1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vera_Pavlovna [14]
4 years ago
15

Draw Lewis structures for each of the following.

Chemistry
1 answer:
Mashutka [201]4 years ago
7 0

Answer:

<em>Structures Attached</em>

Explanation:

1. Total Number of e⁻  in NF₃

Number of e in Nitrogen = 5

Number of e in Flourine = 7

Total Number of e⁻ in NF₃ = 5 + 7(3)

Total Number of e⁻ in NF₃ = 5 + 21

Total Number of e⁻ in NF₃ = 26

2. Total Number of e⁻  in H₂S

Number of e in Hydrogen = 1

Number of e in sulphur = 6

Total Number of e⁻ in H₂S = 1(2) + 6

Total Number of e⁻ in H₂S = 2 + 6

Total Number of e⁻ in H₂S = 8

3. Total Number of e⁻  in F₂

Number of e in Flourine = 7

Total Number of e⁻ in F₂ = 7(2)

Total Number of e⁻ in F₂ = 14

4. Total Number of e⁻  in CO

Number of e in carbon = 4

Number of e in oxygen = 6

Total Number of e⁻ in CO = 4 + 6

Total Number of e⁻ in CO = 10

5. Total Number of e⁻  in SO₂

Number of e in Sulphur = 6

Number of e in oxygen = 6

Total Number of e⁻ in SO₂ = 6 + 6(2)

Total Number of e⁻ in SO₂ = 6 + 12

Total Number of e⁻ in SO₂ = 18

6. Total Number of e⁻  in O₂

Number of e in Oxygen = 6

Total Number of e⁻ in O₂ = 6(2)

Total Number of e⁻ in O₂ = 12

7. Total Number of e⁻  in SF₂

Number of e in Sulphur = 6

Number of e in Flourine = 7

Total Number of e⁻ in SF₂ = 6 + 7(2)

Total Number of e⁻ in SF₂ = 6 + 14

Total Number of e⁻ in SF₂ = 20

8. Total Number of e⁻  in BH₃

Number of e in Boron = 3

Number of e in Hydrogen = 1

Total Number of e⁻ in BH₃ = 3 + 1(3)

Total Number of e⁻ in BH₃ = 3 + 3

Total Number of e⁻ in BH₃ = 6

9. Total Number of e⁻  in CHCl₃

Number of e in Carbon = 4

Number of e in Hydrogen = 1

Number of e in chlorine = 7

Total Number of e⁻ in CHCl₃ = 4 + 1+ 7(3)

Total Number of e⁻ in CHCl₃ = 4 + 1 + 21

Total Number of e⁻ in CHCl₃ = 26

10. Total Number of e⁻  in CS₂

Number of e in Carbon = 4

Number of e in Sulphur = 6

Total Number of e⁻ in CS₂ = 4 + 6(2)

Total Number of e⁻ in CS₂ = 4 + 12

Total Number of e⁻ in CS₂ = 16

11. Total Number of e⁻  in BeCl₂

Number of e in Beryllium = 2

Number of e in Chlorine = 7

Total Number of e⁻ in BeCl₂ = 2 + 7(2)

Total Number of e⁻ in BeCl₂ = 2 + 14

Total Number of e⁻ in BeCl₂ = 16

12. Total Number of e⁻  in HCN

Number of e⁻ in Hydrogen = 1

Number of e⁻ in Carbon = 4

Number of e⁻ in Nitrogen = 5

Total Number of e⁻ in HCN = 1 + 4 + 5

Total Number of e⁻ in HCN = 10

13. Total Number of e⁻  in C₂H₂

Number of e⁻ in Carbon = 4

Number of e⁻ in Hydrogen = 1

Total Number of e⁻ in C₂H₂ = 4(2) + 1(2)

Total Number of e⁻ in C₂H₂ = 8 + 2

Total Number of e⁻ in C₂H₂ = 10

14. Total Number of e⁻  in SiO₂

Number of e⁻ in Silicon = 4

Number of e⁻ in Oxygen = 6

Total Number of e⁻ in SiO₂ = 4 + 6(2)

Total Number of e⁻ in SiO₂ = 4 + 12

Total Number of e⁻ in SiO₂ = 16

15. Total Number of e⁻  in H₂O₂

Number of e⁻ in Hydrogen = 1

Number of e⁻ in Oxygen = 6

Total Number of e⁻ in H₂O₂ = 1(2) + 6(2)

Total Number of e⁻ in H₂O₂ = 2 + 12

Total Number of e⁻ in H₂O₂ = 14

16. Total Number of e⁻  in SO₂⁻

Number of e in Sulphur = 6

Number of e in Oxygen = 6

Total Number of e⁻ in SO₂⁻ = 6 + 6(2)

Total Number of e⁻ in SO₂⁻ = 6 + 12

Total Number of e⁻ in SO₂⁻ = 18

17. Total Number of e⁻  in CH₃OH

Number of e⁻ in Carbon = 4

Number of e⁻ in Hydrogen = 1

Number of e⁻ in Oxygen = 6

Total Number of e⁻ in CH₃OH = 4 + 1(3) + 6 + 1

Total Number of e⁻ in CH₃OH = 4 + 3 + 6 + 1

Total Number of e⁻ in CH₃OH = 14

18. Total Number of e⁻  in NO₃

Number of e⁻ in Nitrogen = 5

Number of e⁻ in Oxygen = 6

Total Number of e⁻ in NO₃ = 5 + 6(3)

Total Number of e⁻ in NO₃ = 5 + 18

Total Number of e⁻ in NO₃ = 23

19. Total Number of e⁻  in ClO

Number of e⁻ in Chlorine = 7

Number of e⁻ in Oxygen = 6

Total Number of e⁻ in ClO = 7 + 6

Total Number of e⁻ in ClO = 7 + 6

Total Number of e⁻ in ClO = 13

20. Total Number of e⁻  in CH₂O

Number of e⁻ in Carbon = 4

Number of e⁻ in Hydrogen = 1

Number of e⁻ in Oxygen = 6

Total Number of e⁻ in CH₂O = 4 + 1(2) + 6

Total Number of e⁻ in CH₂O = 4 + 2 + 6

Total Number of e⁻ in CH₂O = 12

You might be interested in
Determine the molar solubility of pbso4 in pure water. ksp (pbso4) = 1.82 x 10-8.
Misha Larkins [42]
Use the ICE table approach as solution:

           PbSO₄   --> Pb²⁺ + SO₄²⁻
I             -                 0          0
C           -                +s         +s
E           -                  s          s

Ksp = [Pb²⁺][SO₄²⁻]
1.82×10⁻⁸ = s²
Solving for s,
s = <em>1.35×10⁻⁴ M</em>
5 0
4 years ago
Which rule for assigning oxidation numbers is correct?
LenKa [72]
Hydrogen is usually –1. This is INCORRECT. The oxidation number for H is +1.
Oxygen is usually –2. This is CORRECT.
A pure group 1 element is +1. This is INCORRECT. It does not follow. This will depend on the other elements and the overall charge.
A monatomic ion is 0. This is INCORRECT. Diatomic ion is 0.
4 0
3 years ago
Read 2 more answers
A sample contains 16 mg of polonium-218. After 12 minutes, the sample will contain 1.0 mg of polonium-218. What is the half life
mariarad [96]

Answer:

Half-life = 3 minutes

Explanation:

Using the radioactive decay equation we can solve for reaction constant, k. And by using:

K = ln2 / Half-life

We can find half-life of polonium-218

Radioactive decay:

Ln[A] = -kt + ln [A]₀

Where:

[A] could be taken as mass of polonium after t time: 1.0mg

k is Reaction constant, our incognite

t are 12 min

[A]₀ initial amount of polonium-218: 16mg

Ln[A] = -kt + ln [A]₀

Ln[1.0mg] = -k*12min + ln [16mg]

-2.7726 = - k*12min

k = 0.231min⁻¹

Half-life = ln 2 / 0.231min⁻¹

<h3>Half-life = 3 minutes</h3>

5 0
3 years ago
Cl2+2KI→2KCl + I2 is an example of which type of reaction?
Gnoma [55]

Answer:

single replacement

Explanation:

3 0
3 years ago
It refers to the length of the entire path the object travelled
kiruha [24]

Answer:

Path length is the overall distance traveled following the path of where the object travel. ... Displacement is the distance from the starting point of the object to its final point irregardless where it travels.

3 0
3 years ago
Other questions:
  • Can someone help me to answer these Chemistry questions?
    10·1 answer
  • The combustion of 1 mol ch4 releases 803 kj of energy. how much energy is produced from the combustion of 5.00 mol ch4.
    12·1 answer
  • Given the following reaction: 3D(g) + E(g) + 2F(g) → 5G(g) + 4H(g)
    6·1 answer
  • Resonance structures differ by ________. resonance structures differ by ________. number of atoms only number of electrons only
    7·1 answer
  • Which tools should be used to record the most complete data about gas
    5·1 answer
  • Table C: Known Specific Heat Values for Common
    8·2 answers
  • What is the ending substance of? <br><br> C + H2O → CO + H2
    11·2 answers
  • a 20.5g sample of cleaning detergent contains 8.61g of NH40H.CALCULATE the percentage composition of nitrogen in the cleaning de
    7·1 answer
  • The most stable conformation of the following compound has
    12·1 answer
  • Some species of moose are adapted to eat grasses and flowers in the summer and pine needles in the winter.
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!