which scientist would study the force of a car running into a wall? A) A chemist B) A physicist C) A botanist D) A biologist

A methanol-water mixture is to be flash distilled at 1 atm. If the feed is 25 mole %methanol, what are the liquid and vapor comp

frozen [14]

Answer:

Explanation:

Given that:

The distillation is carried out at a pressure of 1 atm

The feed harbors 25% mole of methanol (z)

The total moles of feed is usually 100 moles

In the system, we have both methanol and water

Using the total mole balance for the distillation column.

Fz = Lx + Vy

where;

F = amount of feed

z = mole fraction of ethanol (in feed)

L = amount of liquid product out of the column

V = amount of vapor product out of the column

x = mole fraction of methanol out of the liquid

y = mole fraction of methanol out of the vapor.

SO;

(a)

If all the feed is vaporized, then the vapor will likely have the same composition as the feed.

(b)

If no vaporization of the feed takes place, then the bottoms moving out of the column contains the same composition as the feed.

(c)

If 1/3 of the feed is vaporized; then 2/3 of the feed is at the bottom.

The balance equation would be:

$Fz = (\dfrac{2}{3}F) x + (\dfrac{1}{3}F)y \\ \\ z = \dfrac{2}{3}x+\dfrac{1}{3}y$

Replacing z = 0.25; we have:

$0.25 = \dfrac{2}{3}x+\dfrac{1}{3}y$

0.75 = 2x + y

(d)

If 2/3 of the feed is vaporized;

Then:

$Fz = (\dfrac{1}{3}F) x + (\dfrac{2}{3}F)y \\ \\ z = \dfrac{1}{3}x+\dfrac{2}{3}y$

replacing z = 0.25

$0.25 = \dfrac{1}{3}x+\dfrac{2}{3}y$

0.75 = x + 2y

6 0

2 years ago

At what height above the earth surface will the gravitational acceleration be 5m

slamgirl [31]

Answer: -

The acceleration due to gravity at height r = a = GM/r²

Rearranging

r² = GM /a

= (6.674 x 10⁻¹¹ x 5.972 x 10²⁴ ) / 5

r = 8.917 x 10⁶ m

r = 8917 Km

Now Radius of earth = 6371 Km

So height = 8917 - 6371 = 2546 Km

8 0

3 years ago

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

3 years ago

If the H+ in a solution is 1x10^-1 mol/L what is the OH-

damaskus [11]

PH scale is used to determine how acidic or basic a solution is.
we have been given the hydrogen ion concentration. Using this we can calculate pH,
pH = - log[H⁺]
pH = - log (1 x 10⁻¹ M)
pH = 1
using pH can calculate pOH
pH + pOH = 14
pOH = 14 - 1
pOH = 13
using pOH we can calculate the hydroxide ion concentration
pOH = - log [OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 10⁻¹³ M
hydroxide ion concentration is 10⁻¹³ M

8 0

3 years ago

Suppose a gas starts with a volume of 4.52L, temperature of 23 C, and pressure of 102,00 Pa. If the volume changes to 4.83L and

ivanzaharov [21]

Answer:

The answer to your question is P2 = 84.16 kPa

Explanation:

Data

Volume 1 = V1 = 4.52 L Volume 2 = V2 = 4.83 l

Pressure 1 = P1 = 102 kPa Pressure 2 = P2 = ?

Temperature 1 = T1 = 23°C Temperature 2 = T2 = -12°C

Process

1.- Convert the temperature to °K

Temperature 1 = 23 + 273 = 296°K

Temperature 2 = -12 + 273 = 261°K

2.- Use the Combined Gas law to solve this problem

P1V1/T1 = P2V2/T2

-Solve for P2

P2 = P1V1T2 / T1V2

-Substitution

P2 = (102)(4.52)(261) / (296)(4.83)

-Simplification

P2 = 120331.44 / 1429.68

-Result

P2 = 84.16 kPa

8 0

3 years ago