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belka [17]
3 years ago
5

List three quantum numbers you know

Chemistry
2 answers:
dangina [55]3 years ago
5 0

Answer:

The main quantum numbers are

1) Principal quantum no.

2) Azimuthal quantum no.

3)Magnetic quantum no.

4) Spin quantum no.

Serhud [2]3 years ago
3 0

Answer:principle quantum number

Angular quantum number

Magnetic quantum number

Spin quantum number

Explanation:

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Question 6 of 10
lesya [120]
The answer is B. Molecules move more quickly as temperature increases.

When Allmond molecular motion stops, that is considered absolute zero. That does not mean that it cannot get colder, disapproving A.

C is just wrong.

D says when molecular motion stops the temperature STARTS to decrease, it was decreasing before it got there.
7 0
3 years ago
one molecule of chlorophyll contains 137 atoms. how many of these atoms come from the metal magnesium?
Svetlanka [38]
The chemical formula for chlorophyll is C55H72O5N4<span>Mg. Only 1 of the 137 atoms comes from magnesium.

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7 0
2 years ago
Select the true statements regarding these resonance structures of formate.? Each carbon-oyxgen bond is somewhere between a sing
son4ous [18]

Answer : Option 1) The true statement is each carbon-oxygen bond is somewhere between a single and double bond and the actual structure of format is an average of the two resonance forms.

Explanation : The actual structure of formate is found to be a resonance hybrid of the two resonating forms. The actual structure for formate do not switches back and forth between two resonance forms.

The O atom in the formate molecule with one bond and three lone pairs, in the resonance form left with reference to the attached image, gets changed into O atom with two bonds and two lone pairs.

Again, the O atom with two bonds and two lone pairs on the resonance form left, changed into O atom with one bond and three lone pairs. It concludes that each carbon-oxygen bond is neither a single bond nor a double bond; each carbon-oxygen bond is somewhere between a single and double bond.

Also, it is seen that each oxygen atom does not have neither a double bond nor a single bond 50% of the time.

7 0
3 years ago
Read 2 more answers
What is the molarity (M) of the following solutions?
Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

                                      = 27 + 3(17) = 27 + 51

                                      = 78 g/mole

Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

4 0
3 years ago
3. What is the energy of a photon whose frequency is 5.2 x 1015 Hz? Use the equation: E = hxv
anzhelika [568]

Answer:

3. 3.45×10¯¹⁸ J.

4. 1.25×10¹⁵ Hz.

Explanation:

3. Determination of the energy of the photon.

Frequency (v) = 5.2×10¹⁵ Hz

Planck's constant (h) = 6.626×10¯³⁴ Js

Energy (E) =?

The energy of the photon can be obtained by using the following formula:

E = hv

E = 6.626×10¯³⁴ × 5.2×10¹⁵

E = 3.45×10¯¹⁸ J

Thus, the energy of the photon is 3.45×10¯¹⁸ J

4. Determination of the frequency of the radiation.

Wavelength (λ) = 2.4×10¯⁵ cm

Velocity (c) = 3×10⁸ m/s

Frequency (v) =?

Next, we shall convert 2.4×10¯⁵ cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

2.4×10¯⁵ cm = 2.4×10¯⁵ cm × 1 m /100 cm

2.4×10¯⁵ cm = 2.4×10¯⁷ m

Thus, 2.4×10¯⁵ cm is equivalent to 2.4×10¯⁷ m

Finally, we shall determine the frequency of the radiation by using the following formula as illustrated below:

Wavelength (λ) = 2.4×10¯⁷ m

Velocity (c) = 3×10⁸ m/s

Frequency (v) =?

v = c / λ

v = 3×10⁸ / 2.4×10¯⁷

v = 1.25×10¹⁵ Hz

Thus, the frequency of the radiation is 1.25×10¹⁵ Hz.

8 0
2 years ago
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