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IgorLugansk [536]
3 years ago
8

Explane factor affecting stability of alkene?

Chemistry
2 answers:
egoroff_w [7]3 years ago
8 0

Answer:

The stability of alkene can be determined by measuring the amount of energy associated with the hydrogenation of the molecule. Since the double bond is breaking in this reaction, the energy released in hydrogenation is proportional to the energy in the double bond of the molecule.

Explanation:

n chemistry, an alkene is a hydrocarbon that contains a carbon–carbon double bond. The term is often used as synonym of olefin, that is, any hydrocarbon containing one or more double bonds. However, the IUPAC recommends using the name "alkene" only for acyclic hydrocarbons with just one double bond; alkadiene, alkatriene, etc., or polyene for acyclic hydrocarbons with two or more double bonds; cycloalkene, cycloalkadiene, etc. for cyclic ones; and "olefin" for the general class — cyclic or acyclic, with one or more double bonds.

lions [1.4K]3 years ago
7 0

Answer:

Alkenes have substituents, hydrogen atoms attached to the carbons in the double bonds. The more substituents the alkenes have, the more stable they are. Thus, a tetra substituted alkene is more stable than a tri-substituted alkene, which is more stable than a di-substituted alkene or an unsubstituted one.

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What quantity is the same among atoms of the same element?
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Neutrons or Neutral atoms

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Neutral atoms of each element contain an equal number of protons and electrons. The number of protons determines an element's atomic number and is used to distinguish one element from another.

8 0
2 years ago
Will the calculated Molarity of NaOH be too high or too low or unaffected if the following happen: when you answer the question,
vodomira [7]

Answer:

Explanation:

The result will be affected.

The mass of KHP  weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).

Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.

During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.

Thus, NaOH will appear at a higher concentration than it actually is.

7 0
3 years ago
3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15
shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}

(b) Mass of H

\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

\text{Moles of C = 1400  mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016  + 16.00) u  = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00  \approx 5

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

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