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ankoles [38]
3 years ago
10

Hydrates that have a low vapor pressure and remove moisture from air are said to be ___. Question 8 options: effloresce hygrosco

pic solvation anhydrous
Chemistry
1 answer:
-Dominant- [34]3 years ago
4 0

Answer:

Hygroscopic

Explanation:

An hygroscopic substance is one that absorbs moisture from the atmosphere and becomes wet. Their ability to remove water from air is less than that of deliquescent substances. Most of the solid hygroscopic substances forms pasty substances and not solutions like the deliquescent compounds.

Examples are sodium trioxonitrate(v), copper(ii) oxide e.t.c

Efflorescence compounds gives off their water of crystallization to the atmosphere.

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Determine the wavelength of the energy that needs to be absorbed for a 3p electron in chlorine to be promoted to the 4s subshell
bazaltina [42]

Answer:

The wavelength of the energy that needs to be absorbed  = 52.36 nm

Explanation:

For this study;

Let consider the Rydgberg equation from Bohr's theory of atomic model:

i.e.

\dfrac{1}{\lambda} = R_H (Z^*)^2( \dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})

where

Z* = effective nuclear charge of atom = Z - σ = 6

n₁ = lower orbit = 3

n₂ = higher orbit = 4

R_H = Rydyberg constant = 1.09 × 10⁷ m⁻¹

λ = wave length of the light absorbed

∴

\dfrac{1}{\lambda} = 1.09 \times 10^7}(6)^2( \dfrac{1}{3^2}-\dfrac{1}{4^2})

\dfrac{1}{\lambda} = 1.09 \times 10^7}(36)( \dfrac{1}{9}-\dfrac{1}{16})

\dfrac{1}{\lambda} = 392400000\times0.0486111111

\dfrac{1}{\lambda} =19075000

\lambda = \dfrac{1}{19075000}

\lambda = \dfrac{1}{1.91\times 10^7 \ m^{-1}}

\lambda = 5.236 \times 10^{-8} m

\lambda = 52.36 \times 10^{-9} m

\lambda = 52.36\  n m

Therefore, the wavelength of the energy that needs to be absorbed  = 52.36 nm

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Answer:

white car

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