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Ierofanga [76]
2 years ago
6

A compound with the formula C6H12 may or may not be a saturated hydrocarbon.

Chemistry
1 answer:
astraxan [27]2 years ago
6 0

A compound with the formula C6H12 is not considered a Saturated hydrocarbon.

Why is C6H12 isn't considered a Saturated hydrocarbon?

The ring's presence demonstrates that it is unsaturated. Keep in mind that the general formula for aliphatic hydrocarbons, CnH2n+2, serves as the foundation for its saturation. A chemical is unsaturated if it does not meet this requirement.

Example:

Hexane (C6H14)

C = 6; H = 14 = 2(6) + 2

resulting in hexane becoming saturated.

Cyclohexane(C6H12)

C = 6 and H = 12 do not equal 14 (x)!

cyclohexane is an unsaturated molecule as a result.

Cycloalkanes have the general formula C2H2n as well.

Hence, the given statement is false.

Learn more about the hydrocarbons here,

brainly.com/question/17578846

# SPJ4

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Answer:

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7 0
3 years ago
For the reaction N2(g) + O2(g)2NO(g) H° = 181 kJ and S° = 24.9 J/K G° would be negative at temperatures (above, below) K. Enter
12345 [234]

ΔG deg will be negative above 7.27e+3 K.

<u>Explanation:</u>

  • The ΔG deg with the temperature can be found using the formula and the formula is given below
  • ΔG deg = ΔH deg - T ΔS deg
  • Given data, ΔH deg = 181kJ and ΔSdeg=24.9J/K
  • -T ΔS deg will be always negative and ΔG deg = ΔH deg  will  be  positive and ΔG deg will be negative at relatively high temperatures and positive at relatively low temperatures
  • solving the equation and substitute  ΔGdeg=0
  • ΔGdeg = ΔHdeg - T ΔSdeg
  • T= ΔHdeg/ΔSdeg
  • T=181 kJ / 2.49e-2 kJK-1
  • By simplification we get
  • T=7.27 × 10^3 K.
  • Therefore, Go will be negative above   7.27 × 10^3 K
  • Since ΔG deg = -RT lnK, when ΔGdeg < 0, K > 1 so the reaction will have K > 1 above 7.27 × 10^3 K.
  • ΔG deg will be negative above 7.27e+3 K.

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7 0
3 years ago
2A(g) + B(l) ⇌ 3C(aq) + D(s)
makvit [3.9K]

Answer:

B

[(0.75)^3(0.25)]÷[(0.50)^2(0.75)]

Explanation:

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8 0
3 years ago
Which of the following is a molecular formula for a compound with an empirical formula of CH2O and a molar mass of 150. g/mol.
trapecia [35]

Answer:

C₅H₁₀O₅

Explanation:

Let's consider a compound with the empirical formula CH₂O. In order to determine the molecular formula, we have to calculate "n", so that

n = molar mass of the molecular formula / molar mass of the empirical formula

The molar mass of the molecular formula is 150 g/mol.

The molar mass of the empirical formula is 12 + 2 × 1 + 16 = 30 g/mol

n = (150 g/mol) / (30 g/mol) = 5

Then, we multiply the empirical formula by 5.

CH₂O × 5 = C₅H₁₀O₅

7 0
4 years ago
How many neutrons does this atom have?
Marizza181 [45]

Answer:

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8 0
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