Two independent variables could change at the same time, and you would not know which variable affected the dependent variable
<span>global wind patterns, rotation of the earth, shape of ocean basins.</span>
Answer:
a. Cyclohexanone
Explanation:
The principle of IR technique is based on the <u>vibration of the bonds</u> by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is <em>a specific energy that generates a specific vibration</em>. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.
Now, we must remember that the <u>lower the wavenumber we will have less energy</u>. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.
If we look at the structure of all the molecules we will find that in the last three we have <u>heteroatoms</u> (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of <u>resonance structures</u> which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.
The molecule that fulfills this condition is the <u>cyclohexanone.</u>
See figure 1
I hope it helps!
It would be a,b,c as the answer. hope this helps!
Answer:
Molecular weight of the compound = 372.13 g/mol
Explanation:
Depression in freezing point is related with molality of the solution as:

Where,
= Depression in freezing point
= Molal depression constant
m = Molality


m = 0.26
Molality = 
Mass of solvent (toluene) = 15.0 g = 0.015 kg

Moles of compound = 0.015 × 0.26 = 0.00389 mol

Mass of the compound = 1.450 g

Molecular weight = 