Chemistry, 02.12.2019 08:31 help1572

Draw lewis structures for each of the following.
1. nitrogen trifluoride, nf3
2. hydrogen sulfide, h2s
3. fluorine, f2
4. carbon monoxide, co
5. sulfur dioxide, so2
6. oxygen, o2
7. sulfur difluoride, sf2
8. boron trihydride, bhz
9. chloroform, chcl3
10. carbon disulfide, cs2
11. beryllium chloride, becl2
12. hydrogen cyanide, hcn
13. acetylene, c2h2
14. silicon dioxide, sio2
15. hydrogen peroxide, h2o2
16. sulfate, so2-
17. methanol, ch3oh
18. nitrate, no3
19. chlorite, cio,
20. formic acid, ch2o

Answers: 1

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Answers

Answer from: AshIeigh

Structures Attached

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 = 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

Draw lewis structures for each of the following. 1. nitrogen trifluoride, nf3 2. hydrogen sulfide, h

Answer from: Quest

n2(g) + 3h2-> 2nh3(g) this is the balanced equation

note the mole ratio between n2, h2 and nh3. it is 1 : 3 : 2 this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note: the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

Answer from: Quest

b. examine and describe cells

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Draw lewis structures for each of the following.
1. nitrogen trifluoride, nf3
2. hydroge...