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svetoff [14.1K]
2 years ago
15

Write the electron configuration for the following elements:

Chemistry
1 answer:
vazorg [7]2 years ago
8 0

Answer:

a.Carbon [He]2s22p2

b. Neon [He]2s22p6

c. Sulfur [Ne]3s23p4

d.Lithium [He]2s1

e. Argon [Ne]3s23p6

f. Oxygen [He]2s22p4

g.Potassium [Ar]4s1

h. Helium 1s2

This table is available to download as a PDF to use as a study sheet.

NUMBER ELEMENT ELECTRON CONFIGURATION

1 Hydrogen 1s1

2 Helium 1s2

3 Lithium [He]2s1

4 Beryllium [He]2s2

5 Boron [He]2s22p1

6 Carbon [He]2s22p2

7 Nitrogen [He]2s22p3

8 Oxygen [He]2s22p4

9 Fluorine [He]2s22p5

10 Neon [He]2s22p6

11 Sodium [Ne]3s1

12 Magnesium [Ne]3s2

13 Aluminum [Ne]3s23p1

14 Silicon [Ne]3s23p2

15 Phosphorus [Ne]3s23p3

16 Sulfur [Ne]3s23p4

17 Chlorine [Ne]3s23p5

18 Argon [Ne]3s23p6

19 Potassium [Ar]4s1

20 Calcium [Ar]4s2

21 Scandium [Ar]3d14s2

22 Titanium [Ar]3d24s2

23 Vanadium [Ar]3d34s2

24 Chromium [Ar]3d54s1

25 Manganese [Ar]3d54s2

26 Iron [Ar]3d64s2

27 Cobalt [Ar]3d74s2

28 Nickel [Ar]3d84s2

29 Copper [Ar]3d104s1

30 Zinc [Ar]3d104s2

31 Gallium [Ar]3d104s24p1

32 Germanium [Ar]3d104s24p2

33 Arsenic [Ar]3d104s24p3

34 Selenium [Ar]3d104s24p4

35 Bromine [Ar]3d104s24p5

36 Krypton [Ar]3d104s24p6

37 Rubidium [Kr]5s1

38 Strontium [Kr]5s2

39 Yttrium [Kr]4d15s2

40 Zirconium [Kr]4d25s2

41 Niobium [Kr]4d45s1

42 Molybdenum [Kr]4d55s1

43 Technetium [Kr]4d55s2

44 Ruthenium [Kr]4d75s1

45 Rhodium [Kr]4d85s1

46 Palladium [Kr]4d10

47 Silver [Kr]4d105s1

48 Cadmium [Kr]4d105s2

49 Indium [Kr]4d105s25p1

50 Tin [Kr]4d105s25p2

51 Antimony [Kr]4d105s25p3

52 Tellurium [Kr]4d105s25p4

53 Iodine [Kr]4d105s25p5

54 Xenon [Kr]4d105s25p6

55 Cesium [Xe]6s1

56 Barium [Xe]6s2

57 Lanthanum [Xe]5d16s2

58 Cerium [Xe]4f15d16s2

59 Praseodymium [Xe]4f36s2

60 Neodymium [Xe]4f46s2

61 Promethium [Xe]4f56s2

62 Samarium [Xe]4f66s2

63 Europium [Xe]4f76s2

64 Gadolinium [Xe]4f75d16s2

65 Terbium [Xe]4f96s2

66 Dysprosium [Xe]4f106s2

67 Holmium [Xe]4f116s2

68 Erbium [Xe]4f126s2

69 Thulium [Xe]4f136s2

70 Ytterbium [Xe]4f146s2

71 Lutetium [Xe]4f145d16s2

72 Hafnium [Xe]4f145d26s2

73 Tantalum [Xe]4f145d36s2

74 Tungsten [Xe]4f145d46s2

75 Rhenium [Xe]4f145d56s2

76 Osmium [Xe]4f145d66s2

77 Iridium [Xe]4f145d76s2

78 Platinum [Xe]4f145d96s1

79 Gold [Xe]4f145d106s1

80 Mercury [Xe]4f145d106s2

81 Thallium [Xe]4f145d106s26p1

82 Lead [Xe]4f145d106s26p2

83 Bismuth [Xe]4f145d106s26p3

84 Polonium [Xe]4f145d106s26p4

85 Astatine [Xe]4f145d106s26p5

86 Radon [Xe]4f145d106s26p6

87 Francium [Rn]7s1

88 Radium [Rn]7s2

89 Actinium [Rn]6d17s2

90 Thorium [Rn]6d27s2

91 Protactinium [Rn]5f26d17s2

92 Uranium [Rn]5f36d17s2

93 Neptunium [Rn]5f46d17s2

94 Plutonium [Rn]5f67s2

95 Americium [Rn]5f77s2

96 Curium [Rn]5f76d17s2

97 Berkelium [Rn]5f97s2

98 Californium [Rn]5f107s2

99 Einsteinium [Rn]5f117s2

100 Fermium [Rn]5f127s2

101 Mendelevium [Rn]5f137s2

102 Nobelium [Rn]5f147s2

103 Lawrencium [Rn]5f147s27p1

104 Rutherfordium [Rn]5f146d27s2

105 Dubnium *[Rn]5f146d37s2

106 Seaborgium *[Rn]5f146d47s2

107 Bohrium *[Rn]5f146d57s2

108 Hassium *[Rn]5f146d67s2

109 Meitnerium *[Rn]5f146d77s2

110 Darmstadtium *[Rn]5f146d97s1

111 Roentgenium *[Rn]5f146d107s1

112 Copernium *[Rn]5f146d107s2

113 Nihonium *[Rn]5f146d107s27p1

114 Flerovium *[Rn]5f146d107s27p2

115 Moscovium *[Rn]5f146d107s27p3

116 Livermorium *[Rn]5f146d107s27p4

117 Tennessine *[Rn]5f146d107s27p5

118 Oganesson *[Rn]5f146d107s27p6

Explanation:

I hope it's help

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4.41 g of propane gas (C3H8) is injected into a bomb calorimeter and ignited with excess oxygen, according to the reaction below
gayaneshka [121]

Answer :  The heat of the reaction is -221.6 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter

q_{rxn}=-q_{cal}

q_{cal}=c_{cal}\times \Delta T

where,

q_{rxn} = heat released by the reaction = ?

q_{cal} = heat absorbed by the calorimeter

c_{cal} = specific heat of calorimeter = 97.1kJ/^oC=97100J/^oC

\Delta T = change in temperature = (T_{final}-T_{initial})=(27.282-25.000)=2.282^oC

Now put all the given values in the above formula, we get:

q_{cal}=(97100J/^oC)\times (2.282^oC)

q_{cal}=221582.2J=221.6kJ

As, q_{rxn}=-q_{cal}

So, q_{rxn}=-221.6kJ

Thus, the heat of the reaction is -221.6 kJ

6 0
3 years ago
If you had 240 L container at 479 k and 300 kpa, what would the volume be if you changed the conditions to STP
nydimaria [60]

Answer:

The answer to your question is V2 = 434.7 l

Explanation:

Data

Volume 1 = V1 = 240 l                             Volume 2 = ?

Temperature 1 = T1 = 479°K                   Temperature 2 = T2 = 293°K

Pressure 1 = P1 = 300 KPa                      Pressure 2 = P2 = 101.325 Kpa

Process

1.- Use the combined gas law to solve this problem

                P1V1/T1 = P2V2/t2

-Solve for V2

               V2 = P1V1T2 / T1P2

2.- Substitution

                V2 = (300)(240)(293) / (479)(101.325)

3.- Simplification

                V2 = 21096000 / 48534.675

4.- Result

                V2 = 434.7 l

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3 years ago
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MariettaO [177]

Answer:

c.the slightly positive oxygen on one water molecules is attracted to a slightly negative hydrogen on another molecule of water

5 0
3 years ago
N2+3H2-2NH3 how many moles of hydrogen are needed to completely react with 2.0 miles of nitrogen
forsale [732]
The correct answer is 3 moles of nitrogen

Hope this helped :)
7 0
3 years ago
Please help I’m really stuck:/
iris [78.8K]

Answer:

i would say two

Explanation:

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6 0
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