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iogann1982 [59]

4 years ago

7

When water droplets in a cloud combine, become too heavy, and fall to the ground as snow, sleet or hail, we are experincing what

?

1 answer:

Elza [17]4 years ago

8 0

That would be known as precipitation.

You might be interested in

How many mL of 0.25 M NaOH would have 0.18 moles of NaOH?

8_murik_8 [283]

Answer:

720 mL

Explanation:

To answer this problem we'll use the <em>definition of molarity</em>:

Molarity = moles / liters

With that in mind we can <u>calculate the volume in liters</u>:

0.25 M = 0.18 moles / liters

liters = 0.72 L

Finally we<u> convert liters to mL</u>:

0.72 L * 1000 = 720 mL

Thus 720 mL of 0.25 M NaOH would have 0.18 moles of NaOH.

5 0

3 years ago

2.5 moles of gas occupies 50 L, how many moles of the gas would occupy 100L?

goldenfox [79]

Answer:

5

Explanation:

2.5M=50L

X =100L

2.5MX100L

________ =250M/L

5OL 50L

=5 Moles

6 0

3 years ago

A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 x 10−5 as Ka for ac

Anika [276]

Answer:

a. pH = 5.22

b. Acidic.

c. pH = 5.14

Explanation:

a. It is possible to find the pH of a buffer using Henderson-Hasselbalch equation (H-H equation):

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka (For acetic acid = 4.74), [A⁻] is molar concentration of conjugate base (Acetate salt) and [HA] concentration of the weak acid (Acetic acid).</em>

Replacing:

pH = 4.74 + log₁₀ [0.600M] / [0.200M]

<em>You use the concentration of the acetic acid and sodium acetate because you're adding equal volumes, doing the ratio of the species the same</em>

<em />

<h3>pH = 5.22</h3><h3 />

b. As the solution has a pH lower that 7.0, it is considered as a <em>acidic solution.</em>

<em></em>

c. When you add HCl to the buffer, the reaction is:

CH₃COO⁻ + HCl → CH₃COOH + Cl⁻

<em>Where acetate ion reacts with the acid producing acetic acid.</em>

As you have 0.200L of the buffer, 0.100L are of the acetate ion and 0.100L of the acetic acid. Initial moles of both compounds and moles of HCl added are:

CH₃COO⁻: 0.100L ₓ (0.600mol / L) = 0.0600 moles

CH₃COOH: 0.100L ₓ (0.200mol / L) = 0.0200 moles

HCl: 3.0mL = 3x10⁻³L ₓ (0.034mol / L) = 0.00010 moles HCl

The moles added of HCl are the same moles you're consuming of acetate ion and producing of acetic acid. Thus, moles after the reaction are:

CH₃COO⁻: 0.0600 moles - 0.0001 moles = 0.0509 moles

CH₃COOH: 0.0200 moles + 0.0001 moles = 0.0201 moles

Replacing in H-H equation:

pH = 4.74 + log₁₀ [0.0509moles] / [0.0201moles]

<h3>pH = 5.14</h3>

<em />

8 0

3 years ago

What is the electronic configuration of all the elements

koban [17]

Answer:

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:

3 0

3 years ago

An organic compound is composed of 38.7% C, 9.70% H, 51.6% O. The compound has a molecular formula mass of 62.0g/mol.

Tcecarenko [31]

<span>B)<span>C2H6O<span>2
</span></span></span>
First, convert each percentage to grams: 38.7g, 9.70g, and 51.6g.
Next, calculate the number of moles of each element, based on the number of grams given.
C = 3.23 mol
H = 8.91 mol
O = 3.23 mol
Set up the ratio of moles of each element:
C3.34H9.70O3.23. Convert the decimals to whole numbers by dividing by the smallest subscript, 3.23.
The empirical formula is CH3O.
Now, compute the formula mass, which is 31. Finally, divide the molecular mass by the formula mass, 62/31 = 2. Multiple the subscripts by 2 to get the molecular formula.

5 0

3 years ago

Read 2 more answers