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bearhunter [10]
2 years ago
14

What is the molecular formula for a monocyclic hydrocarbon with 14 carbons and 2 triple bonds?

Chemistry
2 answers:
liq [111]2 years ago
7 0

The molecular formula for a monocyclic hydrocarbon with 14 carbons and 2 triple bond is C₁₄H₂₀

<h3>Molecular formula</h3>

A formula that gives the number of atom of each element present in a one molecule or a compound.

<h3>Monocyclic hydrocarbons</h3>

The name of the saturated hydrocarbons formed by the name attaching the perfix cyclo to the name of acyclic unstaturated hydrocarbon

The molecular formula for a monocyclic hydrocarbon with 14 carbon and 2 triple bonds is C₁₄H₂₀

Learn more about the molecular formula on

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Serggg [28]2 years ago
5 0

The molecular formula to monocyclic hydrocarbon is    C_nH_{2n}CnH_{2n} = C_{14}H_{28}C_{14}H_{28}

<h2>What is monocyclic hydrocarbon?</h2>
  • Monocyclic hydrocarbons that are aromatic in their lowest-lying triplet state but antiaromatic in their singlet initial state.
  • The nature and quantity of each element's atoms in a molecular compound are indicated by the substance's molecular formula.
  • The molecular formula is frequently the same as the formula.
  • Methane's empirical formula is CH_4, which is also its molecular formula because it only has one carbon atom.
  • The hydrocarbon is known to contain two triple bonds. For each triple bond, four hydrocarbons are lost as a result.

Generally speaking, a monocyclic hydrocarbon's molecular formula is C_nH_{2n}CnH_{2n} = C_{14}H_{28}C_{14}H_{28}

Learn more about hydrocarbons here:

brainly.com/question/27399070

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Given the following equation: 2 K3N ---&gt; 6 K + N2 How many moles of N2 can be produced by letting 12.00 moles of K3N react?
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Read 2 more answers
The total volume of seawater is 1.5 x 10²¹ L. Seawater contains approximately 3.5% sodium chloride by mass. At that high of a co
garri49 [273]

Answer:

There are 5.408\times 10^{22} grams contained in all the seawater in the world.

Explanation:

At first let is determinate the total mass of seawater (m_{sw}), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:

m_{sw} = \rho_{sw}\cdot V_{sw}

Where:

\rho_{sw} - Density of seawater, measured in grams per liters.

V_{sw} - Volume of seawater, measured in liters.

If V_{sw} = 1.5\times 10^{21}\,L and \rho_{sw} = 1030\,\frac{g}{L}, then:

m_{sw}=\left(1030\,\frac{g}{L} \right)\cdot (1.5\times 10^{21}\,L)

m_{sw} = 1.545\times 10^{24}\,g

The total mass of sodium chloride is determined by the following ratio:

r = \frac{m_{NaCl}}{m_{sw}}

m_{NaCl} = r\cdot m_{sw}

Given that m_{sw} = 1.545\times 10^{24}\,g and r = 0.035, the total mass of sodium chloride in all the seawater in the world is:

m_{NaCl} = 0.035\cdot (1.545\times 10^{24}\,g)

m_{NaCl} = 5.408\times 10^{22}\,g

There are 5.408\times 10^{22} grams contained in all the seawater in the world.

8 0
4 years ago
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