Answer: C
Explanation:
A. Shows 3-Hexyne (NOT 2-HEXYNE)
B. Shows 7 carbons (too many) (NOT 2-HEXYNE)
C. Shows a triple bond (yne) and 6 carbons and it's on the second carbon (2-HEXYNE)
D. Shows two substitent on the second carbon but the triple bond is on the 3rd carbon so it's 2,2-dimethyl-3-heptyne (NOT 2-HEXYNE)
An amide is less reactive to nucleophilic acyl substitution than an acid chloride because more electron density is donated to the carbonyl by nitrogen.
<h3>
What is electron density?</h3>
In quantum chemistry, electron density, also known as electronic density, is a metric for the likelihood that an electron will be found at a microscopic portion of space surrounding a specific point. The likelihood of finding an electron at a particular position near an atom or molecule is represented by electron density. In general, areas with a high electron density are where the electron is most likely to be located. The attractive forces between the electrons and the nuclei in a molecule are what hold the nuclei together in molecule, hence the electron density is essential to the bonding and geometry of a molecule. The repulsions between the nuclei and the electrons function as an antagonist to these attractive forces.
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The stock solution has a volume of 0.208L. When diluted, a stock solution with a concentration of 0.30 m clo2 becomes a 0.20 m stock solution with a volume of 0.050.
A three-dimensional space's occupied volume is measured. It is frequently measured numerically with SI-derived units (such the cubic metre and litre) or with other imperial units (such as the gallon, quart, cubic inch). Volume and length are related in how they are defined (cubed).
A highly concentrated stock solution, often prepared as a 10x concentrated solution, can be described as a stock solution since it can be diluted to create a working solution. In order to prepare complex solutions with numerous constituents, stock solutions can also be utilized as a component.
0.3M*v = 0.125M*0.50L
V=0.208L
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For a hydrocarbon, the combustion reactions are the following:
C + O₂ --> CO₂
H₂ + 1/2 O₂ --> H₂O
The molar mass of CO₂ is 44 g/mol while C is 12 g/mol. Let's solve for amount of C in hydrocarbon.
Mass of C = (14.1 g CO₂)(1mol/44g)(1 mol C/1 mol CO₂)(12 g/mol) = 3.845 g C
So, that means that the mass of hydrogen is:
Mass of H = 4.4 - 3.845 = 0.555 g
Moles C = 3.845/12 = 0.32042
Moles H = 0.555/1 = 0.555
Divide both by the smaller value, 0.32042.
C: 0.32042/0.32042 = 1
H: 0.555/0.32042 = 1.732
We have to get an answer that is closest to a whole number. Let's try multiplying both with 4.
C: 1*4 = 4
H: 1.732*4 = 6.93≈7
<em>Thus, the empirical formula is C₄H₇.</em>
