Answer:
1) The overlap of the p orbitals of the carbon-carbon π bond would be lost
Explanation:
Unlike simple bonds, a double bond can not rotate, since it is not possible to twist the ends of the molecule without breaking the π bond.
In the structure of but-2-ene present in the attachment, we can see the two isomers, <em>cis</em> and<em> trans</em>. These isomers cannot be interconverted by rotation around the carbon-carbon double bond without breaking the π bond.
Answer:
Molecular formula is C₂₆H₃₆O₄
Explanation:
The compound is 75.69 % C, 8.80 % H and 15.51 % O. This data means, that in 100 g of compound we have 75.69 g, 15.51 g and 8.80 g of, C, O and H, respectively. We know the molar mass of the compound, so we can work to solve the moles of each element.
In 100 g of compound we have 75.69 g C, 15.51 g O and 8.80 g H
In 412 g of compound we would have:
(412 . 75.69) / 100 = 311.8 of C
(412 . 15.51) / 100 = 63.9 g of O
(412 . 8.80) / 100 = 36.2 g of H
Now, we can determine the moles of each, that are contained in 1 mol of compound.
312 g / 12 g/mol 26 C
64 g / 16 g/mol = 4 O
36 g / 1 g/mol = 36 H
Molecular formula is C₂₆H₃₆O₄
The density of it would be 57. Do the math and apply the formula to it and you would get this answer.
Answer:
n=N/NA
n= 3.754×10²³/6.02×10²³
n= 6.24 s
Explanation
since there is number of molecules, make use of Avogadro's constant to get number of moles.
This is because iodine is nonpolar. You also saw that the iodine was less soluble inethanol and acetone than it was in carbon tetrachloride. Ethanoland acetone are more polar thancarbon tetrachloride. Iodine was more soluble in them than it was in water because they are less polar than water.
