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3 years ago

I'm currently doing a Castle Learning part teacher and it's graded

Chemistry

1 answer:

kipiarov [429]3 years ago

4 0

See the image below.

An excited electron is in a high-energy state.

When it drops to the lower-level ground state, it must get rid of this excess energy by emitting it as a quantum of light.

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In order to participate in a hydrogen bond, a hydrogen atom must be covalently bonded to one of three elements. What are they?

Lyrx [107]

Oxygen, Nitrogen and Fluorine

Explanation:

In order to participate in a hydrogen bond, a hydrogen atom must be covalently bonded to one of the oxygen, nitrogen and fluorine.

Hydrogen bonding is a type of dipole - dipole attraction between two specie.
It is an intermolecular force of attraction.
This bond type forms when hydrogen is bonded to a more electronegative atom.
These atoms are oxygen, nitrogen and fluorine
These species draws the electron shared more to themselves and leaves a partial positive charge on the hydrogen and a negative charge on them.
The electrostatic force of attraction between the hydrogen of one molecule and the O/N/F of another molecule cause the formation of hydrogen bonds.

learn more:

Hydrogen bonding brainly.com/question/10602513

#learnwithBrainly

5 0

3 years ago

The water lattice: a. is formed from hydrophobic bonds. b. causes ice to be denser than water. c. causes water to have a relativ

Alexus [3.1K]

Answer: Option (d) is the correct answer.

Explanation:

It is known that water loving bonds are called hydrophilic bonds and water hating bonds are called hydrophobic bonds. Since, water is a polar solvent and it is only able to dissolve polar molecules and not non-polar molecules.

Lattice of water and hydrogen bonding are responsible for the various properties of water like cohesion, adhesion, heat of vaporization etc.

Thus, we can conclude that water lattice excludes non-polar substances.

7 0

3 years ago

Nai(aq)+hg2(no3)2(aq)→ express your answer as a chemical equation. identify all of the phases in your answer. enter noreaction i

Komok [63]

Balanced chemial equation:

2NaI(aq)+Hg2(NO3)2(aq) →Hg2 I2 (s) + 2 NaNO3 (aq)

You can see it better if I use latex:

$2NaI(aq)+Hg_2(NO_3)_2(aq) =Hg_2 I_2 (s) + 2 NaNO_3 (aq)$

As per the phases this is the interpretation:

The symbols (aq) stands for aquous meaning that the compound is dissolved in water.

The symbol (s) stands for solid, meaning tha the compound precipitate and is not dissolved in water.

7 0

4 years ago

Calculate the wavelength, in nanometers, of the light emitted by a hydrogen atom when its electron fallsfrom the n = 7 to the n

Helga [31]

Answer:

λ=2167.6 nm

The wavelength of light emitted is 2167.6 nm.

Explanation:

We recall that Eₙ= $\frac{-2.18*10^{-18} J}{n^{2} }$

since there was transition from n7 to n=4 we will first calculate the change in the energy i.e ΔE

ΔE=E₄-E₇

ΔE= $-2.18*10^{-18} J(\frac{1}{4^{2} } -\frac{1}{7^{2} } )$

ΔE=-9.1760*10^-20 J

Now:

|ΔE|=Energy of photon=h*v=h*c/λ

λ=h*c/|ΔE|

λ= $\frac{6.63*10^{-34}J.s*3.00*10^{8}m/s }{9.1760*10^{-20} J }$

λ=2.1676*10^-6 m

λ=2167.6*10^-9 m

λ=2167.6 nm

The wavelength of light emitted is 2167.6 nm.

5 0

3 years ago

For the vaporization reaction Br2(l) → Br2(

oksian1 [2.3K]

The temperature at which the process be spontaneous is calculated as follows

delta G = delta H -T delta S

let delta G be =0

therefore delta H- T delta s =0

therefore T= delta H/ delta S
convert 31 Kj to J = 31 x1000= 31000 j/mol

T=31000j/mol /93 j/mol.k =333.33K

3 0

3 years ago