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timama [110]
2 years ago
5

How many molecules in each sample? 64.7 g N2 83 g CCl4 19 g C6H12O6

Chemistry
1 answer:
lilavasa [31]2 years ago
3 0

Answer:

  • 1.39x10²⁴ molecules N₂
  • .25x10²³ molecules CCl₄
  • 6.38x10²² molecules C₆H₁₂O₆

Explanation:

First we <u>convert the given masses into moles</u>, using the <em>compounds' respective molar mass</em>:

  • 64.7 g N₂ ÷ 28 g/mol = 2.31 mol N₂
  • 83 g CCl₄ ÷ 153.82 g/mol = 0.540 mol CCl₄
  • 19 g C₆H₁₂O₆ ÷ 180 g/mol = 0.106 mol C₆H₁₂O₆

Then we multiply each amount by <em>Avogadro's number</em>, to <u>calculate the number of molecules</u>:

  • 2.31 mol N₂ * 6.023x10²³ molecules/mol = 1.39x10²⁴ molecules
  • 0.540 mol CCl₄ * 6.023x10²³ molecules/mol = 3.25x10²³ molecules
  • 0.106 mol C₆H₁₂O₆ * 6.023x10²³ molecules/mol = 6.38x10²² molecules
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This question is about iron three of the raw material added to a blast furnace used to extract iron from henatite are coke hemat
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Answer: The fourth material that is added to the blast furnace is HOT AIR which provides OXYGEN for used for combustion of carbon (Coke).

Explanation:

Iron is the second most abundant metal found in the earth's crust after aluminium. It is not found in the free metallic state but are extracted from rocks which are rich in iron that contains other materials. These are known are iron ores and the most common iron ores are haematite ( Fe2O3).

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Question 24 (1 point)
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Sugar is a covalently linked compound with many atoms. Water is able to dissolve sugar. What does this suggest about the atoms o
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il63 [147K]

Answer:

(i) ΔU = 116 J

(ii) ΔU = 289 J

(iii) ΔU = 1 KJ

(iv) ΔU = 0 J

(v) ΔU = 3.25 KJ

Explanation:

first law:

  • ΔU = Q + W

(i) W = 153 J;  Q = - 37 J ( Q ( - ), losing friction )

⇒ ΔU = 153 - 37 = 116 J

(ii) W = 289 J; Q = 0 ( insulated)

⇒ ΔU = W = 289 J

(iii) Q = 1 KJ , W = 0 ( isovolumetric process)

⇒ ΔU = Q = 1 KJ

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∴ ΔT = 0° ( isothermal )

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(v) isobaric ( constant pressure )

⇒ ΔU = Q + W

∴ Q = 15.6 KJ

∴ W = - ∫ P dV = - P ΔV;  W (-) the system performs a job and the volume increases

.

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⇒ ΔU = 15.6 KJ + ( - 12.35 KJ )

⇒ ΔU = 3.25 KJ

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