPJ4

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What atomic or hybrid orbitals make up the sigma bond between Cl and F in chlorine trifluoride, ClF3
Umnica [9.8K]

Answer:

sp³d¹ hybridization

Explanation:

Given Cl as central element with three F substrates ...

The VSEPR structure indicates 5 hybrid orbitals that contain 2 diamagnetic orbitals (non-bonded e⁻-pairs) and 3 paramagnetic orbitals (single, non-paired electron for covalent bonding with fluorine) giving a trigonal bypyrimidal parent with a T-shaped geometry.

Valence bond theory predicts the following during bonding:

Cl:[Ne]3s²3p²p²p¹3d⁰

=> [Ne]3s²p²p¹p¹d¹

=> [Ne]3(sp³d)²(sp³d)²(sp³d)¹(sp³d)¹(sp³d)¹

giving 3 ( [Cl](sp³d) - [F]2p¹ ) sigma bonds and 2 non-bonded pairs on Cl.

Note the following images:

Non-bonded electron pairs are in plane of parent geometry and Fluorides covalently bonded to central element Chloride forming the T-shaped geometry.

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Calculate the silver ion concentration in a saturated solution of silver(i) sulfate (ksp = 1.4 × 10–5).
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Answer:

= 0.030 M

Explanation:

We can take x to be the concentration in mol/L of Ag2SO4 that dissolves  

Therefore;  concentration of  Ag+ is 2x mol/L and that of SO4^2- x mol/L.

Ksp = 1.4 x 10^-5

Ksp = [Ag+]^2 [SO42-]

      = (2x)^2(x)

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Thus;

4x^3  = 1.4 x 10^-5

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molar solubility = 0.015 M

But;

[Ag+]= 2x

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