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3 years ago

Identify which sets of Quantum Numbers are valid for an electron. Each set is ordered (n,ℓ,mℓ,ms).

Chemistry

1 answer:

skad [1K]3 years ago

3 0

3,2,0,1/2

1,0,0,1/2

3,2,2,1/2

4,2,1,1/2

2,1,-1,-1/2

THEY ARE VALID

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Calculate the oxidation number of the iodine (I) in each compound: HIO4 = I2 = NaI = HIO3 =

Makovka662 [10]

1) in periodic acid (HIO₄), iodine has oxidation number +7, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 4 · (-2) = 0.
x = +7.

2) in molecule of iodine (I₂), iodine has oxidation number 0, because iodine is nonpolar molecule.

3) in sodium iodide (NaI), iodine has oxidation number -1, sodium has oxidation number +1:
+1 + x = 0.
x = -1.

4) in iodic acid (HIO₃), iodine has oxidation number +5, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 3 · (-2) = 0.
x = +5.

8 0

3 years ago

Read 2 more answers

Consider a solution that is 0.05 M HCl. Your goal is to neutralize 1 L of this solution (i.e. bring the pH to 7). You also have

Ilia_Sergeevich [38]

Answer:

The volume of NaOH required is - 0.01 L

Explanation:

At equivalence point ,

Moles of $HCl$ = Moles of NaOH

Considering :-

$Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}$

Given that:

$Molarity_{NaOH}=5\ M$

$Volume_{NaOH}=?\ L$

$Volume_{HCl}=1\ L$

$Molarity_{HCl}=0.05\ M$

So,

$Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}$

$0.05\times 1=5\times Volume_{NaOH}$

$Volume_{NaOH}=\frac{0.05\times 1}{5}=0.01\ L$

<u>The volume of NaOH required is - 0.01 L</u>

3 0

3 years ago

Why do people use block and teckle systems to move heavy objects?

yaroslaw [1]

H. using a pulley system can reduce the load force, over a greater distance.<span />

8 0

3 years ago

A solution is prepared by dissolving 0.23 mol of hypochlorous acid and 0.27 mol of sodium hypochlorite in water sufficient to yi

finlep [7]

Answer:

hypochlorite ion

Explanation:

The hypochlorous acid, HClO, is a weak acid with Ka = 1.36x10⁻³, when this acid is in solution with its conjugate base, ClO⁻ (From sodium hypochlorite, NaClO) a buffer is produced. When a strong acid as HCl is added, the reaction that occurs is:

HCl + ClO⁻ → HClO + Cl⁻.

Where more hypochlorous acid is produced.

That means, the HCl reacts with the hypochlorite ion present in solution

3 0

3 years ago

Draw an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+. (Assume standard conditions).

babunello [35]

Answer:

-1.05 V

Explanation:

A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.

Oxidation half equation;

Sn(s) ------> Sn^2+(aq) + 2e

Reduction half equation:

Mn^2+(aq) + 2e ----> Mn(s)

Cell voltage= E°cathode - E°anode

E°cathode= -1.19V

E°anode= -0.14 V

Cell voltage= -1.19 V - (-0.14V)

Cell voltage= -1.05 V

8 0

3 years ago