_____ is the measurement of the density of a substance as compared to water.

Write the electron configuration for the following elements:

vazorg [7]

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

8 0

2 years ago

Sophia and her friend Nora have the same unknown solid. Sophia has progressed a little faster with the purification process, but

Katarina [22]

Answer:

Nora used a minimal volume of hot solvent, this hot mixture was filtered to remove insoluble impurities. She had to perform a rapid filtration with a minimum of evaporation in a short-stemmed funnel previously heated in an oven, and provided with a pleated filter to increase the filtration rate, so in this way she prevented the product from crystallizing prematurely on the funnel stem.

If with this technique does not reach a pure substance, the process can be repeated by heating the solvent, changing the funnel or looking for a solvent with high dissolution power of the substance.

Explanation:

The solid products obtained in a reaction are usually accompanied by impurities that must be removed in order to dispose of the desired product in the highest possible purity. The most suitable method for removing impurities that contaminate a solid is by crystallization either in a pure solvent or in a mixture of solvents. This procedure is given the generic name of recrystallization.

3 0

3 years ago

Calculate the partial pressure of phosgene that is in equilibrium with a mixture of 0. 025 atm of carbon monoxide and 0. 003 5 a

saw5 [17]

According to the reaction equation:
by using ICE table:
CO(g) + Cl2(g) ↔ COCl2(g)
initial 0.025 0.0035 0
change -X -X +X
Equ (0.025-X) (0.0035-X) X

when:
Kp = P(COCl2)/P(CO}*P(Cl2) when Kp= 0.2
by substitution:
0.2 = X / (0.025-X)(0.0035-X) by solving this equation for X
∴ X = 1.74x10^-5

4 0

3 years ago

What fraction of the gas phase molecules initially present for 1.0 atm in the chamber are present at 1.0×10−10 torr?

vivado [14]

The fraction of gas phase molecules is calculated by the division of final pressure to the initial pressure.

Fraction = $\frac{P_{final}}{P_{initial}}$ (1)

Here, initial pressure = 1.0 atm

final pressure = $1.0\times 10^{-10} torr$

First, convert torr into atm

1 atm = 760 torr

final pressure = $\frac{1.0\times 10^{-10}}{760}$

= $1.316\times 10^{-13}atm$

Now, put the value of initial and final pressure in formula (1)

Fraction = $\frac{1.316\times 10^{-13}atm}{1.0 atm}$

= $1.316\times 10^{-13}$

Thus, fraction of the gas phase molecules = $1.316\times 10^{-13}$ $\simeq 1.3\times 10^{-13}$

7 0

3 years ago

In aerobic cellular respiration, if four molecules of pyruvic acid enter steps two, the formation of acetyl CoA and three, the K

BabaBlast [244]

Answer:

The aerobic cellular respiration of the glucose where glucose is converted to energy via four steps as follows

1. Glycolysis (glucose break down to pyruvic acid)

2. Link reaction

3. Krebs cycle

4. Electron transport chain, or ETC

The four pyruvic acid produces Four ATP, twenty NADH, and four $FADH_{2}$ molecules

Explanation:

When four pyruvic acid enters step two of the aerobic cellular respiration, they are converted by Oxidative decarboxylation into acetyl-CoA, four molecules of NADH and four molecule of CO2 are formed. This process is otherwise called the link reaction or transition step, because it connects or links the Krebs cycle and glycolysis.

From the chemical reactions involved in cellular respiration of one glucose molecule, from two pyruvic acid molecules we have 2 ATP molecules, 10 NADH molecules, and 2 FADH2 molecules

Hence from four pyruvic acid molecules we have that the acetyl-CoA produced from the four pyruvic acid enters the the Krebs cycle and forms four ATP molecules, twenty NADH molecules, and four $FADH_{2}$ molecules.

5 0

3 years ago