On the Moon the gravity is approximately 1/6 of that on Earth. If an object has a mass of 60

Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an

Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 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

Predict the boiling point of water at a pressure of 1.5 atm.

Lina20 [59]

Answer:

100.8 °C

Explanation:

The Clausius-clapeyron equation is:

$ln\frac{P_{1} }{P_{2}} =$ -Δ $\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}} )$

Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'

Isolating for T2 gives:

$T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}$

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C

7 0

3 years ago

. What classification should this reaction have? Cu + 2AgNO3 Cu(NO3)2 + 2Ag synthesis decomposition single replacement double di

Monica [59]

What classification should this reaction have?
Cu + 2AgNO₃ ⇒ Cu(NO₃)₂ + 2Ag

single replacement

8 0

3 years ago

What kind of force acts between magnetic poles if the poles are alike?

Alexxx [7]

A forcs of attraction

3 0

3 years ago

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