Two precursor alkenes
H₃C CH₃
I I
H₂C=C-CH-CH₃ 2,3-dimethyl-1-butene
H₃C CH₃
I I
H₃C-CH=CH-CH₃ 2,3-dimethyl-2-butene
alkane
H₃C CH₃
I I
H₃C-CH-CH-CH₃ 2,3-dimethylbutane
H₃C CH₃ H₃C CH₃
I I I I
H₂C=C-CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
H₃C CH₃ H₃C CH₃
I I I I
H₂C-C=CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
Answer:
1.52atm is the pressure of the gas
Explanation:
To solve this question we must use the general gas law:
PV = nRT
<em>Where P is pressure in atm = Our incognite</em>
<em>V is volume = 50.5L</em>
<em>n are moles of gas = 3.25moles</em>
<em>R is gas constat = 0.082atmL/molK</em>
<em>And T is absolute temperature = 288.6K</em>
To solve pressure:
P = nRT / V
P = 3.25mol*0.082atmL/molK*288.6K / 50.5L
P = 1.52atm is the pressure of the gas
I suppose it false, since the oxidation involves the loss or removal of the electrons such forth it does not gain electrons.
Answer : Option 1) The true statement is each carbon-oxygen bond is somewhere between a single and double bond and the actual structure of format is an average of the two resonance forms.
Explanation : The actual structure of formate is found to be a resonance hybrid of the two resonating forms. The actual structure for formate do not switches back and forth between two resonance forms.
The O atom in the formate molecule with one bond and three lone pairs, in the resonance form left with reference to the attached image, gets changed into O atom with two bonds and two lone pairs.
Again, the O atom with two bonds and two lone pairs on the resonance form left, changed into O atom with one bond and three lone pairs. It concludes that each carbon-oxygen bond is neither a single bond nor a double bond; each carbon-oxygen bond is somewhere between a single and double bond.
Also, it is seen that each oxygen atom does not have neither a double bond nor a single bond 50% of the time.
Answer:
The correct answer is b.
Explanation:
The quantum number n specifies the energetic level of the orbital, the first level being the one with the least energy. As n increases, the probability of finding the electron near the nucleus decreases and the orbital energy increases.
In the case of atoms with more than one electron, the quantum number l also determines the sublevel of energy in which an orbital is found, within a certain energy level. The value of l is designated by the letters s, p, d, and f.
Have a nice day!
