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Lubov Fominskaja [6]
3 years ago
13

Give the quantum number set for one electron in the 3p sub level of a sulfur (S)

Chemistry
1 answer:
vesna_86 [32]3 years ago
3 0

Answer:- Atomic number for sulfur is 16 and it's electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^4. Here, there are total for electrons in 3p and the set of quantum numbers for these 4 electrons would be as..

For the first electron of 3p-

n = 3, l = 1, ml = -1 and ms = +(1/2)

for the second electron of 3p-

n = 3, l = 1, ml = 0 and ms = +(1/2)

for the third electron of 3p-

n = 3, l = 1, ml = +1 and ms = +(1/2)

and for the fourth electron of 3p-

n = 3, l = 1, ml = -1 and ms = -(1/2)

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A 34.57 mL sample of an unknown phosphoric acid solution is titrated with a 0.127 M sodium hydroxide solution. The equivalence p
Bas_tet [7]

<u>Answer:</u> The concentration of unknown phosphoric acid solution is 0.034 M

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_3PO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=3\\M_1=?M\\V_1=34.57mL\\n_2=1\\M_2=0.127M\\V_2=28.2mL

Putting values in above equation, we get:

3\times M_1\times 34.57=1\times 0.127\times 28.2\\\\M_1=\frac{1\times 0.127\times 28.2}{3\times 34.57}=0.034M

Hence, the concentration of unknown phosphoric acid solution is 0.034 M

7 0
3 years ago
PLEASE hurry
gizmo_the_mogwai [7]

Answer:

The answer is A.

Explanation:

8 0
3 years ago
Read 2 more answers
Bismuth oxide reacts with carbon to form bismuth metal:
kvv77 [185]

(a) 3.33 mol Bi

Explanation:

Step 1: Convert 283 g to moles

Bi Molar Mass - 208.98 g/mol × 2 = 417.98 g/mol

O Molar Mass - 16.00 g/mol × 3 = 48.00 g/mol

283 g Bi₂O₃ ÷ 465.98 g/mol = 0.0607 mol Bi₂O₃

Step 2: Find the conversion from Bi₂O₃ to Bi

1 mole of Bi₂O₃ equals 2 moles of Bi

Step 3: Use Dimensional Analysis

1.66316 mol  ·   = 3.32632 mol Bi

3.32632 mol Bi ≈ 3.33 mol Bi

What is bismuth oxide used for?

Bismuth oxides have been considered as an interesting material because of their dielectric properties and have been used for applications such as optical coatings, metal-insulator-semiconductor capacitors and microwave-integrated circuits.

What is bismuth used for in everyday life?

Bismuth finds its main uses in pharmaceuticals, atomic fire alarms and sprinkler systems, solders and other alloys and pigments for cosmetics, glass and ceramics. It is also used as a catalyst in rubber production

Learn more about bismuth oxide:

brainly.com/question/6367198

#SPJ4

3 0
1 year ago
How is a scientific law different from other laws in society?
Lemur [1.5K]

Answer:

The answer will be A

Explanation:

A scientific law is based on repeated experiments or observations, that describe or predict a range of natural phenomena. A scientific law is something that is unexpected

7 0
3 years ago
Read 2 more answers
Hydrazine, N2H4, is a good reducing agent that has been used as a component in
Taya2010 [7]

Answer:

Explanation:

Data Given

Formula of hydrazine = N₂H₄

Lewis structure of hydrazine (N₂H₄) =?

Solution:

  • Hydrazine (N₂H₄) contains 2 Nitrogen atoms and 4 Hydrogen atoms
  • Nitrogen atom have 5 valence electrons
  • Hydrogen have 1 valance electron
  • So it have 14 total number of valance electrons

                     N₂ H₄ = 2 (5) + 4(1) = 14 no. o electrons

  • it is a stable compound and obey octet rule for nitrogen and duplet rule for hydrogen by sharing of electrons.

To draw the Lewis Structure

  • Put the 2 nitrogen in center and draw the Hydrogen on the outside
  • Put the 2 hydrogen on out side of each of the nitrogen
  • Then draw the 2, 2 valence electrons between atoms that form chemical bond
  • So we will draw total 14 electrons in the structure
  • then we count electrons around each atom of nitrogen and hydrogen that it have 8 and 2 electrons to complete its octet and duplet respectively  

      Lewis Structure of  N₂ H₄ in attachment

7 0
3 years ago
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