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3 years ago

What is happening at the atomic level to give rise to the observed energies in the flame tests and in the emission spectra

Chemistry

1 answer:

irina1246 [14]3 years ago

7 0

Answer:

See explanation

Explanation:

According to the Bohr model of the atom, electrons are found in energy levels. Energy is absorbed or emitted when an electron moves from one energy level to another.

During flame test, electrons absorb energy and move to higher energy levels; they quickly return to ground state and emit the energy previously absorbed as a photon of light. This is seen as the colour imparted to the flame by the metal.

The emission spectrum tells us about the range of wavelengths emitted by an atom or compound when it is excited. At an atomic level, the electrons are moved to higher energy levels and as they return to ground state, they emit the various wavelengths that comprise the emission spectrum of any particular substance.

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Matthew added a sample of sodium chloride to water and

viktelen [127]

Answer:

11.45

Explanation:

The ph is just 14-pOH so

14-2.55=ph=11.45

3 0

2 years ago

My high school chemistry is learning this and I don't understand it. "Heat is measured by the following equation:

tangare [24]

Answer:

you didn't ask a question so here is your explanation.

Explanation:

Q = mc∆T. Q = heat energy (Joules, J) m = mass of a substance (kg) c = specific heat (units J/kg∙K) ∆ is a symbol meaning "the change in"

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2 years ago

51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0

3 years ago

Р + 02 = P4O6 balance the equation

sergejj [24]

To balance a chemical equation, both sides should have equal mass, or in other words both sides should have same number of atoms as to follow the conservation of mass rule.

P + O₂ = P₄O₆

LHS:

Number of Phosphorus atoms = 1 atom

Number of Oxygen atoms = 2 atoms

RHS:

Number of Phosphorous atoms = 4 atoms

Number of Oxygen atoms = 6 atoms

Also since P₄O₆ has the most number of atoms we will make the LHS equalize to P₄O₆.

Difference between Phosphorous atoms in LHS to RHS = 3

Since phosphorous is a monatomic we need 3 phosphorous atoms extra

Difference between Oxygen atoms is LHS to RHS = 4

But Oxygen is diatomic, so we need 4/2 = 2 Oxygen molecules

Now lets see if it is balanced

P + 3P + O₂ + 2O₂ -------> P₄O₆

4P + 3O₂ ------> P₄O₆

LHS:

Phosphorous atoms = 4 atoms

Oxygen atoms = 3 × 2 = 6 atoms

RHS:

Phosphorous atoms = 4 atoms

Oxygen atoms = 6 atoms

LHS = RHS

Therefore the balanced equation is 4P + 3O₂ = P₄O₆

Happy to help :)

If you need more explanation or help in any other question, feel free to ask

3 0

2 years ago

3 A specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state.

Masja [62]

When a specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state, this emitted energy can
be used to determine the "identity of the element".

5 0

3 years ago