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Ket [755]
4 years ago
5

A neutral atom in the ground state of Sulfur has its outer most valence electron in which orbital? f d

Chemistry
1 answer:
Natasha_Volkova [10]4 years ago
3 0

<u>Answer:</u> The outermost valence electron enters the p orbital.

<u>Explanation:</u>

Valence electrons are defined as the electrons which are present in outer most orbital of an atom.

Sulfur is the 16th element of the periodic table having 16 electrons.

Electronic configuration of sulfur atom is 1s^22s^22p^63s^23p^4

The number of valence electrons are 2 + 4 = 6

These 6 electrons enter s-orbital and p-orbital but the outermost valence electron will enter the p-orbital.

Hence, the outermost valence electron enters p orbital.

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A beaker of nitric acid is neutralized with dilute aqueous solution of calcium hydroxide.Write a net ionic equation for this rea
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Complete balanced equation: 2HNO₃  +  Ca(OH)₂  →  Ca(NO₃)₂  +  2H₂O

Ionized equation (with spectator ions): 
   2H⁺  +  2NO₃⁻  +  Ca²⁺  +  2OH⁻   →  Ca²⁺  +  2NO₃⁻  +  2H₂O

By eliminating the ions that are the same of both sides of the equation (spectator ions):
    
2H⁺  +   2OH⁻   →    2H₂O    [Net Ionic Equation]
 
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What is the formula for percent yield
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The sodium salt, NaA, of a weak acid is dissolved in water; no other substance is added. Which of these statements (to a close a
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8 0
3 years ago
The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. combustion of
Olin [163]

Answer:

            Empirical Formula  =  C₃H₆O₁

Solution:

Data Given:

                      Mass of Ethyl Butyrate  =  3.61 mg  =  0.00361 g

                      Mass of CO₂  =  8.22 mg  =  0.00822 g

                      Mass of H₂O  =  3.35 mg  =  0.00335 g

Step 1: Calculate %age of Elements as;

                      %C  =  (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

                      %C  =  (0.00822 ÷ 0.00361) × (12 ÷ 44) × 100

                      %C  =  (2.277) × (12 ÷ 44) × 100

                      %C  =  2.277 × 0.2727 × 100

                      %C  =  62.09 %


                      %H  =  (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.00335 ÷ 0.00361) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.9279) × (2.02 ÷ 18.02) × 100

                      %H  =  0.9279 × 0.1120 × 100

                     %H  =  10.39 %


                      %O  =  100% - (%C + %H)

                      %O  =  100% - (62.09% + 10.39%)

                      %O  =  100% - 72.48%

                      %O  =  27.52 %

Step 2: Calculate Moles of each Element;

                      Moles of C  =  %C ÷ At.Mass of C

                      Moles of C  = 62.09 ÷ 12.01

                      Moles of C  =  5.169 mol


                      Moles of H  =  %H ÷ At.Mass of H

                      Moles of H  = 10.39 ÷ 1.01

                      Moles of H  =  10.287 mol


                      Moles of O  =  %O ÷ At.Mass of O

                      Moles of O  = 27.52 ÷ 16.0

                     Moles of O  =  1.720 mol

Step 3: Find out mole ratio and simplify it;

                C                                        H                                     O

             5.169                                10.287                              1.720

       5.169/1.720                       10.287/1.720                     1.720/1.720

               3.00                                   5.98                                   1

                  3                                      ≈ 6                                     1

Result:

         Empirical Formula  =  C₃H₆O₁

8 0
4 years ago
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