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nataly862011 [7]
3 years ago
8

What is a possible quantum number set for an electron in the 2s orbital of a beryllium (Be) atom?

Chemistry
1 answer:
Artist 52 [7]3 years ago
7 0

Answer is: 2,0,0,±1/2.

1) n = 1. The principal quantum number (n) is one of four quantum numbers which are assigned to each electron in an atom to describe that electron's state.

2) l = 0. The azimuthal quantum number is a quantum number for an atomic orbital that determines its orbital angular momentum and describes the shape of the orbital.

3) ml = 0. Magnetic quantum number specify orientation of electrons in magnetic field and number of electron states (orbitals) in subshells.

Magnetic quantum number (ml) specifies the orientation in space of an orbital of a given energy and shape . Magnetic quantum number divides the subshell into individual orbitals which hold the electrons, there are 2l+1 orbitals in each subshell.  

4) The spin quantum number, ms, is the spin of the electron; ms = +1/2 or -1/2.

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Create a model of the atoms of a substance moving through the solid, liquid, and gas states.
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5 0
2 years ago
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Which of the following is NOT true regarding Rutherford's Gold Foil experiment?
erastova [34]

Answer:

The area around the nucleus must be of low mass.

Explanation:

Rutherford`s experiment showed that there are some positive charges in the center of the atoms, and because they are all together, they will give a great mass to the atom.

It was quite different from Thomson`s experiment, in which it was thought that the negative charges were mixed with the positive charges, around the atom (like a Pudding Model). In Rutherford`s experiment, because the direction of beta particles, it was the prediction of the positive nucleus.

Hope this info is useful.

8 0
3 years ago
Any help would be appreciated. Confused.
masya89 [10]

Answer:

q(problem 1) = 25,050 joules;  q(problem 2) = 4.52 x 10⁶ joules

Explanation:

To understand these type problems one needs to go through a simple set of calculations relating to the 'HEATING CURVE OF WATER'. That is, consider the following problem ...

=> Calculate the total amount of heat needed to convert 10g ice at -10°C to steam at 110°C. Given are the following constants:

Heat of fusion (ΔHₓ) = 80 cal/gram

Heat of vaporization (ΔHv) = 540 cal/gram

specific heat of ice [c(i)] = 0.50 cal/gram·°C

specific heat of water [c(w)] = 1.00 cal/gram·°C

specific heat of steam [c(s)] = 0.48 cal/gram·°C

Now, the problem calculates the heat flow in each of five (5) phase transition regions based on the heating curve of water (see attached graph below this post) ...   Note two types of regions (1) regions of increasing slopes use q = mcΔT and (2) regions of zero slopes use q = m·ΔH.

q(warming ice) =  m·c(i)·ΔT = (10g)(0.50 cal/g°C)(10°C) = 50 cal

q(melting) = m·ΔHₓ = (10g)(80cal/g) 800 cal

q(warming water) = m·c(w)·ΔT = (10g)(1.00 cal/g°C)(100°C) = 1000 cal

q(evaporation of water) =  m·ΔHv = (10g)(540cal/g) = 5400 cal

q(heating steam) = m·c(s)·ΔT = (10g)(0.48 cal/g°C)(10°C) = 48 cal

Q(total) = ∑q = (50 + 800 + 1000 + 5400 + 48) = 7298 cals. => to convert to joules, multiply by 4.184 j/cal => q = 7298 cals x 4.184 j/cal = 30,534 joules = 30.5 Kj.

Now, for the problems in your post ... they represent fragments of the above problem. All you need to do is decide if the problem contains a temperature change (use q = m·c·ΔT) or does NOT contain a temperature change (use q = m·ΔH).    

Problem 1: Given Heat of Fusion of Water = 334 j/g, determine heat needed to melt 75g ice.

Since this is a phase transition (melting), NO temperature change occurs; use q = m·ΔHₓ = (75g)(334 j/g) = 25,050 joules.

Problem 2: Given Heat of Vaporization = 2260 j/g; determine the amount of heat needed to boil to vapor 2 Liters water ( = 2000 grams water ).

Since this is a phase transition (boiling = evaporation), NO temperature change occurs; use q = m·ΔHf = (2000g)(2260 j/g) = 4,520,000 joules = 4.52 x 10⁶ joules.

Problems containing a temperature change:

NOTE: A specific temperature change will be evident in the context of problems containing temperature change => use q = m·c·ΔT. Such is associated with the increasing slope regions of the heating curve.  Good luck on your efforts. Doc :-)

5 0
3 years ago
determine the frequency and wavelength (in nm) of the light emitted when the e- fell from n=4 and n=2
Lostsunrise [7]

Answer:

Frequency = 6.16 ×10¹⁴ Hz

λ = 4.87×10² nm

Explanation:

In case of hydrogen atom energy associated with nth state is,

En =  -13.6/n²

For n = 2

E₂ = -13.6 / 2²

E₂ = -13.6/4

E₂ = -3.4 ev

Kinetic energy of electron = -E₂ = 3.4 ev

For n = 4

E₄ = -13.6 / 4²

E₄ = -13.6/16

E₄ = -0.85 ev

Kinetic energy of electron = -E₄ = 0.85 ev

Wavelength of radiation emitted:

E = hc/λ = E₄ - E₂

hc/λ = E₄ - E₂

by putting values,

6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev   - (-3.4ev )

6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J

λ = 4.87×10⁻⁷ m

m to nm:

4.87×10⁻⁷ m ×10⁹nm/1 m

4.87×10² nm

Frequency:

Frequency = speed of electron / wavelength

by putting values,

Frequency = 3×10⁸m/s /4.87×10⁻⁷ m

Frequency = 6.16 ×10¹⁴ s⁻¹

s⁻¹ = Hz

Frequency = 6.16 ×10¹⁴ Hz

3 0
3 years ago
What other missions had been to Mars, and what did they accomplish?
lesya692 [45]

Answer:

they stayed alive?

Explanation:

easy

5 0
1 year ago
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